International
L
1959-1999
TRANSLATED BY JANOS PATAKI, ANDRAS STIPSITZ, CSABA SZABO

ISTVAN REIMAN
International Mathematical Olympiad
1959-1999
Translated by Janos Pataki, Andras Stipsitz, Csaba Szabo
© 2001 by Wimbledon Publishing Company. All rights reserved.
No part of this book may be reproduced or transmitted in any form
or by any means electronic or mechanical, including photocopying,
recording, or any information storage or retrieval system without
written permission from Wimbledon Publishing Company,
except for the inclusion of quotations in a review.
ISBN 189885 548 X
Produced in the UK
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A
Anthem Press
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Translation from the Hungarian language edition
Nemzetkozi Matematikai Diakolimpiak
by Istvan Reiman
Copyright © Typotex Ltd.
All rights reserved

Preface
This volume contains the problems from the first forty IMO-s, from 1959
to 1999.
The chronicle of the IMO (International Mathematics Olympiad) starts with
the initiative of the Romanian Mathematics and Physics Society: in July 1959 on
the occasion of a celebration the Society invited high school students from the
neighbouring countries to an international mathematical competition. The event
proved to be such a success that the participants all agreed to go on with the
enterprise. Ever since, this competition has taken place annually (except for 1980)
and it has gradually transformed from the local contest of but a few countries into
the most important and comprehensive international mathematical event for the
young. Only seven nations were invited for the first IMO, while the number of
participating countries was well beyond eighty for the last event; wherever
mathematical education has reached a moderate level, sooner or later the country has
turned up at the IMO.
The movement has had a significant impact on the mathematical education
of several participating countries and also on the development of the gifted. The
aim of a more proficient preparation for the IMO itself has launched the
organization of national mathematical competitions in many countries involved. As
the crucial component of successful participation, the preparation for the contest
has enriched the publishing activity in several countries. Math-clubs have been
formed on a large scale and periodicals have started. Even though the
competition certainly brings up some pedagogical problems, if the educators regard the
competitions not as ultimate aims, but as ways to introduce and endear pupils to
mathematics, then their pedagogical benefit is undeniable.
The administration of the competitions has not changed that much; the
larger scale has obviously necessitated certain modifications but the actual contest
is more or less as it used to be. The participating countries are invited to
delegate a group of up to six students who are attending high school at the year
of the contest or had just finished their secondary school studies. Three
problems are posed each day over two consecutive days and the students have to
produce written solutions in their native tongue. There are two delegation
leaders accompanying each team; one of their tasks is to provide an oral translation
of their students' work into one of the official languages—by now this has been
almost exclusively English—for a committee of mathematicians from the host
country. Together with this group of coordinators they eventually settle the
score the solutions are worth; the highest mark is seven points for each problem.
The contestants are then ranked according to their total scores; the awarding of
3

the prizes has been administered according to the following principle: half of
the participants are given a prize: namely the proportion of the gold, silver and
bronze medals is 1:2:3 respectively.
The occasional professional problems are handled by the international jury
formed by the leaders of the participating delegations; their most important and
difficult task is to select the six problems for the actual contest, to formulate
their official text and to prepare rough marking schemes for each of them. The
organizers ask for proposals from the participating countries well in advance; in
due course they produce a list of approximately twenty to twenty five problems
from those suggested and the jury selects the final six from this supply.
There are almost two hundred problems in this book and they provide a full
image of the challenge the students had to cope with during these forty years.
One cannot claim that every single one of them is a pearl of mathematics but
their overwhelming majority is interesting and rewarding; together they more or
less cover the usual syllabus-chapters of elementary mathematics. When doing
the selection, the jury usually tries to choose from the intersection of the
respective curricula of the participating countries; considering that there are more than
eighty of them this is not an easy job, if not impossible. The reader might notice
that there are no problems at all from the theory of probability, for example, and
complex numbers hardly show up.
From the retrospect of more than forty years one can certainly conclude that
the IMO movement has had a significant role in the history of the second half of
twentieth century mathematics. There are quite a few highly ranked
mathematicians who started their career at an IMO; even at this point, however, we have to
emphasize, that an eventual fiasco at the IMO or any other mathematical contest
whatsoever usually has no implications at all about the mathematical potential of
a well prepared student.
A careful reader will certainly realize that quite a few problems in this book
are in fact simplifications or particular cases of more profound mathematical
results; apart from the intellectual satisfaction of actually solving these problems,
the discovery of this mathematical background and the knowledge gained from
it can be the ultimate benefits of a high level study of this book.
May 2001 Istvan Reiman
4

International Mathematical Olympiad
Problems
1959.
21n + 4
1959/1. Prove that the expression —- is irreducible for every positive
intern+3
ger n.
1959/2. Determine the real solutions of the following equations.
a) y/x + y/2x — 1 + у х — \j2x — 1 = v2,
b) у х + y/2x - 1 + у ж - V2x -1 = 1,
c) у х + \/2x — 1 + y/x — y/2x — 1=2.
1959/3. Suppose that χ satisfies
(1) α cos x + bcosx + c = 0.
Form a quadratic equation in cos 2x whose roots are the same values of x. Apply
your result for the special case α = 4, b = 2, с = — 1.
1959/4. Construct a right triangle with a given hypotenuse, if we know that the
median corresponding to the hypotenuse is equal to the geometric mean of the
two adjacent sides.
1959/5. A point Μ is moving on the interval AB. Squares AMCD and BMEF
over the subintervals AM and MB (on the same side of the line AB) are
constructed The circumcircles of these squares intersect each other in the points Μ
and N. Verify that the lines AE and ВС pass through N. Show that for any
choice of Μ the line MN passes through a fixed point. Find the locus of the
midpoints of the intervals joining the centres of the two squares.
1959/6. The planes Ρ and Q intersect in the line p. A and С are points of the
planes Ρ and Q respectively, neither of them is on p. Construct that symmetric
trapezium ABCD (with AB || CD) for which its vertices В and D are on the
planes Ρ and Q and ABCD admits an incircle.
5

Problems
1960
1960/1. Determine all three digit numbers which are equal to 11 times the sum
of the squares of their digits.
1960/2. Determine the real solutions of the inequality
Ax2
{\-y/\+2x)'
<2ж + 9.
1960/3. The hypotenuse ВС = a of the right triangle ABC has been divided into
η equal intervals with η an odd integer. Let h denote the altitude corresponding
to the hypotenuse; and the central interval subtends an angle a at A. Show that
4nh
tan a = —= .
(nz — l)a
1960/4. Construct the triangle ABC if its two altitudes ma and m&
(corresponding to the vertices A and B) and the median corresponding to A is given.
1960/5. For a given cube ABCDA'B'C'D' let X be a point of the face diagonal
AC, and Υ a point of B'D'.
a) Find the locus of the midpoints of the intervals XY for all possible
choices of X and Y.
b) Consider the point Ζ e XY satisfying the equality ZY = 2XZ and
determine the locus of these points for all choices of X and Y.
1960/6. A cone of revolution has an inscribed sphere tangent to the base of the
cone (and to the sloping surface of the cone). A cylinder is circumscribed about
the sphere so that its base lies in the base of the cone. Let V\ and V} denote the
volume of the cone and the resulting cylinder, respectively.
a) Show that V\ is not equal to Vi-
V\
b) Determine the smallest possible value of k = —, and for this minimal к
V2
construct the half angle of the symmetric cone.
1960/7. The parallel sides of a symmetric trapezium are of length a and b, while
its altitude is m.
a) Construct the point Ρ on the symmetry axis of the trapezium which is on
the Thales circles over the legs of the trapezium.
b) Determine the distance of Ρ from one of the parallel sides.
c) Under what assumption does such Ρ exist?
6

1961
1961
1961/1. Solve the following system of equations for x, у and z:
(1) x + y + z = a,
(2) x2 + y2 + z2 = b2,
(3) xy = z2,
where a and b are given real numbers. What conditions must a and b satisfy for
x,y and ζ to be all positive and distinct?
1961/2. Let a, b and с be the sides of a given triangle while ί is its area. Show
that
(1) a2 + b2 + c2>4tV3
When does equality hold?
1961/3. Solve the equation
cosn χ — sinn χ = 1
where η is a positive integer.
1961/4. Ρ is a point inside the triangle P1P2P3. The intersections of the lines
P\P, P2P and Ρί,Ρ with the opposite sides are denoted by Q\, Q2 and Q3,
respectively. Show that there is one among the ratios
PXP P2P P3P
PQi' PQ2' PQ3
which is not less, and one which is not more than 2.
1961/5. Construct a triangle ABC if the length of the two sides AC = b and
AB = c and the acute angle ΑΜΒ = ω is given — here Μ is the midpoint of
ВС. Show also that the problem admits a solution if and only if
ω
b tan— <c<b.
2 ~
1961/6. Let ε be a given plane and A, B,C three non-collinear points on one side
of ε. Suppose furthermore that the plane determined by these points is parallel to
ε. Let A', B' and C' be three arbitrary points on ε. The midpoints of A A', BB'
and CC' are denoted by L, Μ and N respectively. The centre of gravity of the
triangle LMN is denoted by G. (Those triples A', B1, C1 for which L, Μ, Ν do
not form a triangle, are disregarded.) Determine the locus of G for any possible
choice of the triple А', В', С on the plane ε.
7

Problems
1962
1962/1. Determine the smallest possible positive integer χ whose last decimal
digit is 6, and if we erase this last 6 and put it in front of the remaining digits,
we get four times x.
196^/2. Determine all real χ satisfying
y/3 — χ — Vx + 1 > -.
1962/3. The cube ABCDA'B'C'D' with upper face ABCD and lower face
A'B'C'D1 {AA! || BB' \\ CC' \\ DD') is given. A point X runs along the
perimeter of ABCD (in the direction given by the above order) with constant speed,
while a point Υ does the same (with equal speed) along the perimeter of the
square B'C'CB. X and Υ start in the same instant from A and Β', respectively.
Determine the locus of the midpoint Ζ of the interval XY.
1962/4. Solve the following equation:
cos χ + cos 2x + cos2 3x = 1.
1962/5. Three distinct points А, В and С on a circle к are given. Construct the
point D on the circle for which the quadrilateral ABCD admits an incircle.
1962/6. Let R and r denote the radii of the circumcircle and the incircle of an
isosceles triangle. Show that the distance d between the centres of the two circles
is
(1) ( d = y/R(R-2r).
1962/7. There are five spheres which are tangent to all extended edges of a
tetrahedron SABC. Show that
a) SABC is a regular tetrahedron;
b) conversely: a regular tetrahedron admits five spheres with the properties
described above.
8

1963
1963
1963/1. Determine the real solutions of the following equality (p denotes a real
parameter)
л/ж2 — ρ + 2γ£2 — \-x.
1963/2. Given a point A and a segment Б С in the 3-dimensional space,
determine the locus of those points, P, for which the angle IAPX is a right angle
for some X on the segment ВС.
1963/3. Consider a convex n-gon with equal angles and with consecutive sides
a\, 0,2) · · ·) an satisfying
(1) a\ > a2 > · · · > an-
Show that under the above conditions we have
(2) ai=a2 = ... = an.
1963/4. Determine the values x\, x2, хз, #4, £5 satisfying
(1)
(2)
(3)
(4)
(5)
where у is a given parameter.
1963/5. Show that
(1) cos-
Ж5 + Ж2 = 2/Ж1,
xi+x3 = yx2,
х2 + Х4 = ухз,
Щ+Х5=УХ4,
x4 + xi=yx5
2π 3π
— cos — + cos —-
7 7
1
2"
1963/6. Five students, Л, Б, C, D and £? were placed 1 to 5 in a contest.
Someone made the initial guess that the final result would be the order ABCDE,
but — as it turned out — this person was wrong on the final position of all the
contestants; moreover no two students predicted to finish consecutively did so.
A second person guessed DAECB, which was much better, since exactly two
contestants finished in the place predicted, and two disjoint pairs predicted to
finish consecutively did so. Determine the outcome of the contest.
9

Problems
1964
1964/1. a) Find all positive integers η for which 7 divides 2n — 1.
b) Show that there is no positive integer η for which 7 divides 2n +1.
1964/2. Let a, b and с denote the lengths of the sides of a triangle. Show that
(1) a2(-a + b + c) + b2(a - b + c) + c2(a + b-c)< 3abc.
1964/3. Let a, b, с denote the lengths of the sides of the triangle ABC. Tangents
to the inscribed circle are constructed parallel to the sides. Each tangent forms a
triangle with the other two sides of the triangle, and a circle is inscribed in each
of these three triangles. Find the total area of all four inscribed circles.
Consider the tangents of the incircle which are parallel to the sides. These
tangents give rise to three subtriangles of ABC, consider the incircles of these
subtriangles. Determine the sum of the areas of the four incircles.
1964/4. Each pair from 17 scientists exchange letters on one of three topics.
Prove that there are at least three scientists who write to each other on the same
topic.
1964/5. Five points on the plane are situated so that no two of the lines joining a
pair of points are coincident, parallel or perpendicular. Through each point lines
are drawn perpendicular to each of the lines through two of the other four points.
Give the best possible upper bound for the number of intersection points of these
orthogonals, disregarding the given 5 points.
1964/6. Let ABC Ό be a given tetrahedron and D\ the centroid of the face
ABC. The parallels to DD\ passing through the vertices А, В and С intersect
the opposite faces in A\, B\ and C\, respectively.
a) Show that the volume of ABCD is one-third the volume of A\ B\ C\ D\.
b) Is the result valid for any choice of D\ in the interior of ABC!
1965
1965/1. Find all χ in the interval [0,2π] which satisfy
(1) 2 cos χ < |\/l + sin2ic —-v/1 -sin2a;| < л/2.
1965/2. The coefficients of the system of equations
a\\X\ +a\2X2 + a\3x3 =0,
a2\X\+ a22%2 + а2ЪхЪ = °>
a3iXi+a32X2 + a33x3=Q
10

1966
are subject to the following constraints:
a) ац, a/22 and a-33 are all positive,
b) all other coefficients are negative,
c) the sum of coefficients in each equation is positive.
Verify that the only solution of the system is
1965/3. The length of the edge AB in the tetrahedron ABCD is a, while the
length of CD is b. The distance between the skew lines AB and CD is d, the
angle determined by them is ω. The tetrahedron is divided into two parts by a
plane ε parallel to AB and CD. We also know that к times the distance between
AB and ε equals the distance between CD and ε. Determine the ratio of the
volumes of the parts of the tetrahedron.
1965/4. Find all sets of four real numbers x\, X2, %з, Х4 such that the sum of
any one and the product of the other three is 2.
1965/5. The triangle OAB has angle l АО В acute. Μ is an arbitrary point in
OAB different from O. The points Ρ and Q are the feet of the perpendiculars
from Μ to О A and OB, respectively. Determine the locus of the orthocentre Η
of the triangle OPQ if Μ is
a) on AB;
b) in the interior of OAB.
1965/6. For η > 3 points in the plane denote the maximal distance of pairs of
points by d. Prove that at most η pairs of points are of distance d apart.
1966
1966/1. Problems А, В and С have been posed in a mathematical contest. 25
competitors solved at least one of the three. Amongst those who did not solve
A, twice as many solved В as C. The number of competitors solving only A
was one more than the number of competitors solving A and at least one other
problem. The number of competitors solving A equalled the number solving just
В plus the number of competitors solving just C. How many competitors solved
just ΒΊ
1966/2. Let a, b, с denote the sides of a triangle, while the opposite angles are
denoted by α, β, 7. Prove that if
(1) a + b = tan— (a tan α + Ь tan/3)
11

Problems
then the triangle is isosceles.
1966/3. Prove that a point in the space has the smallest sum of distances to
vertices of a regular tetrahedron if and only if it is the centre of the tetrahedron.
1966/4. Prove that
1 1 1
(1) — + + ... + - =cot£-cot2n:r.
sin 2x sin 4x sin 2nx
for any positive integer η and any real χ (with x^—r, k = 0, 1, 2, ..., η and A
2K
an arbitrary integer).
1966/5. Solve the system of equations
\a\ — ai\x2 + Ial — аъ\хЪ + Ια1 — 0-4^4 = 1,
\d2 — d\ \x\ + |a-2 — аз|ж3 + \a2 ~ Q-4|^4= 1,
|аз — a\ \x\ + |аз — ai\x2 + Ια3 — α4|^4 = 1,
|α4 — a\ \x\ + |a4 — a^Xi + \a/\. — a^\x^ =1,
where a\, a^ аз, ci4 denote four distinct reals.
1966/6. Take any points K, L, Μ on the sides AB, ВС, С A of the triangle
ABC. Prove that at least one of the triangles MAL, KB Μ and LCK has area
at most fourth the area of ABC.
1967
1967/1. The parallelogram ABCD has AB = a, AD = 1, angle lDAB = a and
the triangle ABD is acute. Prove that the circles К a, Kb, Kc and Kd of radius
1 centered at А В, С and D cover the parallelogram if and only if
(1) α <cosa + v3 sin a.
1967/2. Prove that a tetrahedron with just one edge of length greater than 1 has
volume at most -.
8
1967/3. Let k, τη, η be positive integers such that m + к +1 is a prime greater
than (n + 1) and let cs = s(s + 1) ( s = 1, 2, ...). Prove that the product
(1) (cm+l ~ ck)(cm+2 ~ ck) " · · · - (cm+n ~ C>k)
is divisible by
(2) cic2...cn.
1967/4. AqBqCq and A'B'C are given acute triangles. Construct the triangle
ABC with the largest possible area which is circumscribed around AqBqCq (i.e.,
12

1968
AB contains Cq, ВС contains Aq and С A contains Bq) and is similar to A'B'C'
(А, В, С correspond to A', B' and C).
1967/5. Consider the sequence {cn} given as
cl =a\ +a2 + · · · +a8
2 2 2
c2 = a\ + a2 + · · · + a8
сп = а™ + а2+.. · + αξ,
for a\, α/}, · · ·; 0,8 reals, not all zero.
Given that an infinite number of {cn} is zero, find all η for which cn = 0.
1967/6. In a sports contest a total of m medals were awarded over η > 1 days.
On the first day one medal and - of the remaining medals were awarded. On the
second day two medal and - of the remaining medals were awarded, and so on.
On the last day the remaining η medals were awarded. How many medals and
over how many days were awarded?
1968
1968/1. Show that there is a unique triangle whose side lengths are consecutive
integers and one of whose angles is twice another.
1968/2. Find all positive integers χ for which
(1) p(x) = x2-\0x-22,
where p(x) is the product of the decimal digits of x.
1968/3. Prove that the system
αχγ+bxi +c = X2,
ax2 + bx2 + с = хз,
ах^ + Ьхп + с = х\
(α, b, c are real with a^O)
I. has no real solution if (b — l)2 — 4ac < 0;
II. has one real solution if (b — 1) — 4ac = 0;
III. and has more than one real solution once (b — 1) — Aac > 0.
13

Problems
1968/4. Prove that every tetrahedron has a vertex whose three edges have the
right lengths to form a triangle.
1968/5. Let / be a real-valued function defined for all real numbers, such that
for some a > 0 it satisfies
(1)
f(x + a)=^+yjf(x)-(f(x))2.
I. Prove that / is periodic, i.e., there exists a positive real b such that
f(x + b) = f(x)
holds for every x.
II. Give an example of such a non-constant / for a= 1.
1968/6. Let [x] denote the greatest integer not larger than χ (the "integer part"
of x). For every positive integer η evaluate the sum
(1)
n+1
+
n + 2'
+ ...+
2k+\
+
1969
1969/1. Prove that there are infinitely many positive integers a such that
4
z = n +a
is not a prime for any positive integer n.
1969/2. Let
- cos(a2 + x) cos(a3+a0 cos(an + :r)
(1) f(x) = cos(ai+x) + 1 + κ-φ + ···+ 2η_λ ,
where αχ, a2, · · ·, an are real constants and a; is a real variable. Prove that if
f(xl) = f(x2) -0 tnen x2 ~ x\ -πιπ f°r some integer τη.
1969/3. For each к = 1, 2, 3, 4, 5 find necessary and sufficient conditions on α > 0
such that there exists a tetrahedron with к edges of length α and (6 — &) edges
of length 1.
1969/4. Let С be an interior point of the semicircle к over AB and D is the foot
of the perpendicular from С to AB. The circle k\ is the incircle of ABC, the
circle fc2 touches CD, 1)Л and fc while k$ touches CD, D5 and k. Show that
&i, &2 апс* &з have another common tangent apart from AB.
1969/5. Given η > 4 points on the plane (no three collinear), prove that there are
TL — 3
at least [ _ ) convex quadrilaterals with vertices amongst the given points.
14

1970
1969/6. For given real numbers х\,Х2^У\^У2^\^2 satisfying x\ >0, x2 >0,
Х\У\— z\ > 0 and #22/2 — z2 > 0, prove that
8 1 1
(1) .■„.,,. o< o +
(χ1+Χ2)(2/1+2/2)-(^1+^2)2 xm-ζΐ Х2У2~4'
Give necessary and sufficient conditions for equality.
1970
1970/1. Μ is a point on the side AB of the triangle ABC. Let r\, r2 and r
denote the radii of the incircles of AMC, BMC and ABC, respectively. ρ\, ρ2
and ρ stands for the radii of the excircles of the triangles AMC, BMC and ABC
Г] Гп Τ
(corresponding to sides AM, Β Μ and AB), respectively. Prove that = -.
Q\ Q2 Q
1970/2. Real numbers χι (i = 0,1,..., n) with 0 < χι < b and xn > 0, жп_1 > 0
are given. If xnxn_\ ... x\Xq represents the number An base a and Bn base b
whilst xn_\ .. .x\Xq represents An_\ base a and Bn_\ base b, then prove that
a > b holds if and only if
(1) An-1<Bn-1
А Я
1970/3. The real numbers oq» аЬ α2·> ■ · ·, an, ■ · ■ satisfy
(1) 1 = clq < a\ < a2 < ·.. < an < ...
We define the sequence b\, b2, ..., bn, ... as
(2) 6" = Σ(ΐ-—)4=·
I. Prove that 0 < bn < 2 holds for all n.
II. Given с satisfying 0 < с < 2, prove that we can find clq, a\, ..., an, ...
(satisfying (1)) so that infinitely many of the corresponding bn are greater than с
1970/4. Find all positive integers η such that the set {η, η + 1, η + 2, η + 3, η + 4,
η+ 5} can be partitioned into two subsets so that the product of the numbers in
each subset is equal.
1970/5. In the tetrahedron ABCD the angle IB DC is a right angle and the foot
of the perpendicular from D to ABC is the intersection of the altitudes of ABC.
Prove that
(1) (AB + BC + С A)2 < 6(AD2 + BD2 + CD2).
When do we have equality?
15

Problems
1970/6. Given 100 coplanar points (no three collinear), consider all triangles
with vertices among the given points and prove that at most 70 % of these
triangles have all angles acute.
1971
1971/1. Show that the following statement is true for η = 3 and 5 and false for
all other η > 2:
"For any real numbers a\, a2, ..., an the inequality
(αϊ - α2)(αχ - a3)... (a\ - an) + (a2 - a\ ){a2 - a3).. .(a2-an) +...
... + (an- αχ){αη - a2)... (an - an_\) > 0.
holds".
1971/2. Let Pi be a convex polyhedron with vertices A\, A2, ..., Ag. Let Pi be
the polyhedron obtained from Pi by a translation that moves A\ into Αι (i = 2, 3,
..., 9). Show that at least two of the polyhedra Pj, P2, ..., P9 have an interior
common point.
1971/3. Prove that we can find an infinite set of positive integers of the form
{2n — 3} (where η is a positive integer) every pair of which are relatively prime.
1971/4. All faces of the tetrahedron ABCD are acute triangles. Let Χ, Υ, Ζ and
Τ be points in the interiors of the segments AB, ВС, CD and DA, respectively;
and consider the closed path XYZTX.
a) If
(1) LDAB + LBCD 4 LABC + LCD A,
then prove that none of the closed paths XYZTX has minimal length.
b)If
(2) LDAB + IBCD = I ABC + LCD A,
then prove that there are infinitely many shortest paths XYZTX, each with
length
2,4Csin-,
2'
where a = LB AC + LCAD + LDAB.
1971/5. Prove that for every positive integer m we can find a finite set of points
S in the plane such that for any point A of S, there are exactly m points in S at
unit distance from A.
16

1972
1971/6. Let
A =
α21 a22
aln\
a2n
J
\ an\ an2 ■ · · CLnn
be a square matrix with all a^ nonnegative integers. For each i,j with aij=0
we have
(1) Ojl + a^2 + .. . + a%n + &\j + o>2j + · · · + anj > n·
Prove that the sum of all the elements in the matrix is at least
2
У*
1972
1972/1. Given any set of ten distinct numbers in the range 10, 11, ..., 99, prove
that we can always find two disjoint subsets with the same sum.
1972/2. Given η > 4, prove that every cyclic quadrilateral can be dissected into
η cyclic quadrilaterals.
1972/3. Let m and η be nonnegative integers. Prove that
(2m)!(2n)!
m\n\{m + ri)\
is an integer. (According to our conventions 0! = 1).
1972/4. Find all positive real solutions (жь^^З)^)^) of the system
(1) (x\ -хъхъ)(х\-хъхъ)<Ъ,
(2) (x\-X4X\){x\-X4X\)<0,
(3)
(4)
(5)
{x\ - X$X2){x\ - £5^2) < °,
(#4 — x\x^){x^ — x\x^) < 0,
2 ?
(#5 — ΧϊΧδ){χ\ — X2xa) 5: 0.
1972/5. Let / and g be two real valued functions defined on the real line
satisfying
(1)
fix + y) + fix -y) = 2f(x)g(y).
Suppose that f(x) is not identically zero and \f(x)\ < 1 for all x. Show that
\g(y)\ < 1 for all y.
17

Problems
1972/6. Given four parallel planes, prove that there exists a regular tetrahedron
with a vertex on each plane.
1973
1973/1. Let OP[, OP^, ..., OP^ be unit vectors in a plane. РъР2,...,Рп all
lie on the same side of a line through O. Prove that if η is odd, then
|α^+ό^ + ...+α^|>ι,
where \OM\ denotes the length of the vector OM.
1973/2. Can we find a finite set of non-coplanar points, such that given any two
points, A and B, there are two others, С and D, with the lines AB and CD
parallel and distinct?
1973/3. Let a and b be real numbers for which the equation χ + ax +bx + ax +
1=0 has at least one real solution. Find the least possible value of α + b .
1973/4. A soldier needs to sweep a region of the shape of an equilateral triangle
for mines. The detector has an effective radius equal to half the altitude of the
triangle. He starts at a vertex of the triangle. What path should he follow in order
to travel the least distance and still sweep the whole region?
1973/5. Let G be a set of non-constant functions /. Each / is defined on the real
line and has the form f(x) = ax + b for some real a, b.
a) If /, g e G, then so is g о f e G, where g о f(x) = g (f(x)).
b) If / is in G, then so is the inverse / .
rHx>x—b-
a a
c) Every / in G has a fixed point. In other words we can find Xf such that
f(xf) = xf.
Prove that all the functions in G have a common fixed point.
1973/6. Let cli,cl2, ... ,an be positive reals, and q satisfies 0<q<l. Find
b\,b2,--.,bn such that:
a) ak <bk, (fc = l, 2, ..., n);
h)q<b-^-<-,(k=l,2,.r.,n-l)·
h q
c) bi+b2 + .-. + bn<- (ai+a2 + ..- + an).
\-q
18

1974
1974
1974/1. Three players А, В, С play the following game: There are three cards
each with a different positive integer p, q and r where p<q<r. In each round
the cards are randomly dealt to the players and each receives the number of
counters on his card.
After two or more rounds, A has received 20, Б 10 and С 9 counters. In
the last round В received the largest number of counters.
Who received q counters in the first round?
1974/2. Let А, В and С denote the vertices of a triangle. Prove that there is a
point D on the side AB of the triangle ABC, such that CD is the geometric
mean of AD and DB if and only if
С
(1) sin A sin В < sin2 —.
- 2
1974/3. Prove that
is not divisible by 5 for any nonnegative integer n.
1974/4. An 8 χ 8 chessboard is divided into ρ disjoint rectangles (along the lines
between squares), so that
a) each rectangle has the same number of white squares as black squares
b) If щ denotes the number of white squares in the г-th rectangle, then
a\<a2< · ■. <ap holds.
Find the maximum possible value of ρ and all possible a\, a^ ■ · ■> ap se~
quences.
1974/5. Determine all possible values of
„abed
V — ι ι ι
d+a+Ъ a+b+c b+c+d c+d+a
for arbitrary positive reals a, b, c, d.
1974/6. Let P(x) be а поп constant polynomial with integer coefficients. Let η
be the number of distinct integers k, where (P(k)) = 1. Prove that
(1) n(P) - deg(P) < 2,
where degP denotes the degree of P(x).
19

Problems
1975
1975/1. Let x\ > x^ > ... > xn and y\ > yi > ... > yn be real numbers. Prove
that if {zi} is any permutation of the {yi}, then:
η η
(1) Y^{xi-yi)2<Y^{xi-Zi)2.
1975/2. Let a\ < a^ < α-з < · · · be an infinite sequence of positive integers.
Prove that there are infinitely many elements of this sequence that can be
written in the form
Q"m = %CLp "r y&q ι
with x, у positive integers and p^q.
1975/3. Given any triangle ABC, construct external triangles ABR, BCP,
CAQ on the sides, so that
IPBC=ICAQ = 45°
IBCP = IQCA = 30°
IABR=IBAR=\5°.
Prove that IQRP = 90° and QR = RP.
1975/4. Let A denote the sum of the decimal digits of 44444444, and В be the
sum of the decimal digits of A. Find the sum of the decimal digits of B.
1975/5. Can you find 1975 points on the circumference of a unit circle such that
the distance between each pair is rational?
1975/6. Find all polynomials P(x,y) in two variables such that:
I. for every real numbers t, x, y, P(tx,ty) = tnP(x,y), where η is a positive
integer, i.e. Ρ is a homogenous polynomial of degree n.
II. For every real a, b, c, P(a + b,c) + P(b + c,a) + P(c + a,b) = 0.
III. P(1,0) = 1.
1976
1976/1. A plane convex quadrilateral has area 32 cm , and the sum of two
opposite sides and a diagonal is 16 cm. Determine all possible lengths for the other
diagonal.
1976/2. Let Px(x) = x2 - 2; Pj(x) = Px (Ρ0_ι(χ))·, j = 2,3,.... Prove that for
every n, Pn(x) = x has η distinct real roots.
20

1977
1976/3. A rectangular box can be completely filled with unit cubes. If one places
as many cubes as possible, each with volume 2, in the box, with their edges
parallel to the edges of the box, one can fill exactly 40% of the box. Determine
the possible dimensions of the box. (V2= 1.2599...).
1976/4. Determine the largest number which is the product of positive integers
with sum 1976.
1976/5. Consider the following system of ρ equations with q = 2p unknowns:
a\\X\ +ai2%2 + · · · + CL\g%q = 0,
a2\X\ + α22%2 + ■■· + CL2qxq = 0,
dpi X\ + ap2x2 + · · · + CLpqXq = 0;
where ац e {0, 1, -1} (i = l, 2, ..., p, j = l, 2, ..., q).
Prove that the system of equations has a solution x\, X2, · ■ ·, xq satisfying
the following properties:
a) x\, Χ2, ■ ■ ■, хд are integers;
b) hot all xj are 0 (j = 1,2,..., q);
c) \xj\<q (j=l, 2, ..., q).
5
1976/6. The sequence щ,щ,... is defined as follows: uq = 2, щ = -, ип+\ =
un(un_i —2) — щ(п=1,2,...). Prove that
2n-(-l)n
un = 2 3 (n=l, 2, ...).
1977
1977/1. Construct equilateral triangles ABK, BCL, CDM, DAN on the
inside of the square ABCD. Show that the midpoints of KL, LM, MN, NK
and the midpoints of AK, BK, BL, CL, CM, DM, DN, AN form a regular
dodecahedron.
1977/2. In a finite sequence of real numbers the sum of any seven consecutive
terms is negative, and the sum of any eleven consecutive terms is positive.
Determine the maximum number of terms in the sequence.
21

Problems
1977/3. Given an integer η > 2, let Vn be the set of integers 1 + kn for к a
positive integer. A number m e Vn is called indecomposable if there is no number
p,q£Vn such that pq = m.
Prove that there is a number r in Vn which can be expressed as the product
of indecomposable members of Vn in more than one way (decompositions which
differ solely in the order of factors are not regarded as different).
1977/4. Define
(1) f(x) = 1 — a cos χ — b sin χ — A cos 2x — В sin 2x
where a, b, A and В are real constants. Suppose that f(x) > 0 for all real x.
Prove that
(2) a2 + b2<2, A2 + B2<1.
1977/5. Let a and b be positive integers. When (a +b ) is divided by (a + b),
the quotient is q, the remainder r.
Find all pairs (a, b), such that
(1) q2 + r = \911.
1977/6. The function / is defined on the set of positive integers and its values
are positive integers. Let us assume that
(1) f(n + D>f(f(n))
for every positive n.
Prove that for every positive η
f(n) = n.
1978
1978/1. For m and η positive integers, n>m> 1, the last three decimal digits
of 1978m is the same as the last three decimal digits of 1978n. Find m and η
such that m + n has the least possible value.
1978/2. Ρ is a point inside a sphere. Three mutually perpendicular rays from Ρ
intersect the sphere at points U, V and W. Let Q denote the vertex diagonally
opposite Ρ in the parallelepiped determined by PU, PV, PW. Find the locus
of Q for all possible sets of such rays from P.
1978/3. The set of all positive integers is the union of two disjoint subsets:
F = {f(l),f(2),...,f(n),...} and
G = {g(l),g(2),...,g(n),...}
22

1979
where
/(1)</(2)< ... </(n)< ... and g(l)<g(2)<...<g(n)<...,
(1) 9(n) = f(f(n)) + l for every η > 1.
Determine /(240).
1978/4. In the triangle ABC we have AB = AC. A circle is tangent internally to
the circumcircle of the triangle and also to sides AB, AC at P, Q respectively.
Prove that the midpoint of PQ is the centre of the incircle of the triangle.
1978/5. Let {α&} be a sequence of distinct positive integers (k = 1, 2, ..., n, ...).
Prove that for every positive integer η
η η ι
(1) Σρ>Σ[
k=\ к к=\ л
1978/6. An international society has its members from six different countries.
The list of members has 1978 names, numbered 1,2,..., 1978. Prove that there
is at least one member whose number is the sum of the numbers of two members
from his own country, or twice the number of a member from his own country.
1979
1979/1. Let ρ and q be positive integers such that
ρ л 1 1 1 1 1
q 2 3 4 1318 1319
Prove that 1979 divides p.
1979/2. A prism with pentagons А\А2АЪА^А$, and BiB2B3B4B5 as top
and bottom faces is given. Each side of the two pentagons and each of the 25
segments AiBj, (г, j = 1, 2, ..., 5) is coloured red or green. Every triangle whose
vertices are vertices of the prism and whose sides have all been coloured has two
sides of a different colour. Prove that all 10 edges of the top and bottom faces
have the same colour.
1979/3. Let k\ and k2 be two circles on the plane and let A denote one of their
points of intersection. Starting simultaneously from A, two points, P\ and P2
move with constant speed, each traveling along its own circle in the same sense.
The two points return to A simultaneously after one revolution. Prove that there
is a fixed point Ρ in the plane such that, at any time, the distance from Ρ to the
moving points are equal.
23

Problems
1979/4. Given a plane π, a point Ρ in the plane and a point Q not in the plane.
Find all points R of the plane π such that the ratio
QP + PR
(1) -qbT
is maximal.
1979/5. Find all real numbers b for which there exist nonnegative real numbers
x\, X2, X3, X4, £5 satisfying
5 5 5
(1) Y^kxk = b, Y^k3xk=b2, Y^k5xk = b3.
k=\ k=\ k=\
1979/6. Let A and Ε be opposite vertices of an octagon. A frog starts at vertex
A. From any vertex except Ε it jumps to one of the two adjacent vertices. When
it reaches Ε it stops. Let an be the number of distinct paths of exactly η jumps
ending at E. Prove that
O2n-l=0, a2n = -j=(xn-1-yn-1) (n=l, 2, 3, ...)
where χ = 2 + л/l and у = 2 — л/2.
1980
In 1980 there was no International Mathematical Olympics. The
Organisation of the Teachers of Mathematics, Physics and Chemistry of Finland (MAOL)
invited the teams of England, Hungary and Sweden to compare their abilities.
Here we present the problems of that competition.
1980/1. Let α, β, and 7 denote the angles of the triangle ABC. The
perpendicular bisector of AB intersects ВС at the point X, the perpendicular bisector of
AC intersects it at Y. Prove that tan/3-tan7 = 3 implies BC = XY.
Show that this condition is not necessary for ВС = XY, and give a sufficient
and necessary condition.
1980/2. Define the numbers a0> ab · · ·. an in the following way:
1 a?
ao = o> ak+\ =ak + — (n>l, fc = 0, 1, ..., n-l).
2 η
1
к
2 ',
Prove that
(1) 1--<αη<1
η
1980/3. Prove that the equation
(1) xn + l=yn+\
24

1981
where n is a positive integer not smaller then 2, has no positive integer solution
in χ and у for which χ and η +1 are relatively prime.
1980/4. Determine all positive integers η such that the following statement
holds: In the inscribed convex polygon A\A2.. ■ A2n if the pairs of opposite
sides
(ΑιΑ2,Αη+ιΑη+2), (A2A3,An+2An+3), ..., {Αη_λΑη,Α2η_ιΑ2η)
are parallel, then the sides
are parallel as well.
1980/5. In a rectangular coordinate system we call a line parallel to the χ axis
triangular if it intersects the curve with equation
4 3 2
y = x^+px +qx +rx + s
in the points А, В, С and D (from left to right) such that the segments AB, AC
and AD are the sides of a triangle.
Prove that the lines parallel to the χ axis intersecting the curve in four
distinct points are all triangular or none of them is triangular.
1980
1980/6. Find the digits left and right of the decimal point in the decimal form of
the number
(л/5+л/з)'
1981
1981/1. Let Ρ be a point inside the triangle ABC and D, E, F are the feet of
the perpendiculars from Ρ to the lines ВС, С A, AB, respectively. Find all Ρ
which minimise:
ВС СА AB
( } ~PD + PE + PF'
1981/2. Consider all subsets of size r of the set Hn = {\, 2, ... , n}, where
1 < r < n. Each subset has a minimal element, let F(n, r) denote the arithmetic
mean of these elements. Prove that
ч n + 1
F(n,r) = —— .■
r +1
9 9
1981/3. Determine the maximum value of m + η where m and η are integers
satisfying m, η e {1, 2, ..., 1981} and
9 2 2
(1) (n —nm — m ) =1.
25

Problems
1981/4. a) For which η > 2 is there a set of η consecutive positive integers such
that the largest number in the set is a divisor of the least common multiple of
the remaining η — 1 numbers?
b) For which η > 2 is there exactly one set having this property?
1981/5. Three circles of equal radii have a common point О and lie inside a
given triangle. Each circle touches a pair of sides of the triangle. Prove that the
incentre and the circumcentre of the triangle are collinear with the point O.
1981/6. The function f(x,y) satisfies
(1) f(0,y) = y + l,
(2) f(x + l,0) = f(x,l),
(3) f(x + \,y + l) = f(x,f(x + l,y))
for every integer χ and y. Find /(4,1981).
1982
1982/1. The function f(n) is defined on the positive integers and takes on non-
negative integer values. For all n, m
(1) f(m + n) — f(m) — f(ri) = 0 or 1,
(2) f(2) = 0, /(3)>0,
(3) /(9999) = 3333.
Determine /(1982).
1982/2. A non-isosceles triangle A1A2A3 has sides щ opposite to Ai. Mi is the
midpoint of side щ and T{ is the point where the incircle touches side щ.
Denote by Si the reflection of Ti in the interior bisector of angle Ai. Prove
that the lines M\S\, M^S^ and M^S^ are concurrent.
1982/3. Consider the infinite non increasing sequence {xi} of positive reals such
that xq = 1.
a) Prove that for every such sequence there is an η > 1, such that
2 2
(i) sn = -^ + -i + ..
X\ X2
b) Find such a sequence for which
2 2
x\ x2
2
.+ n_1 > 3,999
2
..+ n~l <4.
for all n.
26

1983
1982/4. Prove that if η is a positive integer such that the equation
о 2 3
x — Зху +у =п
has a solution in integers χ, у, then it has at least three such solutions. Show that
the equation has no solutions in integers for η = 2891.
1982/5. The diagonals AC and CE of the regular hexagon ABCDEF are
divided by inner points Μ and N respectively, so that:
AM CN
(1) ^с=Ш=г-
Determine r, if Β, Μ and N are collinear.
1982/6. Let 5 be a square with sides length 100. Let L be a path within S which
does not meet itself and which is composed of line segments AqA\, A\A2, ■..,
Αη_λΑη, where A0^An.
Suppose that for every point Ρ on the boundary of S there is a point of L
of distance from Ρ no greater than 1/2. Prove that there are two points X and
Υ of L such that the distance between X and Υ is not greater than 1 and the
length of the part of L which lies between X and Υ is not smaller than 198.
1983
1983/1. Find all functions / defined on the set of positive reals which take
positive real values and satisfy:
(i) f(xf(y))=yf(x)
for every positive χ and y. Show that
(2) /(ж)->0, if z-^oo.
1983/2. Let A be one of the two distinct points of intersection of two unequal
coplanar circles C\ and C^ with centres Οχ and O^ respectively. One of the
common tangents to the circles touches C\ at P\, C^ at Pi while the other
touches C\ at Qi and C^ at Qi- Let M\ be the midpoint of P\Q\ and M2 be the
midpoint of P2Q2· Prove that Ю\А02 = ΙΜλΑΜ2.
1983/3. Let a, b, с denote pairwise coprime positive integers. Prove that
2abc — ab — be — ca
is the largest integer which cannot be expressed as
(1) xbc + yca + zab,
where x, y, ζ are non negative integers.
27

Problems
1983/4. Let ABC be an equilateral triangle and Ε the set of all points contained
in the three segments AB, ВС and С A (including А, В and C). Determine
whether, for every partition of Ε into two disjoint subsets, at least one of the
two subsets contains the vertices of a right-angled triangle.
1983/5. Is it possible to choose 1983 distinct positive integers, all less than or
equal
sion?
equal to 10 , no three of which are consecutive terms of an arithmetic progres
1983/6. Let a, b, с be the length of the sides of a triangle. Prove that
(1) a2b(a - b) + b2c(b -c) + c2a(c - a) > 0.
1984
1984/1. Prove that
7
(1) 0 < xy + у ζ + zx — 2xyz < —,
where x, у and ζ are nonnegative real numbers for which
(2) x + y + z = l.
1984/2. Find one pair of positive integers, a, b, such that:
(1) ab(a + b) is not divisible by 7;
(2) (a + b)7 -a1 -b1 is divisible by 77.
1984/3. In the plane two different points О and A are given. For each point
X-=fO on the plane denote by ω(Χ) the measure of the angle between О A and
OX in radians counterclockwise from О A (0 < ω(Χ) < 2π). Let C(X) be the
ω(Χ)
circle with centre О and radius OX + . Each point of the plane is coloured
by one of a finite number of colours. Prove that there exists a point Υ for which
ω(Υ) > 0 such that its colour appears on the circumference of the circle C(Y).
1984/4. Let ABCD be a convex quadrilateral with the line CD tangent to the
circle on diameter AB. Prove that the line AB is tangent to the circle on diameter
CD if and only if Б С and AD are parallel.
1984/5. Let d be the sum of the lengths of all the diagonals of a plane convex
polygon with η > 3 vertices. Let ρ be its perimeter. Prove that:
2<i ΓηΊ Γη + 11
(1) n-3< —< - — 2.
Ρ L2.
28

1985
1984/6. Let a, b, c, d be odd numbers, such that:
(1) 0<a<b<c<d,
(2) ad = be,
(3) a + d = 2k, b + c = 2m
holds, where к and m are integers. Prove that a= 1.
1985
1985/1. A circle has centre on the side AB of the cyclic quadrilateral ABCD.
The other three sides are tangents to the circle. Prove that
AD + BC = AB.
1985/2. Let η and к be relatively prime positive integers with к <n. Each
number in the set Μ = {1,2,3,..., η — 1} is coloured either blue or white, such that:
(a) for each i in M, both i and n — i have the same colour;
(b) for each г in Μ not equal to k, both i and \i — k\ have the same colour.
Prove that all numbers in Μ must have the same colour.
1985/3. For any polynomial P(x) = aQ + a\x + ...+ а^хк with integer coefficients
let ω(Ρ) denote the number of odd coefficients, and let Qi(x) = (l +х)г, where
г = 0, 1,2, ....
Prove that if i\,i2, ·.., in are integers such that 0 < i\ < %i < ... < in, then
(1) ^(Qi1+Qi2 + --- + Qin)>^(Qii)·
1985/4. Given a set Μ of 1985 distinct positive integers, none of which has a
prime divisor greater than 26. Prove that Μ contains a subset of 4 elements
whose product is the 4th power of an integer.
1985/5. The circle &i with centre О passes through the vertices A and С of the
triangle ABC and intersects the segments AB and ВС again at distinct points
К and N, respectively. The circumcircles к of ABC and ki of KBN intersect
at exactly two distinct points В and M. Prove that Ю\МВ = 90°.
1985/6. For every real number x\ construct the sequence x\, X2, · · ·, Xn, where
V Π J
for every η > 1. Prove that there exists exactly one value of x\ which gives
0 < xn < xn+i < 1
for every positive n.
29

Problems
1986
1986/1. Let d be any positive integer not equal to 2, 5 or 13. Show that one can
find distinct a, b in the set {2,5,13, d} such that ab — 1 is not a perfect square.
1986/2. Given a point Ρ in the plane of the A1A2A3 triangle. Define As = As_^
for s > 4.
Construct a series of points Pq, P\, P2, ... such that Pk+\ is the image of
Pk under a rotation with centre Ak+\ through an angle —120° (k = 0, 1,2, ...).
Prove that if Ρ\98β = Ρο,tnen tne triangle A1A2A2 is equilateral.
1986/3. To each vertex of a regular pentagon an integer is assigned, so that the
sum of all five numbers is positive. If three consecutive vertices are assigned the
numbers x, y, ζ respectively, and у < 0, then the following operation is allowed:
x, y, ζ are replaced by x + y, —y, z + y respectively. Such an operation is
performed repeatedly as long as at least one of the five numbers is negative.
Determine whether this procedure necessarily comes to an end after a finite number
of steps.
1986/4. Let А, В be adjacent vertices of a regular n-gon (n > 5) with centre O.
A triangle XYZ, which is congruent to and initially coincides with OAB, moves
in the plane in such a way that Υ and Ζ each trace out the whole boundary of
the polygon, with X remaining inside the polygon. Find the locus of X.
1986/5. Find all functions / defined on the non-negative reals and taking non-
negative real values such that:
(a) f(x · f(y)) · f(y) = f(x + y) for every non negative χ and у;
(b)/(2) = 0;
(c) /O)^0, if 0<x<2.
1986/6. Given a finite set of points in the plane, each with integer coordinates. Is
it always possible to colour the points red or white so that for any straight line L
parallel to one of the coordinate axes the difference (in absolute value) between
the numbers of white and red points on L is not greater than 1?
1987
1987/1. Let pn(k) be the number of permutations of the set 5 = {1, 2, ..., n}
(n > 1) which have exactly к fixed points. Prove that the sum from к = 0 to η of
kpn(k) is n\.
30

1988
1987/2. In an acute-angled triangle ABC the interior bisector of angle A meets
ВС at L and meets the circumcircle of ABC again at N. From L
perpendiculars are drawn to AB and AC, with feet К and Μ respectively. Prove that the
quadrilateral AKNM and the triangle ABC have equal areas.
1987/3. Let x\, x2, ..., xn be real numbers satisfying
(1) x2+x2, + ...+x2l = l.
Prove that for every integer к > 2 there are integers щ (ί= 1, 2, ..., η), not all
zero, such that |a^| < к — 1 for all i, and
ι , .(fe-l)v^
(2) |α1χ1+α2^2 + ··· + αη^η1 < τη_, ·
1987/4. Prove that there is no function / from the set of non-negative integers
into itself such that
(1) f(f(n)) = n+1987
for all n.
1987/5. Let η be an integer greater or equal to 3. Prove that there is a set of
η points in the plane such that the distance between any two points is irrational
and each set of 3 points determines a non-degenerate triangle with rational area.
1987/6. Let η be an integer greater or equal to 2. Prove that if к + к + η is prime
/71 /л
—, then кг + k + n is prime for all integers к
such that 0 < к < η - 2.
1988
1988/1. Consider two coplanar circles of radii R and r (R>r) with the same
centre. Let Ρ be a fixed point on the smaller circle and В a variable point on the
larger circle. The line BP meets the larger circle again at C. The perpendicular
I to BP at Ρ meets the smaller circle again at A (if / is tangent to the circle at
Ρ then ,4 = P).
(i) Find the set of values of ВС2 + С A2 + AB2.
(ii) Find the locus of the midpoint of AB.
1988/2. Let η be a positive integer and let A\, A2, ..., A2n+\ be subsets of a
set B. Suppose that
a) each Ai has exactly 2n elements,
b) each AiDAj (1 < i < j < 2n+ 1) contains exactly one element, and
c) every element of В belongs to at least two of the A{.
31

Problems
For which values of η can one assign to every element of В one of the
numbers 0 and 1 in such a way that each Ai has 0 assigned to exactly η of its
elements?
1988/3. A function / is defined on the positive integers by
(1) /(1) = 1, /(3) = 3,
(2) f(2n) = n,
(3) f(4n+l) = 2f(2n+l)-f(n),
(4) /(4n + 3) = 3/(2n + l)-2/(n)
for all positive integers n.
Determine the number of positive integers n, less than or equal to 1988, for
which f(n) = n.
1988/4. Show that the set of real numbers χ which satisfy the inequality
70 к 5
(i) y_^.>£
is a union of disjoint intervals, the sum of whose lengths is 1988.
1988/5. ABC is a triangle right-angled at A, and D is the foot of the altitude
from A. The straight line joining the incentres of the triangles ABD, ACD
intersects the sides AB, AG at the points K, L respectively. S and Τ denote the
areas of the triangles ABC and AKL respectively. Show that S > 2Γ.
1988/6. Let a and b be positive integers such that ab+\ divides a +b . Show
that
a2 + b2
(1)
is the square of an integer.
ab+l
1989
1989/1. Prove that the set {1,2,..., 1989} can be expressed as the disjoint union
of subsets A\, A2, · · ·, ^117 in such a way that each Ai contains 17 elements
and the sum of the elements in each Ai is the same.
1989/2. In an acute-angled triangle ABC, the internal bisector of angle A meets
the circumcircle again at A\. Points Βγ and C\ are defined similarly. Let Aq be
the point of intersection of the line AA\ with the external bisectors of angles В
and С Points Bq and Co are defined similarly. Prove that the area of the triangle
AqBqCq is twice the area of the hexagon AC\BA\CB\ and at least four times
the area of the triangle ABC.
32

1990
1989/3. Let η and к be positive integers and let 5 be a set of η points in the
plane such that no three points of S are collinear, and for any point Ρ of S there
are at least к points of S equidistant from P. Prove that
к < - + V2n.
2
1989/4. Let ABCD a convex quadrilateral such that the sides AB, ВС, AD
satisfy AB = AD + ВС. There exists a point Ρ inside the quadrilateral at a distance
h from the line CD such that AP = h + AD and BP = h + BC. Show that
Vh Vad л/вс'
1989/5. Prove that for each positive integer η there exist η consecutive positive
integers none of which is a prime or a prime power.
1989/6. A permutation x\, x2, ..., X2n-l> x2n °f tne set 1, 2, ..., 2n — 1, 2n
where η is a positive integer is said to have property Ρ if \xi — xi+\ \ = η for at
least one i in {1, 2, ..., In — 1}. Show that for each η there are more
permutations with property Ρ than without.
1990
1990/1. Chords AB and CD of a circle intersect at a point Ε inside the circle.
Let Μ be an interior point of the segment EB. The tangents to the circle through
EG
D, Ε and Μ intersects the lines ВС and AC at F and G respectively. Find ——
EF
r AM
in terms of t = —гт7·
AB
1990/2. Take η > 3 and consider a set Ε of In — 1 distinct points on a circle.
Suppose that exactly к of these points are to be coloured black. Such a colouring
is "good" if there is at least one pair of black points such that the interior of one
of the arcs between them contains exactly η points from E. Find the smallest
value of к so that every such colouring of к points of Ε is good.
2n + l
1990/3. Determine all integers greater than 1 such that —^— is an integer.
nz
1990/4. Construct a function from the set of positive rational numbers into itself
such that
(i) №·№) = —
у
for all x, y.
33

Problems
1990/5. Given an initial integer no > 1, two players A and В choose integers щ,
П2, щ, ... alternately according to the following rules: knowing п^ъ A chooses
any integer ri2k+i sucn tnat
Knowing ri2k+i, В chooses any integer ri2k+2 sucn tnat
n2fc+l r
=p
n2k+2
for some prime ρ and integer 1 < r. Player A wins the game by choosing the
number 1990; player В wins by choosing number 1. For which no does
a) A have a winning strategy?
b) В have a winning strategy?
c) Neither player have a winning strategy?
1990/6. Prove that there exists a convex 1990-gon such that all its angles are
equal and the lengths of the sides are the numbers l2, 22, ..., 19892, 19902 in
some order.
1991
1991/1. Given a triangle ABC, let I be the incentre. The internal bisectors of
angles А, В and С meet the opposite sides at А', В', С respectively. Prove that
1 AI-BI-CI 8
(1) -< <—.
v 4 AA1 · BB' · CO ~ 27
1991/2. Let η > 6 be an integer and let a\, a^, ..., Ofc be all the positive integers
less than η and relatively prime to n. If
α2 — α1 - a3 ~ a2 = ■ · ■ = ak — ak-l > 0,
prove that η must be either a prime number or a power of 2.
1991/3. Let S = {1, 2, 3, ..., 280}. Find the smallest integer η such that each
η-element subset of S contains five numbers which are pairwise relatively prime.
1991/4. Suppose G is a connected graph with к edges. Prove that it is possible
to label the edges 1, 2, 3, ..., к in such a way that at each vertex which belongs
to two or more edges, the greatest common divisor of the integers labelling those
edges is 1.
[A graph is a set of points, called vertices, together with a set of edges
joining certain pairs of distinct vertices. Each pair of points belongs to at most
one edge. The graph is connected if for each pair of distinct vertices x, у there
34

1992
is some sequence of vertices x = vq, v\, ..., vm = y, such that each pair Vi, vi+\
(0 < г < т) is joined by an edge.]
1991/5. Let ABC be a triangle and X an interior point of ABC. Show that at
least one of the angles XAB, XBC, PC A is less than or equal to 30°.
1991/6. Given any real number a > 1 construct a bounded infinite sequence xq,
x\, ... such that
(1) \xi-Xj\-\i-j\a>\
for every pair of distinct i, j. [An infinite sequence xq, x\, ... of real numbers
is bounded if there is a constant С such that \xi\ < С for all г.]
1992
1992/1. Find all integers a, b, с satisfying 1 < a < b < с such that (a — 1) · (b — 1) ·
(c — 1) is a divisor of abc — 1.
1992/2. Find all functions / defined on the set of all real numbers with real
values, such that
(i) f(x2 + f(y)) = y+(f(x)f
for all x, y.
1992/3. Consider 9 points in space, no 4 coplanar. Each pair of points is joined
by a line segment which is coloured either blue or red or left uncoloured. Find
the smallest value of η such that whenever exactly η edges are coloured, the
set of coloured edges necessarily contains a triangle all of whose edges have the
same colour.
1992/4. L is tangent to the circle С and Μ is a point on L. Find the locus of all
points Ρ such that there exist points Q and R on L equidistant from Μ with С
the incircle of the triangle PQR.
1992/5. Let 5 be a finite set of points in three-dimensional space. Let Sx, Sy,
Sz be the sets consisting of the orthogonal projections of the points of S onto
the yz-plane, xz-plane, ху-р\зпе respectively. Prove that
\s\2<\sx\\sy\\sz\,
where \A\ denotes the number of points in the set A.
1992/6. For each positive integer n, S(n) is defined as the greatest integer such
that for every positive integer к < S(n), n2 can be written as the sum of к positive
squares.
35

Problems
(a) Prove that S(n) < η — 14 for each η > 4.
(b) Find an integer η such that S(n) = η — 14.
(c) Prove that there are infinitely many integers η such that S(n) = η — 14.
1993
1993/1. Let f(x) = xn + 5xn +3, where n>l is an integer. Prove that f(x)
cannot be expressed as the product of two non-constant polynomials with integer
coefficients.
1993/2. Let D be a point inside the acute-angled triangle ABC such that
LABD = 90° +1 AC В and AC-BD = AD- ВС.
AB-CD
(a) Calculate the ratio —————.
w AC-BD
(b) Prove that the tangents at С to the circumcircles of ACD and BCD are
perpendicular.
о
1993/3. On an infinite chessboard a game is played as follows. At the start η
pieces are arranged in an η χ η block of adjoining squares, one piece on each
square. A move in the game is a jump in a horizontal or vertical direction over
an adjacent occupied square to an unoccupied square immediately beyond. The
piece which has been jumped over is removed. Find those values of η for which
the game can end with only one piece remaining on the board.
1993/4. For three points P, Q, R in the plane define m(PQR) as the minimum
length of the three altitudes of the triangle PQR (or zero if the points are colli-
near). Prove that for any points A, B, C, X
m(ABC) < m(ABX) + m(AXC) + m(XBC).
1993/5. Does there exist a function / from the positive integers to the positive
integers such that /(1) = 2, /(/(n)) = /(n) + n for all n, and /(n)</(n+l) for
all n?
1993/6. There are n> 1 lamps L0, L\t ..., Ln_\ in a circle. We use Ln+k to
mean L&. A lamp is at all times either on or off. Perform steps sq, s\,. .. as
follows: at step s^, if Z^-l ls nt> tnen switch Li from on to off or vice versa,
otherwise do nothing. Show that:
(a) There is a positive integer M(n) such that after M(n) steps all the lamps
are on again;
(b) If n = 2k, then we can take M(ri) = n2 - 1.
(c) If η = 2k + 1 then we can take M(n) = n2 - η +1.
36

1995
1994
1994/1. Let m and η be positive integers. Let a\, ai, ■ ■,, am be distinct elements
of {1, 2, ..., n} such that whenever a^ + clj <n for some г, j (possibly the same)
we have щ + clj = α& for some k. Prove that
a\ + a,2 + · · · + am η + 1
m "~ 2
1994/2. ABC is an isosceles triangle with AB = AC. Μ is the midpoint of ВС
and О is the point on the line AM such that OB is perpendicular to AB. Q is
an arbitrary point on ВС different from В and C. Ε lies on the line AB and F
lies on the line AC such that E, Q and F are distinct and collinear. Prove that
OQ is perpendicular to EF if and only if QE = QF.
1994/3. For any positive integer k, let f(k) be the number of elements in the set
Ak = {к + l, к + 2, ..., 2k] which have exactly three Is when written in base 2.
Prove that for each positive integer m, there is at least one к with f(k) = m and
determine all m for which there is exactly one k.
1994/4. Determine all ordered pairs (m, n) of positive integers for which
mn — 1
is an integer.
1994/5. Let S be the set of all real numbers greater than (—1). Find all functions
/ from 5 to 5 such that f(x + f(y) + xf(y)) = y + f(x) + yf(x) for all χ and y,
f(x)
is strictly increasing on each of the intervals — 1 < χ < 0 and 0 < x.
χ
1994/6. Show that there exists a set A of positive integers with the following
property: for any infinite set of primes, there exist two positive integers mini
and η not in A, each of which is a product of к distinct elements of S for some
k>2.
1995
1995/1. Let А, В, С, D be four distinct points on a line, in that order. The
circles with diameter AC and BD intersect at X and Y. Let Ρ be a point on
the line XY other than Z. The line CP intersects the circle with diameter AC
at С and M, and the line BP intersects the circle with diameter BD at В and
N. Prove that the lines AM, DN, XY are concurrent.
37

Problems
1995/2. Let a, b, с be positive real numbers with abc = 1. Prove that
1 1 13
-^ + ^5 + -^3 >~·
а5(Ъ + с) Ъ5(с + а) c0(a + b) 2
1995/3. Determine all integers η > 3 for which there exist η points A\, A^ · ·.,
An in the plane, no three collinear and real numbers r\, Г2, · · ·, rn such that for
any distinct i,j, k, the area of the triangle AiAjAk is ri+rj+rk-
1995/4. Find the maximum value of xq for which there exists a sequence xq, χγ,
..., χ 1995 of positive reals with xq = £1995 such that for г = 1, 2, ..., 1995
(i) x0 = xm5;
2 1
(n) Xi-i+ =2xi + —
1995/5. Let ABCDEF be a convex hexagon with AB = BC = CD and DE =
EF = FA such that IBCD = IEFA = 60°. Suppose that G and Η are points in
the interior of the hexagon such that LAGB- LDHΕ =120°. Prove that
(1) AG + GB + GH + DH + HE>CF.
1995/6. Let ρ be an odd prime number. How many p-element subsets A of
{1, 2, ..., 2p} are there, the sum of whose elements is divisible by p?
1996
1996/1. We are given a positive integer r and a rectangular board divided into
20 χ 12 unit squares. The following moves are permitted on the board: one can
move from one square to another only if the distance between the centres of the
two squares is y/r. The task is to find a sequence of moves leading between two
adjacent corners of the board which lie on the long side.
(a) Show that the task cannot be done if r is divisible by 2 or 3.
(b) Prove that the task is possible for r = 73.
(c) Can the task be done for r = 97?
1996/2. Let Ρ be a point inside the triangle ABC such that
(1) LAPB - LAC В = LAPC - I ABC.
Let D, Ε be the incentres of triangles APB, APC respectively. Show that AP,
BD and СЕ meet at a point.
1996/3. Let S be the set of non-negative integers. Find all functions / from S
to itself such that
(1) f(m + f(n)) = f(f(m)) + f(n)
for all m, n.
38

1997
1996/4. The positive integers α and b are such that 15a +166 and 16a— 156 are
both squares of positive integers. What is the least possible value that can be
taken by the smaller of these two squares?
1996/5. Let ABCDEF be a convex hexagon of perimeter ρ such that AB is
parallel to DE, ВС is parallel to EF and CD is parallel to FA. Let RA, Rc,
and Re denote the circumradii of triangles FAB, BCD, DEF respectively.
Prove that
(1) Ra + Rc+Re>\.
1996/6. Let p, q, η be positive integers with p + q<n. Let x$, x\, ..., xn be
integers such that xq = xn = 0, and for each 1 < i < n, Xi — x{_\ =p or —q. Show
that there exist indices i < j with (i, j)^(0, ri) such that xi = xj.
1997
1997/1. In the plane the points with integer coordinates are the vertices of unit
squares. The squares are coloured alternately black and white as on a chessboard.
For any pair of positive integers m and n, consider a right-angled triangle whose
vertices have integer coordinates and whose legs, of lengths m and n, lie along
the edges of the squares. Let S\ be the total area of the black part of the triangle,
and 52 be the total area of the white part. Let
f(m, n)=\Si -S2\.
(a) Calculate f(m, n) for all positive integers which are either both even or
both odd.
(b) Prove that f(m, ri) < - max(m, ri) for all m, n.
(c) Show that there is no constant С such that f(m, n)<C for all (m, ri).
1997/2. The angle at A is the smallest angle in the triangle ABC. The points В
and С divide the circumcircle of the triangle into two arcs. Let U be an interior
point of the arc between В and С which does not contain A. The perpendicular
bisectors of AB and AC meet the line AU at V and W, respectively. The lines
BV and CW meet at T. Show that
AU = TB + TC.
1997/3. Let be real numbers satisfying
\χ\ +Χ2 + ■ · -+xn\ = 1
and
\Xi\<—Z- (г=1; 2> ·■■> n)·
39

Problems
Show that there exists a permutation yi of X{ such that
\y\ +2y2 + 3y3 +... + nyn\ < -γ-.
1997/4. An nxn matrix whose entries come from the set 5 = {1, 2, ...
..., 2n — 1} is called silver matrix if, for each i = 1, 2, ..., n, the ith row and
the ith column together contain all elements of S. Show that:
(a) there is no silver matrix for n= 1997;
(b) silver matrices exist for infinitely many values of n.
199775. Find all pairs (a, b) of positive integers that satisfy
,2
-ha
(1) a° =b
1997/6. For each positive integer n, let / (n) denote the number of ways of
representing η as a sum of powers of 2 with non-negative integer exponents.
Representations which differ only in the ordering of their summands are considered
to be the same. For example, /(4) = 4, because 4 can be represented as
4; 2 + 2; 2+1 + 1; 1 + 1 + 1 + 1.
Prove that for any integer η > 3
η2 η2
2~</(2n)<2~.
1998
1998/1. In the convex quadrilateral ABCD, the diagonals AC and BD are
perpendicular and the opposite sides AB and CD are not parallel. The point P,
where the perpendicular bisectors of AB and CD meet, is inside ABCD. Prove
that ABCD is cyclic if and only if the triangles ABP and С DP have equal
areas.
1998/2. In a competition there are a contestants and b judges, where b > 3 is an
odd integer. Each judge rates each contestant as either "pass" or "fail". Suppose
к is a number such that for any two judges their ratings coincide for at most к
contestants. Prove
к b-l
a ~ lb
1998/3. For any positive integer n, let d(n) denote the number of positive
divisors of η (including 1 and n). Determine all positive integers к such that
d(n2)
d(ri)
for some n.
= k
40

1999
1998/4. Determine all pairs (a, b) of positive integers such that (ab +b + l)
divides (a2b + a + b).
1998/5. Let I be the incentre of the triangle ABC. Let the incircle of ABC
touch the sides ВС, С A, AB at K, L, M, respectively. The line through В
parallel to MK meets the lines LM and LK at R and S, respectively. Prove
that the angle RIS is acute.
1998/6. Consider all functions / from the set of all positive integers into itself
satisfying
(1) f(t2f(s))=s(f(t))2
for all s and t. Determine the least possible value of /(1998).
1999
1999/1. Find all finite sets S of at least three points in the plane such that for
all distinct points А, В in S, the perpendicular bisector of AB is an axis of
symmetry for S.
1999/2. Let η > 2 be a fixed integer. Find the smallest constant С such that for
all non-negative reals x\, · · ·, xn:
(1) Σ ХгХ](х} + х])<с( Σ χλ ■
\<i<j<n М<г<п '
Determine when equality occurs.
1999/3. Given an η χ η square board with η even. Two distinct squares of the
board are said to be adjacent if they share a common side, but a square is not
adjacent to itself. Find the minimum number of squares that can be marked so
that every square (marked or not) is adjacent to at least one marked square.
1999/4. Find all pairs (n,p) of positive integers, such that ρ is a prime, n<2p
and (p — l)n + 1 is divisible by np~ .
1999/5. The circles C\ and Ci lie inside the circle C, and are tangent to it at Μ
and N, respectively. C\ passes through the centre of C^· The common chord of
C\ and C2, when extended, meets С at A and B. The lines MA and MB meet
C\ again at Ε and F. Prove that the line EF is tangent to C^·
1999/6. Determine all functions / : R —»■ R such that
(1) fix- f(y)) = f (f(y)) + xf(y) + fix) -1
for all x, у in R. [R is the set of reals.]
41

I

Solutions
1959.
21n + 4
1959/1. Prove that the expression is irreducible for every positive
14n + 3
integer n.
First solution. If d divides two integers, then it divides their multiples,
their sum and difference. Since
3(14n + 3)-2(21n + 4)=l,
a common divisor of the numerator and the denominator also divides 1; for this
reason the fraction cannot be simplified.
а с
Second solution. Notice that if - = e + - (a, b, с, e integers) then c = a — be
b b
and a = c + be, consequently the common divisors of a and b also divide c; and
similarly the common divisors of с and b divide a. For this reason - can be
simplified if and only if - can be simplified; and the same holds for - and -.
b b a
Since
21n + 4 Λ 7η+1
= 1 +
14n + 3 14n + 3
and
14n + 3 л 1
= 2 +
7n+l 7n+l'
and the last term cannot be simplified, the same is true for our original expression.
1959/2. Determine the real solutions of the following equations.
a) yjx + \/2x — 1 +y x — y/2x — 1 = v2,
b) yjχ + y/2x -l+yjx- y/2x -1 = 1,
c) γ χ + \j2x — 1 + γ χ — \j2x — 1 = 2.
Solution. First of all notice that
x + V2x-l = - (l + л/2ж — l) and χ - y/2x - 1 = - (l - \j2x - l) .
(Here we have to assume χ > -.) Let В denote the expression on the left hand
side of the equation:
B = ^(\l + V2x~^i\ + \l-V2x-l^
43

Solutions
1959/2
Since for χ > — we have 1 + \j2x — 1 > 0, the absolute value sign can be omitted.
The second expression is nonnegative if
(1) l>V2aT-T.
Since both sides of (1) are nonnegative, this inequality is equivalent to
1 > 2x — 1, i.e. χ <l.
According to this,
1 - V2x - 1 > 0 if x<\, and 1 - y/2x -КО if χ > 1.
Consequently, if - < χ < 1 then
(2) Б = — (l + V2x -1 + 1 - >/2z-l) = л/2,
and for χ > 1
(3) Б = — (l + V2x - 1 + y/2x - 1 - l) = л/4ж - 2.
Based on these observations
a) is satisfied if - < χ < 1;
b) holds if \/Ax — 2- 1, i.e. x- -; consequently b) does not admit any solution;
c) holds if \/4x — 2 = 2, in other words x = -; this satisfies assumption (3) hen-
3
сеж = - solves c).
2
1959/3. Suppose that χ satisfies
(1) acos2 x + bcosx + c = 0.
Form a quadratic equation in cos 2x whose roots are the same values of x. Apply
your result for the special case a = 4, b = 2, c= —1.
_ _ . _. о l+cos2a; , a l+2cos2a;+cos22a; ,
Solution. Since cos χ = , and so cos χ = , by
ordering and squaring (1) we can achieve that in the resulting expression only
cos χ and cos χ appear:
b cos χ = —a cos χ — c,
b2 cos2 χ = a2 cos χ + 2ac cos2 x + c2.
9 A.
Substitute cos χ and cos x:
о 1 + 2 cos 2x + cos 2x о 1+cos 2a; о
α2 + (2ac - b1) + c2 = 0.
4 2
44

1959/4
1959.
As a corollary we obtain
(2) a2 cos2 2x + (2a2 + Aac - 2b2) cos 2x + (a2 + Aac - 2b2 + Ac2) = 0;
this is exactly the equation we were aiming for. With the given substitutions
a = 4, b = 2, с = — 1 we get:
4 cos2 χ + 2 cos χ — 1 = 0,
16 cos2 2x + 8 cos 2x - 4 = 0,
or in a simpler form:
(3) 4 cos2 2ж+ 2 cos 2a;-1=0.
Remark. Notice that cos a; and cos 2a; are roots of the same
quadratic equation. The reason for this is that from (3) we derive that
- lib л/5
cosa; = , hence χ = 12° or 144°, consequently 2a; =144° and 288°, but
cos 288° = cos(360° - 288°) = cos 72°.
1959/4. Construct a right triangle with a given hypotenuse, if we know that
the median corresponding to the hypotenuse is equal to the geometric mean of
the two adjacent sides.
Solution. According to the theorem of Thales, the median (corresponding
to the hypotenuse) is equal to half of the hypotenuse, which is equal to the radius
of the circumcircle of the triangle. Denote the two adjacent sides by a and b, the
hypotenuse by 2R and the median by R. According to the assumptions we have
(1) ab = R2.
The altitude corresponding to the hypotenuse will be denoted by m. Expressing
twice the area of the right triangle in two different ways we get
(2) R2 = ab = 2R
m,
implying
R
m= —.
2
Now the construction goes as follows (see Figure 1959/4.1): considering the
Thales circle over the hypotenuse AB = 2R, we intersect this with the line paral-
lei to the hypotenuse of distance —. Any intersection of the circle and the line
will provide a candidate for the vertex С (the four intersections obviously give
four congruent triangles). The triangle we got by this construction satisfies the
D
assumption of the problem since (1) and (2) follow from m= —.
45

Solutions
1959/4
Figure 59/4.1
Remark. Notice that the angle /.COT in the right triangle ОТ С (see Figure
1959/4.1) is equal to 30°, consequently the acute angles of ABC are equal to
15° and 75°. Based on this observation other constructions are also possible.
1959/5. A point Μ is moving on the interval AB. Squares AMCD and
BMEF over the subintervals AM and MB (on the same side of the line AB)
are constructed The circumcircles of these squares intersect each other in the
points Μ and N. Verify that the lines AE and ВС pass through N. Show that
for any choice of Μ the line MN passes through a fixed point. Find the locus
of the midpoints of the intervals joining the centres of the two squares.
Solution. If Μ is the midpoint of AB then C = E = N, hence the first
statement of the problem is obvious: MN = f is the perpendicular bisector of
AB, hence the common point of the lines MN (if such a point exists) should lie
on /. Suppose now that AM 4MB.
First we will show that С is the orthocentre of ABE (Figure 1959/5.1). In
any case, EM is an altitude, and AC is orthogonal to BE since the diagonals
of the two squares are parallel. Consequently С is the intersection of two
altitudes, hence is the orthocentre. Now it follows that ВС is also on the altitude, so
ВС is orthogonal to AE. Let N' denote the intersection of AE and ВС. Since
both diagonals AC and BE form right triangles with N', this point is on both
the Thales circles corresponding to the diagonals, consequently N' is on the cir-
46

1959/5
1959.
Figure 59/5.1
cumcircles of the squares. This, however, means that N1 is equal to the common
point of the two circles: N' = N, hence AE and ВС in fact pass through N.
According to the famous connection between the angles at circumference
and the central angles, the angle LCNM is 45°, hence NM is the bisector of
the right angle of the triangle ANB. The circumcircle of AN В (denoted by k)
coincides with the Thales circle of the interval AB, hence it is independent of
the choice of M. Since the bisector at N intersects к in the midpoint Ρ of the
arc AB (not containing Ν), Ρ is independent of the choice of.M, hence all lines
MN pass through P.
Let Q denote the point of intersection of AC and BE. Since Q is the third
vertex of the isosceles right triangle ABQ constructed over AB, it is independent
of the choice of M. If the centres of the squares are denoted by K\ and K^ then
MKiQK\ is a rectangle. Hence the midpoint F of the diagonal K\Ki is also
the midpoint of QM, so F is on the median XY of ABQ parallel to AB.
Consequently the locus we are looking for is the open interval XY: for
arbitrary point F in XY there is a unique point Μ in AB (the point specified
47

Solutions 1959/5
by twice the vector QF), and according to the above said the midpoint of the
interval joining the centres of the squares determined by Μ is exactly F.
1959/6. The planes Ρ and Q intersect in the line p. A and С are points
of the planes Ρ and Q respectively, neither of them is on p. Construct that
symmetric trapezium ABCD (with AB || CD) for which its vertices В and D are
on the planes Ρ and Q and ABCD admits an incircle.
Solution. Since AB and CD are parallel lines in two distinct, intersecting
planes, they are necessarily parallel to p. The parallels to ρ through Л in Ρ (and
through С in Q) will be denoted by e\ and e2 respectively; the construction will
be performed in the plane determined by e\ and e2 (Figure 1959/6.1). Let d
Figure 59/6.1
denote the distance of e\ and e2, and start with the trapezium constructed on
Figure 1959/6.2. The orthogonal projection of С on e\ will be denoted by C'.
According to the theorem on the tangent intervals to a circle, AC' = AD follows.
Based on this information the trapezium can be constructed in the following
manner:
The circle of centre A and radius AC' intersects e2 in D. Reflecting AD
to the perpendicular bisector t of the interval CD we get ВС. Now ABCD
satisfies the requirements, since (according to its construction) it is a symmetric
trapezium, and since the sum of opposite sides are equal, it admits an incircle.
In case AC' < d, the problem has no solution. If AC' = d, there is a
unique solution and in this case the trapezium turns out to be a square. Finally, if
48

1960/1
1960.
Figure 59/6.2
AC' > d, the circle with centre A and radius AC1 intersects e^ ш two Pomts,
consequently in this case the problem has two solutions (see the trapezium ABC Ό
on the figure).
1960.
1960/1. Determine all three digit numbers which are equal to 11 times the
sum of the squares of their digits.
Solution. Denote the digits by a, b and с (α^Ο). According to the
assumption
(1)
This implies that
100a + 10b + с=Ща2 + Ъ2 + с2).
(99а+1Щ + (а-Ь + с) = Ща2 + Ь2 + с2).
Now the right hand side and the first term on the left hand side is divisible by
11, hence so is a — b + с Since — 8<a — b + c<\S, we conclude that a — b + с is
equal either to 0 or to 11.
In the first case b = a + c; substituting this expression into (1) we get
Ша+Ща + с) + с=Ща2 + (а + с)2 + с2).
After ordering, the quadratic equation
(2) 2a2 + (2c - 10)a + (2c2 - c) = 0
follows. Since the first two terms of this expression are even, the third term
should be even as well. This, however, implies that с is even. Equation (2)
49

Solutions
1960/1
admits integer solutions if and only if its discriminant
4(-3c2-8c + 25)
is a square. It is not hard to check that this is possible only in case c = 0.
Substituting c = 0 in (2) we get 2α2- 10α = 0, and since a^O, we have α = 5. This
implies
b = a + c = 5,
hence we get 550 for the integer we are seeking for; this number, in fact, satisfies
the assumptions of the problem.
Let us now turn to the second case, when b = a + c— 11. After substituting
and ordering (1) we get the expression
(3) 2a2 + (2c-32)a + (2c2-23c+131) = 0.
This time it shows that с cannot be even. The discriminant admits the form
4(-3c2 + 14c-6);
for odd с it is a square only in case c = 3. Substituting this into (3),
2α2-26α + 80 = 0
follows. By solving this quadratic equation we get that a = 5 or a = 8; based on
(3) this implies b = — 3 and 6 = 0 respectively. Obviously (since b is the middle
digit of the positive integer we are searching for), only the latter figure might
serve as a solution of the original problem. This provides the second solution
803, and an easy check shows that this number satisfies the assumptions of the
problem. In conclusion, the two solutions are 550 and 803.
1960/2. Determine the real solutions of the inequality
Ax2
(\-у/\ + Щ
<2ж + 9.
Solution. The expression on the left hand side is defined if its denominator
is nonzero, i.e., if x^0; moreover if the expression under the square root is
nonnegative, implying 1+2ж>0, х> —. Multiply both the numerator and the
denominator of the fraction by И + л/1+2ж) :
Αχ1 (1 + \Л+2ж)
^ = ^<2ж + 9,
Ах1
2л/1+2я;<7,
4 + 8ж<49,
45
χ<γ.
50

1960/3
1960.
Since all the manipulations above can be reversed, the solution set of the equation
can be given by the inequalities
-\<x<0,
45
0<x<—.
о
1960/3. The hypotenuse BC = a of the right triangle ABC has been
divided into η equal intervals with η an odd integer. Let h denote the altitude
corresponding to the hypotenuse; and the central interval subtends an angle a
at A. Show that
4nh
tana:
(n2-l)a
First solution. We can assume that η > 1. The midpoint of ВС will be
denoted by F, while the two endpoints of the interval containing F are Ρ and
Q; let furthermore AP=p and AQ = q. (See Figure 1960/3.1.) According to the
a
theorem of Thales, AF = -. In the triangle APQ the interval AF is exactly the
Figure 60/3.1
median corresponding to PQ and PQ = —, hence
η
a2 l/ 2 „ι a2 .
τ=4 2i,+2"-^··
and this implies that
(1) _ '"- 2* ·
Determine twice the area of APQ in two different ways and conclude
о о (n2 + l)az
pz + qz=y
(2)
ah
pq sin a = —.
η
51

Solutions
1960/3
Apply the law of cosines for the triangle APQ; now (according to (1)) we
get
1 / 2 2 a2\ 1 i(n2 + l)a2 a2\ (n2 - \)a2
2n2 n2 / 4n2
Forming the ratio of appropriate sides of equations (2) and (3) we receive the
expression
pq sin a ah ■ An Anh
tan a = = -z = — ,
pq cos a n(nl— X)al (nl— X)a
which concludes the proof.
Second solution. A simpler solution can be given by using vector calculus.
The vectors pointing form A to the points of the hypotenuse will be denoted by
lower case bold face letters corresponding to the label of their endpoints. Since
f corresponds to the midpoint of the hypotenuse, we know that
f_b + c
2 '
consequently
„ —* b + c b-c (n + l)b + (ra —l)c
p = f+FP = —- + —— = ,
2 2n 2n
and similarly
„ —^ b + c c-b (n —l)b + (n + l)c
2 2n 2n
Since b and с are orthogonal, the above observation provides
_ {n2 - l)(b2 + c2) _ {η2 ~ \)a2
4n2 4n2
which leads to
pq sin a pq sin a ah-An Anh
tana; =
pq cos a pq n(n2-l)a2 (n2-l)a2
(by applying (2)). The proof is now complete.
1960/4. Construct the triangle ABC if its two altitudes ma and ть
(corresponding to the vertices A and B) and the median corresponding to A is given.
Solution. Suppose that the triangle is constructed. Denote the feet of ma
and тъ by Ρ and R. A' stands for the midpoint of ВС and the foot of the
orthogonal to AC from A' is Q (Figure 1960/4.1). Notice that A'Q = — (since
it is the line joining midpoints in the triangle BRC); furthermore the angles
52

1960/5
1960.
Figure 60/4.1
ΑΡΑ' and AQA1 are both right angles, consequently Ρ and Q are on the Thales
circle over AA'.
These observations already dictate the method of construction: construct the
TTLh
circle к with diameter AA' = sa; now the circles with radius — around A' and
with radius ma around A intersect к in Q and P, respectively. The point of
intersection of the lines AQ and A'P is C, while the reflection of С to A' gives B.
The resulting triangle ABC contains AA' as its median; AP = ma is its altitude
TTLh
(corresponding to A) and the altitude corresponding to В is equal to 2 = га&.
Consequently, we verified that the resulting triangle solves the problem.
The construction can be carried out once the intersections exist, i.e., if ma <
TTLh
sa, — < sa and AQ is not parallel to A'P. In general, there are two solutions
since the circles we used in the construction intersect the circle к in two points.
TTLh
In case ma = sa> —— the points A' and Ρ coincide, A'P is tangent to the
circle and the solution is a unique isosceles triangle.
TTLh
If ma <sa = — then Q = A, now AQ becomes tangent.
TTLh
If ma = sa = —, there is no solution since the two tangents are parallel.
TTLh
Finally, if sa > — = ma, there is a unique solution since for one choice of
intersections the lines AQ and A'P are parallel.
1960/5. For a given cube ABCDA'B'C'D' let X be a point of the face
diagonal AC, and Υ a point of B'D'.
a) Find the locus of the midpoints of the intervals XY for all possible
choices of X and Y.
53

Solutions
1960/5
b) Consider the point Ζ e XY satisfying the equality ZY = 2XZ and
determine the locus of these points for all choices of X and Y.
Solution. In solving the problem we will use the following result:
Suppose that α, β and 7 are three parallel planes in the three-space, and 7
is between a and β; suppose furthermore
that it divides the distance between a and
β as a: b. If X and Υ are on the planes
a and β respectively, then the locus of
those points Ζ which divide XY as a: b
(more precisely, those which satisfy XZ:
ZY = a:b) coincides with 7 (see Figure
1960/5.1).
Our first aim is to find the locus
of those points which divide XY (as
given in the problem) according to the
ratio a:b. The two faces ABCD and
A'B'C'D' of the given cube are in
parallel planes, hence (according to the result
quoted above) the points Ζ we are looking for are elements of a plane 7 parallel
to these faces and dividing their distance as a: b (Figure 1960/5.2).
Figure 60/5.1
Figure 60/5.2
54

1960/6
1960.
А, С, В' and D' can be regarded as the vertices of a tetrahedron inscribed
in our cube. Since a tetrahedron is a convex solid, any point of an interval joining
its two points is contained by its interior or boundary. Consequently the intervals
XY and so the points Ζ are in the interior or on the boundary of this tetrahedron.
In conclusion, the locus we are searching for is a subset of the intersection of
the plane 7 and the tetrahedron described above.
It is easy to see that this intersection is a rectangle. This follows from the
fact that for two intersecting planes the intersection of them with a third plane
parallel to their intersection line is a union of two parallel lines; for the planes
AB'D' and CB'D' these are parallel to B'D1, and for B'AC and D'AC to
AC. So the opposite sides of the intersection are parallel; the adjacent ones are
orthogonal since the diagonals B'D' and AC are parallel to them and they happen
to be orthogonal to each other.
In the figure the four vertices of this rectangle are denoted by P, Q, R,
S. Next we will show that every interior and boundary point of this rectangle
belongs to the locus we are searching for. For this, consider Ζ in the rectangle.
A plane is determined by the line parallel to AC passing through Ζ together
with the line AC. The intersection of this plane with the tetrahedron defines a
triangle; its vertex Υ opposite to AC is on the diagonal B'D'. The line Υ Ζ is
contained by the plane determined by the triangle ACY, therefore it intersects
AC in a point X. Since Ζ is in 7, according to the result quoted above we have
XZ: ZY = a:b, hence Ζ belongs to the locus we are looking for.
In summary, the locus asked by the problem is a rectangular section of the
tetrahedron AB'CD' with a plane parallel to the faces ABCD and A'B'C'D'
dividing their distance as a: b.
In part a) of the problem a = b, hence 7 is the mid parallel plane, the vertices
of the resulting rectangle are the midpoints of the edges AB', B'C, CD' and
D'A. In part b) the vertices of the rectangle are equal to the points dividing the
above listed edges as 1:2.
We just note here that in part a) the resulting rectangle is actually a square,
since its edges are equal to half of the diagonal of the cube, and the four vertices
of the square are the centres of four faces of the cube.
1960/6. A cone of revolution has an inscribed sphere tangent to the base
of the cone (and to the sloping surface of the cone). A cylinder is circumscribed
about the sphere so that its base lies in the base of the cone. Let V\ and V2
denote the volume of the cone and the resulting cylinder, respectively.
a) Show that V\ is not equal to V^.
b) Determine the smallest possible value of k= —, and for this minimal к
V2
construct the half angle of the symmetric cone.
55

Solutions
1960/6
First solution. Consider a plane containing the symmetry axes of the
cone, the sphere and the cylinder. Take the intersection of the configuration with
this plane. The resulting configuration consists of an isosceles triangle, its incircle
and a square circumscribed
about the circle (Figure 1960/6.1).
The base of the triangle is equal
to the diagonal 2R of the base
circle of the cone, while the
radius of its incircle equals the
radius r of the sphere.
Recall that the radius of the
incircle in a triangle is equal to
the ratio of the area and half of
the circumference of the given
triangle. Denoting the altitude
of the triangle (i.e., the cone) by
m we get
Rm
r
Figure 60/6.1
R + л/R2 + m2
This implies r + m2 = R(m — r), hence
Since V\ =
2 2
rm π
3(m —2r)
R2 =
and У~2 = 2г π,
2
rm
m — 2r
k=^=
m
V2 6(mr — 2r2) 5 (z. _ 2iL V
now the value of к is minimal if the denominator is maximal. Working with the
denominator we have
-ΐ2((Η)2-ά)=-ΐ2(Η)4;
3 4
we conclude that its maximum is -, hence the minimum of к is - and it is
4 3
r 1
attained when — = -, i.e., when m = 4r.
m 4
Based on these observations,' the angle of the cone can be constructed by
the following way: construct the two tangents to a circle of radius r from a point
which is of distance 3r from the centre of the circle.
56

1960/7
1960.
Notice that the above result already contains the solution of part a): since the
4
minimum value of fc is -, it never takes the value 1, so V\ = V2 is not possible.
Second solution. Using the notations of Figure 1960/6.1 and denoting the
angle of the apothems and the base plane by 2a we get
r = R tan a, m = R tan 2a,
hence the volumes of the cone and the cylinder are
i?37r-tan2a Tr _ _^ 3
Vi = , V2 = 2жЯ5 tan^ a,
respectively. This implies that
V\ tan 2a 2 tan a
V2 6 tan3 a 6 tan3 a{\ — tan2 a)
1 1
3(-tan4a + tan2a) _3 (tan2 α _ 1)
2 3
4
Having fc expressed in the above form we immediately see that k>- (since the
3
maximum of the denominator is -). This fact already implies that k = l, i.e.,
V\ - V2 is not possible.
9 1 л/2
For к minimal we have tan a = -, and since a is an acute angle, tan a = ——
2 tan а л j- _, . .
and tan 2a = — = 2ν 2. This implies
1 — tan2 α
tan φ = tan(90° - 2a) = cot 2a =
2>/2'
and φ can be easily to constructed from this identity.
1960/7. The parallel sides of a symmetric trapezium are of length a and b,
while its altitude is m.
a) Construct the point Ρ on the symmetry axis of the trapezium which is on
the Thales circles over the legs of the trapezium.
b) Determine the distance of Ρ from one of the parallel sides.
c) Under what assumption does such Ρ exist?
Solution. Let us denote the trapezium by ABCD; its base AB = a is
intersected by the symmetry axis in Q, while CD = b intersects the axis in S. For
a point Ρ on the axis the angles LAPD and LB Ρ С are right angles if Ρ is
on the Thales circles over the edges AD ana ВС (Figure 1960/7.1); with this
observation at hand the construction of Ρ is straightforward.
57

Solutions
1960/7
A
m
1
i
-x
' I
к /
χ 1
\ 1
D
*
n\
/ )
1
Ъ
x^
с
Q
Figure 60/7.1
В
Let PQ = x, then PS = m — x. CSP and PQB are similar right angled
triangles (since their acute angles are angles with orthogonal sides). This similarity
implies that
(m
b a
x):- = -:x,
from which it follows that
2 °Ь
x —mx + — =0,
4
The resulting two values of χ are equal to the two distances of Ρ from the
parallel sides. (Notice that their sum is exactly т.) Р exists if the expression
9 9
under the square root is nonnegative, i.e., m > ab. In case m > ab, there are
two points satisfying the requirements of the problem. These two points (as the
construction already shows) are symmetric to the line joining the midpoints of
the legs.
If m -ab then x = —, hence the Thales circle is tangent to the axis of
symmetry; it is fairly easy to see that in this case the trapezium admits an incircle.
1961.
1961/1. Solve the following system of equations for x, у and z:
(1) x + y + z = a,
(2) x2 + y2 + z2 = b2,
(3) %y = z2,
58

1961/1
1961.
where a and b are given real numbers. What conditions must a and b satisfy for
x,y and ζ to be all positive and distinct?
Solution. Subtract (2) from the square of (1) and (according to (3)) subs-
titute xy by ζ :
2xy + 2xz + 2yz = a —b,
2z2 + 2z(x + y) = a2 -b2.
Now substitute x + y with a — ζ (as shown by (1)) and get
2z2 + 2az-2z2 = a2 + b2.
a = 0 implies 6 = 0, hence from (2) we get x = y = z = 0, which — according to
our additional hypothesis — is not a solution. Consequently we may assume that
a ^0, so
2
a2~b2 2 (a2 ~ b2)2
ζ
2a ' 4a2
Now x + y and xy can be expressed in terms of a and b as follows:
a2 + b2 л 2 (a2~b2)2
(4) x + y = a — z = and xy = z= ~ .
У 2а У 4a2
This yields a quadratic equation in it:
4a2u2 - 2a(a2 + b2)u + (a2 - b2)2 = 0,
which is solved by χ and y. The solutions are
• ul,2 = —2 (2a^2 + &2)=t 74α2(β2 + b2)2 ~ 16α2(α2 - b2)A =
(5)
= -ί (a2 + b2 ± y/(3a2 - b2)(3b2 - a2)\ .
Hence the solutions of the system (under the hypothesis a^O) are:
a2-b2
х = щ, у = и2, z = —- ,.
la
a2-b2
x = u2, у = щ, z = —^>—>
la
and these expressions obviously satisfy the equations.
Let us examine what are the necessary and sufficient conditions for the
solutions to exist. (1) implies a > 0; for ζ > 0 we also need a > \b\. (4) implies that
if the solutions exist then both χ and у are positive, χ and у are distinct if the
discriminant in (5) is positive; since a> |6|, this positivity implies \b\y/3>a. In
this case neither χ nor у is equal to z: if, for example, x = z then (according to
59

Solutions
1961/1
(3)) we have у = ζ so χ = у, which is impossible. In summary, the system admits
solutions with the desired properties if
|Ь|<а<|Ь|>/3.
Remark. Let us sketch the geometric background of the problem: (1) is
simply the equation of a plane intersecting the axes in points of distance a from
the origin, consequently its distance from the origin is —=. (2) is the equation of
a sphere of radius |b| around the origin; the two objects have common points if
and only if |b| > —=, i.e., a < |6|v3. (3) is the equation of a cone with the origin
as its vertex.
1961/2. Let a, b and с be the sides of a given triangle while t is its area.
Show that
(1) a2 + b2 + c2>AtVb
When does equality hold?
First solution. Suppose that the largest angle of the triangle ABC is at C.
The foot of the altitude m at С is Г, which is an inner point of the interval AB.
Let χ denote AT {Figure 1961/2.1).
c-x Τ x A
Figure 61/2.1
9 9
Apply the area-formula cm = It and express a and b using the Pythagorean
theorem:
a2 + b2 + c2 - 4tVl = (m2 + (c - x)2) + (m2 + x2) + c2- 2\fbcm =
= 2c2 + 2m2 + 2x2 - 2cx - 2sfbcm = - [(c - 2x)2 + (сл/З - 2m)2] > 0,
providing the solution of the problem.
с сл/3
Equality holds only if χ = - and m = , meaning that the triangle is equi-
lateral.
60

1961/2
1961.
Second solution. (1) is equivalent to
(a2 + b2 + c2)2 >48t2 = 3 ■ I6t2.
To demonstrate this, apply an appropriate version of the formula of Heron:
I6t2 = -a4 -b4-c4 + 2a2b2 + 2b2c2 + 2c2a2.
With this observation we only need to verify that
a4 + b4 + c4 + 2a2b2.+ 2b2c2 + 2cV > -3a4 - 364 - 3c4 + 6a2b2 + 6b2c2 + 6c2a2,
which is equivalent to the inequality
(a2 - b2)2 + (b2 - c2)2 + (c2 - a2)2 > 0.
This latter inequality, however, obviously holds and equality holds if and only if
a = b = c.
Third solution. Let us invoke the formula of Heron again. From the
inequality between the arithmetic and geometric means we have
( ν ш ^f(s-a) + (s-b) + (s-c)Y s3
(s - a)(s - b)(s - c) < I 1 = —,
hence
(2) t = Js(s - a)(s - b)(s -c)<J —
27 Зх/3
and equality holds only for equilateral triangles. This last inequality implies
„ а- о /2s\2 ^/a + b + c\2 ^a2 + b2 + c2 2 ,2 2
W3<3^yJ =3^—j— J <3 = a2 + b2 + c2,
and again equality holds only in case a = b = c.
Fourth solution. Apply the area formula 2t = ab sin 7 and c2 = a + b2 —
2abcosj, which is a direct consequence of the law of cosines:
a2 + b2 + c2 - At V3 = 2a2 + 2b2 - 4ab I —- sin 7 + - cos 7
= 2a2 + 262 - 4ab sin(7 + 30°) >
> 2a2 + 262 - 4ab = 2(a - b)2 > 0.
Equality holds if a = b, j = 60°, i.e., when the triangle is equilateral.
Fifth solution. Our inequality can be substantially refined; for this we use
the following identity valid for positive reals x,y,z:
(3) л/3хуг(х + y + z)<xy + yz + zx.
61

Solutions
1961/2
Now start with the obvious inequality
900
0 < (xy — yz) + (yz — zx) + (zx — xy) .
After squaring and ordering we get
3x2yz + 3xy2z + 3xyz2 < x2y2 + y2z2 + z2x2 + 2x2yz + 2xy2z + 2xyz2,
3xyz(x + y + z)< (xy + yz + zx) ,
which already implies the statement we aimed to prove. Equality holds when
x = y = z.
Apply now (3) with x = s — a, y = s — b, z = s — c, x + y + z = s and simplify
with the formula of Heron expressing
tv3 = y/3(s — a)(s — b)(s - c)s <(s- a)(s -b) + (s- b)(s -c) + (s- c)(s — a)
= 3s2 - s(2a + 2b + 2c) + ab + bc + ca =
= -s2 + ab + bc + ca=-(-(a + b + c)2+ Aab + Abe + Аса) =
= Ua2 + b2 + c2 - (a - b)2 - (b-c)2 - (c-a)2).
This gives
(4) AtV3<a2 + b2 + c2-(a-b)2-(b-c)2-(c-a)2,
with equality if a = b = c. Now (1) obviously follows from (4).
Sixth solution. This last solution reveals the origin of the problem and
A'
Figure 61/2.2
62

1961/2
1961.
shows that it is obviously true for those triangles which admit an angle not less
than 120°. Let the angle of the triangle ABC at A not less than 120° and let
ВС-a. Construct the equilateral triangle А'ВС with sides a. The radius of the
a
circumcircle of this triangle is —= (Figure 1961/2.2). Put the triangle ABC to
V3
the side of ВС opposite to A'. Then A is on or inside the circumcircle of the
equilateral triangle, hence the altitude corresponding to A is at most half of the
radius, so the area of ABC satisfies
1 a a2
t < -a 7= = —=.
2 2л/3 4>/3
This implies M\fb < a2, so AtVb <a2 + b2 + c2 is valid for ABC.
In case the angles of ABC are all less than 120°, then it contains a point /
in its interior (the so-called isogonal point) from which the angles IAIB, LAIC
and IB IC are all equal to 120° {Figure 1961/2.3). According to the previous
case now the triangles IAB, IBC and
1С A have areas not more than
4>/T
a
and —■=, respectively, hence for
4л/3 4л/3!
the area of ABC we have
t<
a
+
+
4х/3 4\/3 4\/3
implying (1). Equality can hold if the
small triangles are isosceles, i.e., when
/ is the centre of the circumcircle,
meaning that ABC is equilateral.
Figure 61/2.3
Remarks. Because of the fundamental importance of (1) in trigonometry it
has relations with many other well-known inequalities. Using those, many more
alternative solutions would be possible to give, here we merely list a few relevant
connections. (1) was published by R. Weitzenbock in 1919, the refinement in (4)
is due to P. Finsler and H. Hadwiger (1937).
1. By constructing equilateral triangles over the sides a, b and с inside the
given triangle, the centres of those form an equilateral triangle with side square
-(a2 + b2 + c2
6
Aty/3).
2. For the sum of the squares of the sides we have:
a2 + b2 + c2 = 4i(cot a + cot β + cot 7).
63

Solutions 1961/2
This implies that (1) is equivalent to the cotangent inequality
cot a + cot β + cot 7 > у/3.
3. An inequality due to F. Goldner states
4i>/3 < ^3(a2b2 + b2c2 + c2a2) <a2 + b2 + c2.
4. If r, ra, гь and rc are the radii of the incircle and the excircles then (4)
is equivalent to
r(ra + rb + rc)>tV3.
5. Equation (2) in the third solution means that among triangles with a fixed
circumference 2s the equilateral triangle admits maximal area; the value of this
2 2
S S 7Г
maximum is —;= = —-cot—. The generalization of this fact also holds: among
Зх/З 3 3 ё 6
n-gons of fixed circumference 2s the regular n-gon has maximal area, and the
S 7Г
value of this maximum is equal to —cot—. Based on this, the method of the
η η
third solution shows that if the sides of an n-gon are equal to a\, a^·, ■ · ·, an and
its area is t then
at + ao +... + at > 4t cot -.
1 L n~ η
1961/3. Solve the equation
cosn χ — sinn χ = 1
where η is a positive integer.
Solution. Suppose first that n= 1. In this case our equality is
cos χ — sin χ = v2 ( —= cos χ — sin χ ) = \fl cos ( — + χ 1 = 1,
\y/2 л/2 / \4 /
/π \ s/г
cos(- + xj = -,
π
from which x = 2kit and х = (Ак + Ъ)— follows. (Here and in the following к
stands for an arbitrary integer.)
Let us consider the case η > 2 now. Rewrite the equation as
cos χ — sin χ = cos χ + sin ж,
sin2 x{\ + sinn_2 x) + cos2 ж(1 - cosn_2 x) = 0.
Factors in the products on the left hand side are all nonnegative, hence the terms
are nonnegative. Their sum can be equal to zero only if they are equal to zero:
(1) sin2£(l + sinn-220 = 0.
64

1961/4
1961.
(2) cos2£(l-cosn-2) = 0.
In case (1) there are two possibilities:
a) sina; = 0, b) sinn x = —l.
a) In this case χ = к-к. In (2) then we have cos x = \, consequently (2) holds
if and only if cosn_ x = 1. For η even this always holds, for η odd, however,
only if cos £ = 1, i.e., х = 2кж.
7Г
b) This can happen only for η odd, and then sin:r = — 1, i.e., х = (Ак + Ъ)—.
For these χ we have cos χ = 0, hence (2) always holds.
In summary we got
in case η even: x = kir,
7Γ
in case η odd: x = 2kn and x = (4k + 3)—.
1961/4. Ρ is a point inside the triangle P1P2P3. The intersections of the
lines P\P, P2P and P^P with the opposite sides are denoted by Q\, Q2 and
Q3, respectively. Show that there is one among the ratios
PXP P2P P3P
~PQ[' ~PQ~2' ~PQl
which is not less, and one which is not more than 2.
First solution. Consider the parallel to P2P3 passing through the centre of
gravity (denoted by S) of the triangle. This line intersects Ρ\Ρ2 in X\ and Ρ\Ρί,
Figure 61/4.1 Figure 61/4.2
in Y\ {Figure 1961/4.1). Since X\Y\ divides the median P\F as 2:1, it divides
all intervals joining P\ with an arbitrary point of P2P3 in the same manner. X\Y\
65

Solutions
1961/4
divides the triangle into a triangle P\X\Yi and a trapezium X\У1Р3Р2; we will
call these domains the triangle- and trapezium-domain corresponding to P\.
PXP
If Ρ is in the triangle-domain then
domain then
PQ\
PQ\
< 2, and if it is in the trapezium-
> 2. Similar consideration holds for the two other vertices P2
and P3.
In order to prove the statement, we only need to show that for an
arbitrary choice of Ρ it belongs to one of the triangle- and one of the trapezium-
domains. By studying Figure 1961/4.2 it is easy to verify that Ρ\ΡιΡ-$ is covered
by both the triangle- and the trapezium-domains, and this observation concludes
our proof.
Second solution. Following the lines of the first solution given above, we
show that Ρ is contained by both a triangle- and a trapezium-domain. It is easy
to see that Ρ is contained by the triangle-domain corresponding to P] if and only
if 44L > i and it
is in the trapezium-domain of P\ if and only if
P\Q\ $ P\Q\
{Figure 1961/43). Notice, however, that the ratio of PQi and P\Q\ is equal
to the ratio of the altitudes of the triangles PP2P3 and P1P2P3 corresponding
to P2P3. Since the two triangles share their base, this latter ratio is equal to
the ratio of their area. Let t1? t2 and t3 denote the area of the triangles PP2P3,
PP3P1 and PP1P2, respectively. For the
area t of P1P2P3 we obviously have t\ +
Ϊ2+Ϊ3 = ί, hence
1
<-
-3
PQi
P\Q\
Since
t\ PQ2 t2 РЯъ
t' P2Q2 t ' P3Q3
t\ t2 ίτ t
— + — + — = - = 1,
t t t t
_*3
t
Figure 61/4.3
among the three ratios there is one which
is not more and one which is not less than
-, concluding the solution.
1961/5. Construct a triangle ABC if the length of the two sides AC = b
and AB = с and the acute angle AM Β = ω is given — here Μ is the midpoint
of ВС. Show also that the problem admits a solution if and only if
ω
6 tan — < c< b.
2 -
66

1961/5
1961.
Solution. Let us start with the triangle already constructed (Figure
1961/5.1). The interval AB = c subtends an angle ω at the point M. Let us
denote the midpoint of AB by F. The length of FM is equal to -, hence Μ is on
the circle of radius - with centre F.
2
s
s
Figure 61/5.1
Now the method of the construction is as follows: Consider the circle к of
points such that the interval AB = с subtends an angle ω at them and intersect к
with the circle s of radius - centered at F. Their point of intersection is exactly
M. By measuring the length of BM twice on the line BM passed Μ we get С
The resulting triangle ABC obviously satisfies the assumptions since AM is a
median, hence AC = 2FM = b and /ΑΜΒ = ω.
The condition of
solvability is exactly that the
circles к and s do
intersect. Since ω is an
acute angle, the Thales
circle over AB is entirely
inside k. Consequently the
circle s intersects к only
if its radius is at least -,
2
i.e., b > с For the
existence of the intersection
we also need that the
farthest point of s from A B
A
cot ω/2
F c/2
Figure 61/5.2
В
67

Solutions
1961/5
is not farther away than the farthest point of к from AB. These distances from
AB are - and - cot — {Figure 1961/5.2), consequently the condition for solva-
bility is
b с ω ω
- < - cot —, hence b tan — < c.
2 ~ 2 2 2 ~
In conclusion, the necessary and sufficient condition we are looking for is
ш
b>c>b tan —.
2
Once this assumption is satisfied, we have two solutions (in Figure 1961/5.1 the
second solution is indicated by dashed lines).
1961/6. Let ε be a given plane and А, В, С three non-collinear points on
one side of ε. Suppose furthermore that the plane determined by these points is
parallel to ε. Let A, B1 and C' be three arbitrary points on ε. The midpoints
of AA', BB' and CC' are denoted by L, Μ and N respectively. The centre
of gravity of the triangle LMN is denoted by G. (Those triples А', В', С' for
which L, Μ, Ν do not form a triangle, are disregarded.) Determine the locus
of G for any possible choice of the triple А', В', С' on the plane ε.
Solution. Consider the centre of gravity of ABC as origin (Figure
1961/6.1). A vector pointing from О to a point denoted by an upper case letter
will be labeled by the same lower case letter. From the definition of the centre
of mass it follows that
a + b + c = 0.
The given data imply that
, a + a' b + b' c + c'
1=-^— > ni= , n=-—-,
2 2 2
hence the vector pointing to the centre of gravity of LMN is
1 /a + a' b + b' c + c'\ 1 , , 1 /a' + b' + c'\
8=з1^ + "Т + т) = 6((а + Ь + С) + (а+Ь+С))=21-3-)·
Let S' denote the centre of gravity of the triple А', В', С'. (The notion of centre
of gravity is extended for collinear points as follows: the centre of gravity of A',
a' + h' + c'
В1, С' is the endpoint of the vector .) According to the above said
g = -s', hence G is in the plane ε' we get by applying the enlargement with ratio
- centered at the origin. In other words, ε1 is the plane which is parallel to ε and
2 '
divides the distance of О and ε by two.
Next we have to show that any point of ε' can be got in this way. Let G be
an arbitrary point of ε' and denote the intersection of OG and ε by S'. Choose
68

1962/1
1962.
Figure 61/6.1
А1 ее in such a way that AA' does not pass through G, hence the midpoint L
of AA' is not equal to G. Fix a line passing through В disjoint both from LG
and A'S', intersecting ε in B'. The midpoint of BB' is denoted by M. Choose
C' in such a way that S' becomes the centre of gravity of the triangle defined
by A', B' and C'. Finally, let N denote the midpoint of CC'. According to the
above said, the vector pointing to the centre of gravity of LMN is half of s',
consequently it points to G. L, Μ and N are not collinear — otherwise G would
be on the line containing the three points, but this fact contradicts the choice of
L and M.
1962.
1962/1. Determine the smallest possible positive integer χ whose last
decimal digit is 6, and if we erase this last 6 and put it in front of the remaining
digits, we get four times x.
First solution. Let the digits of the originally η-digit number χ denoted
by an_i, an_2, ..., αχ, α0 = 6, so
χ = 10η_1αη_ι + 10n-2an_2 + ... + ΙΟαι + 6
According to the problem
Ax = 6 · 10η_1 + (ΐΟη-2αη_! + 10η~3αη_2 + ... + 10α2 + αϊ)

Solutions
1962/1
χ — 6
The expression in the parenthesis clearly equals to , therefore
4ζ = 6·10η-1 + ^.
This implies
40x = 6-10n+a;-6,
Ш = 2(10п-1).
This shows that 10n — 1 (a number containing only 9) must be divisible by 13;
by inspecting these numbers we see that 999 999 is the first which is divisible by
13:
999999 = 13-76923,
therefore
£ = 2-76923 = 153 846,
and this χ solves the problem.
Second solution. It is easy to see that if the last digit of a number is 6,
then four times this number has 4 as its last digit, so if χ is of the form ... 6
then Ax admits the form ... 4. According to the problem Ax has been created by
erasing the last digit of x, consequently χ must be of the form ... 46. Now the
last two digits of Ax is easy to determine: Ax =... 84. Using this principle we
can determine more and more digits of both χ and Ax:
X
... .46
. . . 846
.. 3 846
. 53 846
153 846
Ax
.... 84
... 384
..5 384
. 15 384
615 384
This scheme now shows that 153 846 satisfies the assumptions of the
problem, and it is the smallest such number.
1962/2. Determine all real χ satisfying
a/3 — χ — Vx +1 > -.
2
Solution. By ordering the inequality we get
у/Ъ — х > - + Vx+l.
The expressions appearing here are defined only in the interval —1 < χ < 3;
moreover in this interval all terms in the inequality are positive. Therefore the above
70

1962/3
1962.
inequality is equivalent to its square
1
3-x> - + X + 1 + Vx + l.
4
7
Vx + l < - —2x.
4
Since the left hand side is nonnegative, the right hand side is necessarily positive;
for this reason
(1) ~l~X<\
has to be satisfied. For such χ squaring the inequality is a reversible
transformation, hence
49 о
x+1 < — -7ж + 4ж ,
16
(*-i)2>g,
from which
^i ·31 ^i ^31
ж > 1 + ——- or χ < 1 —
8 8
follows. Only the latter one obeys assumption (1); since we only performed
equivalent transformations, the solution set of our inequality is equal to
■1<ж<1-
8
V31 7 7
We just note here that because of 1 — < -, the inequality χ < - is automa-
8 8 8
tically satisfied.
1962/3. The cube ABCDA'B'C'D' with upper face ABCD and lower
face A'B'C'D' (AA! || BB' || CC' || DD') is given. A point X runs along the
perimeter of ABCD (in the direction given by the above order) with constant
speed, while a point Υ does the same (with equal speed) along the perimeter of
the square B'C'CB. X and Υ start in the same instant from A and B',
respectively. Determine the locus of the midpoint Ζ of the interval XY.
Solution. We will use the following fact: for AB and CD arbitrary unit
intervals with X (and Y) moving on them from A to В (from С to D,
respectively) with equal speed, the midpoint F of XY sweeps the line segment joining
the midpoints of the segments AC and BD.
71

Solutions
1962/3
In order to verify this statement, let a denote the vector pointing to A while
с points to C. If ei and CD = e2 (Figure 1962/3.1) then the position of X
and Υ is determined by the vectors
x = a + Aei, y = c + Ae2 (0<λ<1),
hence the vector f pointing to F is equal to
„ a + с ч ei + eo
(i) f= —+A^'
Consequently, F is on the interval with starting point the midpoint of AC (λ = 0)
Figure 62/3.1
and direction — . The endpoint of this interval is exactly the midpoint of
BD, since in case λ = 1 we have
(a + ei) + (c + e2)
2
and this vector points to the midpoint of BD. Each point of this interval is the
midpoint of an interval XY, since for each f (as given in (1)) one can find the
appropriate χ and y.
Now the motion described by the problem can be subdivided into four
portions, each corresponding to the case of a pair of intervals. These individual cases
are taken care of by our argument above (Figure 1962/3.2). Call the midpoint of
BD, ВС' and AB' by P, Q and R respectively. The four portions of the motion
of X and Υ can be listed as follows:
the interval run by X the interval run by Υ
1. AB, , B'C,
2. ВС, С'С,
3. CD, CB,
4. DA, BB'.
72

1962/4
1962.
Figure 62/3.2
According to our starting argument, the interval swept by the midpoint is
RQ in the first case, QC in the second, С Ρ in the third and finally PR in the
fourth case. All these points belong to the locus we are looking for. In summary,
the locus is equal to the quadrilateral PCQR. By considering the orthogonal
projection to the face ABCD one can show that PCQR is a parallelogram;
since both QC and СР are equal to the half of the diagonal of this face, the
parallelogram is in fact a rhombus. Since to diagonals of adjacent faces starting
from the same vertex of the cube determine an angle of 60°, the acute angle of
our rhombus is 60° as well.
1962/4. Solve the following equation:
о о о
cos χ + cos 2x + cos 3x = 1.
Solution. Use the following substitutions:
9 9
cos 2a; = 2cos x — 1, cos 3x = cos x(4 cos x — 3),
and order the equation:
cos2 χ + (2 cos2 χ - l)2 + cos2 x(4 cos2 χ - 3)2 = 1,
cos2 x(8 cos4 χ - 10 cos2 χ + 3) = 0,
9 9 9
cos x(2 cos χ — 1)(4 cos a: — 3) = 0,
cos χ cos 2a; cos 3a; = 0.
73

Solutions
1962/4
In conclusion, the solution set consists of those χ for which cos x, cos 2a; or
cos 3a; vanishes. Since cosa; = 0 implies cos 3a; = 0, it is sufficient to determine
the solutions of cos 2x = 0 and cos 3a; = 0.
cos2a; = 0, if 2a; = (2fc + l)^, x = (2k + l)j,
cos3a; = 0, if 3x = (2k + l)-, x = (2k + l)^,
2 6
where к is an arbitrary integer. Since all the transformations we have performed
are reversible, the set of values determined above is the solution of the problem.
1962/5. Three distinct points А, В and С on a circle к are given. Construct
the point D on the circle for which the quadrilateral ABCD admits an incircle.
First solution. Recall that a quadrilateral admits an incircle if and only if
the sum of opposite edges are equal.
Suppose that D is a point of the circumcircle of the triangle ABC
which is on the arc AC not containing B, and for which AB + CD = BC + AD
(i.e., ABC Ό admits an incircle, see
Figure 1962/5.1). If AB = BC then CD = AD
also holds consequently D is the midpoint
of the arc AC, and since ABCD is a kite,
it admits an incircle.
Without loss of generality we may
assume that AB>BC, hence AB - ВС =
AD - CD > 0. Measure CD from D onto
DA and call its endpoint P. The triangle
CDP is isosceles with the third angle
being equal to 180° — β, hence the angles on
its base are equal to —, and so the angle
Figure 62/5.1
APC is 180c
β
2
Based on the above said, the construction of D proceeds as follows:
construct the circle such that AC subtends angle 180° — —, and consider the arc op-
posite to B. Intersect this arc with the circle of radius AB — ВС and centre A,
and call the point of intersection P. By construction AP - AB — ВС. Now the
line determined by AP intersects the circumcircle of ABC in the point D we
are seeking for.
Next we show that the quadrilateral ABCD admits an incircle. Since
LCDA= 180° - β and LCPA= 180° - |, the angle LCPD is equal to |, hence
74

1962/5
1962.
LPCD- —. Consequently С DP is an isosceles triangle, so CD = DP, implying
that AD = AP + CD = AB-BC + CD. This shows that AB + CD = AD + BC,
which means that ABCD admits an incircle.
The existence of the construction depends on the existence of P. This point,
however, always exists, since the circle such that AC subtends an 180°
is inside the circumcircle; moreover the circle with radius AP = AB — ВС and
centre A always intersects the inner arc since AP = AB — ВС < AC according
to the triangle inequality applied to ABC. Consequently the problem admits a
unique solution.
Second solution. In this solution
we utilize the fact the if a quadrilateral
admits an incircle then the bisectors
intersect each other in a unique point К
(the centre of the incircle), see Figure
1962/5.2.
Suppose that the quadrilateral has
been constructed and the angles of it at
the vertices А, В and С are α, β and 7,
respectively, a + 7 =180° since ABCD
admits a circumcircle; because LAKB -
180c
α + β
and 1ВКС = \Ш
β + Ί
we have
IAKC = 360°
180c
α + β^
- 180е
β + Υ
Figure 62/5.2
2β + α + η
= 90° + /3.
2 ) Χ 2 ) 2
If /3 = 90° then ABC is a right angled triangle and D is the reflection of В to its
hypotenuse AC.
If β < 90° then the centre point К is the intersection of the angle bisector
at В with the circle such that AC subtends angle 90°+/3. Having constructed
K, the incircle of the quadrilateral can be easily constructed since it is tangent to
В A and ВС. Now the tangents of the incircle passing through A and С intersect
each other in D.
The resulting quadrilateral obviously admits an incircle — we need to show
that it also admits a circumcircle. According to the construction the sum of the
three angles at К is
360° = 90°+β+( 180°
implying
β + Ί
+ 180°-
β + α
= 450°-
α + 7
α + 7 = 180°,
75

Solutions
1962/5
consequently ABCD admits a circumcircle, i.e., D is on the circumcircle of
ABC.
Finally, if β > 90° the construction has to be modified only in one point: we
need to consider the circle such that AC subtends angle 270° — β. It is
straightforward that in all cases the construction yields a unique solution.
Remark. The quadrilateral under consideration is a so-called bicentric
quadrilateral since it admits both an incircle and a circumcircle. Such quadrilaterals
have many interesting properties, for example, two data among the radii of the
incircle, the circumcircle and the distance of their centre determine the third one.
1962/6. Let R and r denote the radii of the circumcircle and the incircle
of an isosceles triangle. Show that the distance d between the centres of the two
circles is
(1) d = ^R(R-2r).
First solution. Let ABC be an isosceles triangle (AC = ВС), and let О be
the centre of its circumcircle while К is the centre of its incircle. The bisector
at С intersects the circumcircle for the second time in the midpoint Q of the arc
AB; the foot of the orthogonal from К to AC is Τ (Figure 1962/6.1).
Figure 62/6.1
ι
We start with the observation that QA = KQ. This can be shown
ex. ύ
by demonstrating that AKQ is an isosceles triangle: IAKQ = — + — and
76

1962/6
1962.
LQAB = LQCB = j-, hence LQAK = — + -}-, verifying that AKQ is an isosceles
triangle.
Let OK = d. Based on the formula regarding the power of К to the circum-
circle we have
(2) KQ-KC = R2-d2.
Using the similarity of the right angled triangles CAQ and CTK we get
r =KC=KC
U QA CQ 2R'
implying
2Rr = QA-KC,
and since QA = KQ, this shows
2Rr = KQ-KC.
Comparing it with (2) we conclude that
2Rr = R2-d2,
d = jR(R-2r).
Second solution. We claim that (1) holds for any triangle. We prove it by
vector methods.
Let us fix the origin at the centre of the circumcircle; now the vectors
pointing to the vertices of the triangle are a, b and c. By definition we have
|a| = |b| = |c| = R. It is known that the vector pointing to the centre of the incircle
is equal to
аа + bb + cc _aa + bb + cc
a + b + c 2s
where a, b and с stand for the lengths of the sides of the triangle. According to
our notation we have that |k| = d Consider the square (with inner product!) of
the equality
2sk = аа + bb + cc.
This leads to
(2) 4s2 d2 = (a2R2 + b2R2 + c2R2 + 2abab + 2bcbc + 2caca),
now applying of a — b| = c, i.e., (a — b) = с yields
2#2-2ab = c2,
which is equivalent to 2ab = 2R2 — c2. Similarly, we have 2bc = 2R2 — a2 and
9 9
2ca = 2i? — b . Substitute these into (2) and conclude:
4s2d2 = R2(a + b + c)2 - abc(a + b + c) = R2 · 4s2 - abc ■ 2s,
77

Solutions
1962/6
giving
2s
According to the radius formula abc = 4tR, t = rs, and so
d2 = R2-2Rr, d = y/R2-2Rr,
concluding the solution.
Remarks. 1. Equation (1) was discovered by L. Euler; its important
consequence is the so-called radius-inequality:
R>2r,
and equality holds if and only if d = 0, i.e., in case of an equilateral triangle.
2. (1) is in close connection
to Poncelet's prisms.
3. A slight modification of
our first solution also gives that
(1) holds for all triangles. If ABC
is not an isosceles triangle,
hence К and О are not on the
same diameter of the circumcircle,
we should consider the diameter
QC', where Q is the midpoint of
the arc AB; therefore the
bisector С К passes through Q (Figure
1962/6.2). Now the proof should
be modified at (3) by substituting
CQ with C'Q and in the previous
line we should invoke the
similarity of C'AQ and CTK.
Q
Figure 62/6.2
1962/7. There are five spheres which are tangent to all extended edges of
a tetrahedron SABC. Show that
a) SABC is a regular tetrahedron;
b) conversely: a regular tetrahedron admits five spheres with the properties
described above.
Solution. If a sphere is tangent to the extended edges of a tetrahedron, then
the faces intersect the sphere in circles which are tangent to the extended edges;
consequently in plane sections the resulting circles are the incircles or excircles
of the faces of the tetrahedron.
78

1962/7
1962.
Two touching circles of a triangle cannot be on the same sphere and an
excircle can share a sphere with an incircle only in case they touch the edge at
the same point (Figure 1962/7.1).
Figure 62/7.1
If we add the fact that two tangent circles lying in different planes uniquely
determine the sphere containing them and that this sphere contains all circles
tangent to these two circles, we conclude that one of the five spheres (call it G)
contains the four incircles of the faces, while the remaining four spheres contain
one incircle and three excircles tangent to this circle. (Recall that two circles are
said to be tangent if in their common point their tangents coincide.)
In order to show that SABC is regular, it is enough to verify that any face
of it is an equilateral triangle. Let us consider the face ABC. Mark the points on
the edges of the tetrahedron where G intersects them. On edges starting from a
common vertex these marked points are of equal distance from the vertex (since
the tangents to a sphere from an external point are of equal length). In Figure
1962/7.2 these equal intervals are denoted by the lower case letters corresponding
to the labels of the vertices.
79

Solutions
1962/7
Figure 62/7.2
Let Gs be the sphere containing the incircle of ABC and the containing
excircles of the other three faces. It intersects the extended edges SA, SB and
SC in the points Χ, Υ and Z, respectively. Since the length a of the tangent
interval from A to the incircle of ABC is the same as the tangent interval from A
to Gs, we have AX = a and similarly BY = b, CZ = c. Using the above principle
the tangent intervals from S to Gs are equal, consequently SX - SY = SZ, hence
s + 2a = s + 2b = s + 2c,
implying a = b = c. This shows that the sides of ABC are equal, hence it is an
equilateral triangle.
Now if SABC is a regular tetrahedron with centre К and the radius of
the incircle is equal to r, then by denoting the radii of the incircles of the faces
by ρ we have that the distance of К from any edge is equal to r' = yr2 + ρ2
(Figure 1962/7.3). Therefore the sphere of radius r' and centre К is tangent to
all extended edges.
Similarly, if the centre of the sphere touching the face ABC of the
tetrahedron SABC from outside is Ks and its radius is rs, furthermore the radius of
the excircle of ABC is ρ3, then the distance of Ks from the edges is equal to
r's = Jr2 + ρ2. Consequently, the sphere with centre Ks and radius r's is tangent
80

1962/7
1962.
Figure 62/7.3
to all edges. This shows the existence of five spheres with the desired properties
for a regular tetrahedron.
Remarks. 1. Studying Figure 1962/7.2 a little longer, we might notice that if
the tetrahedron admits a sphere which is tangent to the edges in their inner points
then the sum of opposite edges are equal (on the figure this sum is a + b + c + s).
One can prove that this condition is sufficient for the existence of a sphere tangent
to the edges.
2. Similar observations can be made regarding the existence of spheres
which are external tangents of the edges: the necessary and sufficient
condition for the existence of one such sphere is that the difference of any base edge
and the facing edge should be constant. (Here base edges are the edges of the
triangle which is tangent to the sphere from inside.) For tetrahedrons with equal
faces this condition is fulfilled, hence they admit four external tangent spheres,
but the inner tangent exists only in case the tetrahedron is regular.
3. Our proof showed that the existence of the inner tangent sphere and the
external tangent sphere tangent to the edges of the face ABC from inside imply
that ABC is equilateral and SA, SB, SC are equal. In a similar vein it follows
that if the external tangent sphere to the face SBC exists then SA = AB = AC.
This however implies that all edges of the tetrahedron are equal, consequently
it is regular. According to this observation the existence of the inner and two
external tangent spheres already implies that SABC is regular.
81

Solutions
1963/1
1963.
1963/1. Determine the real solutions of the following equality (p denotes
a real parameter)
x1 — p + 2yfxz — l = x.
Solution. Since χ is the sum of two nonnegative numbers, χ > 0 obviously
holds; moreover we need x2 > ρ and χ > 1 for the square roots to make sense.
Under these assumptions the equality
4^x2-pjx2- l=p + 4-4x2
is equivalent to our starting one.
The left hand side is nonnegative, hence ρ + 4 — Ax > 0 has to be satisfied,
• , · ? P + 4
implying χ < ——.
With this assumption, another squaring yields
8(2-p)z2 = (p-4)2,
and this equation is equivalent to our starting one.
Now p-^2, since otherwise the left hand side vanishes while the right hand
side does not. This shows that
(1) 32 = £ζί£
Next we examine the possible values of ρ for which our previous (necessary)
assumptions on χ are satisfied.
ο (ν- 4)2 ν2
a) x2 > 1, i.e. ζ \ > 1 or ζ > 0, implying ρ < 2.
8(2-ρ) 8(2-ρ)
о (р-4)2 (Зю-4)2
b) xL > ρ, i.e. — > ρ or — > 0, implying ρ < 2.
8(2-ρ) 8(2-ρ)
ϋ + 4 (ю-4)2 ю + 4 Зр(р-^) 4
The common part of these assumptions is
4
0<p<-.
In conclusion, for these values of ρ the equation admits a unique solution of the
form
1 _ 4-p
while for other values of ρ the equation has no solution at all.
82

1963/2
1963.
1963/2. Given a point A and a segment ВС in the 3-dimensional space,
determine the locus of those points, P, for which the angle LAPX is a right
angle for some X on the segment ВС.
Solution. Let X be an arbitrary point of ВС. If one leg of the right angle
contains A and the other one X, then the vertex Ρ of the right angle is on the
sphere with AX as its diameter, and all the points of this sphere satisfy the
condition of the problem.
Once X runs through the points of ВС, the corresponding spheres sweep
out the locus Μ asked by the problem. (*)
Figure 63/2.1
In the following we would like to give a better description of this set M.
Let Gb and Gq denote the balls with diameters AB and AC, respectively. We
would like to show that Μ is equal to the set of points which are not exterior to
one and not interior to the other one.
Let us first examine a plane section of this set; let the plane S contain our
three given points А, В and C. (Notice that if the three points are not collinear
then S is determined by these points.) The centres of the spheres are on the
interval joining the midpoint B1 of AB with the midpoint C' of AC; the line
B'C' = t turns out to be a symmetry axis for each sphere in M, moreover by
rotation around t any point of Μ might be transported into S. Let the circle кв
be the plane section of Gb with S while the intersection of S and Gq is denoted
by kc (Figure 1963/2.1).
In order to prove our statement, it is enough to verify that the main circles
of the spheres in Μ fill up the domain we get by deleting the common interior
83

Solutions
1963/2
points from the union of the discs кв and kc· In the following this domain will
be denoted by T. The other intersection point of кв and kc will be denoted by
A' (notice that it might coincide with A), in this notation t is the perpendicular
bisector of AA'. Since any circle in S with centre on t which passes through A
also passes through A', we conclude that the main circles under consideration all
pass through A and A'.
It is easy to see that the points of к в and kc belong to M. Let now Q be an
inner point of T, for example in the interior of кв. The centre К of the circle к
passing through A, A' and Q is on t, and the endpoint X of its diameter is on the
interval BA'. This last assertion follows from the fact that the angle I ABA' is
less than the angle IAXA', consequently Q is on a circle such that its diameter
is an interval joining a point X of ВС with A. This observation now shows that
Q belongs to M. For similar reasons any circle with diameter AX belongs to T.
We just note here that (after slight modifications) our arguments are also
valid in the case when А, В and С are collinear.
Remark. In fact, the solution of the problem could be terminated at the
point indicated by (*), since by that point we answered the question raised in the
problem. Since a locus can be described in many different (equivalent) ways, the
expectations for solutions of such problems always contain indeterminacy.
1963/3. Consider a convex n-gon with equal angles and with consecutive
sides a\,CL2,... ,an satisfying
(1) . a\ > a,2 > · · · > an-
Show that under the above conditions we have
(2) a\ — a2 - · · · = an-
Solution. The equality of the angles imply that the sides of the n-gon are
parallel to the sides of a regular n-gon. We have to prove that the n-gon itself is
regular. Let n = 2k or η = 2k + 1 and
an = b\,an_i =b2,... ,an_k+i = bk,an_k = bk+i.
Denoted the inner bisector at the common vertex of a\ and an = b\ by / (Figure
1963/3.1).
The endpoint of ak coincides with one of the endpoints of bk if η is even;
for η odd the edge ak+\ = bk+\ is orthogonal to /. For this reason the orthogonal
projection of 01,02,... ,ο& to f coincides with the orthogonal projections of
b\, έ>2) · · · j Η- The projections of o; (bi) will be denoted by a[ (Ь[); so
(3) a\ + o2 +... + a'k = b\ + b'2 +... + b'k.
84

1963/4
1963.
Figure 63/3.1
Since the angle of щ and b{ with / are equal, according to (1) we have
(4) ai>bi, hence а[>Ь[.
Now from (3) it follows that
(α,1-ί/1) + (4-6^) + ...+ (4-^ = 0;
since the terms in the parentheses are all nonnegative, their sum is zero only
in case all terms are equal to zero. This shows a\ =b[, implying a\ =b\ =an,
consequently in (1) we have equality. This concludes our solution.
Remark. The statement of the problem can be visualized in the following
way: consider a regular n-gon with side a\. Starting from a\ construct the n-gon
given by the array (1); now if there is a place in (1) where > is valid, then from
that point our polygonal path is inside the regular n-gon, and will never close to
form an n-gon.
1963/4. Determine the values x\, x2, хъ> x4> x5 satisfying
(1) Х5 + Х2 = УХ\,
(2) xi+x3=yx2,
85

Solutions
1963/4
(3) x2 + x4 = yx3,
(4) x3+x5=yx4,
(5) x4 + xi=yx5
where у is a given parameter.
Solution. Express x$ and x3 from (1) and (2):
(6) x5 = yxi-x2,
(7) x3=yx2-x\.
Substitute (6) into (5) and get
(8) x4 = (y2-l)xi-yx2.
After substituting (8) and (7) into (3) and ordering the equation we get
(9) (y2 + y-l)(xl-x2) = 0.
Substituting (6), (7) and (8) into (4) we see that
(10) (y2 + y-l)((y-l)xl-x2) = 0.
If y2 + y- 1=0, i.e.
-l±>/5
y = -
2 7
then (9) and (10) is satisfied for arbitrary x\ and x2, since all our previous
transformations are reversible, these values uniquely determine x3, χ4, x$.
If у2 + у - 1 7^0 then (9) and (10) gives
(11) xx-x2 = 0,
{y-\)xi -x2 = 0,
implying
(12) (y-2)Xl=0.
For у = 2 the value of χ ι = x2 can be arbitrary; if x\ = x2 = с then £3 = £4 = £5 = с
Finally if у ^2 then x\ =0 from (12) and x2 = 0 from (11) implying that all
unknowns in the original system are equal to 0.
In summary:
1. If y = (— 1 ± v5)/2 then x\ =a, x2 = b, x3 = yb — a, X4 = (y — \)a — yb =
—y(a + b), х$ = уа — b, for a, b arbitrary reals.
2. If у = 2 then x\ = x2 = x3 = X4 = X5 = с for an arbitrary real с
3. Finally if is у none of the above three values then x\ = x2= x3= £4= £5= 0.
86

1963/5
1963.
1963/5. Show that
(1)
π 2π 3π 1
cos — — cos —- + cos —- = -,
7 7 7 2
7Γ
First solution. Introduce ct='— (so 7α = π). Consider the equilateral
triangle OAB inside the angle α with vertex О and then consider the isosceles
triangles ABC and BCD such that ОЛ = AB = ВС = CD = 1 is satisfied (see
Figure 1963/5.1).
Figure 63/5.1
According to theorems regarding the sum of angles and external angles of
a triangle, we conclude that the base angles of the above triangles are a, 2a and
3a, respectively. Moreover the angles of the triangle COD on CD are equal to
3a, hence it is an isosceles triangle, implying OC = OD. But
OC=l+2cos2a,
OD = 2 cos a + 2 cos 3a,
hence
1 + 2 cos 2a = 2 cos a + 2 cos 3a,
which obviously implies (1).
Second solution. By considering the identity
5π / 2π\
COS -r-=COS 7Γ
7 V V
(1) can be rewritten as
= — cos
2тг
У
(2)
π 3π 5π 1
cos — + cos — + cos — = -,
7 7 7 2
This identity, however, is connected to geometric properties of the regular
heptagon. The symmetry axis of the regular heptagon with unit sides passing through
A bisects the opposite side. The angle between the side and the perpendicular m
7Γ
to the axis in A is — = a, hence the projection of this side on m is cos a (Figure
87

Solutions
1963/5
α = π/7
cosoc
Figure 63/5.2
1963/5.2). Since consecutive sides can be got by rotating the previous side by
2a, the angles of the next two sides with m are 3a and 5a. Consequently the
lengths of the projections are cos 3a and cos 5a, this latter one being negative.
The sum of these three oriented projections (signs are assigned according to the
orientation) is equal to half of the side (see the figure), hence:
cos α + cos 3a + cos 5a = -,
2'
which is exactly what we wanted to demonstrate.
Third solution. The equality will be proved in the form given by (2).
Denote the left hand side of (2) by К and apply
2 sin χ cos у = sin(cc + y) — sin(y — x).
7Г
Using the notation α = — and taking sin a = sin 6a into account we get
К · 2 sin α = 2 sin α cos α + 2 sin α cos 3a + 2 sin a cos 5a =
= sin 2a + sin 4a — sin 2a + sin 6a — sin 4a = sin 6a = sin a,
implying
4-
88

1963/6
1963.
Remark. All three solutions above generalize to the proof of the following
statement:
If a = —^— (k = l, 2, 3, ...) then
2k +1 v
cosa + cos3a + .. . + cos(2k — \)a = -.
1963/6. Five students, A, B, C, D and Ε were placed 1 to 5 in a contest.
Someone made the initial guess that the final result would be the order ABCDE,
but — as it turned out — this person was wrong on the final position of all the
contestants; moreover no two students predicted to finish consecutively did so.
A second person guessed DAECB, which was much better, since exactly two
contestants finished in the place predicted, and two disjoint pairs predicted to
finish consecutively did so. Determine the outcome of the contest.
Solution. For simplicity, we assume that the competitors are vertices of a
graph on five points, and two vertices are connected by an oriented edge if and
only if the competitor corresponding to the starting vertex of the edge finished
right before the student corresponding to the end vertex of the edge. Now any
outcome can be represented by an oriented path containing four edges. We also
label each vertex with its place.
B2 B5
Figure 63/6. la, b
ABCDE corresponds to G\ {Figure 1963/6.1) while DAECB corresponds
to G2. We have to find a graph G which
a) does not share an edge or label with G\ and
b) shares two edges and two labels with G2.
According to our conventions, if G2 and G share an edge and the labels
on one of its endpoints are equal in the two graphs then the labels on the other
endpoints are equal as well. Furthermore: G2 and G cannot share two adjacent
edges since in that case none of the three endpoints can have equal labels. Two
89

Solutions
1963/6
СЗ А2
Dl E4
Figure 63/6.1 с, d
C4 A3
Figure 63/6.1 e, f
adjacent edges cannot be common in G^ and G even if the labels are different,
because in that case the labels differ at least in four positions, which contradicts
our assumption.
Notice that the common labels of G and Gi are the endpoints of one of
their common edges. Since among the remaining three vertices there is a
common edge with different labels, the common edge connects vertices either with
positions 1 and 2 or with 4 and 5. The second guess corresponds to the situation
in which one of the two above edges is correct and the position of the remaining
ones are cyclically permuted.
In conclusion the final result (based on the second guess) is among:
a : DACBE, β: DAB EC,
η-.AEDCB, δ: ED ACВ.
In case a and 7 the graphs G and- G\ share vertices with equal labels (E5 and
Al), in β they share a common edge (AB). Consequently these do not satisfy the
assumptions of the problem, δ however provides a solution, hence the outcome
of the competition was EDACB.
90

1964/2
1964.
1964.
1964/1. a) Find all positive integers η for which 1 divides 2n — 1.
b) Show that there is no positive integer η for which 1 divides 2n + 1.
Solution. After examining particular cases, one can see that divisibility of
2n ± 1 by 7 is connected to divisibility of η by 3. Let η = 3 k + r (r = 0, 1,2); then
2n = 23k+r = 2r-8k=2r(8k-\k) + 2r = 7E + 2r,
where Ε stands for some integer, hence
2n-l=7£ + 2r-l and 2n + 1 =7£ + 2r + 1.
From this, if
r = 0, 2n-l=7£, 2n + l=7£ + l,
r = l, 2n-l=7£ + l, 2n + l=7£ + 3,
r = 2, 2n-l=7£ + 3, 2n + l=7£ + 5,
This means that 2n — 1 is divisible by 7 if and only if η is divisible by 3
and 2n + 1 is not divisible by 7.
1964/2. Let a, b and с denote the lengths of the sides of a triangle. Show
that
(1) a2{-a + b + c) + b2(a -b+c) + c2(a + b-c)< 3abc.
First solution. Let us start with the following obvious inequality (the
second factors, according to the triangle inequality, are all negative):
(b-c)2(a-b-c)<0,
(c-a)2(b-c-a)<0,
(a-b)2(c-a-b)<0.
Adding these inequalities we get:
a (b — с — a + с — a — b) + b (a — b — c + с — a — b) + с (a — b — c + b — с — a) —
— 2bc(a — b — c) — 2ca(b — c — a) — 2ab(c — a — b) =
= 2a2(-a + b + c) + 2b2(a -b + c) + 2c2(a + b-c)- 6abc < 0,
from which (1) directly follows.
Second solution. After multiplication and ordering we get
-a3 -b3 - c3 + a2b + ab2 + a2c + ac2 + b2c+ be2 - 2abc < abc;
factoring the left hand side we get
(2) (—a + b + c)(a — b + c)(a + b — c)<abc,
91

Solutions
1964/2
and this inequality is equivalent to the one we started with. Since in this last
expression both sides are positive, it is equivalent to its square:
(-a + b + c)2(a - b + c)2(a + b - c)2 < a2b2c2,
(a2 -(b- c)2)(b2 - (c - a)2)(c2 - (a - b)2) < a2b2c2.
Now the terms on the left hand side are smaller (not larger) than the
corresponding terms on the right hand side — verifying this last, hence our starting
inequality.
Third solution. After ordering (1) divide both sides by 2abc:
a(b2 + c2- a2) + b(a2 + c2- b2) + c(a2 + b2 - c2) < 3abc,
b2 + c2-a2 a2 + c2-b2 a2 + b2-c2 3
+ + < -.
2bc 2ac 2ab 2
The quotients on the left hand side are equal to the cosines of the angles of the
triangle, as it is shown by the law of cosines, hence the above inequality becomes
3
cos a + cos β + cos 7 < -.
This latter inequality is, however, the famous cosine inequality which is valid
for all triangles, and (according to the above said) it is equivalent to (1).
Fourth solution. We show that (1) is satisfied by an arbitrary triple of
nonnegative numbers a, b and с Since their role in the inequality is symmetric,
we can assume that
0<c<b<a.
In this case
9 9 9
a (b + c — a) + b (c + a — b) + c (a + b — c) — 3abc =
999
= a(ab + ac—a —bc) + b(bc + ba — b —ac) + c(ac + cb —с —ab) =
- —a(a — b)(a — c) + b(b — c)(a — b) — c(a — c)(b — c) <
< -a(a - b)(b - c) + b(b - c)(a -b)- c(a - c)(b -c) =
= -(a - b)2(b - c) - c(a - c)(b - c) < 0.
Equality holds in case a = b = c or a = b and с = 0.
Remarks. 1. Solutions 1.-3. also show that equality holds only in case a =
b = c, i.e., for an equilateral triangle.
2. The fact that (1) is equivalent to (2) — one of the most fundamental
inequalities in trigonometry — indicates that our inequality has deep connections
with numerous other geometric inequalities, as it was pointed out by the third
solution. Another connection can be found by writing (2) in the form
8s(s — a)(s — b)(s — c)<s· abc,
92

1964/3
1964.
and applying Heron's formula together with the identities t = rs, abc = 4tR where
r and R denote the radii of the incircle and the circumcircle and t is the area of
the triangle. Now this rewriting yields
8i2 < s · AtR,
2r s <s -rsR,
2r<R,
the famous radius inequality.
TOO
3. Adding 2(a +b +c ) to the form of the inequality we found in the
second solution we get
(a2 + b2 + c2)(a + b + c)<2(a3+b3 + c3) + 3abc,
still equivalent to (1). This form admits the following generalization: for a,i non-
negative,
9 ? ?
< (η — \)(a^ + α2 + · · · + an) + nala2
a
n·
1964/3. Let a, b, с denote the lengths of the sides of the triangle ABC.
Tangents to the inscribed circle are constructed parallel to the sides. Each tangent
forms a triangle with the other two sides of the triangle, and a circle is inscribed
in each of these three triangles. Find the total area of all four inscribed circles.
Consider the tangents of the incircle which are parallel to the sides. These
tangents give rise to three subtriangles of ABC, consider the incircles of these
subtriangles. Determine the sum of the areas of the four incircles.
Solution. Let the tangent of the incircle parallel to Б С be denoted by
B'C (where B' is
on AB and C' is on
AC). Let r\
denote the radius of the
incircle of AB'C,
while the radius and
area of the
incircle of ABC is
denoted by r and t.
Finally, ma is the
altitude of ABC
corresponding to ВС =
a is ma (Figure
1964/3.1).
Figure 64/3.1
93

Solutions
1964/3
The triangles ABC and AB'C' are similar, hence the ratio of the radii of
the incircles coincides with the the ratio of the altitudes:
r\ ma - 2r ( 2r \
— = , ri=r[l .
r ma \ maJ
Since It = ama = 2rs, ma = , this shows
a
( a\ r(s-a)
(1) ri=r[l--\ = .
v( s — U)
Similar reasoning for the radii of the two other incircles provide r^ =
s
v(s — b)
and Г3 = . The sum of the four areas is equal to
s
T = 7r(r2 + rf+rl + r^) =
2 2
= ^r(s2 + (s-a)2 + (s-b)2 + (s-c)2) = '^r(a2 + b2 + c2).
sl sz
2 t2 (s-a)(s-b)(s-c)
Since r - — = 5 we get
sz s
7r(s - a)(s - b)(s - c)(a2 + b2 + c2)
~ Ρ "
Remark. An interesting corollary of (1) states that the sum of the radii of
the incircles of the three subtriangles is equal to the radius of the incircle of the
original triangle:
(л а л Ь . C\
rl+r2+r3-M 1 ■" 1 !" 1 =r·
V s s s/
1964/4. Each pair from 17 scientists exchange letters on one of three
topics. Prove that there are at least three scientists who write to each other on the
same topic.
Solution. Consider a graph with 17 vertices and associate a vertex to each
scientist. Two vertices are connected by an edge if the corresponding scientists
exchange letters. The edge is red if they discuss the first topic, it is blue if
the second, and it is green if they exchange letters on the third topic. We have
to show that the graph contains a monochromatic triangle, i.e., there are three
vertices connected by edges of the same colour.
First we will prove a useful lemma, sometimes also called Ramsey's
Theorem (see our remark after the solution). The lemma says the following: Suppose
that the edges of the complete graph on six vertices are coloured by two colours.
Then the graph contains a monochromatic triangle.
94

1964/4
1964.
Suppose that the colours we use are blue and green. Fix a vertex A. Among
the five edges starting from A there are at least three of the same colour, say
blue, joining A with B\, Bi and Б3. If one of the edges of the triangle ΒχΒιΒ^
is blue (for example В\В^) then AB\B2 is a monochromatic triangle. If none
of the edges in B\BiB^ is blue then all of them are green, hence we have found
a monochromatic triangle again.
Let us now return to our original problem. Fix a vertex A of the graph on
17 vertices. There are at least 6 edges among the 16 edges starting from A which
are decorated by the same colour, say red. Let the endpoints of these 6 edges be
called B\, B2, B^, B4, Б5 and B^. If one of the edges connecting these vertices
(say B1B2) is red, then AB\Bi is monochromatic, so we are done.
If there are no red edges joining any two of B\,..., B^ then all edges are
blue or green, hence the lemma proved above shows that there is a
monochromatic triangle among them. This last observation concludes the solution.
Remark. 1. The statement does not remain true for a graph with 16 vertices:
such a complete graph admits a colouring with three colours such that no triangle
is monochromatic.
2. The problem is one of the Ramsey-type theorems of graph theory; it
admits many generalizations and applications.
3. The following generalization directly relates to our problem: let сц =3
and an = nan_\ — (n — 2) for η > 1. If the complete graph on an vertices is
coloured by η colours then it admits a monochromatic triangle.
For η = 2 this theorem is simply the lemma we used above, for n = 3 it is
exactly the problem. The proof goes by induction: for η = 2,3 the statement —
as we verified above — is true. Suppose that it is true up to η — 1. Now consider
the complete graph on an vertices coloured by η colours and fix a vertex A of it.
The an — 1 =nan_\ — n+ 1 edges starting from it are also coloured by η colours,
so there are at least an_\ which are coloured by the same colour.
Let the endpoints of the an_\ vertices of the same colour (say red) starting
from A be denoted by B\, Β2,. ■. Ba _χ. If there is a red edge among them (say
between Bi and В к), then ABiB^ is a red triangle, and we proved the assertion.
If there is no red edge among B\, B2, · · ·, Дхг _λ, then it is a subgraph on an_\
vertices coloured by η — 1 colours, so — according to the inductive hypothesis
— it admits a monochromatic triangle.
It can be shown that
an = [n!e] +1.
(See also the remark at problem 1978/6.)
95

Solutions
1964/5
1964/5. Five points on the plane are situated so that no two of the lines
joining a pair of points are coincident, parallel or perpendicular. Through each
point lines are drawn perpendicular to each of the lines through two of the other
four points. Give the best possible upper bound for the number of intersection
points of these orthogonals, disregarding the given 5 points.
Solution. Denote the five given points by A\, A^ Αί,, Α4. and Л5;
these points determine I J = 10 lines connecting them. Fixing Ai, the number of
lines joining the remaining four points is equal to I J = 6, hence there are 6
orthogonals starting from Ai, from the 5 given points this adds up to 30 lines.
/30N
The maximal number of points of intersections of these 30 lines is I
435.
Let us now determine how much this quantity has to be reduced because of
possible coincidences and by disregarding the 5 given points.
a) The three orthogonals to AiAj~ are parallel, so we have to reduce the
number of points of intersection for each pair of points by 3, meaning the deduction
of 30.
/5\
b) There are I J = 10 triangles with vertices from the A^s; there are
three lines concurring in the orthocentres of these triangles. For this reason each
triangle reduces by 2, altogether by 20 the number of points of intersection.
c) There are 6 orthogona, passing through a given point *. the Q -15
points of intersection of these lines do not count, hence we have to reduce the
total number of points of intersection by 5 · 15 = 75.
The sum of the above reductions is 30 + 20 + 75 = 125, consequently there
are at most 435 — 125 = 310 points of intersection.
Remark. The above reasoning did not show that the maximal number of
points of intersection is equal to 310; for this we should provide a set of 5 points
for which the number of appropriate intersection points of the orthogonals is
eventually equal to 310. This can be done, but the description is very time- and
space-consuming.
The original form of the problem — as it was given in the competition —
went as follows: Determine the maximum number of intersections .... Right
before the competition the members of the jury realized that the full solution
would consume too much time, hence they verbally instructed the competitors
96

1964/6
1964.
that "the verification of maximum is not needed". At the end of the day, they
expected roughly the same amount of work we presented in the solution above.
1964/6. Let ABCD be a given tetrahedron and D\ the centroid of the face
ABC. The parallels to DD\ passing through the vertices А, В and С intersect
the opposite faces in A\, B\ and C\, respectively.
a) Show that the volume of ABCD is one-third the volume of A\BiC\D\.
b) Is the result valid for any choice of D\ in the interior of ABC?
Solution, a) and b) will be shown at once; we will show that the volume
of A\B\C\D\ is three times the volume of ABC Ό for any choice of D\.
Figure 64/6.la Figure 64/6.lb
First we show that if the tetrahedrons ABCD and XYZU are such that the
lines AX, BY, CZ and DU are parallel, D and U are interior points of the
triangles XYZ and ABC, moreover ABC and XYZ are disjoint, then ABCD
and XYZU have equal volumes (Figures 1964/6.la and lb).
Notice that the 4-sided pyramids ABYXD and ABYXU have the same
base and the distances of D and U from ABYX are equal. Therefore the volumes
of ABYXD and ABYXU are equal. The same can be said for the pyramids
BCZYD and BCZYU, and for the pyramids CAXZD and CAXZU. Now if
we take away the first of the pyramids of the above pairs from the ungula of a
prism ABCXYZ, we get the tetrahedron ABCD. By taking away the second
97

Solutions
1964/6
of the pyramid pairs from ABCXYZ, we are left with the tetrahedron XYZU.
This clearly implies that ABCD and XYZU have equal volumes.
Now we turn to the solution of the problem. Fix D\ in ABC and suppose
that DD\ intersects A\B\C\ in D' {Figure 1964/6.2). According to our result
Figure 64/6.2
above the tetrahedrons ABCD1 and AiBiC\D\ have equal volumes. If we
prove that D\ D' is three times as much as DD\ then we verified that the volume of
ABCD is third of the volume of ABCD', hence of the volume of ΑχΒιΟχΌι.
In conclusion it is enough to show that DD' -2DD\.
We will use vector calculus in the verification of this last statement. Let
D be the origin; vectors pointing to given points will be labelled by lower case
letters coinciding with the labels of the points. Since D\ is in the plane given by
А, В and C, we get
(1) di =aa + /3b + 7C,
98

1965/1
1965.
where α + β + η=\, and this decomposition is unique. Since CC\ and DD\ are
parallel, we have cj = c + Adi. Substitute (1) and conclude
(2) C!=Aa:a + A/3b + (l + A7)c.
Since ci is in the plane of a and b, the coefficient of с is 0, hence A = .
7
Substituting this into (2) we get:
7C1 = — aa — /3b.
Similar reasoning gives aa\ = — βЪ — 7c,
/3bi = —aa — 7c.
The three equalities above yield
aai + βЬ\ + 7C1 = —2ά\.
Since a + β + 7 = 1, the endpoint of the vector on the left hand side is in the plane
of A\ B\ C\, while the endpoint of the one on the right hand side is on DD\. This
shows that
d' = -2db
implying DD' = 2DD\, which concludes the proof.
In summary, the volume of the tetrahedron ABCD is third of the volume
of A\B\C\D\ for any allowable choices of D\.
1965.
1965/1. Find all χ in the interval [0,2π] which satisfy
(1) 2 cos χ < |\/l + sin2ic —Vl -sin2a;| < л/2.
Solution. If χ satisfies the inequality then it necessarily satisfies cos χ <
л/2 . . .
—, implying
(2) 1<Х<Ц.
Let us first examine the right hand side inequality of (1):
I\/l + sin2ж — >/l -sin2a;| < y/l.
Since both sides are nonnegative, it is equivalent to its square
2 —2\/l — sin2 2x < 2,
i-e-t0 -Vl-sin^^O.
This latter expression is satisfied for all χ since the left hand side is nonnegative.
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1965/1
Next we study the left hand side inequality of (1):
(3) 2 cos χ < \Vl + sin 2x — y/\ — sin 2x\.
This is obviously satisfied if cos χ is nonpositive since the right hand side
is nonnegative. Consequently
π 3π
(4) -2<Χ<Ύ
belongs to the solution set. For this reason it is enough to consider (3) on the
7Г 7Г 3"7Γ 77Γ
intervals — < χ < — and — < χ < —. Here both sides of (3) are nonnegative,
4 ~ _ 2 2 ~ ~ 4 & >
hence (3) is equivalent to its square
4cos2 χ < 2 - 2\] 1 - sin2 2x = 2 - 2| cos 2x\,
from which 2 cos χ — 1 < — | cos 2x \
and so cos2a; < — | cos2a;|
7Γ 3"7Γ 7Γ
follows. This is true if cos 2x < 0, giving the constraints — < 2x < — or — +
Зтг
2π < 2x < — +2π for χ. These latter conditions transform to
~ ~ 2
π 3π 5π 7π
4 4 4 4
In summary, the solution set of (1) is the interval
π 7π
4 4
1965/2. 77ze coefficients of the system of equations
a\\X\ +а\2Х2 + а\ЪхЪ =®·>
a2\X\ + a22X2 + а2ЪхЪ = °,
α3\Χι + α32Χ2 + α33χ3 =®
are subject to the following constraints:
a) ац, a,22 and a33 are all positive,
b) a// other coefficients are negative,
c) the sum of coefficients in each equation is positive.
Verify that the only solution of the system is
X\ —X2 = X3-^·
100

1965/3
1965.
я\
Q2
Q3
а\2
^22
^32
°·\3
^23
ЯЗЗ
Solution. Suppose that the system admits a solution such that not all χι = 0.
Since for a solution (x\, X2, £3) also (—x\, —X2, —£3) is a solution, we may
assume that one of the a^'s is positive, suppose that the largest one is x\. Consider
the first equation and notice that xi < x\ implies а\2%2 > а\2хъ so:
0=a\\X\ +a\2X2 + CL13x3 >a\\x\ + a\2x\ + a\3x\ =
(αη+α12 + αΐ3)χι>0.
This is, however, a contradiction, so no solution Xi can be different from 0.
Remarks. 1. Both the problem and the solution admits a (suitable)
generalization for a system with η unknowns.
2. The solution also follows from the theorem asserting that such a
system admits a nonzero solution if and only if its determinant vanishes. Hence by
showing that this determinant is positive, we conclude x\ =X2 = x?>=Q·
For this reason add the second and third columns to the first one of the
determinant (these operations keep the value of the determinant unchanged):
D =
here q\, qi and q^ are the sums of the coefficients in the rows, respectively, hence
these are positive. Now we get D = q\(a,22a33 — α23α32) + #2(α32α13 ~а12азз) +
Чз{а\2а23 ~αΐ3α22)> and in this expression the second and the third summands
are positive because of a) and b) respectively. α22 + α23 > a2\ + α22 + α23 > 0
implies 0,22 > —a,23> and similarly 0,33 > — 0,32- Therefore α.22α33 > α23α32> and
so the first summand, and so D is positive.
1965/3. The length of the edge AB in the tetrahedron ABCD is a, while
the length of CD is b. The distance between the skew lines AB and CD is d,
the angle determined by them is ω. The tetrahedron is divided into two parts
by a plane ε parallel to AB and CD. We also know that к times the distance
between AB and ε equals the distance between CD and ε. Determine the ratio
of the volumes of the parts of the tetrahedron.
Solution. In order to get a better feeling about the problem, we pictured
the tetrahedron on Figure 1965/3.1 by the parallelepiped containing it. Now ε
cuts out the parallelogram XYZU from it if we intersect two planes by a third
plane parallel to their intersection line then the resulting intersection lines will
be parallel.
According to the problem the distance between the lower and the upper
faces of the parallelepiped is d, therefore the distance of ε from the lower face
is дДр and from the upper face this distance is ^-. Let XY = ZU = a' and
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Solutions
1965/3
Figure 65/3.1
YZ = UX = b'. By shrinking the edge AB from D with quotient XY:AB
:d=~—r we get XY, hence
k + l
(1)
k+l
XY-.
a
a
a
k + l
= k+l.
a'
Similarly, shrinking CD from В with quotient
bk
dk
(2)
YZ = b' =
k+l
b _k + l
d =
к
k + l
we get
k + V b' к
The plane ε divides the tetrahedron into the solids ABXYZU and
CDXYZU; the area of their common base will be denoted by t while their
kd d
altitudes are -—- and . Their edges parallel to XY and Υ Ζ are a and b,
k+l k+l &
finally their volumes are V\ and Vi respectively. Using (1) and (2) the ratio of
the volumes is
Vry1= kdt J2\a\- dt (2\b) = k{k + 3) k^k + 3)
6(k+l) V a'J 6(k + l)
3k+\
3k + l
Remarks. 1. The solution used a formula expressing the volume of the solids
ABXYZU and CDXYZU. Such solids (also called roof-shapes) can be got by
intersecting an (infinite) triangular prism with two planes in such a way that the
remainder of one face is a parallelogram, becoming the base of the solid (on
Figure 1965/3.2 this is the parallelogram XYZU). The edge parallel to the base
102

1965/4
1965.
Figure 65/3.2
face is called the spine (denoted by AB = д here). The distance of the spine from
the base is the height m of the solid. The sides of the base will be denoted
by a and b (a\\g). The solid ABXYZU can be decomposed into the pyramid
AXYZU and the tetrahedron ABYZ. Notice that the angle ω of the sides a
and b of the base parallelogram is equal to the angle of the edges AB and Υ Ζ.
Since the volume of the roof-shaped solid is simply the sum of the volumes of
the pyramid and the tetrahedron, we get
T„ mabs'mu) mqbs'mu) mab sin ω ( q\
v = о + ——r = 7 (2+-).
3 6 6 V aj
If the area of the base parallelogram is t (so t = ab sin ω), the above equality
implies
V =
tm
2+9-
a.
2. Notice that the ratio of the volumes depends only on k. The reason for
this phenomenon is that two tetrahedrons can be transformed into each other by
an affine transformation. Such a transformation maps parallel objects to parallel
ones and preserves the ratio of collinear intervals and volumes. Therefore the
ratio should be the same for all tetrahedrons.
1965/4. Find all sets of four real numbers x\, x^, £3, £4 such that the sum
of any one and the product of the other three is 2.
Solution. The problem is equivalent to solving the system
X\ +X2X?,%A = 2,
Х2 + Хъх4х\ -2,
χί, + Χ4.χ\Χ2 = 2,
X4 + X\X2%3 =2.
Notice first that xi^O, since for x\=0 the last three equations give а^-^з-
£4 = 2 which do not satisfy the first equation. For a solution (^1,^2,^3,^4) let
103

Solutions
1965/4
Q denote the product of the a^'s. It is easy to see that all four Xi satisfies
Q
x + -=2.
χ
This means that all x^ are roots of χ — 2x + Q = 0, hence the possible values of
Xi are
l+Jl-Q and 1-Jl-Q.
We conclude that there are at most two different values among the a^'s. In the
following we will analyze the various cases.
a) All Xi are equal. Let Xi-k (г = 1, 2, 3, 4), then к + к = 2, so
(k-l)(k2 + k + 2) = 0.
Since the second term has no real roots, the only such solution is x\ =X2 = £3 =
X4 = к = 1 which solves the system.
b) Three a^'s are equal and the fourth is different from them. Let x\ =X2 =
X2 = k, X4 = n and к ^п. The first and the fourth equations now read as
k + k2n = 2
n + k3=2.
The difference of the two equations imply (к — n)(l — k2) = 0; since k^n, this
gives fc =1 and so k = ±l. For fc = l the fourth equation gives n = k which is
excluded. In case k = — \ the fourth equation implies
n— 1 =2, n = 3,
providing xi=X2 = X3 = — l,X4 = 3, and all permutations of these. These numbers
satisfy the original system of equations.
c) Finally we assume that x\=X2 = k, x^=x^ = n and k-=fn. The first and
the fourth equations give
k + kn =2,
n + k n = 2,
and their difference shows (k — n)(l—nk) = 0. Since k^n, we have nk = l.
Substituting this we get k + n = 2, hence η and к are two different solutions
of ж — 2x +1 = (#— 1) =0. Since this equation provides a unique solution
\=k = n, we conclude that c) does not provide further solutions.
In summary, the solutions are
(1,1,1,1); (3,-1,-1,-1); (-1,3,-1,-1); (-1,-1,3,-1); (-1,-1,-1,3).
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1965/5
1965.
1965/5. The triangle OAB has angle l АО В acute. Μ is an arbitrary
point in OAB different from O. The points Ρ and Q are the feet of the
perpendiculars from Μ to О A and OB, respectively. Determine the locus of the
orthocentre Η of the triangle OPQ if Μ is
a) on AB;
b) in the interior of OAB.
Solution. Suppose that Μ is not on the lines О A and OB. Let Υ and X
denote the points of intersection of MP with OB and of MQ with О A (Figure
1965/5.1). Since Μ is the orthocentre of the triangle OXY, the correspondence
Figure 65/5.1
Μ <^H associates the orthocentre of OXY to the orthocentre of OQP. Let us
first examine this correspondence.
Since X, Y, Q, Ρ are on the same circle (hence both XPY and XQY are
right angles), XYPQ admits a circumcircle, hence
/OXY = lOQP = a, LOYX = LOPQ = $.
Therefore the triangles OXY and OQP are similar with ratio
OP
where 7 is the given acute angle. Consequently, by shrinking OXY from О with
cos 7 and then reflecting it to the bisector / of LAOB we get OQP. It means
that the correspondence Μ <-» Η is a similarity with centre O, axis / and ratio
105

Solutions
1965/5
cos 7. It is easy to see that the above said holds in case Μ happens to be on О А
or on OB. Consequently, the image of the triangle OAB (and so the interval AB
contained by it) can be got by shrinking it with ratio cos 7 and then reflecting it
to /. Since this transformation is one-to-one, the locus is simply the image of
the interval AB (the triangle OAB) under the above transformation.
1965/6. For η > 3 points in the plane denote the maximal distance of pairs
of points by d. Prove that at most η pairs of points are of distance d apart.
Solution. Let the points be denoted by P\,P2,...,Pn-\- An interval of
length d joining two points Pi and Pj is called the diameter of the set. The degree
of a point is by definition the number of diameters starting from it. Since the sum
of the degrees is twice the number of diameters, in case the degree of each point
is at most 2 we get that the sum of degrees is at most 2n, yielding that there are
at most η diameters.
We will solve the problem by induction on n: it is obviously true if n = 3
and if all points Ρ\,Ρ\,... ,Pn have degrees at most two. Suppose now that
there is a point, say P\, of degree at least 3 {Figure 1965/6.1). According to the
inductive hypothesis, among the remaining η — 1 points there are at most η — 1
diameters.
Figure 65/6.1
The diameters starting from P\ are on an arc of a circle к of radius d around
P\. Let the extremal points on this arc be called Pi and P3. As P2P3 is not more
than d, the arc P2P3 is not more than 60°. Since the degree of P\ is at least three,
there is a diameter joining it with an inner point of the arc ΡχΡ^, we denote this
point by Pn. Next we show that the degree of Pn is one.
106

1966/1
1966.
Suppose to the contrary that there is another diameter PnPi (besides PnP\)
starting from Pn. Obviously Pi is not on k, and cannot be outside of k, since then
Pi Pi is larger than d. In case Pi is inside k, it cannot lie in the triangle P1P2P3
since this implies PnPi < d; consequently Pi is outside of P1P2P3. This implies
that PnPi intersects P1P2 or P1P3; without loss of generality we can assume
that it intersects P1P3. This shows that P\P2PnPi is a convex quadrilateral, and
according to a famous inequality (see Remark 2.) the sum of the diagonals is
more than the sum of any pair of opposite sides:
PlP2 + PnPi<PiPn + P2Pi, i-e., d<P2Pi.
This, however, contradicts the definition of d, hence Pi does not exist, so the
degree of Pn is one. Let us now delete Pn; according to the inductive hypothesis
there are only η — 1 diameters on the remaining η — 1 points, and now adding
P\ back we get the statement we wanted to prove.
Remarks. 1. The basic combinatorial geometric theorem on which the
problem rests is due to H. Hopf and E. Panwitz, and there are many related problems
to it.
2. We used the following result: if ABCD is a convex quadrilateral then
AB + CD < AC + BD. Let Μ denote the point of intersection of AC and BD.
By summing the triangle inequalities AB <AM + BM and CD<CM + DM
we get the desired statement.
3. The problem admits a generalization into the three-space: η > 3 points in
the space determine at most 2n — 2 diameters.
1966.
1966/1. Problems А, В and С have been posed in a mathematical contest.
25 competitors solved at least one of the three. Amongst those who did not solve
A, twice as many solved В as C. The number of competitors solving only A
was one more than the number of competitors solving A and at least one other
problem. The number of competitors solving A equalled the number solving just
В plus the number of competitors solving just C. How many competitors solved
just B?
Solution. Let us visualize the sets of competitors solving А, В or С by
discs, while the students solving more problems correspond to intersections of
these discs {Figure 1966/1.1).
Now let Xa, Xb, Xc denote the number of those competitors who solved
just А, В or C, Xab denotes the number of those who solved just A and B, and
107

Solutions
1966/1
Figure 66/1.1
likewise Xabc stands for the number of students solving all three problems.
According to the problem
Xa + Xb + Xc + Xab + Xbc + Xac + Xabc = 25,
Xb + Xbc = 2{Xq + Xbc),
Xa = 1 + Xab + Xac + Xabc,
XA + XB+Xc = 2(XB + Xc).
We have to determine the (positive integer) Xb- After ordering we have:
(1) Xa + Xb + Xc + Xab + Xbc + Xac + Xabc = 25,
(2) Xb-2Xc -Xbc =0,
(3) Xa -Xab -Xac- Xabc = 1,
(4) Xa-Xb- Xc = 0.
Now subtract twice the fourth equation from the sum of the first three and
get:
(5) 4XB + XC = 26.
From (2) it follows that XB - 2XC = Xbc > 0, hence XB > 2XC, therefore
(5) implies
26 = 4XB + Xc> 8Xc + Xc = 9Xc-
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1966/2
1966.
Consequently
26
and since Xq is a positive integer, its possible values are 0, 1 and 2. For Xq = 0
and 1 equation (5) does not admit an integer solution for Хв, hence the only
possibility remains Xq-2. This implies
Xb = 6,
hence В has been solved by six competitors.
Remark. From the above solution now it is easy to determine the remaining
unknowns Xa-% and Хвс = ^\ the remaining three unknowns, however, can
take any positive integer values with sum equal to 7:
Xab + Xac + Xabc = 7.
1966/2. Let a, b, с denote the sides of a triangle, while the opposite angles
are denoted by α, β, η. Prove that if
(1) α+ 6 = tan— (a tan α + 6 tan/3)
then the triangle is isosceles.
First solution. We assume that the expression under (1) does make sense;
now express the tangent function as the quotient of sine and cosine and multiply
the equation by the product of the denominators. After ordering we get
sin^ (a sin a bsin/Λ
a + b = φ + -J.
cos ^ V cos α cos ρ /
(Ί Ί\ ( 7 τ\
cos a cos — — sin a sin — 1 + b cos a I cos β cos — — sin β sin — I = 0,
a cos β cos I a + — J + b cos a cos ί β + — J = 0.
Notice that since
we have
H)+H)=180°'
cosip4^J=-cosU + ^J ,
hence (1) is equivalent to
(2) cos la + — J (a cos β — b cos a) = 0.
If cos (a + -) =0 then a + -=90°, and also /3 + -=90° is satisfied. This
shows that α = β, consequently the triangle is isosceles.
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Solutions
1966/2
If the second term vanishes then
α cos/3 = 6 cos a,
i.e.,
(3) a2 cos2 β = b2 cos2 a.
Add the square of the equation a sin β = b sin a (originated from the law of sines)
to (3) and get
a2 sin2 /3 = fr2 sin2 a,
a =o ,
showing again that the triangle is isosceles.
Second solution. There are various geometric interpretations of (1). Let us
consider those two right angled triangles for which one acute angle is a (and β,
respectively), and one adjacent side is a (and b, respectively). The opposite sides
in these triangles are equal to α tan α and btз.nβ, respectively. Put these two
triangles next to each other as shown by Figure 1966/2.1. Now inscribe the two
triangles into a right angled triangle ABC with sides a + b and a tan a + b tan β.
С «tana b tan β
Figure 66/2.1
}
(Since a and b have symmetric roles, we may assume that b > a.)
110

1966/3
1966.
7
Since (1) uniquely determines —, and in the triangle ABC we have
..π^ a + b
tan Ζ ABC =
a tan a + b tan β'
we get that IABC = —. If the common vertex of our original subtriangles is
denoted by M, we have
ΖМЛБ = 90° - ^ - a, and so ΖМЛБ = a- (9О0 - ^) = a - 90° + ^,
ίΜΒΑ=Ί-- (90° -β), andso ZMiL4 = 900-/?-^
depending on whether Μ is on the same side of AB as С or it is on the opposite
side. In both cases:
\ΙΜΑΒ-ΙΜΒΑ\ = \ΙΖΟ°-α-β-Ί\ = 0, hence MABl = lMBA.
Therefore the (possibly degenerate) triangle ABM is isosceles, hence AM =
Β Μ . This shows that in the subtriangles
a2(l +tan2 a) = b2(l +tan2 β),
a2 cos2 β = $ cos2 a,
which is identical to equation (3) in our first solution. Since (3) implied a = b,
the statement follows.
Remark. In the second solution it has been used that a and β are acute. This
is legitimate, since the notation can be chosen to achieve b > a, hence β>α. This
shows that β cannot be obtuse, since in this case 180° — β is acute, implying
90°>180°-/3 = α + 7>α,
and therefore — tan β> tan a, i.e., (since tan/3 is negative) — tan/3= | tan/3| >
tana and (from b>a) also | έ> tan /31 > α tan a. This, however, shows that the
right hand side of (1) is negative, which is impossible since a + b is positive.
In conclusion β and so a are both acute.
1966/3. Prove that a point in the space has the smallest sum of distances to
vertices of a regular tetrahedron if and only if it is the centre of the tetrahedron.
First solution. For a better picture we draw the regular tetrahedron ABCD
with the cube containing it, see Figure 1966/3.1. The line t joining the midpoints
of the edges AB and CD is orthogonal to the two opposite faces of the cube,
hence t is orthogonal to the edges AB and CD.
Let Ρ be an arbitrary point in the space. The reflection of Ρ on t is P' and
the midpoint of PP' — i.e., the foot of the orthogonal from Ρ to t — is Q. Let
111

Solutions
1966/3
Figure 66/3. J
us first study the sum of the distances of Ρ from A and B, which is PA + PB.
Since we applied reflection, we get PB = P'A and so PA + PB = PA + P'A.
The interval QA is a median of PAP1. According to the median inequality (see
Remark 2.), the median of a triangle is shorter than the arithmetic mean of the
two adjacent sides, therefore the sum of the distances from Q to A and В is less
than the same sum from Ρ once Ρ is not on t.
The same holds for the
distances from С and D (since
we only used the fact that the
reflection of A to t is exactly
B, and the same relation holds
for С and D). Therefore if
Ρ is not on t then it cannot
minimize the sum of distances
from the vertices of ABCD.
The distance of Q from
A and В remains unchanged
if we rotate AB around t
until it becomes parallel to CD.
In this case t becomes the me-
Figure 66/3.2 dian of the rectangle ABCD.
Л
D
&
K^^^ \
1 ^^"-ч^\
t
В
112

1966/3
1966.
Let К denote the centre of this rectangle. The sum of the distances between К
and the vertices A, B, C, D is less than the same sum for any other point of t,
since if Q is a point of t different from К (Figure 1966/3.2), then according to
the triangle inequality QA + QC > AC and QB + QD > BD, showing
QA+QB+QC+QD>KA+KB+KC+KD.
Now К is the centre of the circumsphere of the tetrahedron ABCD, hence we
showed that the sum of distances from К is smaller than the corresponding sum
for any other point in the space.
Second solution. This solution rests on the following statement: the sum
of (oriented) distances of any point Ρ in the space from the faces of a regular
tetrahedron is constant, moreover it is equal to the altitude of the tetrahedron.
(The distance of a point from a face
of the tetrahedron is positive if it
is on the same side of the face as
the opposite vertex. For example,
for inner points of the tetrahedron
all distances are positive.)
The proof of this statement
proceeds by computing volumes.
Let the distance of Ρ from the
face ABC of the regular
tetrahedron ABCD be negative, and
positive from all other faces
(Figure 1966/3.3). The tetrahedrons
PABD, PBCD and PACD
together provide a cover of ABCD,
and even more; this additional
piece is exactly the tetrahedron
Ρ ABC. Let us denote the area of
the faces by t, the altitudes by m
and the distances of Ρ from the
various faces by dU, άβ, dc and dp
(in our case dp is negative). Since Figure 66/3
У ABC Ό = VPABD + VpBCD + У Ρ AC Ό — У Ρ ABC-,
multiplying it by 3 we get
tm = tdc + tdA + tds + tdrj,
implying
dA + dB+dc + dr) = m (constant).
113

Solutions
1966/3
A similar method works for points in other space segments, the number of
negative distances varies from 1 to 3.
Now we are ready to prove the statement of the problem. For a point Ρ
in the space dA, άβ, dc, dp denote its distances from the faces, and the feet of
the orthogonals are denoted by PA, Рв, Рс and Pd, respectively. The triangle
inequality implies
PA + PPA>m, i.e. PA + dA>m,
since PA + Ρ Pa > APA > m. This last inequality remains true even if PPA = dA
is negative (Figure 1966/3.4), with equality if and only if Ρ is on the altitude
starting from A. Writing down the above inequalities for all vertices we get
Figure 66/3.4
PA + dA>m, PB + dB>m, PC + dc>m, PD + dD>m.
By summing them we conclude
PA + PB + PC + PD>4m-(dA + dB + dc + dD) = 3m
with equality if and only if Ρ is on all altitudes of the tetrahedron, in other words,
if Ρ is the centre of the tetrahedron.
Remarks. 1. The first solution did not really made use of the regularity of
the tetrahedron. We only used that it admits two edges with equal length and the
interval joining their midpoints is perpendicular to both. This property is satisfied
by a wider class of tetrahedrons.
114

1966/4
1966.
2. The proof of the median inequality used in the first solution can be
verified as follows: Let the reflection of the vertex A of the triangle ABC to the
midpoint of the opposite edge be denoted by A'. Now AA' = 2sa, and the triangle
b + с
inequality for ABA1 gives b + c> 2sa, or equivalently sa < —τ-·
(1)
1966/4. Prove that
1
+
1
sin 2x sin Ax
+ ...+
1
sin 2nx
= cota; — cot 2 x.
λττ
for any positive integer η and any real χ (with χ-4—r, k = 0, 1, 2,
2k
an arbitrary integer).
Solution. For such χ we have the identity
η and λ
cot χ — cot 2x =
cos χ cos 2x cos χ
cos2 χ — sin χ
1
sin a; sin 2a; sin a: 2 sin a; cos a; sin 2a;
In verifying the equality of the problem we just use the above identity over
and over again:
1 1 1
+
sin 2a; sin 4a;
>n—1
+ . . .+
sin 2na;
= (cot χ — cot 2x) + (cot 2a; — cot 4a;) + ... +
+(cot 2n_1 χ - cot 2na;) = cot χ - cot 2na;,
which is exactly the statement we wanted to prove.
У
5(0,1)
1
οΊ
,
/2χ
Рг
2гх/
'i——■"-"""'—
χ
Ъс^-
Λ
-ί-—~~
ρα
χ
Figure 66/4.1
Remark. In case all the angles appearing are acute, the problem has a
geometric interpretation. The line with equation у = 1 is intersected with the lines
passing through the origin with slopes x, 2x,..., 2na; in the points Ρ0,Ργ,...
..., Pn (Figure 1966/4.1); the distances of these points from B(0,1) are cot a;,
cot 2a;, ..., cot2^at. Consequently
Pi+lPi = cot2*a;-cot2i+1a;,
therefore twice the area of the triangle OPi+\ Pi is cot 2га; - cot 2г+1 х, while twice
the area of OPn+\P\ is equal to cot χ — cot2na;.
115

Solutions
1966/4
On the other hand, OP; = ——— and ОЛ-+1 = ^τ-, hence twice the
sin2*ar г+[ sin2*+V
area of OPi+\Pi is
sin 2гх _ 1
sin2*;c-sin2i+1;c sin 2й"1 ж'
Now (1) just says that the area of OPn+\P\ is the sum of the subtriangles
OPi+iPi.
1966/5. Solve the system of equations
\a\ — a,21^2+ \al ~ аз1ж3 + lal ~ a4\ XA~ 1)
la2 — al 1^1 + \a2 ~ аъ\хЪ + \a2 ~ Q-4|^4 = 1,
|аз — a\ \x\ + |аз — Cb2\x2 + \a3 ~ а4|ж4 = 1)
|a4 — aj \x\ + |a4 — 0,21^2 + |сц — а-з|^з = 1,
where a\, (12, 0-3, 04 denote four distinct reals.
Solution. First we will get rid of the absolute value signs. Notice that
by permuting the c^'s the system remains unchanged, so assume that ае<а^<
ah < an (here e к h η is just a permutation of 1 2 3 4). Introduce the notations
q=ae, c2 = ak, c3=ah, cA = an and xe = y\, xk = 2/2> хк~Уъ^ хп=У4> hence q <
c2 < c3 < c4· Now the system has the following form:
(1) . (c2-Ci)2/2 + (c3-Ci)2/3+(c4-cl)2/4 = l>
(2) (C2-q)2/i+ (сз-С2)2/з+(с4-С2)2/4=1,
(3) (c3-ci)yi+(c3-c2)y2+ (c4-c3)y4 = l,
(4) (с4-С1)У1+(с4-С2)г/2 + (с4-сз)УЗ =1·
By taking the difference of each equation with its preceding equation
together with the sum of the first and the last we get
(C2 ~ С] ){y\ - У2 - 2/3 - У A) = °>
(сз - C2X2/1+У2-У3- 2/4) = °,
(C4 - C3X2/1 + 2/2 + УЗ - У4) = 0,
(c4 - q )(2/j + y2 + УЪ + 2/4) = 2>
The Ci's are distinct, so we can divide by their differences, resulting
(5) У\-У2-УЪ-У4 = ®,
(6) У\+У2-У3-У4 = °,
(7) У\ +У2 + УЗ- 2/4 = °,
2
(8) 2/1+2/2 + 2/3+2/4 = ·
С4 - С!
The difference of (6) and (5), and similarly the difference of (7) and (6) shows
2/2 = 2/3=°;
116

1966/6
1966.
and the difference of (8) and (7) together with the sum of (5) and (8) gives
1
У4 = У\= ·
c4-ci
This implies that in case ае<ак<а^< an the solution is
1
Xk=Xh = 0,
JjQ — Jjn —
CLri CLf
"n
Now it is not hard to verify that this is actually a solution of our original system.
1966/6. Take any points K, L, Μ on the sides AB, ВС, С A of the triangle
ABC. Prove that at least one of the triangles MAL, KB Μ and LCK has area
at most fourth the area of ABC.
First solution. In the following we will adopt the convention that if an
interval of length d is cut into two pieces then one will be denoted by dx while
the other one by d(l —x) (0<x< 1).
Let the sides of the triangle be denoted by c, a and b, and the points M,
K, L divide them as cz, c(l — z); ax, a(l — x) and by, b(l — y), respectively (x,
y, ζ are positive reals < 1, see Figure 1966/6.1). Let t denote the area of ABC,
the area of the sub triangle MAL at A is t^, while for the other two Ьв and tc
denote their respective areas.
b(l-y)
ciX-z)
Figure 66/6.1
We need to show that among the ratios
tA te tc
117

Solutions
1966/6
there is one not exceeding -. Notice that
4
tA b{\ — y)cz · sin a
t be sin a
= (1-2/)*,
and similarly
γ = (1-ζ)χ, — = {l-x)y.
Therefore we only need to demonstrate that the products
(1) (1-2/)*, (l-*)s, (l-s)j/
u 1
contain one not more than -.
4
Apply the inequality between the arithmetic and geometric means and
conclude that
2
x(l — x) <
'x+l
χ
2
1
4"
(1 -y)z · (1 - z)x -(1 -x)y = x(l - x) -y(l -y) · z(l - z) < (-) ,
hence not all the terms in the product can be larger than -. This, however,
concludes our solution.
Second solution. Let the midpoints of AB, ВС, С A be denoted by C1,
A', B'. The sides of the subtriangle A'B'C' partition ABC into four subtriangles
with areas equal to fourth of the area of ABC.
If two of the
points K, L, Μ are in one
of the subtriangles then
the statement of the
problem is obvious. Therefore
we can assume (after
possible renaming) that K,
L and Μ are on the
intervals ΒΆ, CB' and
AC1, respectively (Figure
1966/6.2). We will show
that the area of KLM is
at least fourth the area of
ABC, in other words its
area is at least the area of
A'B'C.
Μ
С
Figure 66/6.2
118

1967/1
1967.
By pushing the vertex К of KLM into A' we do not increase the area of
KLM since the altitude belonging to LM did not increase. Therefore
(2) area of KLM> area of A 'LM.
Now the area of A'LM does not increase when we push L into Β', since the
altitude belonging to MA! did not increase, hence
(3) area of A'LM> area of Ά B'M.
Notice, however, that the area of A1 B'M coincides with the area of A'B'C'
(consequently with fourth the area of ABC), so (2) and (3) imply
area of KLM >- · area of А В С.
~ 4
Therefore the sum of the three subtriangles of ABC given by the sides of KLM
3
is at most - the area of ABC, hence there is one with area at most fourth the
area of ABC.
Remarks. 1. Since (1) is unsensitive for cyclic permutations of x, у and z,
we may assume that χ is the largest among x, y, z, hence у < χ, showing
(l-x)y<(l-x)x<-.
2. There are many noteworthy connections between ABC, KLM and the
further subtriangles; here we mention only a few of these
a) one of the triangles MAL, KBM and LCK has area not more than the
area of KLM;
b) the circumference of one of MAL, KBM and LCK is not more than
the circumference of KLM.
1967.
1967/1. The parallelogram ABCD has AB = a, AD = 1, angle I DAB = a
and the triangle ABD is acute. Prove that the circles Кд, Kb, Kq and Kd of
radius 1 centered at А В, С and D cover the parallelogram if and only if
(1) a<cosa + v3sina.
Solution. Let R denote the radius of the circumcircle of ABD (and so of
CDB, which is congruent to it). If К a, Kb, Kd cover ABD, then Kb, Kq,
Kd cover CDB. We will show that R < 1 is a necessary condition for this to
happen.
119

Solutions
1967/1
cos α
a - cos α
Figure 67/1.1
Since ABD is acute, the centre О of its circumcircle is in its interior. We
obviously need R < 1 in order to cover O. This assumption is also sufficient: in
order to show this, choose Ρ in ABD (Figure 1967/1.1).
Ρ is in one of the six right angled triangles we get by considering the radii
О A, OB, ОС together with the orthogonals from О to the sides. Now if Ρ is
in one of the triangles containing D as a vertex, then since DP < DO = R, Kb
covers P.
We only need to show that R < 1 is equivalent to (1).
AD 1
Let /ABD = p. Since R = —.—^ = —:—^, we get that R < 1 is equivalent
1
2sin/3 2sin/3
to sin/3 > -, i.e. to β >30°. Let the foot of the orthogonal from D to AB be
denoted by T. In DTB we have DT = sin а, ТВ = а — cos a and so
n a — cos a
cot/3 =
sin α
In this region β > 30° is equivalent to cot/3 < y/з, hence it is the same as
a — cos a
sin α
<>/3,
or
a< cosa + л/3 sin α.
This is identical to the statement we wanted to prove.
1967/2. Prove that a tetrahedron with just one edge of length greater than
1 has volume at most -.
8
120

1967/2
1967.
Solution. The
main idea of the solution
can be summarized as
follows: Let the longest
edge of the tetrahedron
ABCD be CD, and the
length of the opposite
edge AB be χ < 1. We will
give an estimate for the
volume of the tetrahedron
as a function of x.
Let us first
estimate the altitude CT of the
triangle ABC. Suppose
that on AB, A is the
closest vertex to T. Then
BT>-, and from the
~ 2
right triangle BTC we conclude (Figure 1967/2.1):
A
Figure 67/2.1
(1)
Xх
ct=Jbc2-bt2<Ji-^-.
Similar reasoning gives that the altitude m\ of ABD belonging to AB satisfies
(2) mi <
The altitude CQ = m of the tetrahedron cannot be more than CT, therefore
(3)
m <
1 1
Now for the volume V of the tetrahedron the expression V = - · -AB · m\
m together with (1), (2) and (3) implies
(4)
1
V<%[\-X-\=^x{A-x2)·
1 о
We have to show that V <-, which means x(4 — χ ) < 3 according to (4). This
8
is true, because
3 - x(4 - x1) = (1 - x)(3 - χ - x2) > 0,
since in [0,1] we have 1 — χ > 0 and 3
χ
x2>0.
121

Solutions
1967/2
Remarks. 1. The function x(4 — x2) in [0,1] considers its maximum at χ = 1,
and in a tetrahedron corresponding to this maximum the faces ABC and ABD
are equilateral triangles with sides of unit length, and their planes are orthogonal
to each other.
2. The maxima of the function under (4) cannot be directly determined using
differentiation since the zeros of the derivative are outside of the unit interval
1 о
[0,1]. In fact, the derivative —(4 — 3x ) is positive and monotone increasing in
[0,1] hence it takes its maximum at the endpoint x = l.
1967/3. Let k, m, η be positive integers such that m + k + 1 is a prime
greater than (n + 1) and let cs = s(s + 1) ( s = 1, 2, ...). Prove that the product
(1) (cm+i - ck)(cm+2 - ck) ·... · (cm+n - ck)
is divisible by
(2) c\c2...cn.
Solution. Let Ρ and Q denote the products under (1) and (2). First we will
examine these expressions in detail. A generic term in Ρ admits the form
ca - cb = a(a + 1) - b(b + 1) = a2 - b2 + a - b = (a - b)(a + b + 1).
From this we get that
p = (m-k+ l)(ra + к + 2)(ra - к + 2)(ra + к + 3)... (m - к + n){m + k + n+l) =
= ((m-k+l)(m-k+2)... (m-k+nj) ((m+k+2)(m+k+3)... (m+k+n+l)).
Let us introduce the following notations:
A = (m — к + l)(m — к + 2)... (m — к + ή),
В = (т + к + 2)(т + к + 3). ..(m + k + n+l).
With these notations P = AB. Next we focus on Q:
ρ = 1(1 + 1)2(2+1)...η(η + 1) = (1·2·...·η)(2·3·...·(η+1)) = η!(η + 1)!
In conclusion, we have to prove that
AB A B
n!(n + l)! n! (n+1)!
is an integer.
A B
To achieve this, it is enough to confirm that — and — are both integers.
n! (n+1)!
Now — is an integer, since by its definition A is the product of η consecutive
n!
integers, hence it is always divisible by n! (see also the remark following the
solution).
122

1967/4
1967.
ту
In order to show that is also an integer, notice that it is an integer if and
(n + 1)!
only if —— is an integer, since (according to the problem) m + k + l
(n + 1)!
is a prime larger than n+1. According to the definition of В the expression
(m + k + l)B is the product of n+1 consecutive integers, so it is divisible by
(n + 1)!. This last argument now completes our solution.
Remark. In our solution we used the fact that the product of η consecutive
integers is divisible by n!. It means that if the first term is k +1 then
_ (fc + l)(fc + 2)...(fc + n)
E = 1
n!
is an integer. Notice that
kl(k+l)...(k + n) (n + k)\
Ε = = =
k\n\ k\n\
which (according to basic properties of binomial coefficients) is an integer.
1967/4. AqBqCq and A'B'C' are given acute triangles. Construct the
triangle ABC with the largest possible area which is circumscribed around
AqBqCq (i.e., AB contains Cq, ВС contains Aq and С A contains B$) and is
similar to A'B'C' (А, В, С correspond to A', B' and C').
Solution. First let us inscribe a triangle in A'B'C' similar to AqBqCq
in such a way that its vertices are on the edges B'C', C'A' and A'B'. This is
possible, since for an arbitrary interval CqBq parallel to B'C a triangle similar
to AqBqCq can be constructed on its side opposite to A'. Then it can be shrunk
from A' until the vertex corresponding to Aq will lie on B'C'. For the sake of
simplicity this inscribed triangle will be denoted by AqBqCq as well (Figure
1967/4:1).
We denote the intersection (other than Cq) of the circumcircles к a and кв
of A'BqCq and B'AqCq by M. Since A'B'C and AqBqCq are both acute, Μ
is an inner point of the triangle AqBqCq and hence of A'B'C'. It is on the arcs
BqCq and AqCq of the circles к a and кв not containing A' and B'. Therefore
if the angles of A'B'C' are α, β, 7, then
LBqMCq = 180° - a, LAqMCq = 180° - A
and so
/ЛоМБо = 3600-(1800-а)-(1800-/?) = а + /?=180°-7.
This shows that Μ is on the arc AqBq of the circumcircle of C'BqAq not
containing С.
123

Solutions
1967/4
Figure 67/4.1
Since the rectangles MBqA'Cq, MCqB'Aq and MAqC'Bq admit circum-
circles, we have IA'C0M = IB'A0M= IC'B0M (Figure 1967/4.2). Let φχ and
Ψ2 denote stretching-rotations with centre Μ taking the line A'B1 into B'C' and
B'C' into C'A'. It follows that ψι(Ο0) = Α0 and φ2(Α0) = Β0. F°r an
arbitrary point C\ on A'B' with φ\(0\) = Αι and φ2{Α\) = Β\, the triangles Ло-ВоО)
and ΑγΒγΟγ are similar since they are composed from the similar triangles
CqMAq~CiMAi, AqMBq~AiMB\. Therefore starting from a point C\ on
A'B', infinitely many triangles inscribed in A'B'C' and similar to AqBqCq can
be constructed.
Next we will show that Μ is determined by the triangles A'B'C' and
AqBqCq: For another M* with the same properties we could achieve that sides
of AqBqCq were parallel to corresponding sides of AqBqCq, and so in A'B'C'
there are two similar triangles in similar positions which is clearly a
contradiction.
In conclusion, with the aid of the point Μ we can describe all triangles
inscribed in A'B'C' in the required way. The one with the minimal area will
obviously have the feet of the orthogonals from Μ to the sides as vertices. These
feet are interior points of the sides since A'B'C' is acute and Μ is an inner point.
124

1967/5
1967.
С
В'
Figure 67/4.2
The construction concludes by shrinking the triangle we got until the
resulting AqBqCq will become congruent to the triangle given by the problem.
If AqBqCq was the inscribed triangle with the smallest possible area then the
corresponding circumscribed triangle has the largest possible area.
Remark. Of course the problem has many other solutions. The difficulty in
evaluating such a solution is to find the appropriate amount of precisity when
discussing facts which are obvious when visualized. The solution given above
roughly reflects how much is expected in the competition. For example, the fact
that Μ is in AqBqCq is fairly obvious, but a rigorous proof requires quite an
amount of work.
196775. Consider the sequence {cn} given as
cl = al + a2 + · · · + a8
2 2 2
c2 = a\ + a2 + · · · + a8
for a\, 0,2, · · · j Q-8 reals, not all zero.
Given that an infinite number of {cn} is zero, find all η for which cn = 0.
125

Solutions
1967/5
Solution. If η is even, then the summands defining it are all nonnegative
and contain nonzero terms, hence cn ^0.
Suppose now that η is odd. There are positive and negative reals among ai
otherwise the sum of their (odd) powers could not be zero. Let a\ has the
largest absolute value, while a2 has the largest absolute value among those having
opposite sign than a\. Obviously, a\ and a2 are nonzero. Write cn as
<■> __ -Κ1*©"*©"*-*(ϊ)Ή*
The term — is negative and not more than any other —, and the same holds for
all their odd nth powers. Therefore
(2) 5>1+7^V.
According to (1) now cn is zero only in case £ = 0 since a\^0. If \a\\ ^\a2\ (i-e-
\a2\ < lall) tnen ( — ) converges to zero, hence after some no the expression
a2
η ι
is less than -. In that case the right hand side of (2) is positive; and the
a\ 1
same holds for S and cn. This implies that cn is 0 only in finitely many cases,
showing that la-il^l^l ls impossible, so a2 = — α\, in other words a\ + a2 = 0.
This means that a" + aj = 0 f°r a^ °dd n.
If all аз, щ, ..., ag are zero, then we are done. If there are nonzero reals
among these six numbers then the method given above can be applied again. In
conclusion we get that the a^'s can be paired up to give 0 summands for odd
powers, hence cn = 0 for all odd η and these are the only indices with cn = 0.
1967/6. In a sports contest a total of m medals were awarded over η > 1
days. On the first day one medal and — of the remaining medals were awarded.
On the second day two medal and - of the remaining medals were awarded, and
so on. On the last day the remaining η medals were awarded. How many medals
and over how many days were awarded?
First solution. The number of medals awarded on the fcth day will be
denoted by e^, the number of medals available at the beginning of the same day
is mfc. According to the assumptions
mk-k mk 6k
(1) ek = k + —-— = — + —.
Using mk_\ — mk = ek_\ we would like to find some connection between ek and
4-1'·
mk_i-mk 6 ek_\ 6
ek_i -ek =
7 111
126

1967/6
1967.
(2) ^-^ΊΓ-1·
According to the problem en = n, therefore en_\ = —-—. This shows that η is
6
divisible by 6, so η >6. With p = n — 6 (p > 0) we get
7en In 7p
en -1 = —r ~ l = ~Γ ~ l = ~F + 6'
so (2) implies
en_2=i-j P + 6,en_3 = (-j p + 6,...,ei = l-j p + 6.
e\ is an integer only if ρ = η — 6 is divisible by 6 . This, however, implies
η = 6 since η — 6 < 6 . Therefore, the solution is:
η = 6, ej =β2 = ... = eg = 6 and m = 36.
Second solution. We will count the medals which remained after each day:
Since mk — mk+i = ek, (1) shows
77lfc — /c 6(?7lfc — fc)
mfc+l = mk~ek=mk-k-
7 7
7
For the sake of brevity let q denote -, and express mk as
mk = k + qmk+h
Repeating this process we get
mn =n
mn_i=(n-l) + qn,
mn-2 = (n-2) + q(n-l) + q2n,
m = m\=\+2q + 3q2 +.. .+nqn l.
Writing the sum on the right hand side in closed form we arrive to
(3) m=l+2q + 3q +...+nq = —~2 = —-^-+6.
ι ι
m is an integers only in case η — 6 is divisible by 6n and since η — 6 < 6n
this implies η = 6, showing that e^ = e2 = ... = eg = 6 and m = 36.
Remarks. 1. The proof of the identity in (3) goes as follows
S =l+2q + 3q2+ ... + nqn~l
Sq= q +2q2 + 3q3 + (n - l)gn_1 + ngn
127

Solutions
1967/6
Sq-S = S(q-l) = -(l+q + q2 + ... + qn-l) + nqn =
i+l
= nq
η
qn-\ nqn+i -(n+l)qn + l
implying
S =
nq
n+1
1
(n+\)qn + \
1
(<z-D2
2. We just mention here that the Hungarian sports press dealt with the
IMO only because of the above problem. A magazine blamed the organizers
that by posing such "trivial" problems, the Olympic spirit had been discredited.
The members of the Hungarian team invited the editors to their annual meeting
and asked them to show the solution of this trivial problem. Needless to say, the
answer and the solution has not been arrived at in the past 25 years ...
1968.
1968/1. Show that there is a unique triangle whose side lengths are
consecutive integers and one of whose angles is twice another.
First solution. With the usual notations let β = 2a. The bisector at В
intersects AC in B' (Figure 1968/1.1) and divides it into subintervals of lengths
CB' = and B'A = . (This is because an angle bisector divides the op-
a + c
a + c
В
a+c
a+c
Figure 68/1.1
posite side with the ratio of the adjacent sides.) Since their angles agree, ABC
and BB'C are similar triangles. Therefore
(1) ^ = ^
y J AC ВС
i.e. BC2 = AC-CB'
128

1968/2
1968.
о ab о
(2) α = , bz = a(a + c).
a + c
According to our assumption on the angles b > a holds; this implies that
either b = a+l or b = a + 2:
In the first case
2 1
(a + 1) =a(a + c), c = 2 + —,
a
and since с is an integer, this shows a= 1, and so b = 2 and c = 3; there is,
however, no triangle with these sides.
In the second case b = a + 2, but since a, b, с are consecutive integers, с is
between a and b: c = a + l. Now (2) gives
(α + 2)2 = α(2α+1), α2-3α-4 = 0,
(α+1)(α-4) = 0.
The unique positive solution of this equation is a = 4, implying b = 6 and с = 5. In
conclusion there is at most one triangle satisfying the assumption of the problem.
In order to show that in the triangle of sides 4, 5 and 6 the angle opposite to the
side of length 6 is twice the one opposite to the side of length 4, we denote the
edges as follows: AB = 5, ВС = 4, CA = 6. As we noted towards the beginning
24
of the solution, we have CB' = -—. The triangles ABC and BB'C are similar
(they share an angle at С and the ratio of the adjacent sides). This implies that
LCBB' = LCAB = a, hence the angle at В is 2a.
Second solution. The key ingredient of the above solution is formula (2)
— this formula can be derived using trigonometry as follows: Starting with β =
2a we apply the law of sines:
b sin 2a b2 + c2 — a2
- = —. = 2 cos α = ,
a sin α be
о 9 9 о 9
b c = ab +ac —a, b (c — α) = α(α + c)(c — a).
Since Cyta, it shows
9
b =a(a + c).
Now we complete the solution as above.
1968/2. Find all positive integers χ for which
(1) p(x) = x2-\0x-22,
where p(x) is the product of the decimal digits of x.
129

Solutions
1968/2
First solution. After taking a few examples it is not hard to see that the
number of digits of χ should not exceed 2: if η is the number of digits of χ and
η > 2, then — since the product of the digits is at most 9n:
9n>p(a;) = a;(a;-10)-22>10n-1(10n-1 -10)-22 =
= 10n(10n_2-l)-22>10n,
providing a contradiction. Since p(x) = x — \0x — 22 = x does not admit integer
solution, the number of digits cannot be one either. Thus we get that
x = 10a+ 6,
where 1 < α < 9, 0<b<9. It transforms the equation into
(10α + 6)2-10(10α + 6)-22 = α6,
which is equivalent to
(2) 100a(a-l) + (19a-10)fr + (fr2-22) = 0.
If a > 2, the left hand side is at least
200 + 28Ь + Ь2-22 = Ь2+ 286 + 178,
which is strictly positive. Therefore a= 1, and now (2) shows
b2 + 9b-22 = 0.
The positive solution of this equation is 2, hence x = \2, and this number satisfies
(1)·
Second solution. p(x) = 0 does not admit integer solution, hence p(x) is
positive, so
ж2-10ж-22>0, (ж>0)
implying
(3) χ > 11,86.
If a; is of η digits with first digit equal to A then χ = 10n_1 A + B, with В of η - 1
digits. From this the maximal value of the product of digits of χ is A · 9n~ ,
hence
p(x)<A-9n-1 <\0n-lA<l0n-lA + B = x
so
ж2-11ж-22<0,
implying
(4) ' χ < 12,73.
Based on (3) and (4) we get that 11,86 < χ < 12,73, and since χ is an integer, the
only possible solution is x = 12, which number eventually solves the problem.
130

1968/3
1968.
1968/3. Prove that the system
(1)
о
αχ ι +bx\ +c = X2,
aX2 + bx2 + с = X2,
aju γ. ι ujb^i ι ο — ju ι
(a, b, с are real with a^O)
9
I. has no real solution if (b — 1) — Aac < 0;
II. has one real solution if(b—l) —4ac = 0;
9
III. and has more than one real solution once
Solution. Notice that (b - l)2 - 4ac is equal to the discriminant of the
quadratic equation
(2) f(x) = ax2 + (b-l)x + c = 0.
The ith equation in (1) can be written as
axi+bxi + c = xi+\ (n + l:=l)
(3) f(xi) = axj + bxi-Xi + c = xi+i-Xi',
adding these expressions for г = 1, 2, ..., η we get
(4) f(xl) + f(x2) + ... + f(xn) = 0.
Hence for a solution x\, x2, ..., xn °f (1) the above equality is satisfied.
I. In case (b - l)2 - 4ac < 0 the polynomial f(x) has no real solution, hence
it is either positive or negative for all x, therefore the left hand side of (4) cannot
be 0. This shows that (1) admits no solution in this case.
II. If (b — 1) — Aac = 0 then f(x) is either nonnegative or nonpositive for all
x. Therefore (4) holds only if
f(xi) = f(x2) = ... = f(x4) = 0.
The unique solution of f(x) = 0 is x = , hence the unique solution of (1) is
2a
_\-b
X\ —X2 — · · · — xn — ~^ ·
la
III. Finally if (b- l)2 -4ac>0, then f(x) has two different real roots гц
and y2. Now
х\=Х2 = ... = хп = У1, and х\=Х2 = --- = Хп = У2
both satisfy (1), hence the system admits more than one solution.
131

Solutions
1968/3
Remark. Notice that III. does not prove that in that case the system
admits exactly two solutions; we only showed that there are at least two different
solutions. In fact, in certain cases there are more solutions, for example
о
Χγ —4,5^1 +6 = X2
о
£2 — 4,5x2 + 6 = x\
is solved by the following four pairs:
(1,5; 1,5) (4; 4), (1;2,5) (2,5; 1).
1968/4. Prove that every tetrahedron has a vertex whose three edges have
the right lengths to form a triangle.
Solution. The intervals a, b, с form a triangle if and only if the following
(so-called triangle) inequalities are satisfied:
a<b + c, i.e. — a + b + c>0,
b<c + a, i.e. a — b + c>0,
c<a + b, i.e. a + b — c>0.
If a is the longest among the three intervals then it is enough to check the
first inequality, the other two are automatically satisfied.
Label the vertices of the tetrahedron in a way that the longest (or one of the
longest) edge is be denoted by AB. Apply the triangle inequality for the triangles
ABC and ABD:
-AB + BC + AC>0,
-AB + BD + AD>0.
Adding up these inequalities we get
(-AB + AC + AD) + (-AB + BC + BD)>0.
This means that at least one expression in the parentheses must be nonne-
gative.
If -AB + AC + AD >0 then AB, AC and AD form a triangle, since AB
is (by our assumption) the longest edge in the tetrahedron.
In case -AB + BC + BD >0, the edges AB, ВС, BD form a triangle. In
conclusion the edges of Л or Б form a triangle.
Remark. For tetrahedrons with the property that the two longest edges are
opposite, one can find a vertex whose edges form an acute triangle.
132

1968/5
1968.
1968/5. Let f be a real-valued function defined for all real numbers, such
that for some a > 0 it satisfies
(1) f(x + a)=^ + Jf(x)-(f(x))2.
I. Prove that f is periodic, i.e., there exists a positive real b such that
f(x + b) = f(x)
holds for every x.
II. Give an example of such a non-constant f for a=\.
Before proceeding further we note that
(2) \ < fix) < 1
holds for all χ since fix) is the sum of - and a nonnegative real, and the exp-
ression under the square root is nonnegative only if
/(s)(l-/(s))>0,
which (according to fix) > -) implies fix) < 1.
First solution. I. Since (1) provides a relation between f(x + a) and fix),
it seems plausible to examine the substitution χ ^ χ + a in (I):
fix + 2a) = - + V f{x + a) - (f(x + a))2 =
=\+i\+^f(x) ~(f(x))2 ~(l~ ^m"(/(a0)2)=
= \ + Л + у/fix) - (Я*))2 -\- Я*) + (fix)f ~ y/fix) ~ (fixf =
=\+γ \"m+(/(ж))2=\+γ (/(a°" I)2=\+m ~\=f(x)-
This exactly shows that b = 2a is a period of /.
II. A simple example satisfying the assumptions of the problem can be given
by the following formula (see also Figure 1968/5.1):
fix) =
This function obviously satisfies (1)
1, if2n<x<2n + l,
- if 2n +1 < χ < 2n + 2, η arbitrary integer.
133

Solutions
1968/5
i
1
<> 1/2
-1
0
4
χ
Figure 68/5.1
Second solution. Regard (1) as a quadratic equation with f(x) as the
unknown. Taking its square we get
f(x + a)-^=Jf(x)-(f(x))2
f(x)2-f(x)+(f(x + a)-^J =0
The solutions are provided by the quadratic formula:
1
- + Jf(x + a)-(f(x + a)) ,
f(x + a)-(f(x + a)Y
1
The second root is less than -, hence the unique solution for f(x) is:
f(x)=-+y/f(x + a)-(f(x + a))2.
Now perform the substitution and get
f(x-a) = ±+y/№-{№)2.
According to (1) this, however, shows that
f(x-a) = f(x + a),
hence a final substitution χ -^ χ + a implies
f(x) = f(x + 2a),
consequently b = 2a is a period of /.
Remark. There are many ways to produce an example of a function with the
required properties. A continuous function with period 2 (a= 1) can be given by
the formula
№)=4(i +
πχ
sin
134

1968/6
1968.
1968/6. Let [χ] denote the greatest integer not larger than χ (the "integer
part" of x). For every positive integer η evaluate the sum
~n + 2k^
(1)
n + 1
+
n + 2'
"22-
+ ...+
2k+l
+ . . .
First solution. Notice first that the sum in (1) is finite, since once 2 > η
wehaven<2fc=2fc(2-l) = 2fc+1 -2k, and so
n + 2k
2k+l
< 1, therefore
n + 2k
2k+l
= 0.
First we prove the following statement: for any real χ
(2)
1"
x + —
2
= [2x]-[x].
1
Any real χ can be written either as m + a or as m + - + a, where m is an integer
and 0 < a < -. In the first case
2
[x] =m,
1"
x + -
2
= m, [2x] = [2m + 2a] = 2m,
implying (2). In the second case
[x] = m,
1'
x + -
2
m + 1, [2x] = [2m + 1 +2a] = 2m + 1,
which again implies (2).
Using (2) (and assuming that 2k >ri) now write (1) as:
'n+1
+
n + 2
η Γ
- + -
2 2
22
+
+ ...+
n + 2k
2k+l
' η 1
22 + 2
+ ...+
+ .. .=
η Г
+ -
= [n]
η
2
+
η
2
η
22
2k+l 2
η
+ ...+
2k
+ .. .=
η
2k+\
= [n]
η
2k+l
= [n] = n,
since
η
2k+l
= 0. Therefore the value of the sum in (1) is equal to n.
Second solution. Write η in base two:
η = as ■ 2s + as_\ ■ 2S_1 +... + afc · 2k + ... + a\ -2 + clq,
where щ, = 0 or 1 and as = l. If к > s then к > s +1, 2k > 2S+1 > n, and so
'n + 2k'
n + 2k 2fc+2fc_
2^+1 2k+^ ~ '
2k+l
0.
135

Solutions
1968/6
Therefore we can assume that к < s, hence
n + 2k = as ■ 2s + as_i -2s~l +... + (ak + l)2k +... + ai -2 + a0
n + 2k
(3)
2k+l
(as-2s-k-l+as_l-2s-k-2 + ... + ak+l) + {(ak + l)2-l)+
>-2
-k-V
+ (ak_x-2 / + ... + a1 •2~/c + a0-2~/c_1)
The value of the first expression in (3) is an integer, while the value of the
second one is 1 if ak = 1 and - if ak = 0. The third term can be estimated as
>—2 , „ _o_3 , , „ о—к , „_ о-к—1 /о_2 . л-3 ι л.О~к—1
afc-l2 + α&-22 +... + αι·2 κ + α0·2"
<2_2 + 2-3 + ...<
<
Г2
Τ"
2
2'
Therefore
η + 2Α
2fc+1
as-2s fc ^vi^5 fc 2 + afc+l+^·
Consider this expression for all к between k = 0 and k = s:
= as-2s-l+as_i-2s~2 +
+ CL2-2 + CLI+CL0
= ая · 2
8-2
+ a
s-l
>s—3
•2s J +... + a2 + a\
= as + as_i
= a.
Summing these equalities we get that (1) is equal to
as -2s + as_\2s~ + .. .+2a\ +ao = n.
Third solution. First we prove a useful formula: The cardinality of the set
of those integers in Η = {1, 2, ..., η} which are divisible by 2k but not by 2k+l
is precisely
^n + 2k'
(4)
2k+\
(These are the integers for which their prime decomposition contains 2 on power
k. Partition the elements of Η into Hq, H\, ..., Hk in such a way that Щ
contains those elements of Η for which their prime decomposition contains 2 on
power i. Obviously, the Hi's provide a cover of Η with disjoint subsets.) The
proof of the formula is quite easy: The number of elements in Я divisible by 2k
136

1969/2
1969.
is exactly
η
2^
fc+1
and the number of elements divisible by 2 is
η
2k+l
This
implies that the number of elements of Η divisible 2k but not by 2k+l is
η
2k
η
2k+l
η
Applying (2) for χ = ——- we get
2k+l
П
2k
η
2k+l
η
Г
2^+Γ + 2
n + 2A
2k+l
which proves the statement in (4). Since the disjoint union of the H^s is the set
^n + 21'
Η containing η elements, and there are
2й-1
integers in Щ, we get that
n+1
+
n + 2'
+ ...+
n + 2k
2k+l
= n.
Remark. The first solution applies even for the case of an arbitrary real n;
in this case the value of the sum in (1) is equal to [n].
1969.
1969/1. Prove that there are infinitely many positive integers a such that
4
z = n +a
is not a prime for any positive integer n.
Solution. Let a = 4b where b> 1 is an arbitrary integer. We show that ζ
is not prime:
z = n4 + 4b4 + 4n2b2 - 4n2b2 = (n2 + 2b2)2 - (2nb)2 =
= in2 + 2b2 + 2nb)(n2 + 2b2 - 2nb) = ((n + b)2 + fr2) ((n - b)2 + b2^ .
Since b>\, both factors are greater than 1, hence ζ is not prime.
1969/2. Let
cos(a2 + £) cos(a3+a;) cos(an + a;)
(1) f{x) = cos(ai + x) + + ^2 + · · ■ + —^T] >
where a\, a2, ..., an are real constants and χ is a real variable. Prove that if
f(x\) = f(x2)= 0 then X2 — x\= τηπ for some integer m.
First solution. According to a trigonometric identity we have
cos(a,j + x) = cos щ cos χ — sin ai sin x;
137

Solutions
1969/2
applying this to the left hand side of (1) we get
/ cos 02 coson\
f(x)=[ COS Οι Η + . . . Η — COS X —
( · sino2 sinon\ . _ .
— sin oi л +... л — sin χ = A cos χ — В sin x.
V 2 2"-1 /
A and Б cannot simultaneously vanish, since this would imply f(x) = 0 for all
x, although
ro\ f( \ ι , cos(o2-oi) cos(on-oi)
(2) /(-αι) = 1 + + ... + —χ >
>!_(! 4 + ...+ Μ 1 >0
\2 22 2n-V 271-1
(since cos(oi — a\) > — 1). Let us write the above expression for f(x) in the
following form:
f(x) = A cos χ — В sin χ = ν Л2 + Б2 I
cos ж , sin χ
л/A2 + В2 л/Л2 + В2
Let a real number </? be chosen in such a way that cos φ = —=== and so
VA2 + E2
ту
sin φ = = hold. With this notation at hand
Va2 + в2
f(x) = VA2 + B2(cos χ cos φ — sin χ sin φ) = л/Л2 + В2 cos(x + cp).
Since the distance between two roots of cosine is an integer multiple of π,
for
/(si) = /(s2) = 0
we get
cos(£i +φ) = cos(£2 + </?) = 0,
therefore
(#2 + ¥>) — (#1 + </?) = #2 — x\ - m7r·
The proof is now complete.
Second solution. Let z^ be the complex number with absolute value
2/c-i
and argument o^:
*fc = ^Tl(cos afc +«sin afc) (fc = 1, 2, ..., n),
and let ,
ζ = cos χ + i sin ж.
The real part of a complex number с will be denoted by 3ftc.
138

1969/3
1969.
With these notations Zfcz = ——r(cos(ak + x) + 2sm(ak + x)), and so
2«—ι
(3) f(x) = №(z\z +z2z +.. . + znz) = ^.{z{z\ +Z2 + . ■ . + zn)).
It is not hard to see that z\ + z2 +... + zn ^0, since otherwise
zi = -(z2 + Z3 + ... + zn)
\zi\ = \z2 + z3 + ... + zn\< \z2\ + \z3\ + ... + \zn\
would imply
л 11 1,1
K- + -77 + ... + Г = 1-
2 22 "'" 2n_1 2n~l'
This last inequality is clearly a contradiction, hence we get that z\ + z2 + ■ ■ ■ + zn =
с7^0. Let c = r(cos(/? + isin</?); now (3) implies
/(a;) = $ft(c2) = rcos(a; + <p), (^7^0)·
If /(η ) = /(s2) = 0, then
cos(a:i + y) = cos(£2 + ψ) - 0,
showing that
χ2 + φ — (χΐ+φ) = χ2 — χ\=τηπ (πι integer),
completing the proof.
1969/3. For each k = \, 2, 3, 4, 5 find necessary and sufficient conditions
on a > 0 such that there exists a tetrahedron with к edges of length a and (6 — k)
edges of length 1.
Solution. Let us examine the various values of к one after another.
A)fc=l.
For the tetrahedron ABCD we have AB = a and all the other edges are
of unit length. The midpoint of CD is denoted by F (Figure 1969/3.1). From
л/3
the equilateral triangles CD A and CD В we get AF = BF=—, therefore the
£*
condition of the existence of the isosceles triangle ABF is:
(i) a<2.^- = V3.
The above condition is also sufficient for the existence of the desired tetrahedron
λ/3
since with sides AF = BF = — and AB = a sl triangle ABF can be constructed
2 6
139

Solutions
1969/3
Figure 69/3.1
Figure 69/3.2
in the plane and by constructing an orthogonal to its plane of length DF = FC =
- at F we get a tetrahedron with edges of required lengths.
B) k = 2.
There are two possible choices for the two edges of length a. In case both
of them are on the same face, say AC = BC = a (Figure 1969/3.2) then consider
the midpoint Ρ of AB.
For the existence of ABC the inequality 2a > 1, or equivalently a> — is
л/3 / 1
obviously necessary. Since PD = —— and PC = Ja2 — -, the triangle inequality
PC + PD> CD for the triangle PCD yields
(2)
ι \/з
a τ + ~
4 2
>1.
Using (2) we get the necessary conditions
and
(3)
^u.^wi.^
a2 > 2 - y/З, α > у 2 - л/3.
1
This condition is in accordance with a > -.
2
Similarly, from PD + DC > PC we conclude
λ/з Λ Ι 2 ι
^ + 1>Va 4'
140

1969/3
1969.
and
(4) a2<2 + V3, а<у2 + \/3.
The union of (3) and (4) now provides
(5)
у2-\/3<а<у2 + л/3.
Since CD + PC > PD automatically holds, in case (4) is satisfied, we can
reverse our reasoning and this implies the existence of the triangle PCD. Now
1 1
the orthogonals PA and Ρ Β in Ρ of length - and - give the desired tetrahedron.
If the two edges of length a are skew, for example, AB = CD = a (and all
other edges are of unit length) then consider the midpoint Q of CD. As above,
the triangle inequality for ABQ gives AQ + BQ>AB (Figure 1969/3.3), and
so
(6)
2J 1 —— > a, hence a < ν 2.
This last inequality is the necessary and sufficient condition for the existence
of the triangle ABQ. Once ABQ
is given, the orthogonals at Q with
length DQ = QC = - provide the
tetrahedron. Hence in this case (6)
is the condition we are looking for.
Summarizing the two cases
above, if there are two edges of
length a and four of length 1, then
the necessary and sufficient
condition for the existence of a
tetrahedron with the given edges is that
one of (5) and (6) has to be
satisfied, which is equivalent to
(7) α< yflWi.
C) k = 3.
We will show that for any choice of a there is such a tetrahedron. Choosing
the edges of ABC to be equal to a and the remaining edges are of length 1, the
/я
distance of the centre of ABC from the vertices is ——. This should be less
than the sides,
(8) ^
В
Figure 69/3.3
<1, a<\/3.
141

Solutions
1969/3
Once this inequality is satisfied, a tetrahedron with base ABC and side edges of
unit length exists.
If the base has edges of unit length, then (similarly to the above said) the
necessary and sufficient condition is
(9)
l-л/З l
—-— < a, a>
V3
Since one of (8) and (9) is always satisfied, we get that for к = 3 there is no
restriction on the positive real number a.
D) fc = 4.
This case is essentially the same as the case of к = 2; we just have to invert
the roles of a and 1. This substitutes (7) with
which gives
1<ау2 + л/3,
a > V2 - л/3.
Ε) k = 5.
This case is basically the same as к = 1: again by changing the roles of a
and 1 we get
1 < av3, a >
1
л/3'
In conclusion, the necessary and sufficient condition for the existence of a
tetrahedron with the desired properties is:
k = l: 0<a<\/3
к = 2: 0<а<у2 + л/3
k = 3: 0<α
k = 4: V2-\/3<a
k = 5 : —-= < a.
λ/3
1969/4. Let С be an interior point of the semicircle к over AB and D
is the foot of the perpendicular from С to AB. The circle k\ is the incircle of
ABC, the circle ki touches CD, DA and к while k% touches CD,DB and k.
Show that k\, k^ and k^ have another common tangent apart from AB.
First solution. The centres and radii of k\, k^-, &з will be denoted by
0\, O2, O3 and r\, Г2, Г3, respectively. The tangent points of the circles on
142

1969/4
1969.
Figure 69/4.1
С
Figure 69/4.2
AB are Ex, E2, E3, where Ελ is in the interval E2E3 (Figure 1969/4.1). The
statement of the problem is equivalent to saying that 0\, 02, 03 are collinear:
The reflection of AB in this line provides the other common tangent.
We will use the following notations: AB-c, BC = a, CA = b, AD=p,
BD = q, a + b + c = 2s. The circle k\ divides the sides AB, ВС, С А — as it
is well-known — into intervals of lengths s — a, s — b; s — b, s — c; s~ c, s — a
(Figure 1969/4.2). Since ABC is a right triangle we have s — c = r\. The centre
of к is O, and we assume that the notation is chosen such that a > b is satisfied.
From the right triangle 02Е20 we get
143

Solutions
1969/4
cq = (r2+qf.
We also know that cq = a , therefore a = r2 + q and so
(1) r2 = a-q.
Starting with O3E3O, a similar argument provides
(2) r3 = b-p.
Now (1) and (2) implies
AE2
AE3
and
E2E\ = AE\ - AE2 = (s - a) - (c- a) = s - с
Εχ E3 = АЕЪ — AE\ = b — (s — a) = a + b — s = 2s — c—s = s — c.
This simply means that E\ is the midpoint of E2E3. Once again from (1) and
(2) it follows that
r2 + r3=a + b — (p + q) = a + b — c = 2(s — c) = 2r\,
i.e., r\ is equal to the arithmetic mean of r2 and r3. These last two results show
that in the right trapezium 02E2E303 the median starting at E\ is of length
r\, hence its other endpoint Οχ is the midpoint of 0203. This observation now
completes the solution.
Second solution. We will show that the statement of the problem holds for
any triangle ABC with D an arbitrary inner point of AB and к the circumcircle
of ABC.
In the solution we will use the fact proved in the first solution of Problem
1962/6., according to which if Q is the midpoint of the arc AB of the
circumcircle к of ABC not containing C, then the centre К of the circumcircle is the
point on the angle bisector CQ which satisfies QA = QB = QK.
Now let D be an arbitrary point of AB, and the circle k2 touches AB in E,
CD in F and the circumcircle к in Τ (Figure 1969/4.3).
Τ is the centre of similitude transforming k2 into k\ this enlargement maps
Ε into Q, since in these points the tangents of k2 and к are parallel. R on к
corresponds to F, therefore the enlargement maps EF into QR, thus EF is
parallel to QR. К will stand for \he intersection of EF and CQ. We will show
that К is the centre of the circumcircle of ABC, i.e. К = 0\; for this we have
to verify that QA = QK.
= p — r2=p + q — a = c—a,
= p + r3=b,
144

1969/4
1969.
Q
Figure 69/4.3
From the theorem regarding angles at the circumference it follows that
ITFC = ITEF = ITQR,
therefore their complements are equal as well; these are denoted by double arcs
on the figure:
ICFR = IFEQ = ITCR.
Taking the equality LTQK = ITRC = 7 into account we conclude the
similarity
(3)
QEK ~ RFC ~ RCT
QT
RT
of the corresponding triangles. It follows from the intercept theorem that
QE QE QK
——, while the similarity of QEK and RFC yields -^— = -ττρς· Combining
RF RF RC
these two results we get
(4)
QT QK
RT RC
It implies the similarity of the triangles RCT and QKT since they share an
angle (7) and according to (4) the ratios of the edges adjacent to 7 are equal.
(3) shows that QEK ~ QKT, hence
(5)
QE = QK
QK QT
consequently QK2 = QE ■ QT.
145

Solutions
1969/4
Furthermore, the triangles QEA and QAT are also similar since they share the
angle IAQE, and LQAE= IATQ = 6 since these are both equal to the angle of
AQ and the tangent at Q. From this similarity we deduce
g = f, i-, QA^-QE.QT,
and using (5) it implies QA = QK. Now this latter equality shows that К is the
centre of the incircle k\ of ABC; therefore the intersection of EF and CQ is K.
By repeating the above said for кт, we get that the intersection of CQ with
E'F' is K\ here E' and F' are the points of tangencies of fc3 with AB and CD,
respectively.
Notice that if the centres of
fc2 and &з are Oi and O3, then
О^D and O3D are orthogonal to
EF and S'F', respectively
(Figure 1969/4.4). Since they also
bisect the angles at EDF and
E'DF1, the interval 02D is
orthogonal to O3D. Moreover, since
OjED and DE'Ot, are similar
triangles, their altitudes at _E and Е'
divide the hypotenuse and the
interval О2О3 in the same manner.
O,
Ε D
Figure 69/4.4
This shows that K-0\ is on О2О3, which proves our assertion
Remarks. 1. Notice that our second solution uses only similarities. It is not
true in general that Οχ is the midpoint of О2О3, as it was deduced in the first
solution.
2. Another specialization of the same general problem will be discussed in
Problem 1978/4.
1969/5. Given η > 4 points on the plane (no three collinear), prove that
n — 3
2
there are at least
points.
convex quadrilaterals with vertices amongst the given
First solution. The convex hull of the points contains at least three of
them as vertices; let us denote such a triple by А, В and C. Add two more
points, called X and Υ to the above triple. The line XY does not pass through
vertices of the triangle ABC and intersects at most two of its edges.
146

1969/5
1969.
Suppose that it is disjoint from AB. In this case Α, Β, Χ, Υ form a convex
quadrilateral, since otherwise their convex hull is a triangle, implying that XY
intersects the interval AB.
Therefore for any two points of the given set minus {А, В, С] we can
add two of А, В and С such that the resulting quadrilateral is convex. Since
different point pairs give rise to different quadrilaterals (because their vertices
are different), the number of convex quadrilaterals found in this way is at least
С;1)-
Second solution. We start with the fundamental lemma of the subject:
from five given points (no three collinear) we can always choose four which
form a convex quadrilateral. The proof is similar to the reasoning given above:
if the convex hull of the five points is a pentagon or a quadrilateral then four
vertices of them will provide the desired convex quadrilateral. In case the convex
hull is a triangle, say ABC, then the two inner points X and Υ and the edge of
ABC disjoint from the line XY — say AB — provides the four points forming
a convex quadrilateral (Figure 1969/5.1).
There are I J choices for five-tuples of points among η points, according
. Μ
to the above said each tuple determines a convex quadrilateral. This gives I J
convex quadrilaterals; notice, however, that the same quadrilateral might be
given by many five-tuples. In fact, there are η — 4 such tuples determining the
same quadrilaterals, since the fifth point can be chosen arbitrarily. In conclusion
147

Solutions
1969/5
this argument gives that there are at least
1 In
n-4\5
different convex quadrilaterals with vertices amongst the given points. This
bound is even better than the one required by the problem, since we will show
that
n —3
2
This inequality is equivalent to n(n — l)(n — 2) > 60(n — 4), which can be
verified for η = 5, 6, 7, 8 directly; for η > 9 we have n(n — 1) > 60 and η — 2 >
η — 4, hence n(n — l)(n — 2) > 60(n — 4) obviously follows.
Remark. This problem represents one of the most popular circle of problems
in convex geometry. A generalization of our starting lemma goes as follows:
For any к there exists Z(k) such that if η > Z(k) then for η points in the
plane in general position there are к which form a convex k-gon.
The exact value of Z(k), however, is surprisingly hard to determine. We
only have estimates for it, for example
A long-standing conjecture due to P. Erdos and G. Szekeres asserts that
Z(k) = 2 +1. For fc='4 this is exactly the result we found above, and it is
known that Z(5) = 9.
These kind of problems share many common features with Ramsey-type
problems.
1969/6. For given real numbers Χ\·)Χ2^\^2·>ζ\ιζ2 satisfying x\>0,
X2 > 0, x\y\ — Ζγ > 0 and Х2У2 ~ z2 > ®> Prove inat
ω , —^— .<^+ ι
{x\+X2){y\+y2)-{z\+Z2)2 X\y\-z\ Х2У2~^2
Give necessary and sufficient conditions for equality.
First solution. Let us denote the denominators appearing in (1) by A, A\
and A2, respectively. The following relation will be crucial for us:
(2) А = А1+А2 + х\У2 + Х2У\ -2z\Z2-
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1969/6
1969.
Applying x + y> 2л/ху (x>0, у>0) this implies
A = A1+A2 + —(A2 + z^) + —(A1+zf)-2ziZ2,
x2 x\
x2 x\ \ у χ\ ]jx2i
Since the resulting value is positive, it shows that
(3) A>0,
and (since x + y> 2^/xy) this provides:
(4) A>Al+A2 + ^- A2 + ^A1>A1+A2 + 2Ja1A2 = (Ja~1+Ja:
x2 xx v \v v
It is not hard to see that equality holds only in case
(5) *.J- = W- and ~Аг = -^
Using the above notations (1) can be rewritten as
8 11 . , , Λ 8Α\Α2
— <-— + -—, or equivalently A > - —.
A~ Ax A2 - Αγ+Α2
In order to verify this last inequality (according to (4)) it is enough to show that
[jAi+jAt) >^^~2, .-е. ^ ) .——>А,А2.
Based on the inequality between the arithmetic and geometric means we have
(^Щ>^12 and ^>У1^.
Equality holds if and only if A\ =A2. Together with (5) it gives x\ = x2,
consequently x\ =x2 is necessary for equality. Based on (5) these conditions imply
z\ = z2; now A\=A2 forces y\=y2. But x\ = x2, y\ = y2, z\ = z2 obviously imply
equality in (1), hence we determined the necessary and sufficient condition.
Second solution. With the notations of the above solution we note first
that the discriminants of the quadratic equations
Fi(t) = x\t2 + 2z\t + y\,
F2(t) = x2t2 + 2z2t + y2,
F(t) = Fl(t) + F2(t) = (xl+x2)t2 + 2(zl+z2)t + (yl+y2)
149

Solutions
1969/6
are —4A\, —4A2 and — 4 A, respectively. Note furthermore that if a>0, then
ax2 + bx + c attains its minimum at — — with minimum value — — where D is
2a 4a
the discriminant of the quadratic equation.
This implies
Ал An A
(5) minF1(i) = —, min F2(t) = —, mm F(t) =
X\ X2 X\+X2
Since
(6) min F{t) > min Fx (t) + min F2(t),
(5) gives
А Ал А2
> — + l
X\+X2 X\ %2
Using (3) this inequality is equivalent to
I< \ A
A (xi+X2)(% + %)
8 11
We would like to prove that — < —— + ——; according to the above said it is
A'Ax A2
enough to show that
1 1 8
+ —>
Since all the quantities appearing in the formula are positive, it is equivalent to
A\ + A2) {{ + xj l + \ + x2
) A2) > 8,
\x2 x\) \A2 Αχ) \x\A2 x2Ax)
This inequality trivially holds since the sum of a positive number and its
reciprocal is at least 2; this observation completes the proof of (1).
Equality in this last step holds once x\ = x2 A\ = A2\ in (6) we need that the
minima of F\ and F2 coincide, meaning — = —. Hence x\ = x2 implies z\ = z2
X\ X2
and these equalities together with A\ =A2 yield y\ =y2. In conclusion, equality
holds in (1) if and only if x\ =x2, y\ =y2 and z\ =z2.
1970.
1970/1. Μ is a point on the side AB of the triangle ABC. Let r\, r2
and r denote the radii of the incircles of AMC, BMC and ABC, respectively.
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1970/1
1970.
ρ\, Q2 and ρ stands for the radii of the excircles of the triangles AMC, BMC
and ABC (corresponding to sides AM, BM and AB), respectively. Prove that
r\ Г2 r
Q\ Q2 Q
Solution. The incircle of ABC touches AB at Cq wmle the excircle
touches the same edge at C\ (Figure 1970/1.1). According to a well-known relation
for the tangent intervals we have
(1) AC0 = BCi=s-a.
The centre of the incircle of ABC will be denoted by K, while the centre of the
Figure 70/1.1
excircle touching AB is K'. If lKAC0 = a' and ΙΚ'ΒΟχ=β" then
r = (s — a) tan a', Q = (s — a) tan β",
and so
r _ tana'
( } ^ = tan/3"'
Furthermore, the centres of the excircle of AMC and the incircle of BMC
are denoted by K\ is K2, and /K[MA = ΙΚ2ΜΒ = φ. Applying (2) for AMC,
and then for the circles of BMC we get (Figure 19970/1.2):
r\ tana' 7'2 tan</?
ρ l tan</?' ρ2 tan^"
Multiplying the corresponding sides, (2) implies
r\ T2 tana' tan</? tana' r
ρ Ι ρ2 tan φ tan^" tan^" ρ'
151

Solutions
1970/1
which is exactly what we wanted to prove.
1970/2. Real numbers X{ (i = 0,1,..., n) with 0<Xi<b and xn > 0,
xn-\ >0 are given. If xnxn_\ .. .x\xq represents the number An base a and
Bn base b whilst xn-\ ... x\Xq represents An_\ base a and Bn_\ base b, then
prove that a>b holds if and only if
(1)
Аг-1 ^ Аг-1
<
■™-n ^n
First solution. According to the notations above
An =anxn + an~ xn-\ +.. . + ax\ +xq, Bn = bnxn + bn~ xn-\ +
ίη
+ bx\+XQ,
A.
η
= Ar
_\ — s±n a xn,
B„_:=B„-bnx
"n
η
jn·
Easy to see that (1) is equivalent to
(2) ΑηΒη_1-Αη_χΒη>0.
Perform the substitutions indicated above and get
AnBn— 1 — An_\Bn = AnBn — Anb xn — AnBn + Bna xn =
= anxn(bnxn + bn~lxn__i +.. . + bx\ + xq)-
152

1970/3
1970.
—bnxn(anxn + anxn_i +.. . + ax\ + xq) =
= xnxn-\an~xbn~l(a - b) + xnxn_2an~2bn~2(a2 - b2) +...
... + χηχ\(ώ{αη~1 - bn~l) + xnxo(an - bn).
Since in this last expression the signs of the terms in the parentheses are the
same, (2) is satisfied if and only if a > b.
Second solution. Let us introduce the following notations:
f(t) = xntn + xn_\tn~l +.. . + x\t + x0,
fit) fit)
Since xq, x\, ..., xn-2 are nonnegative and xn-\, xn are positive, for positive
values of t the functions f(t),
fit) . ι . i.i
— ·Εη ' ""η — 1 ' Xn 9 о "г · · · "г Х()'
+п t +1 £
-η
х tn
and " are strictly increasing while g(t) is strictly decreasing.
I\t)
A H
Notice that g(a) = n~ and g(b)= n~ , therefore a>b is equivalent to
An %
gia) < gib), meaning
An-\ Bn-\
An Bn
1970/3. The real numbers clq, a\, ai, ..., an> ... satisfy
(1) 1 = O-O < a\ < a2 ^ · · ■ 5: 0"n < · · ·
We define the sequence b\, b^, ..., bn, ... as
(2) b, = £(l-2*zi)_l
I. Prove that 0<bn<2 holds for all n.
II. Given с satisfying 0 < с < 2, prove that we can find clq, a\, ..., an,
(satisfying (1)) so that infinitely many of the corresponding bn are greater than c.
First solution. I. Since k~ < 1 and therefore 1 ^— > 0, the terms in
ak ak
the sum defining bn are nonnegative, and so bn is nonnegative. One term in the
sum under (2) can be rewritten as
\ ak / \/^k V V ak / V V ak / л/ак
0>k \ У o>k J y/ak-\ V V ak 7 \y/ak_\ y/Ok
• 153

Solutions
1970/3
Applying this estimate for (2) we get
6η<2(4=-4=+^=--?=+···-4=)=2ίι-^=)<2'
Щ Ja\ Ja[ Jai y/On / \ л/ап
proving Part I. of the problem.
II. Choose oq, a\, ... to be a geometric sequence with clq = \ and quotient
1 1
-« (0 < q < 1). Since -=■ > 1, the sequence is increasing, therefore it satisfies (1).
qi gi
Now (2) takes the form
hence
h
-(ΐ-ήΣ^^1^2) * =gd+g)(i-gn)> 2^(1-0-
r—ί l—o
η
fc=l
Since 0 < q < 1, the sequence gn converges to 0, and so 1 — qn converges to
1. This implies that for some no the equation 1 — qn > q is satisfied once η > no-
This shows that
(3) bn>2q3.
Fixing д=Л-, 0<c<2 implies 0<q<l, hence (3) implies that there is n0
such that for п>щ
bn > 2q3 = c,
concluding the solution.
Remark. It is not hard to guess the origin of this problem; with this guess
we get an alternative solution:
A term in (2) can be written as
ak — ak_i
(4)
4
Let us fix clq, a\, ..., α^, ... on the ж-axis. Now (4) becomes the area of
the rectangle with two vertices being equal to ak_\, ak, and the third vertex is
on the curve У-— (Figure 1970/3.1). The value bn is the sum approximating
the area under the curve from below. Therefore bn is less than the area under the
curve from ag t° an> i-e->
an - -,η
(5) bn < j
_з
χ 2dx =
a0
X-*a0
= 2-—=z<2.
Wan
154

1970/4
1970.
Figure 70/3.1
This observation verifies Part I. of the problem. The geometric content of Part II.
is also clear, but its rigorous proof requires a delicate approximation of the
integral. It can be done, but does not provide a simpler result than the one discussed
above.
1970/4. Find all positive integers η such that the set {η, η+1, η + 2, n + 3,
η+ 4, η+ 5} can be partitioned into two subsets so that the product of the
numbers in each subset is equal.
First solution. We will show that there is no η satisfying the requirements
of the problem. Among positive integers every pth is divisible by ρ and
between any two numbers divisible by ρ there are at least ρ — 1 numbers. Therefore
there are no two among the given six numbers divisible by 7. If the partition is
possible, the only prime factors in the numbers are 2, 3 and 5, since the prime
factors in the two sets should be the same.
The prime 5 can appear only in η and η+ 5, hence the prime factorizations
ofn + 1 n + 2, n + 3 and n + 4 may contain only 2 and 3 as prime factors. Two
among the four numbers are odd, hence divisible only by 3. Their difference is
necessarily 2, hence 3 cannot divide both. This contradiction shows that there is
no such n.
Second solution. Suppose that there is a partition of the six numbers with
the required property. There is no multiple of 7 among our numbers since among
six consecutive integers there is at most one divisible by 7, but we need one in
each subset of the partition. Therefore n— 1 is divisible by 7, hence the mod 7
residues are 1, 2, 3, 4, 5, 6. Now the mod 7 residue of the product 6! = 120 of
155

Solutions
1970/4
the six numbers is equal to 6. If the products in the subsets of the partition are
equal, their mod 7 residues coincide as well; let this common value be equal to
ra. The above reasoning shows that mod 7 we have the equality ra =6. The
mod 7 residue of ra can take the values 0, 1,2, 3, 4, 5, 6, the mod 7 residues
of the squares of these are 0, 1,4, 2, 2, 4, 1, respectively. Since 6 is not among
them, the required partitioning is impossible.
Third solution. We will show that if η -=f\ then there is one among the six
numbers containing a prime factor in its prime decomposition which is at least 7.
Since there is at most one such number among the six chosen one, this argument
again shows that the desired partition cannot exist.
The six numbers contain one of the form 6k+ \ and another of the form
6k —I: there is one which is divisible by six and if it is not η or η + 5 then the
assertion is trivial. Now if η is divisible by 6, then n+1 and η+ 5, if n + 5 is
divisible by 6 then η and n + 4 will admit the promised form. 6k +1 and 6k —I
have only odd divisors and 3 is not among them. Only one of the two numbers
is divisible by 5 since their difference is 2 or 4. This shows that the other —
since it is not equal to 1 — contains a prime factor not less than 7. This final
observation concludes the solution.
Remarks. 1. Another way to solve the problem is that we write down all
possible partitions of the six numbers, and after computing the products we show
that the two resulting numbers are never equal.
2. Even this simple problem has deep number theoretic background. In the
second solution we utilized the fact that 6!ξ6ξ-1 (mod 7). This is a special
case of the famous Wilson congruence theorem, which asserts that
(p — 1)! = — 1 (mod p)
holds if and only if ρ is a prime.
3. In our third solution we showed that there exists a prime dividing only
one of the given six numbers. The following theorem — frequently called the
Sylvester-Schur theorem — is closely related to this fact: if η > 2k then I J
admits a prime factor greater than k.
/n\ (n — k + l)(n — k + 2)...(n—l)n ,. ,
Since I = , this means that the ratio is
\kl k\
divisible by a prime ρ greater than k. Obviously the denominator is not divisible
by any prime greater than k. Therefore the product (of к terms) in the numerator
must contain a term which is divisible by p. Since the numerator is the product
of к consecutive numbers, there is at most one divisible by p. In particular this
156

1970/5
1970.
means that if η > 6 and so η + 5 > 2 · 6, then
n(n + l)(n + 2)(n + 3)(n + 4)(n + 5)
is divisible by a prime which is greater than 6; this is exactly what we used in
our third solution.
The Sylvester-Schur theorem implies that for к consecutive numbers
there is a prime dividing exactly one of them. It then implies that к consecutive
positive integers can never by partitioned into two subsets with equal products.
The most well-known special case of the Sylvester-Schur theorem is when
n = 2k: in this case the theorem states that for n> 1 there is a prime number
between η and 2n (n<p<2n). This statement is also called the Chebyshev
theorem.
1970/5. In the tetrahedron ABCD the angle IB DC is a right angle and
the foot of the perpendicular from D to ABC is the intersection of the altitudes
of ABC. Prove that
(1) (AB + BC + С A)2 < 6(AD2 + BD2 + CD2).
When do we have equality?
Some preparatory remarks. In all solutions we show that the edges at D are
perpendicular to each other, hence we have a so-called right angled tetrahedron.
After verifying this, the proof of (1) proceeds as follows: let AB = c, BC = a,
CA = b, DA = ai, DB = b\, DC = c\. The Pythagorean theorem applied to the
faces ABD, BCD, CAD (Figure 1970/5.1) yields
D
В
Figure 70/5.1
157

Solutions
1970/5
(2) а2 = Ъ\ + с\, b2 = c2+a\, c2 = a\ + b\.
After summing these equalities and multiplying both sides by 3 we get
6(a2 + b\ + c2) = 3(a2 + b2 + c2) =
= (a + b + c)2 + (a-b)2 + (b-c)2 + (c-a)2>(a + b + c)2,
which proves (3). Equality holds only in case a = b = c, based on (2) this implies
a\ =b\ =c\.
First solution. Let the intersection of the altitudes in ABC be denoted by
M. DM is perpendicular to the plane of ABC and therefore to С A: CALDM.
We also know that CALBM since BM is an altitude, hence С A is
perpendicular to two (nonparallel) lines in the plane of DBM. This implies that it
is perpendicular to all lines in the plane at hand, in particular CALDB. This
shows that DB is perpendicular to two (nonparallel) lines in the plane of CAD
(namely to С A and DC), hence it is perpendicular to all lines in this plane. This
immediately gives DB1.DA. Changing the roles of В and С we get DCJ-DA,
hence the edges at the vertex D are pairwise perpendicular, and this concludes
the solution.
Second solution. Let a, b, с and d denote the vectors pointing from Μ
to the vertices А, В, С and D. We recall that two intervals are perpendicular if
and only if the corresponding scalar product vanishes.
DM is perpendicular to the altitudes of ABC:
(4) da = db = dc = 0.
MB is perpendicular to AC and Μ С is perpendicular to AB:
(5) b(c-a) = 0, i.e. bc = ab,
and
(6) c(a-b) = 0, i.e. ac = bc.
DB is perpendicular to DC, hence
(b-d)(c-d) = 0, i.e. bc + d2-dc-db = 0,
and so according to (4) we get
(7) bc + d2 = 0.
We want to show that DB and DC are perpendicular to DA. In the spirit
of (7) these statements are equivalent to
ab + d2 = 0, and ac + d2 = 0.
158

1970/6
1970.
These latter equations, however, follow directly from (7), since (5) and (6) show
bc = ab = ac.
Remarks. A tetrahedron with the above property is called an orthocentric
tetrahedron; the altitudes of these solids pass through a common point. In our
case the tetrahedron has an additional special property: it admits a vertex at which
the edges are pairwise perpendicular to each other. This property explains the
name: right angled tetrahedron. In a certain sense such a solid can be interpreted
as the three dimensional generalization of a right angled triangle. For example,
there is a relation reminiscent to the usual Pythagorean theorem: the squares of
areas of the three faces with right angles is equal to the square of the area of the
acute face.
1970/6. Given 100 coplanar points (no three collinear), consider all
triangles with vertices among the given points and prove that at most 70 % of
these triangles have all angles acute.
First solution. First we verify the statement for 5 points (we always
assume that no three are collinear). Consider the convex hull of the given points; it
is a triangle, a quadrilateral or a pentagon.
a) If the convex hull is the triangle ABC (Figure 1970/6. la) then it contains
A
Figure 70/6. la
the remaining two points, D and E, in its interior. The intervals joining D with
the vertices of the triangle form 3 angles, and their sum is 360°. Therefore at
most two of these angles are not acute. This shows the existence of two obtuse
triangles. The same argument applies to E, hence there are at least 4 obtuse
triangles on the given 5 points.
159

Solutions
1970/6
b) If the convex hull is a quadrilateral, denoted by ABCD {Figure
1970/6.lb), then the fifth point Ε is in the interior of, say, the triangle ABC.
Figure 70/6.lb
At least one angle of the quadrilateral is not acute, hence the edges of this angle
determine an obtuse triangle. As before, the intervals Ε Α, ΕΒ, EC determine at
least two obtuse angles, giving rise to further two obtuse triangles. In summary,
we found at least three obtuse triangles.
c) Finally, assume that the convex
c\ у—---^^ hull is equal to the pentagon ABCDE.
Since the sum of the angles in the
pentagon is 540°, it has at least two
obtuse angles (Figure 1970/6.lc). If these
obtuse angles are at neighbouring
vertices A and B, then AEB and CBA
are obtuse triangles. The
quadrilateral EDCB also has an obtuse angle,
and the corresponding triangle is
obtuse again. This shows the existence of at
least three obtuse triangles in this case
again.
If the two obtuse angles are not
neighbouring, say these are at the
vertices A and C, then besides the
triangles at these vertices the quadrilate-
ral ACDE provides an obtuse triangle
(since at least one of its angles is obtuse). Once again we found at least three
obtuse triangles.
Figure 70/6.1 с
160

1970/6
1970.
In conclusion we deduced that among the I J = 10 triangles determined by
the 5 given points there are at least 3 obtuse, in conclusion the number of acute
triangles is at most 7, i.e., at most 70 % of all the triangles.
Now let the number of points be equal to η > 5. We can choose five tuples
from these η points in I J ways, and each such five tuple contains at least three
obtuse triangles. At first glance this seems to give at least 311 obtuse triangles.
Notice, however, that each such triangle is counted with multiplicity, we counted
them in as many five tuples as many extensions of the three points (the vertices)
to a five tuple exist. There are I J such extensions, hence the number of
obtuse triangles we have found in this way is at least
3(g)
Γ2Ύ
ίη\ 3(2)
This shows that the number of acute triangles is at most | I ^-. The ratio
W cf)
of this number and the total number of triangles is
з(5)
(5)
Cf) 3(?) 3
ν 2 ) =1 V5^_ = 1_ o,7 = 70%.
(5) ©C23) 10
Thus we proved the assertion for every η > 5.
Second solution. The idea is the following: we will show that the ratio
of the number of acute triangles and the number of all triangles on η points
decreases once η increases. Since the ratio does not exceed 70 % for η = 5, this
monotonicity shows that the same holds for all η > 5.
Let M(n) denote the maximum number of acute triangles determined by
n>3 coplanar points. Let furthermore A\, A2, ..., An+l denote n+1 points
on the plane (no three are collinear). Delete Ai\ the number of acute triangles
on the remaining η points is denoted by hi. The sum k\ + k2 + - · - + ^n+l w^
give the acute triangles on A\, A2, ..., An+i in a way that, for example, the
triangle AjA^Ai is counted in this sum as many times as many ways we can
add η — 3 points to Aj, A^, Αι from the remaining η — 2. This means that each
such triangle is counted with multiplicity η — 2. The number of acute triangles
determined by A\, A2, ..., An+\ will be denoted by h. Obviously h < M(nVl)
since h corresponds to a specific set while M(n+1) refers to all sets of n + 1
161

Solutions
1970/6
elements. Therefore
k\ + k2 +... + kn+i
= h.
n — 2
By definition we have ki < M(n), hence
ki + k2 + ... + kn+i =(n- 2)h < (n+ l)M(n),
i.e.
n + 1 , r, ч
h< -M(n).
n — 2
This inequality holds for any set A\, A2,..., An+\ on the plane, in particular
for the one in which the number of acute triangles is exactly M(n+1). This
shows
n+1
M(n+1)< -M(n).
n — 2
т^· · ,· , · ι- ι /n+i\ , 1t n-2 /n + l\ /n\ .
Dividing the inequality by and taking = I ) into account
\ 3 / n+1 \ 3 / \3/
we get
M(n + 1) M(n)
fi1) - (5) ·
This last inequality means exactly that the ratio of the acute triangles and the total
number of triangles on η points gives a monotone decreasing function when η
increases.
If n = 5, then — as was shown by the first solution — M(5) = 7, and so
M(5) 7
= — = 0,7. Now (1) implies that for η > 5
10 10
M(n)
< 0,7 = 70%.
1971.
1971/1. Show that the following statement is true for n = 3 and 5 and false
for all other η > 2:
"For any real numbers a\, a2, · · ·, an the inequality
{αχ -α2)(αι -α3)...(α1 -αη) + (α2-αι)(α2-a3).. .(a2-an) +...
... + (an- a\)(an -a2)... (an - an_\) > 0.
t
holds".
162

1971/2
1971.
Solution. For η = 3 we have
(αϊ - a2)(ai - аз) + (a2 - αλ){α2 - a3) + (a3 - а\)(аъ - a2) =
2 2 2
= αγ + a2 + a3 — a\ a2 — a2a3 — a3a\ =
= - ((αϊ - α2Ϋ + (a2 - α3)2 + (a3 - a\ f J > 0,
hence the inequality holds.
For n = 5 the expression is symmetric in the a^'s hence we can assume that
a\ > 0-2 > a3 > a4 > a5- Rewrite the expression as
Ol - α2)[(α1 - аз)(а1 - θ4)(°1 _ a5) - (a2 ~ оз)(°2 ~ α4)(α2 ~ α5)] +
+Ol - аз)(й2 _ аз)(^4 - аз)(а5 _ аз)+
+(<24 - α5)[(αΐ - ^5)(α2 ~ α5)(α3 - as) - (a! - a4)(a2 - a4)(a3 - a4)].
In the first and third terms a\ - a2 > 0 and a^ — а$ > 0. In the square
brackets we have differences of products involving three factors and we
have a\ — a3>a2 —a3, a\—a^>a2 — a^, ai~a^>a2 — a^, and similarly:
a\ — а$ > a\ — 04, a2 — αζ > a2 — a^, a3 — a^>a3— a^. Therefore these sums
give nonnegative numbers, hence the first and third terms in the above sum are
nonnegative. The middle term is simply a product of two nonpositive and two
nonnegative numbers, consequently it is nonnegative. Therefore the expression
is nonnegative once η = 5, hence the statement is verified.
For η = 4 consider a\ = 0, a2 = a3 = a^ = 1; the value of the expression then
becomes —1, hence the inequality in the statement is false.
Now for η > 5 consider the substitution a\ = \, α2 = α3 = α^ = 2, a$ = ag = ...
... = an = 0; the resulting value is —1 again. This last example concludes the
solution of the problem.
Remark. The following rewording of the problem might shed light on its
origin: Let p'(x) be the derivative of the polynomial
p(x) = (x — a\)(x — a2)... (x — a*;) (n > 2).
The inequality
η
EpW>0 (* = l, 2, ...,n)
i=l
holds for any real η-tuple a\, a2, ..., an exactly in case n = 3 and η = 5. Hence
the problem admits a solution based on the relation between polynomials and
their derivatives.
1971/2. Let P\ be a convex polyhedron with vertices A\, A2, ..., Ag. Let
Pi be the polyhedron obtained from P\ by a translation that moves A\ into Ai
163

Solutions
1971/2
(г = 2, 3, ..., 9). Show that at least two of the polyhedra P\, Pi, ..., Pg have an
interior common point.
Solution. Let Q be a point of Pj and apply the translation A\Ai to it
{Figure 1971/2.1), and denote the resulting point by Q'. Fixing the origin at A\,
in vector notation we have A\Q' = A\Q + A\Ai. According to the parallelogram
rule Q' can also be obtained by reflecting A\ to the midpoint С of the interval
AiQ (this holds even in case Q in on A\Aj). Since both Q and Αι are in P\, and
Pi is convex, it follows that С is in P\ as well. This shows that Q' can be got
by applying an enlargement with centre A\ and quotient 2 to C.
Figure 71/2.1
In summary, we showed that by translation we got points which are in the
solid P\ we have got from P\ by the above enlargement.
Let the volume of Ργ be denoted by V; therefore P[ has volume 8V. Since
all the polyhedrons Pj, P2, ..., Pg with volume V are in P[ of volume 8 V, they
cannot be disjoint, hence there are at least two different sharing common interior
points.
Remark. For a polyhedron with eight vertices the statement is usually false
— consider the cube, for example. For all polyhedra with more than 8 vertices,
though, the statement turns out to be true.
1971/3. Prove that we can find an infinite set of positive integers of the
form {2n — 3} (where η is a positive integer) every pair of which are relatively
prime.
Solution. For η = 3, 4, 5 the values of 2n — 3 are 5, 13 and 29, respectively,
and these are relatively prime. We will give a procedure which gives further
elements of the desired form and the resulting numbers are relative primes to the
previously constructed ones.
164

1971/4
1971.
Let αϊ, α2, ..., ak be given with the required properties. Consider
s = αϊ αϊ... dk and take the s + 1 numbers
oO r\\ ryL r\S
z,z,z,...,z.
There are two among them with the same residue mod s, say 2r and 2q (r > q).
Since 2r — 29 = 2<?(2r-<? — 1) is divisible by s and s is a product of odd numbers
(and so itself is odd), s divides 2r~q — 1, i.e., for some integer e we have 2r~q —
1 = es. Define now
ak+l=4es + l=4-2r-s-4 + l=2r-s+2-3.
ak+\ is obviously relatively prime to аг· (i<k), since a common divisor (greater
than 1) of α2· and ak+\ divides both s and 4es + 1, which is a contradiction. Finally
afc+l is greater than any а; (г < к), since it is greater than their product. Now
applying this method an infinite sequence of integers with the desired properties
can be constructed.
Remarks. 1. Our argument can be slightly modified as follows: Let
afc+i=2*<e>+2-3
where φ is the Euler function φ(ή). According to the Euler congruence theorem
2^(s) ξ 1 (mod s) and 2φ^' — 1 = es, where e is an integer; consequently
afc+1=4-2^(s)-3 = 4es+l.
This implies ak+\ > сц (г < к), furthermore ak+\ and щ are relatively prime since
their common divisor (different from 1) divides 4es + 1, which is a contradiction.
2. The same argument applies to numbers of the form {pn — (p+1)} for
some prime p.
3. Our construction followed the line first used by Euclid in proving the
existence of infinitely many primes: for finitely many prime numbers p\,p2, · · ·>
Pk, consider pk+\ =p\P2 .. .pfc + 1. It has to be prime since it is not divisible by
any of the p^s.
1971/4. All faces of the tetrahedron ABCD are acute triangles. Let X,
Υ, Ζ and Τ be points in the interiors of the segments AB, ВС, CD and DA,
respectively; and consider the closed path XYZTX.
a) //
(1) LDAB + LBCD 4 LABC + ICDA,
then prove that none of the closed paths XYZTX has minimal length.
b)//
(2) LDAB + LBCD = LABC + LCDA,
165

Solutions
1971/4
then prove that there are infinitely many shortest paths XYZTX, each with
length
2AC sin -,
2
wherea = lBAC + lCAD + lDAB.
Solution. Cut the tetrahedron along the edges AC, AB, BD and unfold
it into the plane; this construction yields the lattice of the tetrahedron (Figure
1971/4.1). This operation does not change intervals on individual faces, hence the
closed path on the tetrahedron has minimal length if and only if the corresponding
path on the plane has minimal length.
в С
Figure 71/4.1
The union of two faces (since they are acute triangles) form a convex
quadrilateral. If, for example, on the quadrilateral ABDC the intervals XY and Υ Ζ
are not collinear then Υ can be chosen to achieve that Χ, Υ and Ζ are collinear
and Υ is an interior point of ВС. Consequently, in the plane the points of the
shortest path XYZTX are collinear; the same is true if we cut our tetrahedron
along another edge, say AD.
Assuming that Χ, Υ, Ζ and Τ are collinear, we introduce the following
notations:
ΙΑΧΤ = /ΒΧΥ = φι, lBYX = /.CYZ = ip2,
/CZY = lDZT = ip3, ΙΌΤΖ = ΙΑΤΧ = φ4.
With these notations at hand we have
/DAB + /BCD = 2π - (φλ + φ2 + Ψ3 + φ4),
/ABC + ICDA = 2k - (φι +Ψ2 + Ψ3 + Ψα)^
which contradicts (1), hence there is no shortest closed path XYZTX.
166

1971/4
1971.
Suppose now that (2) is satisfied. Let the angles at the vertices А, В, С and
D be denoted by α, β, η and δ, respectively. Applying (2) we get
a+j=lDAB+lDAC+lCAB+lBCD+lBCA+lACD=
= (IDAB + IBCD) + (ICAB + IBCA) + (IDAC+IACD) =
= IABC + ICDA + tt - IABC + 7T - ICDA = 2tt,
therefore
(3) α + 7 = 2π.
The same reasoning shows that (2) implies
(4) β + δ = 2π.
Assume that 7 < π and δ<π. (In the contrary case: (3) and (4) implies that
at least one of the pairs (α, δ), (α, β), (j, β), (j, δ) satisfies the inequality, if this
happens to be (α, β), for example, then cut the tetrahedron and unfold it into the
plane in a way that AB plays the role of "central" edge [played by CD on our
figure].) This shows that the hexagon ABDB'A'C is convex.
Now (2) implies that AB and A'B' are parallel (see also our remark), and
since these intervals are of equal length, we conclude that AA'B'B is a
parallelogram. Hence any closed path mapped into an interval XX' parallel to A A' is
of minimal length, and all these lengths are equal.
Since AC = A'C, the triangle АСА' is isosceles with angle I AC A' equal
to 7, therefore the length of XX' is
XX' = AA' = 2AC sin ^ = 2AC sin (ж - -\= 2AC sin -,
2 V 2/ -2'-
and this is what we wanted to prove.
Remark. The fact that AB and A'B' are parallel can be verified as follows:
according to (2), if ΙΌΑΒ = ω2, ίΒΟΌ = ωλ, lABC = eh lADC = e2 then
ωι+ω2 = ει+ε2,
hence
—s\ + ω\ — ε2 + ω2 = 0.
This means that by rotating a vector parallel to AB with angles —ε\, +ω\, —ε2
and +ω2 respectively, its direction remains unchanged, but applying the same
transformations to the line AB with centres В, С, D, A' one after another, then
it is mapped to A'B'. Since its position has not been changed, we get that AB
and A'B' are parallel.
167

Solutions
1971/5
1971/5. Prove that for every positive integer m we can find a finite set of
points S in the plane such that for any point A of S, there are exactly m points
in S at unit distance from A.
Solution. For m = 1 we can choose S to be the two endpoints of the unit
interval; for m = 2 the three vertices of an equilateral triangle of unit side length
will do. Next we will describe a simple method for producing a set Sm+\
corresponding to the value m +1 from a set Sm satisfying the requirements of the
problem for m.
Translate the points of Sm by a given appropriate unit vector; the resulting
points define S'm. The union of Sm and S'm now provides a set which has the
property we require from Sm+\: for each point of Sm and S'm there is an additional
one which is of unit distance.
Not all unit vectors will provide an appropriate 5m+1, though. Applying
translation by e for the equilateral triangle shown by Figure 1971/5.1 gives a
set of type 53 while doing the same with the vector e' is not suitable for our
purposes, since there are four points of unit distance from A'.
Figure 71/5.1
Let us examine how can we avoid the "bad" translations. A translation is
bad if it maps one point of Sm into another point of the same set, or into one for
which there are more than one points of Sm+i of unit distance.
Construct a circle ki of radius 1 for each Αι e Sm as centre. The points of
intersection of ki with the other unit circles are denoted by P\, P2, ..., Pk\ these
are the points Д· is not supposed to be translated. Therefore we omit AiP\,
A1P2, ..., AiPk from the unit vectors we would like to use in our translation
process. After doing so for every i we exclude finitely many vectors; now all
168

1971/6
1971.
Figure 71/5.2
the remaining ones will provide an appropriate set 5m+i· On Figure 1971/5.2
we denoted by Ρ those points corresponding to the vertices A\, A2, Αί, yielding
unit vectors not suitable for our present purposes.
In conclusion, we showed that starting from any set S\, by iterating the
above process a set Sm for any m with the required properties can be constructed.
1971/6. Let
/au ai2 ··· a\n\
a2\ a22 ··· a>2n
A =
\ anl an2 ■ · · ann'
be a square matrix with all aij nonnegative integers. For each i,j with aij =0
we have
(1) Ojl + a^i + . · · + din + a\j + d2j + .. . + anj > n.
Prove that the sum of all the elements in the matrix is at least
vL
2 ■
First solution. (1) simply means that if there is a position in the matrix
where the corresponding value is zero, then the sum of the elements in that row
and column is at least n. Notice that this assumption remains unchanged after
changing two rows or two columns or rotating the entire matrix by 90°. (This
last operation turns rows into columns and vice versa.)
169

Solutions
1971/6
Add the numbers in each row and column and denote the minimum of the
resulting 2n numbers by s. We can assume that this minimum is attained by the
first row. If s > η then the sum of the elements in the matrix is at least η , hence
the statement is true.
If s < n, then the first row contains at least n — s zeros. Rearrange the matrix
by column changes such that the zeros are at the beginning of the row. In that
columns the sum of the elements — according to (1) — is at least n — s.
Henceforth in the first n — s columns (in the ones surely starting with 0) the
sum of terms is at least (n — s)(n — s) = (n — s) . In the remaining s columns the
sums are at least s columnwise, giving at least s to the total sum. Hence the
sum of all the elements can be estimated by
( <l^ 2 n2 (n-2s)2 n2
(n-s) +s = —+ >^r>
v 2 2 ~ 2
which now solves the problem.
Second solution. Rearrange the matrix using row-row and column-column
changes such that the most possible zeros stand in the upper left end of the main
diagonal. Let us assume that the maximum number of the zeros in the main
diagonal is k. Now partition the matrix into four parts as it is shows by Figure
1971/6.1. The upper к х к minor A has all zeros in the main diagonal, while in
О
о
о
λ
ч
1
с
к
А
О
О
4.
ς —
п-
-к
в
τ
ч
J
D
\
<
п-к
п-к
Figure 71/6.1
I
D there are no zeros — otherwise a row-row or column-column change would
transport an additional zero into the main diagonal.
170

1972/1
1972.
В and С cannot contain zeros which are symmetric to the main diagonal,
otherwise by an appropriate change of two rows a zero can be transported to D.
The sums of the terms in the individual parts will be denoted by s(A), s(B),
s(C) and s(D).
Since D contains no zeros, we get s(D) > (n — k) . Adding up the rows and
columns corresponding to diagonal positions in A (because of (1)) we get at least
kn. In this sum the elements of A appear twice while the elements of В and С
appear once. Therefore
2s(A) + s(B) + s(C)>nk.
Now add the elements of В and С in the following way: add an element in
В first to its reflection to the main diagonal (which is in C). Since there are no
opposite zeros, this procedure provides at least 1 in every position in C, implying
s(B) + s(C)>k(n-k).
Adding 2s(D) > 2(n — k) to these two inequalities we get
2 (s(A) + s(B) + s(C) + s(D)) >nk + k(n -k) + 2(n - k)2 = n2 + (n- k)2.
This shows
2
TL
s(A) + s(B) + s(C) + s(D)>—,
which coincides with the inequality we wanted to demonstrate.
Remark. The estimate cannot be improved as the following example shows:
TL
Let η = 2k, and put — in every position of the main diagonal and zero everywhere
else.
1972.
1972/1. Given any set often distinct numbers in the range 10, 11, ..., 99,
prove that we can always find two disjoint subsets with the same sum.
Solution. A set of ten elements has 210 — 2 = 1022 proper subsets. The sum
of at most 9 numbers, each of two digits is at least 10 and at most
91+92 + ... + 99 = 855;
hence there are at most 846 possible values for such a sum. Therefore by forming
sums containing one, two, three, ... , nine terms from the given ten numbers,
there are two which are equal. If there are common terms in these two sums,
then we can delete them and we still have equality. It cannot happen that during
this procedure one sum becomes zero, since then the same should happen to the
171

Solutions
1972/1
other one, although the two sums are not composed from the same terms. Hence
there are always two disjoint subsets with equal sum.
1972/2. Given η > 4, prove that every cyclic quadrilateral can be dissected
into η cyclic quadrilaterals.
Remarks before the solution. The cyclic property of a quadrilateral depends
only on its angles, more precisely if it admits two opposite angles with sum 180°
then the quadrilateral is cyclic. Therefore if we perform changes on a cyclic
quadrilateral leaving the angles unchanged, the result will be cyclic as well.
Every symmetric trapezium is a cyclic quadrilateral and we can dissect it
into an arbitrary number of cyclic quadrilaterals by lines parallel to its base.
Therefore the statement is obviously true for a symmetric trapezium and hence
for rectangles — so we can disregard them in our subsequent arguments.
First solution. Let the angle of the cyclic quadrilateral ABCD at A be
acute (since IA + 1С = 180°, and not all angles of the quadrilateral are right
angles, such vertex exists). Reflect ABCD to the bisector of IA and then shrink
it until it gets inside ABCD. The resulting quadrilateral will be denoted by
AB'C'D'. According to our remark above it is cyclic (Figure 1972/2.1). The
parallels to AD and AB passing through С intersect CD and ВС in X and
Figure 72/2.1
Y, respectively. B'C'XD is a symmetric trapezium since the angles at B' and
D coincide. The same reasoning shows that D'C'YB is a symmetric trapezium.
Finally, XC'YC is a cyclic quadrilateral since its angles coincide with the angles
of ABCD.
ι
In this way ABCD has been dissected into four cyclic quadrilaterals; now
dissecting one of the symmetric trapeziums further, the number of pieces can be
increased arbitrarily with keeping the condition that all quadrilaterals are cyclic.
172

1972/2
1972.
Second solution. The two obtuse angles of the cyclic quadrilateral
cannot be opposite since the sum of opposite angles is equal to 180°. Let 1С = 7
and ID = 8 two neighbouring obtuse angles. The section A'B' parallel to DC
dissects the original quadrilateral into a cyclic quadrilateral A'B1 В A and a
trapezium A'B'CD. If 7 = 8 then ABCD is a symmetric trapezium, therefore we
can assume (by symmetry) that 7 > 8 (Figure 1972/2.2).
A
Figure 72/2.2
Choose Ρ and P' on CD and A'B', respectively, in such a way that
IDPP' = 8, and then Q and Q' on CB' and A'B' such that ICQQ' = 8. Now
the intervals PP' and QQ' can be translated to intersect each other in an
inner point R of the trapezium. In this case the trapezium A'B'CD is dissected
into three cyclic quadrilaterals: A'P'PD is a symmetric trapezium since DP
and A'P' are parallel and IA'DP= LDPP' = 8\ QCPR is a cyclic
quadrilateral since LCQR-8 is equal to the complement of the opposite angle and finally
B'P'RQ is a cyclic quadrilateral because /B'P'R = 8 is equal to the complement
of the opposite angle.
In this way ABCD has been dissected into four cyclic quadrilaterals and the
symmetric trapezium can be further dissected into as many cyclic quadrilaterals
as one wishes.
Remark. There are numerous different solutions of the problem; the above
two share the advantage that they apply to all cases at once. Most of the other
solutions distinguish cases according to where the centre of the circumcircle of
the cyclic quadrilateral lies.
173

Solutions
1972/3
1972/3. Let m and η be nonnegative integers. Prove that
(2m)!(2n)!
m\n\(m + n)\
is an integer. (According to our conventions 0! = 1).
(1)
First solution. Introduce the notation
(2m)!(2n)!
f(m,ri) =
m\n\(m + ri)\
/(m,0) is an integer for all nonnegative integer m since
(2m)! /2m
m
f(m,0) =
(τη!)2
and the combinatorial interpretation of this expression shows that it is an integer.
The function f(m,ri) satisfies the following equation:
(2) /(m+l,n) + /(m,n + l) = 4/(m,n).
We prove this by substituting the definition of the left hand side:
tf j.i ^/y xn (2m + 2)!(2n)! , (2m)!(2n + 2)!
j(m+ l,n) +j(m, n+ 1) = ——- Η — =
(m+ \)\n\(m + n+ 1)! m!(n+ l)!(m + n+ 1)!
(2m)!(2n)! 4m + 2 + 4n + 2
= 4/(m,n).
т!п!(?тг + п)! m + n + 1
This verifies (2). After substituting n:=n — 1 and reordering the equation we
have
(3) f(m,n) = 4f(m,n— 1) — f(m+ l,n — 1).
Now induction on η implies that f(m,ri) is an integer for every m and n: We
already saw that it holds for n = 0 and arbitrary m. Suppose that f(m,n— 1) is
an integer for all m, now (3) implies that f(m, n) is an integer; this concludes
our first solution.
Second solution. The main idea of this solution is the following: we show
that if a prime ρ divides the denominator then it also divides the numerator,
and in addition the exponent of it in the numerator is at least as much as in the
denominator. This exactly means that the ratio is an integer.
First we prove a statement originally due to Legendre: the exponent of a
prime ρ in the prime factorization of n! is:
η
LPJ
+
η
+
η
Lp3 J
+ . ..+
η
-Pk _
k+\-
(pK<n<pK+l)
This is true, since among the first η positive integers every pth is divisible
by p, hence there are
η
LPJ
such numbers. Every ρ th is divisible by ρ , and
174

1972/3
1972.
there are
lb
_P2.
of those, and so on: there are
Therefore ρ appears in n! with exponent
ρ > η we have
in an infin
ite sum.
η
pk+l
lb
-P.
+
lb
,pk _
η
~~9
_P .
numbers divisible by pk.
+...
lb
-Pk -
. Notice that for
= 0, hence there is no mistake by writing these terms
We will need one more statement: For a, b nonnegative reals
(4) [2a] + [2b] > [a] + [b] + [a + b].
To prove this let a = [a] + a and b = [b] + β where 0<a<l,0</3<l.Ifa + /3<
1 then [a + b] = [a] + [b] and so
[2a] + [2b] = 2[a] + 2[b) + [2(a + β)] > 2[a] + 2[b] = [a] + [b] + [a + b].
In case α + β > 1 we have either 2a > 1 or 2β > 1. Suppose that 2a > 1. Then
[a + b] = [a] + [b] + [α + β] = [a] + [b] + l,
[2α] = 2[α] + [2α] = 2[α] + 1.
This implies
[2a] + [2b] > 2[a] + 1 + 2[b] = [a] + [b] + [a + b],
proving the statement. A similar argument concludes the verification of (4) in
the case 2/3 >1.
Now the exponent of ρ in the numerator of the ratio under examination is
A =
'2m'
. Ρ .
+
'2m'
9
. p .
+ . . .+
"2m'
. Pk .
+
~2n~
. Ρ .
+
~2n'
-P2.
+ ...+
~2n~
.Pk.
Choose к such that pk+l is greater than 2m and 2n.
The exponent of ρ in the denominator is
m
—
I_P J
+
m
—
_p2_
+ ...
m
—
.pk .
+
η
—
LpJ
+
η
_P2_
B= — + -=·
+ .
+ ...+
η
+
+
n + m
LP"J L Ρ J L ρ
+
n + m
...+
n + m
Ρ
к
In order to solve the problem we have to show that A > В which follows
from
2m
pi
+
2n
pi
>
m
pi
+
η
pi
+
n + m
pi
(i=l, 2, ..., k).
This last statement, however, has been proved above at (4), hence the solution is
complete.
Remarks. 1. The problem is originally due the Belgian mathematician E.
Catalan, who got the result as a byproduct when analysing certain functions.
175

Solutions
1972/3
2. There are various generalizations of the problem; for example, on the
Mathematics Olympiads of the United States the following form appeared: Prove
(5m)!(5m)!
that is an integer.
m\n\{3m + n)!(3n + m)!
1972/4. Find all positive real solutions {x\,X2,x3,X4,xs) of the system
(1) {x\-x3X5){x\-x3x5)<0,
(2) (χ2 — X4X\){x3 — x^x\) <0,
(3) {xl-X5X2){xl-X5X2)<0,
(4) {χ1-χιχ3){χ1-χλχ3)<0,
9 9
(5) (x$ — Χ2Χδ){χ\ — X2xa) < 0·
First solution. Perform the multiplications and sum the left hand sides;
their sum will be denoted by B:
D 2 2 2 2 ,22,
B = X^X2 — Х2Хъх5 ~Х\ХЪХ5 + ХЪХ5+
2 2 2 2 2 2
+ XnX3 ΧηΧ^,Χγ XnX/^X\ "г ХлХл τ
2 2 2 2 2 2
+ Х3Хл — ХлХ^Х2 — Χ3ΧζΧ2 "Ί" Х<Хп~^~
2 2 2 2 2 2
+ 3:4^5 —Х5Х\ХЪ —Х^Х\ХЪ +Х\Х3 +
2 2 2 2 2 2
+ XcXi —ХлХ2Х^ — ХсХоХ^"^-ΧθχΑ·
From this we deduce
2Β-Χγ ({χ2 — Χ4) +{x3—xs) ) +
+ x\ {{x3 - x5)2 + (ж4 - ^!)2) +
+ ж3 \(х4 ~ х0 + (х5~ х2) ) +
+ ^4 ((»5 ~ ^2>2 + Ol - ^з)2) +
+ xl ((χι - х3)2 + {χ2 - Χ4)2) ·
We conclude that 2Β is the sum of five nonnegative numbers, therefore В > 0.
On the other hand, summing the original system gives В < 0, hence we get
that В = 0. This, however, can happen only in case all terms in IB are equal to
zero:
x2 ~ XA = x3 ~ x5 - XA ~ xl = x5 ~ x2 = x\ ~ x3 = ®i
giving χι =χ2 = χ3 =χ^ = Χζ.
The fact that x\-x2-x3-x/^-x^-c for arbitrary positive с satisfies the
equality can be seen directly, hence the solutions of the system are given by
x\ = X2 = x3 = X4 = Х5 = c.
176

1972/4
1972.
Second solution. For any positive real с the five tuple
(6) X\ =X2 = %3 =Χ4 = Χζ =C
obviously solves the system. Furthermore, if (2:1,2:2, · · · >xs) ls a solution then
so is I —, —,..., — ). Next we show that the assumption of having a solution
\xi x2 2:5/
of the form different from the above ones leads to contradiction. An additional
solution violates at least one of the following equations:
2:1=2:3, 2:3 = 2:5, 2:5=2:2, 2:2 = 2:4, 2:4 = 2:1.
Since a cyclic permutation of the Xi's leaves the system unchanged, we can
assume that
(7) 2:3 < 2:5.
We divide our argument according to these further solutions satisfy 1) x\.<
2?2 ОГ 2) X\ > X2.
0 0 0
1) Let x\ <x2. In this case x\ — Ж3Ж5 < #2 — £3^5, therefore x\ — 2:32:5 >
О 0 >—»-ч
0, otherwise 2:2 — 2:32:5 >0 holds, which contradicts (1). Hence 2:^ — 2:32:5 <ΓΌ,
or equivalently
(8) x\ < -^/2:32:5 < x$ i.e. 2:i<2:5.
9 9
We cannot have 2:^ — 2:32:5 < 0, since it implies χ γ — 2:32:5 < 0 which again
contradicts (1). Consequently, 2:2 — £3^5 >0 or
(9) х2>^/х~ъХ5>хъ, i-e- 2:3<2:2.
(7) and (8) implies
9 ■ 9
2:5 > 2:12:3, i.e. 2:5 — 2:12:3 >0,
which — according to (4) — now gives 2:4 — 2:12:3 < 0, which provides
2
(10) 2:4 < 2:1X3 < 2:32:5
by applying (8). On the other hand (7) and (9) shows that 2:3 < 2:22:5, which
together with (3) and (9) yields
(11) 2:4 > 2:5^2 > 2:52:3.
Now (10) and (11) shows the desired contradiction.
9 9
2) Now suppose that 2:1 > 2:2· Then 2:1 — 2:32:5 > 2:^ — 2:32:5, and so (1) imp-
9 9
lies that χ^ — 2:32:5 is nonpositive, hence 2:^ < 2:32:5 and so
(12) Х2<^/хъХ5<Х5-
177

Solutions
1972/4
9 9 9
Χγ — χ-^Χζ <0 implies x% — χ3Χζ<0 which again contradicts (1), hence x\ >
χ^Χζ, or equivalently
(13) xx >^/x3x5>x3.
(2) holds only if
X4X1 <max (х2,хз) ·
Since (12) and (7) imply x\ < χ3Χζ and x\ <x3x$, we have
(14) x&x\ <χ3Χζ.
On the other hand, (5) is satisfied only if
X2X4 > min (χ5,χΛ .
Since (7) and (13) proves χϊ- > ΧζΧ3 and x\ > ΧζΧ3, we get
(15) #2^4 ^Х5ХЪ-
Comparing (14) and (15) we get хдх\ < χ^χδ, or equivalently x\ < x^· This,
however, contradicts our starting assumption.
This last line now shows that there is no solution different from the ones
listed in (6).
1972/5. Let f and g be two real valued functions defined on the real line
satisfying
(1) f{x + y) + f{x -y) = 2fix)giy).
Suppose that fix) is not identically zero and \fix)\ < 1 for all x. Show that
\giy)\<\ for ally.
First solution. Since the absolute value of a sum does not exceed the sum
of absolute values, we have
2\fix)\\giy)\ = \fix + y) + fix-y)\<\fix-y)\ + \fix + y)\,
hence one of the following inequalities hold:
\№\Ш\ < \f(x + y)\, \№\\g(y)\ < \f(*-y)\-
This means that for у fixed and for every χ there is an χ ι (in the form of
x\ = χ + у or x\ = χ — y), satisfying
(2) \m\\g{y)\<\f{xi)l
Now apply (2) for x = xq; we may assume that fixo)^0:
\№о)\Ш\<\№0\·
178

1972/5
1972.
By iterating the above reasoning we get a sequence xq, x\, ..., Xk which
satisfies
\f(xk-i)My)\<\fMl
hence
\f(xo)\\g(y)\k <\f(xk)l
where к is an arbitrary positive integer. Since \f(xk)\ < 1, we have
(3) ш^ ШГ
Now using (3) the existence of a number у with \g(y)\ > 1 provides a
contradiction, since the right hand side is fixed while for suitable к the left hand side
can be arbitrarily large.
Second solution. In this solution we will use the fundamental property of
reals that every bounded subset of the reals admits a least upper and a greatest
lower bound. The least upper bound of a bounded set is well-defined, although it
might not belong to the set itself; for example the least upper bound of the real
numbers less than 1 is equal to 1.
Since the set of reals \f(x)\ is bounded, it admits a least upper bound H,
and since f(x) is not identically zero, Η is positive; therefore \f(x)\ <H for
all x.
Applying the formula for the absolute value of a sum we get
2\f(x)\\g(y)\<\f(x + y)\ + \f(x-y)\<2H,
hence
(4) \№\\g(y)\<H.
Now suppose that there exists у$ with \g(yo)\ > 1. In this case (4) implies
that for all χ
showing that the set of values |/(ж)| admits an upper bound Щ strictly less than
the least upper bound H. This is, however, a contradiction, and so \g(y)\ < 1 is
proved for all real y.
Remarks. 1. The condition |/(ж)| < 1 was used only in the assumption that
| f(x) | is bounded.
2. There are functions satisfying the equation; one example is given by
/Or) = sin ж, <70r) = cos:r.
179

Solutions
1972/6
1972/6. Given four parallel planes, prove that there exists a regular
tetrahedron with a vertex on each plane.
First solution. Suppose that the tetrahedron ABCD exists with vertices
A, B, C, D on the parallel planes α, β, η, δ. The planes are on one side of a
and their distances from a are b, с and d, respectively. On Figure 1972/6.1 these
planes are perpendicular to the plane of the figure, hence they appear as parallel
lines.
Figure 72/6.1
β intersects AC in C\ and AD in D\. Since the ratio of the distances
between a—β and a—7 is b: c, the same holds for the ratio of AC\ and AC:
(i) ^ = S.
w ACX b
For the same reason
AD _d
~b'
(2)
ADi
Notice that an enlargement with centre A transforms our regular tetrahedron
into another regular tetrahedron. The planes passing through the vertices are
mapped under this enlargement into planes with the same ratios of distances.
Based on the above observations the existence of the desired solid can be
verified as follows: consider an arbitrary regular tetrahedron ABCD. Fix C\ and
D\ on AC and AD such that they satisfy (1) and (2). Since B, C\ and D\ are
not collinear, they determine a plane β'. Apply now an enlargement with centre
A transforming the distance of A and β' to b and mapping the plane a passing
180

1972/6
1972.
through A and parallel to β'. After the enlargement the distances of С and D
from a become с and d, therefore the parallel planes to a passing through В, С
and D are of distances b, с and d from a. This shows that the desired tetrahedron
exists.
Second solution. The original version of the problem received by the jury
at the competition also asked the edge length of the tetrahedron in terms of the
distances of the parallel planes. The following analytic geometric solution will
answer this question as well.
Let us choose our coordinate system in such a way that the
coordinates of the vertices of the tetrahedron become Л(0,0,0), J3(3s,3s\/3,0),
C{—3s, 3s\/3,0), D(0,2s\/3,2s\/6) (the edge length of this tetrahedron is 6s).
The unit vector η starting at A has coordinates n(p, q, r), so
(3) p2 + q2+r2 = l.
We will use the notations А§ = Ъ, Аб = с, AS = d. The orthogonal
projections of B, C, D to the line of η are denoted by В', C, D', respectively. If we
can choose η and s in a way that AB' = b, AC' = c, AD' = d is satisfied (where
b, c, d are given positive distances) then the planes α, β, η, δ orthogonal to η at
the points А, В, С, D will have the above prescribed distances and so the points
provide the desired regular tetrahedron ABCD.
Now the coordinates of the vectors representing the edges are as follows:
b(3s,3sV3,0) c(-3s,3s\/3,0) d(0,2sVl, 2s\/6).
Since the scalar product with a unit vector is equal to the (signed) length of the
projection to the line determined by the unit vector, we have
nb = 6, nc = c, nd = d,
or in coordinates
3sp + 3\/3sq = b,
—3sp + 3y/3sq = c,
sq + 2v6sr = d.
This system can be easily solved for the unknowns sp, sq, sr and we get:
b — c b + c 3d — b — c
(4) sp= , sq = —γ=, ,- sr =
6 ' бл/3' бл/6
In view of (3) this gives
1 / 9 9 \
(sp)2+(sq)2+(sr)2 = s2 = ^-^((V6(b-c)) +(V2(b + c)) +(3d-b-c)2) =
= —Ц (b2 + c2 + d2 + (b- cf + (c- d)2 + (d-b)2y
181

Solutions
1972/6
In conclusion, the edge length of the tetrahedron is equal to
(5) 6s = -^=y/b2 + c2 + d2 + (b - c)2 + (c- d)2 + (d -b)2.
(4) also shows the coordinates of the unit vector n:
(6)
P =
6s
q =
b + c
γ -
3d — b — c
6йл/6
Consider a tetrahedron ABCD of edge length 6s given by (5) in the
coordinate system and fix the unit vector η with coordinates (p, q, r) given by (6). The
planes α, β,η and δ orthogonal to η passing through the vertices of this
tetrahedron will have the prescribed distances; consequently the desired tetrahedron
exists.
Remark. Notice that the first solution did not use the fact that the tetrahedron
is regular, consequently it applies to any tetrahedron.
1973.
1973/1. Let OP\, OPi, · ■ ·, OPn be unit vectors in a plane. P\, Ρ2,..., Pn
all lie on the same side of a line through O. Prove that if η is odd, then
\OPi+OP2 + ... + OPn\>\,
where \OM\ denotes the length of the vector OM.
First solution. First, observe that if the angle of the vectors a and b is at
most 90°, then
(1)
|a + b|>|a| and |a + b|>|b|.
Figure 73/1.1
Indeed, if a, b, a + b are
the sides of a triangle where the
angle opposite to a + b is at most
90°, then it is the largest angle
of the triangle, so the longest
side is |a + b|. (1) holds even if
one of a or b is the nil-vector,
in this case their angle is 0° (see
Figure 1973/1.1).
Let A and В denote the poiqt of intersection of the line e and the unit circle
with centre O, and order the points Pi from A to B. So the points lie on the arc
between P\ and Pn (see Figure 1973/1.2).
182

1973/1
1973.
А О В
Figure 73/1.2
We extend the statement of the problem with the following condition: the
sum of the vectors lies in the angular domain P\OPn. We prove the statement
by induction: for n = 1 the statement is obvious; assume that it is true for η — 2,
that is, for the vector s = OP2 + OP3 +... + OPn_\
\s\ = \OP2 + OP3 + ... + OP^{\>h
and s lies in the angle P2OPn_\. The vector OP\ +OPn = \ lies along the
bisector of the angle P\OPn, and so by our introductory remark it makes an acute
angle with s. Hence:
|v + s| = \θΡχ +ΟΡ2 + ■ ■ ■ + θΚ\ > |s| > 1.
It is easy to see that v + s lies in the angular domainPi OPn, and this is what we
wanted to prove.
Second solution. Using the notation of the first solution, denote О Pi by
щ (i = 1, 2, ..., n). For arbitrary 1 <i, j <n the vector tj +ti bisects the angle
of tj and ei, and so ej+e^ makes an acute angle with every vector es in the
angular domain bounded by tj and e^. Thus е5(е;+ег)>0. If n = 2k — 1, the
"middle" vector, e^ is in the angles determined by the pairs of vectors (ei, fyk-i),
(β2>β2Α;-2)> ·■·. (Ч-иЧ+l)· So:
ek(ei+e2 +.. . + ek_i+ek+ek+i +.. . + е2к-2 + ^2к-\) =
(2) =ek(e1 +e2k-i) + ek(e2 + e2k-2)+· .- + ek(ek_i +ek+i) + e2k>e2k = l.
On the other hand,
(3) efc(ei + e2 +... + en) < le^I |ei + e2 +... + en| = |e! + e2 +... + enI.
Now (2) and (3) gives
|ei +e2 + . - - + en| > 1.
183

Solutions
1973/2
1973/2. Can we find a finite set of non-coplanar points, such that given
any Wo points, A and B, there are two others, С and D, with the lines AB and
CD parallel and distinct?
Solution. Let ABCD and A'B'C'D' the bottom and the top faces of a
cube. Let the reflection of S, the centre of the cube, to the bottom and the top
faces К and Κ', respectively. The 8 vertices of the cube along with К and K'
make a set of points, Μ satisfying the conditions of the problem. (See Figure
1973/2.1).
SX
С
Indeed, pick two points of M. If S does not lie on the line through them,
then it is parallel to its reflection to S. This reflection is through two points of
M, too, as Μ is symmetric to S. If this line is a main diagonal of the cube,
then it is parallel to one of the segments К A, KB, КС, KD. Finally, KK' is
parallel, for example, to AA!.
Remark. There are several ways to construct a set of points M, satisfying the
conditions of the problem. Our example can be generalized in two ways: Instead
of a cube we can take any regular prism with even many faces. Alternatively,
our example can be realized as the union of two central symmetric hexagons,
AA'K'C'CK and BB'K'D'DK with the common diagonal KK'. Making a
similar construction with e.g. two regular 2n-gons the resulting An —2 points
satisfy the conditions of the problem.
184

1973/3
1973.
The 12 midpoints of the edges of a cube is a good construction, as well.
For a non central symmetric Μ put together 5 cubes of the same size, such that
their centres are collinear and the neighbour cubes have a face in common. The
24 vertices and the centre of the first, second and fifth cubes form a set of 27
points that satisfies the conditions.
1973/3. Let a and b be real numbers for which the equation χ +ax +
\-CLQ
a2 + b2.
9
bx + ax + 1 = 0 has at least one real solution. Find the least possible value of
First solution. As χ = 0 is not the root of the equation, we can divide by
x . After reordering we get
x2 + -^r+alx + -)+b = 0.
xl \ x/
1 о 1 о
Let χ + — = z. Thus χ + —=■ = ζ — 2. The sum of a positive number and its
X X1
reciprocal is at least 2, so
(1) z2 = x2 + -\ + 2>4.
Using the substitutions above, we get
z2 + az + b-2 = 0,
(2) 2
2-z* = az + b.
Applying the Cauchy-inequality to the right hand side of the equation:
(3) (2 - z2)2 = (a-z + b-l)2<(a2 + b2)(z2 + 1),
yielding
(4) α2 + ()2>(2ς^¥=(χ2_2)Λ 3 ^
2 + l \ z2 + V'
9
Both factors of the product on the right hand side is increasing when ζ is
9
increasing, so according to (1) it takes its smallest value at ζ =4:
а2 + Ъ2>\.
9 9 4
Hence the minimum of a + b is -; we have to show that under these conditions
the original equation has a real root. Since in this case there is an equality in (3),
by the Cauchy inequality
a:b = z:l, that is a2 = b2z2,
185

Solutions
1973/3
9 9
must hold, and so a =4b . Hence
2 , г2 сг2 4 г2 4 2 16
α +6 =56 = -, b =-, α =-.
The possible values of the coefficients are:
4 2
If we choose the negative value in both cases, χ = 1 is the root of
4 4 3 2 2 4 . .
x --xJ - -r- -£ + 1=0,
о о 4
and so the least possible value of a +b is -.
Second solution. Let us assume that χ is a positive solution of the
equation. As for every real с the inequality c> — \c\ holds,
A Ί О A Ί О
0 = £ +аж + bx +ax +1 > ж — \a\x — \b\x — |а|ж + 1,
hence
(5) xA + l < \a\x3 + \b\x2 + \a\x = \a\(x3 + x) + -^-2x2.
Now, as (x — 1) > 0, we have
2x2<xA + l.
Since the sign of χ — 1 and χ — 1 are the same, (x — l)(x — 1) > 0 follows, and
so
from (5) we get
that is
3 / 4 л
χ +χ<χ +1;
z4 + l<|a|(z4 + l) + ^(z4 + l),
.SW4
|Ь|>2-2|а|=2(1-|а|).
If |а|<1,
9 9 I |9 I |9 I |9 I |9 / I I 4 \ 4
a^ + 6^= or + \b\ > or + 4 + 4 or-8 a =5 a +-.
ι ι ι ι -ι ι V 5/ 5
thus
(6) α2 + 62>ί
and equality can only hold if \a\ = -.
186

1973/4
1973.
If \a\ > 1, then (6) clearly holds.
Now, if we assume that the original equation has a negative solution, then
the positive χ is a solution of the equation
4 3 9
x — ax +bx — ax+ 1=0;
and, again we get (6). As in the first solution, we can choose the coefficients such
о о 4 9 9
that a + b = - and the equation has a real root. Hence the minimum of a + b
. 4
1S5
Remark. The equation in the problem is a so called reciprocal equation; a
polynomial equation is called reciprocal if along with the root xq the value —,
XQ
(or ) is a root, too. The coefficients of these equations are symmetric (or
XQ
antisymmetric), i.e. the η — k-th and the k-th. coefficients of the polynomial of
degree η are equal (or opposites). The substitutions from the first solution work
even for reciprocal polynomials of degree 9.
1973/4. A soldier needs to sweep a region of the shape of an equilateral
triangle for mines. The detector has an effective radius equal to half the altitude
of the triangle. He starts at a vertex of the triangle. What path should he follow
in order to travel the least distance and still sweep the whole region?
Solution. Let us consider the triangle ABC, with side 1, and
suppose that the soldier starts at A. In order to sweep the other two vertices, his
path must intersect the circles
τη
k\ and ki of radius — with
centre В and C, respectively.
By symmetry we may assume
that he first intersects k\ at
the point P. The shortest path
from Ρ to fc2 nes on the line
PC. Let Q denote the point
of intersection of ki and PC.
The path AP + PC is longer
τη
than AP + PQ by —, hence
AP + PC is the shortest when
AP + PQ is the shortest (see
Figure 1973/4.1).
Figure 73/4.1
First, we are looking for the point Pq on the circle k\, such that APq + PqC
is the shortest. We show that Pq is the point where the altitude of В intersects the
187

Solutions
1973/4
circle. Let e be the tangent of k\ at Pq and pick an arbitrary point Ρ distinct from
Pq on k\. We prove that AP + PC > AP0 + PqC. Let A' denote the reflection of
A to e. Since AA' = m, the points A', Pq, С are collinear and so APq + PqC =
A'C. The line e separates k\ from A thus the point of intersection of AP and e
is a single point M. Clearly, M = i'M, so AP + PC = A'M + MP + PC', the
polygonial path A'MPC is longer than A'C, thus
AP + PC<AP0 + P0C.
Now, if Qq denotes the intersection of fc2 and Pq, the shortest path is: A
Qo.
P(
о
Figure 73/4.2
conditions are satisfied by the A —» Pq
It remains to check that
the whole triangle can be
swept from this path. For
a segment XY the points
not farther than —· cover a
2
"stadium-shaped region" that
is shown of Figure 1973/4.2.
The stadium-regions
belonging to APq and PqQq
cover the ABC triangle, so the
Qo path. (See Figure 1973/4.3).
В
Figure 73/4.3
188

1973/5
1973.
1973/5. Let G be a set of non-constant functions f. Each f is defined on
the real line and has the form f(x) = ax Л-Ь for some real a, b.
a)Iff,geG, then so is go fe G, where g о f(x) = g (f(x)).
b) If f is in G, then so is the inverse f~ .
ι χ b
a a
c) Every f in G has a fixed point. In other words we can find Xf such that
f(xf) = xf.
Prove that all the functions in G have a common fixed point.
Solution. / not being constant means that a^O. From c) we get that if
in f(x) = ax + b, the condition a=l implies 6 = 0, because xj + b = Xf cannot be
true.
ι X b
From b) f~l(x) = and from a)
a a
/o/_1=/ort)=«(
+b = x
a a J
belong to G. Set e(x) = x; as G is non-empty, e(x) is in G\ every real χ is a
fixpoint of e(x).
If G contains only e(x), the statement is obvious. If G has an element / not
equal to e, then it has infinitely many elements as /, / ο /, (/ ο /) ο /, ... are all
distinct functions, because the coefficients of χ are a, a , a , ... and they are all
distinct.
Now, let f(x) = ax + b, g(x) = cx + d (a^l, c^l) be two arbitrary elements
of G. We have to show that if A; is a fixpoint of /, then it is a fixpoint of g, as
well. If A; is a fixpoint of / then
b
ak + b = k, that is k =
1 — a
d
Similarly, the fixpoint of g is , hence to prove our statement we have to
1-е
show that
(i) *= " d
1 — a 1-е
Consider the functions p = f og and q = go f:
p(x) = a(cx + d) + b = acx + ad + b
q(x) = c(ax + b) + d = acx + bc + d.
The composition ρ ο q~l is an element of G, too:
poq~ (x) = ac ( I + ad + b = χ + (—be — d + ad + b).
\ac ac J
189

Solutions
1973/5
Since the coefficient of χ equals 1, —bc — d + ad + b = 0, hence
b(l-c) = d(l-a)
b _ d
1 — a 1-е
and (1) really holds.
Remarks. 1. In this problem, the set G is a group. As for e(x) = x we have
f oe = eo f = f, e is the unit element of the group; in order to show that G is
a group, it only remained to prove that G is associative, i.e if f,g,he G, then
(I ' ° g)°h = f o(goh) which is an easy calculation.
G is called a group of linear mappings of the line. As we showed, for every
f(x) = ax + b in G, f(k) = k, that is ak + b = k, b = k(l —a); hence every element
of G is of the form
(2) f(x) = ax + k(l~a).
2. The following observations show the geometric motivation of the
problem: (2) implies that / is a similarity transformation of the line with centre к and
ratio a. Indeed, let χ be an arbitrary point of the real line. Its distance from к is
χ — k, the the distance of f(x) from к is f(x) — k, reordering (2) we get
fix) ~ к
Γ~ = α'
χ κ
which means that the proportion of the distances of f(x) and χ equal to a
constant a. The geometrical meaning of c) says that there is no translation in G, as
every element has a fixed point. The statement itself is a well known theorem in
geometry: If a set of similarities of the line form a group that contains no
translations, then the elements of the set are central similarities with a given centre
(The theorem holds in higher dimensions, too).
Even the proof of the theorem has some geometric motivation based on
the following: Apply enlargements with proportions a and then c, followed by
1 1
enlargements with proportions - and - to the plane then the resulting objects
а с
do not change size and direction of the objects. As there is no translation, it has
to be the identity transformation.
2. An other geometric correspondence is the following one: G is a set of
lines with equation y = ax + b that are not parallel to the χ axes and have the
following properties:
a) If the lines y = ax + b and y = cx + d belong to G then so does у = acx +
(ad + b).
b) If a line is in G, then so is its reflection to у = χ.
c) Every line in G meets the у = χ line.
Corollary: Every element of G contains a point K(k,k) in common.
190

1973/6
1973.
1973/6. Let αι,α,2,... ,an be positive reals, and q satisfies 0 < q < 1. Find
b\, 625 · · · 5 bn such that:
a) ak < bk, (k=l,2,..., n);
b)g<%i<-,(/c = l,2,...,n-U·
bk Q
c) b\ + 62 + · · · + bn < (a\ + a>i +... + an).
Solution. Condition c) suggests us to construct the numbers bk in the
following way:
Ь\ - a\ + a<iq + a^q + ... + anqn~
2 = aiq + a2 + CL2q+. . . + anq
(1) hi~a\q +d2q + CL3 +.. . + anqn
bn=aiqn~1 + a2qn~2 + CL3qn~3 +.. .+an.
In general
к— 1 η
(2) bk = Y^aiqk-i + ak+ Σ а^~к.
г=\ г=к+\
We show that the construction in (1) satisfies conditions a)-c). a) is obvious as
on the right hand side of (1) there are additional positive summands with ak. for
b) observe that
к п к -ι η
bk+\=Y,ai(ik+l~i+ Σ ^~к~1=яТ,а^к~1+- Σ а^~к·
ν
= 1 i=k+l г=1 ^ i=k+l
AsO
<q<
1 and
Q
bk+1
, we conclude
1 k
<-Y^aiqk~l
к
1
+ -
Q
η
Σ
i=k+l
η
г=1 г=к+1
Thus we proved both inequalities in b).
To get c) sum the the columns of the expressions in (1) ; the sum of the г-th
column:
a>i (V ~l + Ч1~2 + ... + q+l^+ai(q + q2 + ... + q4'^ =
l-qi 1-дп~Л l+q-qi-qn-i+l i+q
= ai I -, + Q—; =ai ; <ai~, ·
1-q 1-q / l-q 1-q
191

Solutions
1973/6
Summing up the η sums of the columns:
Dj + b2 + . .. + bn < {αχ + a2 +... + αη)γ—,
hence c) is satisfied.
Remark. Our proof shows that the щ-s do not uniquely determine the bi-s.
Indeed, there are infinitely many η-tuples for bi-s that satisfy the conditions.
1974.
1974/1. Three players А, В, С play the following game: There are three
cards each with a different positive integer p, q and r where ρ <q < r. In each
round the cards are randomly dealt to the players and each receives the number
of counters on his card.
After two or more rounds, A has received 20, В 10 and С 9 counters. In
the last round В received the largest number of counters.
Who received q counters in the first round?
Solution. Let us assume that they played η rounds (n > 2). In each round
the total of the cards is p + q + r, so the number of counters received altogether
is n(p + q + r). By the assumptions this equals to 20 + 10 + 9 = 39:
(1) n(p + q + r) = 3-13.
Since p> 1, q>2, r>3 and n>2, p + q + r >6, and (1) can only hold if
n = 3 and p + q + r =13. The highest score was 20, so r cannot be smaller than
7, so r>7. As p + q + r > 10 and В received r counters in the last round and
altogether 10 counters, the distribution of B's counters cannot be q + q + r, p +
q + r, and could not receive r counters twice. So the number of B's counters is
p + p + r, so
2p + r=10,
and p> 1, r > 7 implies p= 1, r = 8 and so q = 4.
С had 9 counters, so he could not get r = 8 counters, only 1 or 4. As В
received the card with 1 in the first round, С had to receive the card 4. Thus in
the first round С received the q = 4 counters. Now, we can set the table of the
game:
I. round
II. round
III. round
All
A
в
С
number of counters
8
8
4
20
1
1
8
10
4
4
1
9
192

1974/2
1974.
Remark. The following question was raised at the evaluation of the
solutions: does one have to make the table, i.e. verify the existence of the game, or
was it assumed. (They accepted the latter one). If we do not assume the existence
of the game, we can derive a wrong conclusion. For example, there is no game
that can result 18, 8 and 7 counters, respectively, although even in this case our
method shows that С received the q counters in the first round. The game (and
so the table) does not exist, though.
1974/2. Let A, B and С denote the vertices of a triangle. Prove that there
is a point D on the side AB of the triangle ABC, such that CD is the geometric
mean of AD and DB if and only if
С
(1) sin A sin В < sin2 —.
2
First solution. We have to show that (1) holds if and only if there is a
point D on AB such that
(2) AD-DB = CD2.
Let us assume that such a D exists and CD intersects the circumcircle of the
triangle in Ε (see Figure 1974/2.1). The product of the lengths of the segments
F
Figure 74/2.1
through an inner point of the circle is constant, so AD · DB = CD ■ DE, hence
from (2) we have CD = DE, that is the distance of Ε from AB equals the altitude
at C. Thus the point D exists, if and only if the line e parallel to AB of distance
hc from AB intersects the circumcircle. This intersection exists, if the distance
of F, the midpoint of the AB arc not containing С from AB is at most hc. This
193

Solutions
1974/2
holds if and only if the area of the triangle AFB is at least the area of the triangle
ABC:FAFB>FABC.
Let the length of the diameter of the circumcircle equal to 1. Then, BC-
sin A, AC = sin В and so
Fabc = r sin A sin В sin C.
In the isosceles triangle, AFB we have LAFB = 180° — C, the other two
С
angles equal to —, so its area
1 С
Fafb = £ sin2 уsin ci
and the inequality for the areas is equivalent to
1 1 о С
- sin A sin В sin С < - sin — sin C.
2 ~2 2
that is equivalent to (1). Thus we proved the equivalence of (1) and (2).
В
Figure 74/2.2
Second solution. Let us
assume, again, the existence of the point
D, and suppose that the angle С is
split into Cj, and C2 by CD (see
Figure 1974/2.2). Apply the Law
of Sines to the triangles ADC, and
BDC:
sin A CD sin Б CD
sinCi AD: sinC2 DB
Multiplying the two equations and applying (2) we get:
sin A sin В CD2
(4)
(5)
= 1.
sinQsin^ AD-DB
sin A sin В = sin C\ sin C^ ·
.2 С
As 2 sin Cj sin C2 = cos(Ci - C2) - cos(Ci + C2) and cosC= 1 - 2 sin —, we
conclude
С С
2 sin A sin В = cos(C! - C2) - cos С = 2 sin2 (1 - cos(C! - C2)) < 2 sin2 —,
so we showed that (2) implies (1).
С
Now, if (1) holds, from 2 sin2 — = 1 - cos С it follows that
2 sin Л sin Б < 1 - cos C,
2 sin Л sin В + cos С < 1.
194

1974/2
1974.
Since О < 2 sin A sin В and — 1 < cos C,
-1 < 2 sin A sin В + cos С < 1.
This means that there is an angle φ such that 0 < φ < π and the following
holds:
(6) cos φ = 2 sin A sin В + cos С
As cos</?>cosC we have 0<φ<Ο, introduce the following notations: C\ -
C + φ „ C — φ
2sin AsinB = cos </? — cos С = 2sin —-— sin —-— = 2sinC\ sin C2.
This means that if we split С into C\ and C2 then (5) holds, and hence (3)
implies (4) and so (2), as well. Thus (1) implies (2) and this is what we wanted
to prove.
Third solution. We prove that both (1) and (2) are equivalent to the
following inequality for the sides of the triangle:
a + b<cV2.
Let the bisector of С intersect the circumcircle in F, and the side AB in
G. By the remark referring to Figure 1974/2.1 in the first solution the point D
exists if and only if CF intersects e, i.e. GF > CG.
The triangles CGB and CAF are similar, because two of their angles are
measured by the same arcs. From the equality of the proportion of the sides we
have:
a CF CG + GF , . ^,^,9 , ^.^, ^,π
—— = —— = that is CGz = ab-CG-GF
CG b b
As GF > CG and the equality of the products of the segments of a chord CG ·
GF = AG-GB, we obtain
AG-GB = CG-GF> CG2 = ab-AG-GB,
(7) 2AG ■ GB > ab.
Also, the bisector splits the opposite side into the proportion of the two adjacent
sides, hence we get:
a+b a+b
Substituting this into (7) we obtain:
2abc2
(а + Ъу-
and
195

Solutions
1974/2
(8) a + b<cV2.
As the reverse of our arguments are true, (8) is equivalent to (2).
Now, we prove that (1) and (8) are equivalent, as well. Let us assume that
(1) holds and choose the diameter of the circumcircle to be equal to 1. The Law
of Cosine (2ab(l + cos C) = (a + b)2 — (?) implies:
„ . , . „ „ . о С 4sin2^cos2^ sin2 С sin2 С
2sinAsmB<2sinz — = l . n l = ;—^ = —.
2 2cos2^ 2cos2^ 1+cosC
Substituting o = sin Д b = sinB, с = sin С we get
c2
2ab< -, 2a6(l+cosC) = (a + 6)2-c2<c2,
1 +cosC
and (8) can easily derived from here. The reverse of our arguments hold, so (1)
and (8) are equivalent.
Remark. This last result shows that the point D exists if a + b is not "very
far" from c, since c< a + b <c\/2~ 1,41c. This always holds if С is not acute,
because in this case from the Law of Cosine с >a + b , and so
2c2>2(a2 + 62) = (a + 6)2 + (a-6)2>(a + 6)2.
1974/3. Prove that
n ^η+η 93jb
2/c + l
k=0
is not divisible by 5 for any nonnegative integer n.
First solution. The sum in the problem reminds us to the binomial
theorem. As ν 23 = л/8, by this theorem
Set
k=0
-sm-^-se::)--
So bn denotes our sum, and
(1) (л/8,+ 1)2п+1=ап + 6пл/8.
Similarly, we get
(2) (л/8-1)2п+1 = -ап + 6пл/8.
196

1974/4
1974.
Multiplying the two equalities we conclude:
72n+l=Sb2n-a2n,
(3) 7.49n + a2n = Sb2n.
As an and bn are integers, in case 5 divided bn the last digit of the right
hand side of (3) would equal to 0. The last digit of 49n is 9 or 1, hence the last
digit of 7 ■ 49n is 3 or 7. The possible last digits of (l\ are 0, 1,4, 5, 6, 9, the last
digit of the left hand side cannot equal 0, so 5 does not divide bn.
Second solution. We start with formula (1):
. ап+1+6п+1л/8 = (л/8 + 1)2(п+1)+1 = (л/8 + 1)2п+1(л/8 + 1)2 =
= (an + bnVS)(9 + 2л/8) = (9αη + 16bn) + (2αη + 9bn)Vs.
The numbers in the parenthesis are integers, so
an+i=9an + 16bn
bn+i=2an + 9bn.
As we are interested in the divisibility by 5, from now on we handle these
equations as congruences mod 5
an+\ Ξ4αη + 6η,
(4) bn+i=2an + 4bn.
Repeating the procedure:
an+2 = 4an+i+ bn+i =3an + 3bn,
bn+2 = 2an+i+4bn+i= an + 3bn.
Finally
bn+3 = 2an+2 + 4bn+2 = 3bn.
Thus in the series bn an element is divisible by 5 if and only if the third
preceding element is divisible by 5. From (1) a,Q = bQ = l and from (4) a\ ξ0,
b\ = 1; b2 = 4, so 5 does not divide bo, b\, and 62> hence there is no bn divisible
by 5.
1974/4. An 8 χ 8 chessboard is divided into ρ disjoint rectangles (along
the lines between squares), so that
a) each rectangle has the same number of white squares as black squares
b) If di denotes the number of white squares in the i-th rectangle, then a\ <
CL2 < ■. ■ < dp holds.
Find the maximum possible value of ρ and all possible a\, α2, ..., ap
sequences.
197

Solutions
1974/4
Solution. There are 32 white squares on a chessboard, so a\ +a,2 + ... +
ap = 32. Since a^ > 1, a2 > 2, ..., ap >p,
1 + 2 + .. .+p<a\ +a2 + .. . + ap = 32;
hence the maximal possible value of ρ is 7. Now, we determine all possible ways
of splitting 32 as the sum of 7 positive integers.
аз > 3, but if it was greater, аз + a^. +... + αη > 30 would hold what is
impossible. Therefore a\ = 1, a<i = 2, аз = 3, and a\ + a^ + аз = 6, α^ + αζ + α^ + αη =
26.
If 05 = 5, then a^ = A, and so αβ + αη = 17, hence the possible cases are:
6 + 11, 7+ 10 and 8+ 9. We have:
a) 1+2+3+4+5+6+11, β) 1+2+3+4+5+7+10, η) 1+2+3+4+5+8+9.
If a$ = 6, then 0,4. equals 4 or 5, in the first case ag + αη = 7 + 9, in the second
ag + αη = 7 + 8, so we got:
δ) 1+2+3+4+6+7+9, ε) 1+2+3+5+6+7+8.
a$ > 6 would imply αζ+α^ + αη> 24 which is impossible. So we listed the
arithmetically possible cases. But a) cannot be realized because for 11 white
squares you need a 1 by 22 or a 2 by 11 rectangle. The remaining four cases are
realized on Figure 1974/4.1.
■
ΑΙ
?
?
1
10
■5-
■3·
4
-i
\
ъ
5
о
Ζ
I
9
3·
■4
7
/
t
?
>
ζ,
I
.9.
3·
5
3
<
ι
г
3
>
ζ,
I
7
*
с
I-
Fi'gwre 74/4.1
198

1974/5
1974.
1974/5. Determine all possible values of
a b с d
о —
d+a+b a+b+c b+c+d c+d+a
for arbitrary positive reals a, b, c, d.
Solution. We decrease the sum S if we increase the denominators of the
fractions to a + b + c + d:
a b с d
s>—ι j + —ι 1 + —ι 1 + —ι 1 = 1-
a+b+c+d a+b+c+d a+b+c+d a+b+c+d
We increase S if we decrease the denominator of the first two fractions to a + b,
and the other two to c + d:
a b c d
s<—τ + —z +—1 + —i=2i
a + b a + b c + d c + d
hence
1<5<2.
We show that S attains every value in ]1,2[ so this is its range. First, let
a = b = x and c = d=l. Set the function
2x 2
S = f(x) =
2ж+1 χ+2
lim /(я) = 1 and /(1) = ί
x—>0+ J
(x —> 0+ refers to the limit from the right). As / is continuous in [0,1], it attains
4
every λ where 1 < λ < -.
Moreover, let a = c = x, b = d=l; then for
2x 2
x + 2 2x +1
4
#(1) = - and km g(x) = 2,
J x->0+
4
g(x) is continuous on the [0,1] interval hence it attains every λ, where - < λ < 2.
The range of the sum S is the ]1,2[ interval.
Remarks. 1. We used several times the well known theorem of Bolzano,
asserting that a continuous function attains every value between its values at the
endpoints. Although, we used the argument for semi closed intervals, but we can
close them arbitrarily close to 0 or 1, since because of the existence of the limit
there is a place in [0,1] where, for example, it attains a value arbitrarily close
to 1.
199

Solutions
1974/5
4
The arguments using continuity are not necessary. For every 1 < λ < - we
can find an x, for which f(x) = X. Indeed, consider the equation
2r 2
/(iC) = _£Ξ_ + _Ц = A.
2x + 1 χ + 2
Hence
2(λ-1)ζ2 + (5λ-8)ζ + 2(λ-1) = 0,
and
8-5λ + χ/3(λ-4)(3λ-4)
x =
4(λ-1)
is a positive solution, because by the restrictions to λ, the inequalities λ — 1 > 0,
8 — 5λ > 0 hold and the discriminant is nonnegative, too. So there is a positive
χ such that f(x) = X.
4
Similarly, g{x) = λ has a positive solution for - < λ < 2.
2. Instead of / and g we can use the single function t(x): let a— 1, b = x,
c=l — x, d = {\ — x)x (0<x < 1). Then
S = t(x) =
Since
and t is continuous on [0,1], it attains every λ, where 1 < λ < 2.
1974/6. Let P(x) be a поп constant polynomial with integer coefficients.
Let η be the number of distinct integers k, where (P(k)) — 1. Prove that
(1) n{P) - deg(P) < 2,
where deg Ρ denotes the degree of P(x).
First solution. We start with a widely used observation: if P(x) is a
polynomial with integer coefficients, and b, с are distinct integers then b — c divides
P(b) - P(c).
Indeed, let
P(x) = amxm + am_\xm~ +.. . + a\x + aQ,
then
P(6)-P(c) = am(6m-cm) + am_1(6m-1-cm-1) + ...+a1(6-c).
ι
As b — c divides the right hand side, it also divides the left hand side.
Let m denote the degree of P(x): m = deg(P); we have to show that
(2) n(P)<m + 2.
1
—x2+2x +
lim t(x)-
- +
1
= 2,
X
2
1-х
+ 2 1 +
—χΔ + x + 1
lim t(x) = 1,
χ—Λ
(1
— x)x
x2 + 2
200

1974/6
1974.
Assume to the contrary that n(P)>m + 2, thus n(P)>m + 3. In this case
there are m + 3 distinct integers where the value of P(x) equals +1 or — 1; let us
call them: k\ < k<i < ... < кт+з.
\P(ki) — P(k\)\ equals 0 or 2, on the other hand k^ — k\ > 1, % — k\ > 2,
k4 — k\>3; but we saw that ki — k\ divides P{ki) — P(k\), thus for i>4,
P(ki) — P(k\) has to be equal to 0. Then the m +1 values
P(h), P(fe4), P(fc5), ■··, Р(кт+з)
equal which is impossible as a polynomial of degree m can attain the same value
at most at m places. Since n(P) >m + 2 leads to contradiction so (2).
Second solution. P(x) attains +1 or —1 at the integer χ if and only if
the polynomials P\(x) = P(x)- 1, or P2(x) = P(x) + l have an integer root. We
show that one of P\ and P^ has at most two integer roots.
Assume to the contrary that both P\ and Pi have three integer roots. They
cannot share a root xq in common, because
P(a;o)-l = P(a;o) + l
is impossible.
Let к be the smallest integer root of P\ and P<i e.g. the root of P\, P\ (k) = 0.
There is a polynomial Q(x) with integer coefficients such that
(3) Pl(x) = (x-k)Q(x).
Let a, b and с denote the integer roots of P2(x), P2i°) = Pi(b) = Plic)= 0· The
values Q(a), Q(b), Q(c) cannot equal 0, for example, in case Q(a) = 0 the equation
(3) implies Pi(a) = 0, but P\ and Pi have no root in common.
As
Pl(x) = P2(x)-2 = (x-k)Q(x),
substituting a, b, с respectively, we get
(a - k)Q(a) = (b- k)Q{b) = (c - k)Q(c) = -2,
and so the distinct positive integers a — k,b — k,c — k divide —2, a contradiction,
so Pi and P2 cannot have both 3-3 roots.
The number of roots of P\ and P^ is at most the degree of P(x), that is
deg(P). Thus P(x) can attain +1 and —1 at at most deg(P) + 2 places, so
n(P)<deg(P) + 2,
therefore (1) holds.
Remark. Let d denote the difference n(P) — deg(P). Our problem claimed
that d<2. The truth is that for most polynomials d<0. Using the arguments of
the first proof this is easy to prove for polynomials of degree at most 5.
201

Solutions
1974/6
Indeed, let us assume that deg(P) = m>5 and P(x) attains +1 or —1 at
the places k\ < k^ < . ·. < km+\. Then, for i > 4, ki — k\ divides P(ki) — P{k\)
and so P{k\) = P{k^) = .. - = P(km+i)', on the other hand, for i = l, 2, ..., m-2,
fcm+l — fej >3 and it divides P(km+\) — P(h). This implies P(fcm+1) = P(/c2) =
Р(&з) and so P(x) takes the same value at m+1 different points, so in case
m>5, n(P) < m, thus d < 0.
We can achieve equality if we consider
P(x) = (x-l)(x-2)...(x-m) + l
where (m > 5).
It can be shown that the only polynomials, for which n(P) — deg(P) > 0 are
the following ones (here, a denotes an integer):
P(x) = ±((x- a)2 + (x - a) - l) d = 2,
P(x) = ±((x - a)(x - a- l)(x - a-3) + l) d=\
P(x) = ±((x- a)(x -a- 2)0 - a - 3) +1) d=\,
P(x) = ±(2(x-a)2-l) d=l.
P(x) = ±2(x-a) + l d=l,
P(x) = ±(x + a) d=\.
1975.
1975/1. Let x\ > x<i > ... > xn and y\ > yi > ... > yn be real numbers.
Prove that if{zi} is any permutation of the {yi}, then:
η η
First solution. Squaring the binomials on both sides and omitting the
squares of the Xi-s and yi-s we get the following inequality equivalent to (1):
η η
(2) ΣχίΖί-Σχ№·
г=1 г=1
And this is what we shall prove. In (2) equality holds if the order of the Zi-s
agrees with the order of the y^-s. If this is not the case, let us assume that they
first differ at the k-th place, i.e. y\=z\, y^-z^·, · ■ ·, Ук-\ ~zk-\-> but Vk¥zk·^^
Zk = yr and yk = zs, where r and s are greater than к, г>к and s> к. Using this
notation
^kzk + %sZs = %кУг + ^sVk-
Now, change the left hand side by substituting XkZk + xszs-%kyr + %syk by
ХкУк + ^sVr· Doing this, we increased (did not decrease) the sum on the left
hand side, because
(хкУк + ХзУг) ~ (хкУг + ХзУк) = (хк - Хз)(Ук - Ут) > 0.
202

1975/1
1975.
With this change we achieved that the first к summands of the left and right hand
sides agree. Iterating the procedure, we increase (do not decrease) the left hand
side and we arrive to the right hand side, thus verify (2).
Second solution. We shall prove (2), again. Let us introduce the following
notations: Ak = yx + y2 + ...+yk, Bk = zi +z2 + ... + zk, (k = l, 2, ..., n); and by
convention let Aq = Bq = 0. Using these notations we have:
Ук = Лк-Ак_1 and zk = Bk-Bk_i.
Hence (2) becomes:
η η η η—1
Σ хкУк = Σ Хк(Ак ~ Ак-0 = Σ ХкАк ~ Σ Хк+\Ак =
к=\ к=\ к=\ к=0
п-1
= хп-А-п + 2_j(-Xk ~ xk+VAk-
к=\
Similarly, we have
η η—1
Σ XkZk = ΧηΒη + Σ (Хк xk+l )Bk.
к=\ к=\
Since An and Bn are the sum of the same numbers, An = Bn. Moreover, Ak > Bk
(k = l, 2, ..., n) because Ak is the sum of the к largest numbers from these
numbers. Thus (2) can be written as
n—1 n—1
Y^(Xk ~ xk+l)Bk < ^Ofc xk+l
k=\ k=\
n-\
J2^Xk ~ Xk+\)(Ak ~ Bk) > 0.
k=\
This is obvious as both factors of the summands of the sum are positive.
Remark. The equation used in the proof says that if x\, x2, ..., xn and z\,
z2, ..., zn are real η-tuples, then the sum
η
О = / j X{ Z{
i=\
is maximal if the tuples are ordered on the same way.
With our methods it is easy to show that the sum is minimal, if the ordering
of the z^-s is the reverse of the ordering of the Xi-s.
From these two theorems several well known inequalities can be derived,
as for example the Chebyshev, the Cauchy and the A.M.-G.M. inequalities.
203

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1975/2. Let a\ < ai < 0,3 < ... be an infinite sequence of positive integers.
Prove that there are infinitely many elements of this sequence that can be
written in the form
with x, у positive integers and p^q.
First solution. Let us assume to the contrary that there are only finitely
many elements of the sequence of this form and let as be the last one of them.
Omit the first s element of the sequence (up to as) and denote the remaining
elements by
bh b2, Ъ3, ....
(If no element is of that form, b\ -a\). Now, starting at b2, divide b\ + \ many
elements from the series by b\. There will be two of them, bk and bT (br > bk),
where the remainders agree, so their difference is divisible by b\. Hence
br - bk = χ ■ b\,
where χ is a positive integer, so br = x· b\ + bk. Choosing pi we get
br = x-bi+ybk.
Thus br is of the required form, contradicting our assumption.
Second solution. We shall utilize the following observation (see the
Remark): If a, b, с are positive integers such that с > ab, where a and b are coprime,
then there are positive χ and у which satisfy the equation ax + by = c.
Let di denote the greatest common divisor of a\ and ai (i = 2, 3, ...). Since
a\ has only finitely many divisors, there is a number in the sequence di, d3, ...
that occurs infinitely many times, so in the щ sequence there are infinitely many
elements of the form
a\ =b\d, b<id, b^d, ...
The sequence bi is strictly increasing, so it contains infinitely many elements
greater than b\ 62. The greatest common divisor of a\=b\d and bid is d, implying
that b\ and 62 are coprime. So, by our introductory observation, the equation
bxx + b2y = bk
has a positive integer solution along with the equation
b\dx + b2dy = bkd,
where b\d, bid, bkd are in the sequence a;. There are infinitely many choices for
bk, therefore the statement is proved.
Remark. We prove the observation in the second solution: the integers a, 2a,
3a, .. .,ba give distinct residues mod b. Indeed, in case ia and ja (1 < i < j < b)
give the same residue, then, b divides (j — i)a that is impossible since j — i<b
and a and b are coprime. Hence the numbers a, 2a, ..., ba give all possible
residues mod b, they form a so called complete residue system.
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1975/3
1975.
So, one of them — e.g. xa —, gives the same residue as c, and so b divides
с — ax, hence there is an integer y, such that
c — ax = by.
As χ < b, we have ax<ab<c and с — ax = by > 0, so у > 0 and the equation
ax + by = c has a positive solution x, у.
1975/3. Given any triangle ABC, construct external triangles ABR,
BCP, CAQ on the sides, so that
LPBC=LCAQ = 45°
LBCP = LQCA = 30°
IABR=IBAR=15°.
Prove that IQRP = 90° and QR = RP.
First solution. Choose the notations such that the triangles ABC, BPC,
CQA ARB are labelled counterclockwise (see Figure 1975/3.1).
AC _BC _
~aq~Hp~ '
because the triangles CQA and СРВ are similar. Let r\ denote the rotation-
C
205

Solutions
1975/3
stretching with ratio λ, degree —45° and centre Α, τι the rotation-stretching with
ratio —, angle —45° and centre В. т\ maps Q to C, τι maps С to P, so the
Л
composition of them, τ\τ<ι maps Q to P. τ\τ<ι is a rotation-stretching with ratio
λ · — = 1, angle 2 · 45° =90°, so it is a congruence, thus a rotation by 90°.
Λ
The transformation т\ maps R to a point R', such that the triangles AQC
and ARR' are similar. Thus /.ARR'= 105°, hence ДЛ' is orthogonal to AB.
Since ЛШ? is an isosceles triangle, RR! is the axis of symmetry of the triangle
ABR*, therefore ABR' is an isosceles triangle (it is equilateral as /.BAR' = 60°).
This implies that ARR' and BRR' are congruent triangles, hence τ<ι maps R' to
R. Thus Λ is the fixpoint of τχτ^ and so its centre. Hence t\ti is a rotation by
90° with centre R mapping Q to P. Thus QiZP is an isosceles right triangle and
this is what we wanted to prove.
Second solution. Trigonometry provides another solution.
Applying the Law of Sines to the triangles ACQ and PBC we get
AQ Β Ρ sin 30° 1 CQ С Ρ sin 45° 1
b a sin 105° 2 sin 75°' b a sin 105° ^2 sin 75°'
hence we obtain
(1) AQ = -X-, BP=—£—, CQ= ^ b _, , CP= a
2 sin 75°' 2 sin 75°' ^ χ/2 sin 75°' \/2sin750'
In the isosceles triangle ARB we have
(2) AR = BR=^ °nc .
2 sin 75°
Applying the Law of Cosine to QAR, Ρ BR and PQC using (1) and (2) we get:
QR2 = AQ2 + AR2 - 2AQ ■ AR cos(a + 60°) =
Similarly,
= = (V + c2 - 26c cos(a + 60)) .
4 sin2 75° v )
PR2 = = (a2 + c2- lac cos(/3 + 60°)) ,
4 sin2 75° v >
PQ2 = = (a2 + b2- lab cos(7 + 60°).)
2 sin2 75° v >
We show that the expressions in the latter three parentheses represent the
same value. We will express them with the area of the triangle in the same way,
e.g.:
5 = a2 + 52-2a5cos(7 + 60q) = a2 + 52-2a5i-^cos7- ^3sm^
9 τ9 a2+b2-c2 /- . . a2 + b2 + c2 л /-
(3) =az + bz + V3ob sin 7= + lV3t.
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1975/3
1975.
Clearly, we get the same formula starting with any of the three expressions, so
QR = PR and PQ2 = QR2 + PR2,
therefore PQR is an isosceles right triangle.
Third solution. The following generalization goes deeper into the
background of the problem. Two 4-tuples of points are called similar if there is a
similarity transformation mapping one to the other. If you pick 3-3 corresponding
vertices in two similar 4-tuples, the triangles spanned by them are similar. Two
similar 4-tuples have the same orientation if the corresponding triangles have the
same orientation. Theorem:
If XYZU, X'Y'Z'U', X"Y"Z"U" are similar 4-tuples and Z' = Y,
Y" = X', Z" = X, moreover U' JU, U" JU, then the triangles XX'X", YY'Y",
ZZ'Z" and UU'U" are pairwise similar (see Figure 1975/3.2).
л Υ'
Ζ
Figure 75/3.2
Visually, if there is an arbitrary triangle on the plane (XYX' on the figure)
and we insert three similar 4-tuples on the following way: the XY edge of the
first 4-tuple is the same as the XY edge of the triangle; the Z'X' edge of the
second 4-tuple is the same as the YX' -Z'Y" edge of the triangle; the Y"Z"
edge of the third 4-tuple is the same as the X'X = Y"Z" edge of the triangle,
then the four triangles mentioned in the theorem are pairwise similar.
The similarity of the first three triangles is trivial. Indeed, XY Ζ is
identical to Z"Z'Z and X'Y'Z' is identical to Υ'ΎΎ; it only remains to show that
UU'U" is similar to the first three triangles.
207

Solutions
1975/3
For the proof, we insert our figure to the complex plain with U as the 0;
we denote the complex numbers belonging to the points with the corresponding
lower case letters. We do not distinguish the points from the corresponding
numbers .
Every similarity keeping the orientation mapping the point ρ to p* is given
by a linear function. Let
(4) p* = hp+th
be the transformation mapping XYZU to X'Y'Z'U' and
(5) P* = hP + t2,
the one mapping XYZU to X"Y"Z'U", where k\,t\, k2, t2 are the appropriate
complex numbers.
As u = 0 and (4) maps и to u', and (5) to u", we have u' = k\ -0 + t\=t\ and
u" = k2 · 0 + t2 = t2, so (4) and (5) become
(6) p* = k\p + u', and p* = k2P + u".
Now, apply (6) to the χ —»χ', z—>z' = y, and y^-y" = χ', ζ —» ζ" = χ
correspondences:
(7) x' = kix + u' (9) x' = k2y + u"
(8) y = k\z + u' (10) x = k2z + u"
As χ 7^0, у 7^0, (7) and (9) gives:
x' — u' , x' — u"
χ у
Substituting to (8) and (10) we get:
x'z — u'z + u'x x'z — u"z + u"y
y- ? χ- .
χ у
Combining the two formulas:
xy = x'z — u'z + u'x = x'z — u" ζ + u"y; u'(x — z) = u"(y — z).
Now, set:
^-_f = y-£ (ttVo Voo
и" и
Applying the similarity transformation p* = ki,p + z to и, и', and u", respectively,
we obtain
k^u + z = z,
k^u' + z = y,
k^u" + z = x.
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1975/4
1975.
That is U"U'U is similar to XYZ and this is what we wanted to prove.
Now, if we consider the XYZU 4-tuple where XYZ is an isosceles right
triangle, (the right angle is at Z) and inside take U such that XUY is an isosceles
triangle with LARU - 150°, then we get the configuration of the problem.
Remarks. 1. The method of the first proof gives us the following
generalisation:
If we construct the BPC, CQA, ARB triangles outside to the arbitrary
ABC triangle such that
lPBC = lCAQ = (p,
IBCP = IQCA = 6,
IABR=IBAR=\90° -(φ + δ)\
and AB separates R and С if </? + <5<90°; AB does not separate them if φ +
<5>90°;finally, R is the midpoint of AB if φ+ 6 = 90°, then Ζζ)#Ρ = 2</? and
QR = RP.
In our case φ = 45°, <5 = 30°. This formulation of the problem has several
interesting special cases. For example, if </? = <5 = 45°, then we construct isosceles
right triangles to the AC and ВС sides of the ABC triangle; in this case P, Q
and the midpoint of AB form a isosceles right triangle.
2. A historically interesting special case of the generalisation above: if
<^ = <5 = 30°. The vertices of the isosceles triangles are the centres of the
equilateral triangles constructed above the sides. By the result, they form an
isosceles triangle with an angle of degree 60°, and this is an equilateral triangle. This
problem is due to Napoleon.
An other generalisation is the Napoleon-Barlotti theorem: if we construct
regular n-gons on the sides of an n-gon, the centres of these n-gons form a
regular n-gon if and only if the original n-gon was a regular affine n-gon. (An
n-gon is affine regular if it is the projection of a regular n-gon. eg. every triangle
is affine regular; the affine regular quadrilaterals are the parallelograms.)
1975/4. Let A denote the sum of the decimal digits of 4444 , and В be
the sum of the decimal digits of A. Find the sum of the decimal digits of B.
Solution. Let N denote the number 4444 , and С the sum of the
decimal digits of B. As the mod 9 residue of a number is the same as the residue
of the sum of its digits, N gives the same residue mod 9 as А, В and C. 4444
gives residue 7, so N = 9k + 7, for some positive integer k. The powers of 9k + 7
are of the form: (9k + if = 9/q + 4, (9k + if = 9k2 +1, (9k + 7)4 = 9/c3 + 7, so the
mod 9 residues are periodic, 7, 4, 1, 7, ... If the exponent is 3m +1, 3m+ 2, 3m
(
209

Solutions
1975/4
the mod 9 residues are 7, 4, 1, respectively. Since 4444 is of the form Зга +1,
N gives 7 mod 9 and 7 is the residue of А, В and C.
As N < 100004444 = 1044444 = 1017776, TV has at most 17776 decimal
digits. Thus, A <9-17776= 159984. The first digit of Л is at most 1, hence
В < 1 + 5 ■ 9 = 46. Among the first 46 number the sum of the digits is the
largest in case of 39. Thus С < 12, but С gives residue 7 mod 9, and so С = 7.
Remarks. 1. TV has 16211 digits, Л = 72 601, Б =16 finally, С = 1 .
2. We can get the mod 9 residue of N using congruences:
7V" = 44444444 = 74444 = (_2)3'1481+1 ξ7·(-8)1481 =7-11481 =7
1975/5. Can vow jfmd 7975 points on the circumference of a unit circle
such that the distance between each pair is rational?
First solution. We show that we can choose η points on the unit circle
such that the distance of any two of the points is rational (n > 1).
We shall utilize the existence of an arbitrary small angle with rational
tangent. Indeed, let in the ABC right triangle 1 be the length of the ВС leg and N
(positive integer) the length of the AC hypotenuse, LB AC = φ. Then tan φ = —
is rational; increasing N, we can make tan φ and φ arbitrarily small.
7Γ Ci
Now, choose the angle a such that a < with tan — rational. In this
n— 1 2
case sin a and cos a are rational, too, as
2 tan 2 l-tan2f
sin a = ^— and cos a = T—,
l+tan2f l+tan2f
moreover, sinna and cosna are rational, as well. This can be proved by
induction using that
sin(n + \)a = sin na cos a + cos na sin a,
cos(n + \)a - cos na cos a — sin na sin a.
For η = 1 the statement is true, and if it is true for n, the above formulae prove
it for η + 1.
Now, consider the points P\, Pi, P3, · ·., Pn on the unit circle such that
the angle between two neighbouring points is 2a (see Figure 1975/5.1). These
points are distinct, because (n — 1) ■ 2α < 2π. The length of the arc between Pi
and Pk (k > i) is (k — i)· 2a, hence
PiP^-2 sin(fc — i)a,
and as sin α is rational, the length of PiPk is rational, too.
210

1975/5
1975.
Ρ,
Second solution. Construct right triangles over the segment Pj Pn as
hypotenuse of length 2, with legs:
ΡΛ =
Ak
PkPn =
2(k2 -1)
Clearly, РгР^ + РкР2 = 4, so these
triangles exist, moreover their Pk
vertices are on the unit circle
constructed over PxPk. The points Pb P2,
..., Pn satisfy the conditions of the
problem, because P] Pk and PfcPn are
rational by definition. Now, applying
Ptolemy's theorem for 1 < i < к < η
(see Figure 1975/5.2) vje get:
(fc = 2, 3, ..., n— 1).
Figure 75/5.2
Р1Рк-РпРг = Р\Рп-РгРк+Р\Рг-РпРк,
and
№ = J (i>i Pk ■ PnPi -Pi Pi- PnPk)·
As the right hand side is the product and sum of rational numbers, PiPk is
rational, too.
Remark. The theorem implies that there are N points in the plane such that
no 3 of them are collinear, but the distance of any 2 of them is an integer. Indeed,
211

Solutions
1975/5
if the distances in the construction are
Pi P2 El
q\' Я1 " Qr
enlarging the plane by q\qi... qr, the distances become integers.
It is surprising, however, that there cannot be found infinitely many points
on the plane such that the distance of any two is an integer. To show this, let
AB = с, ВС = а, С A = b in the ABC triangle, where a, 6, с are integers. We show
that there are only finitely many points D on the plane such that its distance is
integer from the vertices A, B, and C.
From the triangle inequality we have:
\DB - DC\ < a, \DA - DB\ < с
The set of points, D where \DB — DC\ = k (constant) is a hyperbola with
foci В and С (0 < к < a), or the perpendicular bisector of ВС (к = 0), or the ВС
line, omitting the segment ВС {к-a). Since a is an integer, к has to be 0, 1,
2, ..., a so D is situated on one of the two lines or a — 1 hyperbolas. Similarly,
\DA — DB\ < с implies that D is lies on one of the two lines or с — 1 hyperbolas.
Two hyperbolas or pair of lines intersect in at most four points. Thus there are
finitely many positions for D, and it is impossible to find infinitely many points
with integer distances under the conditions of the problem.
1975/6. Find all polynomials P(x, y) in two variables such that:
I. for every real numbers t, x, y, P(tx,ty) = tnP(x,y), where η is a positive
integer, i.e. Ρ is a homogeneous polynomial of degree n.
II. For every real a, b, c, P(a + b,c) + P(b + c,a) + P(c + a,b) = 0.
III. P(1,0)=1.
Solution The general form of a homogeneous polynomial of degree η is
(1) P(x,y) = anxn + an_ixn-ly + an_2xn~2y2 + ... + a1xyn-l+a0yn.
(1) (or property I) implies that P(0,0) = 0. If the degree of P(x, y) is 1, then Ρ
is of the form апх + а$у, and from II and III we have x = 2y. We show that in
case n> 1, for every real x, the identity P(—x,x) = 0 holds. Applying property
II by substituting a = b = x and c=—2x gives:
P(2x,-2x) + P(-x,x) + P(-x,x) = 0,
((-2)n+2)P(-x,x) = 0,
' P(-x,x) = 0.
This says that if χ + у = 0, than
(2) P(x,y) = 0.
212

1975/6
1975.
Now, assume that x + y^O. Applying II with c= 1 — a — 6 we obtain:
(3) Ρ(α + 6,1-α-6) + Ρ(1-α,α) + Ρ(1-6,6) = 0.
Substituting 6 = 0 we get:
P(o, l-o) + P(l-o,o) + l=0
Hence
P(l-o,o) = -P(o, l-o)-l.
As (3) is symmetric in a and 6 we have
Р(1-Ь,Ь) = -Р(Ь,1-Ь)-1.
Substituting to (3) and reordering the expression we get
(4) Р(а + Ь,1-а-Ь) = Р(а,1-а) + Р(Ь,1-Ь) + 2.
Introducing Q(z) = P(z, l—z) + 2, (4) gives
Q(a + 5) = Q(a) + Q(6),
where Q(z) is a polynomial of one variable. From III. we get Q(l) = P(l, 0) + 2 =
3, and with induction, Q(n) = 3n, because Q(n + l) = Q(n) + Q(l) = 3n + 3 = 3(n +
1). The values of Q(z) and 3z agree at infinitely many places, hence
Q(z) = 3z, and so P(z, 1 — z) = 3z — 2.
This gives ^
Р(Х1у) = (х+уГр(*^) = (х + уГр(*1-^) =
\x+y x+y/ \x+y x+y/
= (х + уГ(—-2)=(х + УГ-\х-2у),
\x + y J
and this satisfies (2) if χ + y = 0, as well.
The polynomial obviously satisfies I and III. Condition II holds, as
(a + b + c)n-\a + b-2c) + (a + b + c)n~\b + c-2a) + (a + b + c)n-\c + a-2b) = 0.
Thus the only polynomial satisfying the conditions of the problem:
P(x,y) = (x + yT-\x-2y).
Remark. Using additional properties of the polynomials (e.g. continuity)
there are different ways to solve the problem. We sketch one of these
possibilities:
If n> 1, by (2) P(x, —x) = 0. Fix χ and consider P(x,y) = Px(y) that is a
one variable polynomial of y. As P(x, —x) = 0, x + y is a factor of Px(y):
P(x,y) = (x + y)Ql(x,y),
213

Solutions
1975/6
where Q\(x, y) is a homogeneous polynomial of two variables.
0 = Р(а + Ъ + с) + Р(Ъ + с,а) + Р(с + а,Ъ) =
= (a + b + c)(Qi(a + b, c) + Qi(b + c, a) + Qi(c + a,b)),
so in case a + b + c^O, Q\(x,y) satisfies II.
In case x + y^O, Q\(x,y) satisfies (1), because
Q1(tx,ty) = ^^=tn-1Ql(x,y),
t(x + y)
that is Q\(x, y) satisfies the conditions I—III. Now, using the continuity of P(x, y)
this holds if a + b + с = 0, or χ + у = 0. Applying the same procedure to Q\(x, y),
we can factor out x + y from Q\(x,y) until we get the polynomial Qi(x,y) of
degree 1, that has to be of the form χ — 2y, thus
P(x,y) = (x + y)n-\x-2y).
1976.
1976/1. A plane convex quadrilateral has area 32 cm , and the sum of two
opposite sides and a diagonal is 16 cm. Determine all possible lengths for the
other diagonal.
Solution. Let ABCD be the convex quadrilateral and AB = a, CD = c,
AC = f, BD = e (see Figure 1976/1.1). Let FABC denote the area of the ABC
triangle. By the assumptions a + e + c=16.
Figure 76/1.1
Twice the area of a triangle is smaller or equal than the product of its sides,
and is equal to it if and only if the two sides are orthogonal. Thus
(1) 2-32 = 2FABD+2F'BDC<ae + ec = (a + c)e = (16-e)e,
hence
(2) 0>e2-16e + 64 = (e-8)2,
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1976/2
1976.
Figure 76/1.2
)
and this holds only if e = 8 and a + c=8. In case of equality in (1) and (2) AB
and CD are orthogonal to the BD diagonal (see Figure 1976/1.2). Thus the only
possible length of the other diagonal is:
/ = У(а + с)2 + е2 = л/82 + 82 = 8\/2.
1976/2. Let Px{x) = x2-2; Pj(x) = Pi(Pj_l(x)); j = 2, 3, .... Prove that
for every n, Pn(x) = χ has η distinct real roots.
First solution. As the degree of the polynomial Pj is twice the degree of
Pj-l, (j = 2, 3, ...), the polynomial Pn(x) — x has degree 2n and has at most
2n real roots. Let us examine the roots in the interval [—2,2]; its points can be
written in the form χ = 2 cos t. As cos 2t = 2 cos t — 1,
(1) Px (x) = Pi (2 cos 0 = 4 cos21 - 2 = 2 cos It.
Using induction we get
(2) PnO) = Pn(2cosi) = 2cos2ni.
Indeed, for n-\ our statement is the same as (1). Assume that Pn_\(x) =
2cos2n_1i, then
Pn(x) = (Pn_l(x))2-2 = 4cos22n-lt -2 = 2(2 cos2 2k~4 -l) = 2cos2ni,
and we proved (2).
We can write the equation in the form
cos2ni = cosi
and hence
2nt = t + 2/С7Г or 2nt = -t + 2&7Γ, к arbitrary integer
and so
(3)
*1 =
2k\ π
)n
1
and F2 =
2&2π
2n + l
2.15

Solutions
1976/2
If in case t\ we have k\ = 0, 1, ..., 2 — 1 and in case t>i, &2 = 1, 2, ..., 2n ,
then we get the 2n roots of the form £ = 2cosi. The values corresponding to
t\ are all distinct, and their cosine values are distinct, too, because the possible
values of t\ are in the interval [0, π] where the cos χ function is strictly
increasing. The same holds for the t^ values. It only remains to show that no t\ and
^2 values agree. This would mean that there are integers k\ and ki such that
2k\Ti _ 2k2ir
2n - 1 ~ 2n +1'
fcl(2n + l) = fc2(2n-l)
held, as 2n + 1 and 2n — 1 are coprime and 2n +1 cannot divide k<i- Hence the 2n
roots given in (3) are all distinct and so these are all the roots of the equation.
Second solution. The problem does not ask to determine the roots, only to
prove their existence. Observe that if Pn_\(x) attains the values —2, 0, 2 then at
the same places the values of
Pn(x) = {Pn-i(x)f-2
are 2, —2, 2, respectively.
Using induction we prove the following property of Pn(x): from —2 to +2,
the polynomial Pn(x) attains the values +2 and —2 alternating, it attains +2 at
2 +1 many places and —2 at 2n many places.
For n=l our statement is true as P1(-2) = 2, Pl(0) = -2, Pl(2) = 2. Now,
assume that it holds for Pn_\{x), that is from -2 to +2 it attains +2 at 2n_2 + 1
many places and —2 at 2n many places, alternating. These points split the
interval between the first and last examined points into 2n parts. Pn_\{x)
equals to —2 at one end, and to +2 at the other end of these intervals. Because of
the continuity of Pn_\{x), it has a root in the interval. Now, joining these roots
to the examined points, the values attained at these places are
2, 0, -2, 0, 2, -2, ..., -2, 0, 2,
respectively. Between every (2, —2) and (—2,2) pair there is a 0, hence there are
2 0-s in the series. By our introductory remark, the values of Pn(x) at these
0 places are
2, -2, 2, -2, 2, -2, ..., 2, -2, 2,
1 1
that is, there are 2n +1 2-s and 2n~ —2-s, and we proved our statement.
The statement of the problem easily follows, as these 2n points split the
interval [—2,2] into 2n subintervals that can only intersect at the endpoints. At
one end of these intervals Pn{x) — χ is positive and at the other and it is negative
(except for χ = 2). Hence, inside these intervals Pn{x) — x has a root that is a
216

1976/2
1976.
solution of the equation Pn(x) = x. In the last interval, at one end the value is
negative, at the other end non-negative, so there is a root, too. Thus Pn(x) — χ
has at least 2n distinct real roots, and it cannot have more roots since 2n is the
degree of Pn(x) — x.
Remarks. 1. The last step of the second solution says the following: as Pn{x)
attains the value 2 at 2 +1 places and the value —2 at 2n many places in
[—2,2], the curve y = Pn(x) and the line y-x intersect in at least 2n points, at
the roots of the equation Pn(x) = x (see Figure 1976/2.1).
У = Рп(х)
y=x
Figure 76/2.1
2. Rewriting Pn(x) = χ to (Pn_\(x)) — 2 = χ we get
Pn_l(x) = ±V2 + x;
continuing the procedure we get:
Pn_20) = ±V 2 ± л/2 + х
(2)
Pl(x) = ±^2±yj2±...±V2 + ^ = x2-2
χ
= ±\2±J2±...±V2 + x.
This last form is, in fact, 2n equations because that is the number of the possible
combinations of the signs in the square roots. We get a solution to our equation
if we fix a combination of the signs in (2). This means that the values χ = 2 cos t,
t given in (1), satisfy the equation (2) for some combination of the signs. This
217

Solutions
1976/2
fits the well-known formula that if the values of a\, 0*2,..., an are +1 or — 1,
then
2 sin αϊ + —-— + =— +... + — =
V 2 22 2n 7 4
αιγ2 + α2γ2 +.. .+ αην2.
1976/3. Λ rectangular box can be completely filled with unit cubes. If one
places as many cubes as possible, each with volume 2, in the box, with their edges
parallel to the edges of the box, one can fill exactly 40% of the box. Determine
the possible dimensions of the box. (v2= 1.2599 .. .)·
Solution. Let a, b, c, denote the sides of the triangle and ε the side of the
cube of volume 2 (ε =2). From the approximation we get
(1)
- = 1.25<ε<1.2857... = -
4 7
When we fill exactly 40% (2/5-th) of the box with cubes of volume 2, then
they can be pushed together to a rectangular block or we could insert more cubes
of edge ε into the box.
■"a"
Along the edge of the box of length a we can insert at most
cubes, hence the volume of the filling is
many
"a"
_ε_
"6"
_ε_
"с"
_ε_
2 - - -
hence
abc
= abc■ —,
= 5.
J__/CL t"n —
η '
П
m
(2)
[f] [ί] [f]
Our problem is to determine the solutions of
Here is a table for the values of t.
n·
η
ίη
2
2
2,00
3
3
2
1.50
4
4
3
1.33
5
5
3
1.67
6
3
2
1.50
7
7
5
1.40
8
4
3
1.33
Now, it is convenient to make some upper and lower estimates for tn. First
considering (1):
(3)
η 5
ε
218

1976/4
1976.
on the other hand,
zn — ε
η
ε
<ε
[?] + 1
= ε 1 +
Now, if n > 8,
(4)
Γη]
_ε _
>
Γ8Ί
_ε_
[?] [?] \ [;]/'
= [6,34] = 6, and so
/ 1\ 9 7 3
For η > 2 in the table, tn<-. Thus, if a, b and с are greater than 2,
*а*ь*с<(з) =4,62... < 5,
contradicting (2); this also holds for η > 8; thus at least one of a, b and с equals
2. Let a = 2, then ία = 2 and (2) becomes
(5)
5
tbtc=y·
(3) implies that none of Ц and ic can equal 2 and (4) implies that none of them
can be greater then 7, because this gives
3 5 5 /3\2 5
^с<г-т = г, or i6ic< - <-
2 3 2
contradicting (5). Hence
(6)
Since
3<6,c<7.
abc = 5
a
ε
5 divides b or c, so (6) implies that 5 = 5 or c = 5; assume that 6 = 5, so t& = - then
3
(5) implies tc= - and hence the possible values of с are c = 3 or с = 6.
The possible dimensions are (2,3,5) or (2,5,6). These rectangular blocks
satisfy the conditions of the problem. The volume of the first one is 30 and one
can place 1-2-3 = 6 cubes into it, the volume of the second one is 60 and one
can place 1 · 3 -4= 12 cubes in it, the proportion of the volumes is 40% in both
cases.
1976/4. Determine the largest number which is the product of positive
integers with sum 1976.
First solution. As 1976 can be decomposed to a sum of integers only
finitely many ways, there is a decomposition where the number of the summands
219

Solutions
1976/4
is maximal. Let a\, 02,..., on denote the summands in this decomposition; thus
a\ +a,2 +.. ■ + on = 1976. Let M = a\a2 ... on.
There is no 1 among the o^-s as omitting щ = 1 and taking o^ +1 instead
of a^ we do not change the sum but increase the product as 1 · o^ < a^ + 1; thus
сц > 2.
We also show that щ <5: if, to the contrary, o^ >5, then splitting o; into
the sum of o^ = 2 and a'( = щ—2 we do not change the sum, but increase M,
since
aiai = 2(щ — 2) = 2сц — 4 > сц.
So, аг equals 2, 3 or 4. 4 can be replaced by 2 + 2, changing neither the sum,
nor the product, so we may assume that there are only 2-s and 3-s in the sum.
There cannot be more then two 2-s in the sum, as replacing 2 + 2 + 2 by 3 + 3
the sum remains the same and the product becomes greater.
Our result is the following: if we split an arbitrary integer into a sum such
that the product of the summands is maximal, there are at most two 2-s among
the summands, the rest equal 3. Since 1976 = 3-658 + 2, a\=2, 02 = 03 = ...=
o658=3andsoM = 2-3658.
Second solution. Again, assume that from the decompositions of 1976 the
product Μ = ο^θ2 ... on is maximal, and first assume that η > 5.
a) there cannot be two щ-s with difference greater than 1 as if o^ — oj > 1,
then substituting a!k = o^ — 1, a'j = a,j +1 the sum does not change, but the product
increases:
akCLj = (ak — 1)(oj + 1) = a>kaj + ak ~ aj ~ 1 > akaj-
β) As for к τβ
\к>кЪ
(proof by induction), we can replace αι + αι + щ by щ times 3 not changing the
sum but increasing the product if ai ^3.
Since η > 5, a) implies that one of the summands has to occur at least 3
times and by /3) this has to equal 3. The other summand (that occurs at most
twice) is 4 or 2. By the equality of their sum and product a 4 can be substituted
by two 2-s, hence in the maximal decomposition there can occur only 3-s and at
/ГСО
most two 2-s. As 1976 = 3 ■ 658 + 2, for η > 5 the maximal product is 3 · 2.
If η < 5, by the A.M.-G.M. inequality the product cannot be larger than
(—) ·but
(i^)"<2.3«8
for every possible n.
220

1976/5
1976.
Remarks. 1. In general, if a positive integer is of the form 3k+ r (r = 0,
1, 2), then splitting it into a sum the product of its summands is maximum 3k,
22-3k~l, 1-Ък for r = 0, 1, 2, respectively.
2. If we want to split a number into the sum of к integers, then applying the
arguments of our methods in case n = kt + r (0 < r < k), the maximal product is
(t+iytk-r.
3. The most general form of the problem is the following: Split the real
number η into a sum such that the product of the summands is maximal. Here, the
only problem is to determine the number of the summands, because the A.M.-
G.M. inequality tells us that they have to be equal. Using calculus one can show
that in case of a maximal product the number of summands equals к =
η
e
(e
the Euler-constant e = 2.7182818 ...), or k+ 1 and so the maximal value is
G)" - Ш"
As —, or -—- is approximately e, one can say that in case of a maximal product
the summands equal approximately e. This is the deeper reason of our results as
3 is the integer closest to e.
1976/5. Consider the following system of ρ equations with q = 2p
unknowns:
a\\x\ +ai2X2 + · · · + a\qxq = 0:
a2\Xi+a22X2 + ·· - + a2qxq=0,
a>p\X\ +0>p2x2 + · · - + apqxq = 05
where a^ G {0, 1, -1} (i = 1, 2, ..., p, j = l, 2, ..., q).
Prove that the system of equations has a solution x\, X2, · · ·, xq satisfying
the following properties:
a) icj, X2, ■ · ·, xq are integers;
b) not all Xj are 0 (j = 1, 2, ..., q);
c) \xj\<q (j = h 2, ..., q).
Solution. Let us suppose that substituting the two different g-tuples of
integers
(щ,и2,...,ид) and (vi,V2,...,vq)
221

Solutions
1976/5
where \uj\<p, \vj\<p (j = l, 2, ..., q) into the i-th equation we get the same
values yi, (i = 1, 2, ..., p). Then
(щ -vi,u2-v2,...,uq-vq)
satisfy the equations and the conditions a), b), c): a) and b) directly follow from
the choice of the щ-s and v^-s and c) holds because \щ — V{\ < \щ\ + \vi\ < 2p-q.
Therefore to solve the problem we only have to show the existence of the
щ-s and Vi-s.
In such a g-tuple (щ, u2,...,uq) as — ρ<щ<р, there are 2p+\ choices
for щ so there are
(2p+l)q = (2p+l)2p = (4p2 + 2p+l)p.
g-tuples. By our definition
yi = ail ui+ai2u2 + ... + aiquq (г = 1, 2, ..., ρ),
hence
Ы <\щ1\\щ\ + \щ2\\и2\ + ■ · - + \o<iq\\uq\ <
<\щ\ + \u2\ + ... + \uq\ <pq,
so yi can attain at most 2pq +1 different values, so the number of the possible
p-tuples for (y\, yi-, ■ · ■, yp) is at most
(2pq+l)p = (4p2 + l)p.
But, as
(4p2 + 4p+lf>(4p2 + l)p,
There are two distinct g-tuples producing the same results; and this is what we
wanted to prove.
5
1976/6. The sequence uq,u\,... is defined as follows: uq = 2, щ = -,
9
un+l =un(un-\ —2) — u\ (n= 1, 2, ...). Prove that
2n-(-l)n
un = 2 3 (n=l, 2, ...).
Solution. Construct the first few elements of the series:
11111 1
щ = \ + -, ui=2 + -, u2 = 2+-, ^3 = 8 + -, щ = 32+ —, ii5 = 2048 +
V l 2' Δ 2' J 8' ч 32' J 2048
They can be rewritten in the form
^3 = 2' + ^з, u4 = 2:) + ^, u5 = 2ii + ^TT.
222

1977/1
1977.
So we formulate the following conjecture:
(1) un = 2/(n) + 2~/(n)
where
2n-(-l)n
(2) f(n)= Kj—-.
The conjecture holds for n = 0, 1, ..., 5. We prove the statement by
induction. Assume that (1) and (2) holds for until some integer n. By the definition of
the sequence
un+1 = (2'W + 2-'<»>) (2Wrb-D + 2-2f(n-l^ _ 5 =
(3) =2/W+2/(n-l) + 2-/(n)+2/(n-l) + 2/(n)-2/(n-l) + 2-/(n)-2/(n-l) _ 5
2"
Examining the exponents according to (2) we have:
/(n) + 2/(n-l) = i(2n-(-l)n + 2n-(-l)n-1.2) =
= ^(2n+1-(-l)n+1) = /(n+l),
/(n)-2/(n-l) = i(2n-(-l)n-2n + (-l)n-1.2) =
= ^(-(-Dn + (-lf"1-2) = i((-l)n+1+2(-lf+1)=(-lf+1
3
Using this for (3) we get:
so we proved our conjecture.
f(n) is always an integer, because 3 divides 2n-(-l)n; but 2~^n^ < 1,
hence
[un] =
2n-(-l)n
2/W + 2~/(n) = 2^n^ = 2 3
1977.
1977/1. Construct equilateral triangles ABK, BCL, С DM, DAN on the
inside of the square ABCD. Show that the midpoints of KL, LM, MN, NK
and the midpoints of AK, BK, BL, CL, CM, DM, DN, AN form a regular
dodecahedron.
Solution. We can observe lots of symmetries of our object; the
combination of them can lead to the proof in several ways. We choose one of these
approaches (see Figure 1977/1.1).
223

Solutions
1977/1
D С
Figure 77/1.1
The equilateral triangle ABK is the rotated image of the triangle BCL by
90°, hence В К and CL are perpendicular. Moreover В К lies on the
perpendicular bisector of CL intersecting it in X, so X is the midpoint of CL. Similarly,
AK and DN intersect in Υ, the midpoint of DN. Let Ζ denote the midpoint of
LK, Ζ is on the BD diagonal. Let О denote the centre of the square. L and Μ
are symmetric to the diagonal AC, so the midpoint U of LM, lies on AC and
LM is orthogonal to AC.
To prove our theorem it is enough to show that ZOY and XOY are
congruent equilateral triangles (the central triangles of the dodecahedron), because
the other two central triangles are the reflections of one of these triangles to the
diagonals of the square. The quadrangle KLMN is a square as the rotation by
90° about О fixes it, thus ZO = UO.
If we join a vertex of a triangle inside the square (e.g: L) with the nearest
vertices {A and D), then the base angles of the isosceles triangle LAD are 15°,
because the base angles of the isosceles triangle ABL are 75°.
In the congruent triangles ACL and BDN we have AL = BN, and XO
is the median parallel to AL in the first, YO is the median parallel to BN in
the second triangle, so XO = YO. As AL and BN include angles of 15°—15°
with the direction of AD, the vertex angle of the isosceles triangle XOY is
30°. As INВD = 30°, the similarly oriented angle LYOZ is 30°, as well. ZO =
224

1977/2
1977.
LU = AL/2, because one of the angles of the right triangle, AUL is 30° and as
YO = AL/2, ZOY and XOY are congruent equilateral triangles.
1977/2. In a finite sequence of real numbers the sum of any seven
consecutive terms is negative, and the sum of any eleven consecutive terms is positive.
Determine the maximum number of terms in the sequence.
First solution. We show that such a sequence cannot have 17 or more
terms, but there is a sequence with 16 terms. Let us assume that the sequence
has 17 terms: a\, α-2,..., α\η\ order them by the following scheme:
a\ α-2 α-3 ... αη
CL2 Q-3 СЦ ... <2g
аз сц α,ζ . .. ад
а11 а12 а13 ··· а17·
The sum of the terms is positive in every column and negative in every row.
If we sum up all the numbers of the table by the rows, the sum is positive, if by
the columns, it is negative, a contradiction. So we can have at most 16 terms in
the sequence. That is possible by the following example:
8, 8, -21, 8, 8, 8, -21, 8, 8, -21, 8, 8, 8, -21, 8, 8.
Thus the maximum number of the terms is 16.
Second solution. The advantage of the following solution is that it enables
us to construct infinitely many examples with maximal number of terms.
Let Si denote the sum of the first i elements, sq = 0. The conditions of the
problem say that
(1) si+j-Si<0, i.e. Si>si+j (г = 0, 1, 2, ...)
and
(2) si-si_n>0, i.e. s;>s;_n (г = 11, 12, ...)
If the sequence has 17 elements, then iterating the two inequalities we get:
i
(3) s7 > s14 > s3 > 510 > s17 > s6 > s13 > s2 > s9 > s16 >
>S5>Sn>Sl >58>515>54>5П >S0 = °>
that is a contradiction, because sj < 0.
Now, let us assume that the sequence has only 16 terms and start the chain
of inequalities at sg ( denoted by an arrow in (3)), and continue from sq as
follows:
(4) s0 > 57 > «η > 53 >S10·
225

Solutions
1977/2
Now the chain is interrupted as none of (1) and (2) apply. Give arbitrary
values to the s^- s satisfying (3) and (4), e.g: 12 > 11 > 10 > ..., and construct
the table:
si s2 S3 s4 55 56 s7 s8 s9 s10 sn sn s13 s14 s15 s16
5 10 -3 2 7 12 -1 4 9 -4 1 6 11 -2 3 8
Since a,i = Si — Sj_i, the table produces:
5, 5, -13, 5, 5, 5, -13, 5, 5, -13, 5, 5, 5, -13, 5, 5.
As we have infinitely many choices for the Si-s, we can construct infinitely
many sequences with 16 terms satisfying the conditions of he problem.
Third solution. We show that the assumption that the sequence has 17
terms leads to a contradiction.
Choose arbitrary 11 — 7 = 4 consecutive terms in the sequences. As the
sequence has 17 terms, these 4 terms are the first or last 4 terms of an 11 tuple. As
the sum of the 11 terms is positive and the sum of the complementary 7 terms
is negative, the sum of the four terms has to be positive; thus the sum of any 4
consecutive terms is positive.
Now, choose arbitrary 7 — 4 = 3 consecutive terms. There is a 7 tuple in the
sequence such that these 3 terms are the first or last 3 terms of a a 7 tuple . The
sum of the 7 terms is negative, the sum of the complementary 4 terms is positive,
so the sum of the 3 terms has to be negative.
Now, choose arbitrary 4 — 3 = 1 term. This has to be positive as this is the
complement of 3 terms in a 4-tuple where the sum of the 3 terms is negative and
the sum of the 4-tuple is positive. Thus every term has to be positive so there is
no sequence satisfying the conditions of length 17.
As shown in the previous solutions, there are sequences of length 16
satisfying the conditions.
Remarks. 1. With the method of the first solution it can be shown that if in
a sequence the sum of any consecutive η terms is negative and the sum of any
consecutive ρ terms is positive then the sequence has at most n + p—1 terms.
2. The method of the third solution reminds us to the Euclidean algorithm.
Indeed, if the sum of any consecutive η terms is negative and the sum of any
consecutive ρ terms is positive, then the sequence cannot have more than n +
p — gcd(n,p) terms, where gcd stands for the greatest common divisor. If — as
in our problem — η and ρ are coprime, then the sequence has at most n + p — 1
terms.
226

1977/3
1977.
1977/3. Given an integer n>2, let Vn be the set of integers l + kn for
к a positive integer. A number mEVn is called indecomposable if there is no
number p,q EVn such that pq = m.
Prove that there is a number r in Vn which can be expressed as the product
of indecomposable members of Vn in more than one way (decompositions which
differ solely in the order of factors are not regarded as different).
First solution. First, we collect a few simple observations about the
numbers in Vn:
a) 1 is not in Vn.
β) The products of two members of Vn is in Vn: if k\n+l 6 Vn and kin +
1 G Vn, then
(kin+l)(k2n+l) = (kik2n+ki +/c2)n+l eVn.
7) the product of kin — 1 and k^n — 1 is also in Vn:
(k\n— 1)(к2П— \) = {к\к2П — к\ —k2)n+l. (n>2).
The beginning of the list of the members of Vn in increasing order is:
n+1, 2n+l, 3n+l, ..., (n+l)n+l, (n + 2)n+l=(n+l)2.
We have to find a number in Vn that splits into the product of
indecomposable numbers in two different ways in Vn.
Let a = n- 1, b = 2n-l. By β) and 7) we have that a2, b2, ab, (ab)2 are
in Vn:
1 1
ab· ab-a -b
are two decompositions of (ab) . These two decompositions are different as e.g.
a =ab implies a = b that is a contradiction.
We show that a , b , ab are indecomposable in Vn.
9 9
a is indecomposable, because it is smaller than , (n +1) , the smallest
decomposable number in Vn.
9 9
lib = (In — 1) is decomposable, then one of its factors has to be η + 1,
because (2n+ l)2 > (In — l)2 and 2n+ 1 is the second smallest term of Vn. Hence
in case b were decomposable, then
9 9
b =(2n —1) =(n+l)(cn + l) (с positive integer)
possible, and so
c + 5
n = - .
4-c
This is an integer only if c = 3 and η = 8. So if n^8, 62 is indecomposable.
227

Solutions
1977/3
In a decomposition of ab = (n — l)(2n— 1) both factors cannot be greater
than (n+ 1) as (2n+ 1) > (n — l)(2n — 1), hence if a6 can be decomposed, it is
done in the following form:
(n-l)(2n-l) = (n + l)(dn+l)
where d is a positive integer. Hence
d + 4
П=2^'
that holds in case of d= 1 and η = 5. Thus ab is decomposable if η ^5.
We solved the problem if n^5 or n^8. We show that the statement holds
in these latter cases, as well.
In case η = 5, 3 -5 +1 = 16, 15 -5 + 1 =76, 72-5 + 1 = 361 are
indecomposable numbers in V5 because they do not have two divisors of the form 5k + 1, but
762 = 76·76=16·361,
so 76 G V5 is decomposed in V5 in two different ways.
If η = 8, 8 · 8 + 1 = 65, 3 ■ 8 + 1 = 25, 21 · 8 +1 = 169 are indecomposable
numbers in Vg because they do not have two divisors of the form Sk + 1, but
652 = 65 -65 = 25 169,
and so 65 G V% is decomposed in two different ways.
Second solution. The following solution refers to the origins of the
problem, but it uses a deep theorem of number theory, namely Dirichlet's theorem:
In the arithmetic progression an + b where a, b are coprime integers there are
infinitely many primes.
We shall utilize the theorem through the following statement: there are three
primes of the form nk — 1:
p\=nq-l, p2 = nr-l, p3 = ns-l,
where q, r, s are positive integers. None of these numbers is in Vn as, for
example,
nq—l=nk+l
implies n(q — k) = 2, and it is impossible since η > 2.
ίΡΐίΡ2> ίΡ2ίΡ3> Ρ3Ρ1 ancl Ръ are m Ki> but they are indecomposable in Vn since
they have no prime divisors in Vn. р\Р2Ръ е ^n can ^e decomposed in Vn in at
least two ways:
(Р\РЪ)(Р2РЪ) = (Р\Р2)РЪ·
The two decompositions are different sinceρχρ^ =ρ^ would imply p\=p?). Hence
we proved our statement.
228

1977/3
1977.
Remarks. 1. The condition η >2 is necessary, because in case of η = 2, Vn
is the set of odd primes; here, the indecomposable numbers are the primes and
if an odd number is a product of two primes, it can be done in a unique way.
2. Using Dirichlet's theorem is a very strong theorem in the second solution,
but the jury accepted it as a full solution. The third solution will use a simple
special case of the theorem.
Third solution. First, we prove three lemmas:
Lemma 1. There are infinitely many primes that are not of the form kn + 1
(n, к positive integers, η > 1 a fix integer).
Let us assume to the contrary that there are only finitely many primes of this
kind, ti,t2,...,ts and let U = nt\t'i ...ta — l. None of the t{-s divide U, hence
all its prime factors are of the form kn + 1; the product of these primes is of the
form kn+l, but U is not. Thus there are infinitely many primes not of the form
kn+1.
Lemma 2. There is an integer a, such that there are infinitely many primes
of the form кп + а (к positive integer).
According to Lemma 1. there are infinitely many primes of the form kn + r,
where 1 < r < n. The possible values of r are 2, 3, ..., η — 1, hence there is r = a,
such that there are infinitely many primes of the form kn + a. Let Ρ denote the
set of these primes.
Lemma 3. There is a positive exponent a <n such that aa is of the form
kn + 1, where a is the positive integer defined in Lemma 2.
First, observe that there are two powers in the sequence a, a ,..., a that
give the same residue mod n, e.g., ααι and aai («2 > cq). Then η divides aai —
ααι =ααΐ(αα2_αι — 1). a and η are coprime as the elements of Ρ are divisible
by the greatest common divisor of a and n, so it has to equal 1. So η divides
aa2~ai — \t hence for a2 — a\=a, aa — 1 = kn (k positive integer), thus
(1) aa = kn+l.
a is greater than 1, because a= 1 implies a = kn+1 that is impossible since
a<n, thus α > 2. Let и denote the smallest of the as satisfying (1) (u>2);
PliPli ■ ■ - iPu be arbitrary primes in P. v~P\P2 · · -Pu is °f the form An + au,
where A is a positive integer. By the choice of u, au is of the form Bn + 1 (B
positive integer), hence p = Cn+ 1 (C positive integer), so ρ is in Vn.
ρ is indecomposable in Vn, because in the other case the product of a few
(let's say s) of the primes p\,P2,---,Pu, was m Vn, i-e- °f the form kn+1. This
product is of the form Dn + as (D positive integer), that is as gives residue 1
mod η contradicting the definition of u. Thus the product of any и elements from
Ρ is in Vn and it is indecomposable.
??Q

Solutions
1977/3
f(x) = 1 — να2 + b2 ( . cos χ + , sin χ ) —
Finally, let pi,p2> · · · >iV> <?i»?2> ■ · · > 9« be primes in P. Their product
PlP2 · · ■ PuQl Я2---Яи = (P\Pl ■ ■ ■ Pu)(q\ Я2---Яи) =
= (qiP2P3 · · · Pu)(pi qiqi··· qu)
factors into the product of indecomposables in at least two ways.
1977/4. Define
(1) f(x)=l — acosx — bsmx — Acos2x — Bsm2x
where a, b, A and В are real constants. Suppose that f(x) > 0 for all real x.
Prove that
(2) a2 + b2<2, A2 + B2<1.
Solution. If a = b = A = B = 0, then (2) clearly holds. Assume that at least
one of a, b, and A, B is non-zero. Then
b
b2 Va2 + b2
-Va2+b2 , = cos 2x + - sin 2x .
WA2 + B2 VA2 + B2 J
Let a and 2/3 denote the angles of the unit vectors e I —==, ■—=== I and
V у a2 + b2 Va2 + b2/
Ε ( ζ, ζ Ι respectively. Using the above notations
\VA2 + B2 VA2 + B2/
f(x) = 1 - Va2 + b2 cosO - a) - Va2 + B2 cos 2{x - /3),
and this holds even if a2 + b2 = 0, or A2 + B2 = 0. As f(x) > 0,
fW) + /Ο + β) = 1 - Jo2 + b2 cos(/3 - a) - VA2 + B2+
+ 1 + Ja2 + b2 cos(/3 - a) - ^A2 + B2 = 2( 1 - VA2 + B2) > 0,
Hence
A2 + B2<1.
Moreover,
f(a + j)+f(a-j) = l- \la2 + b2 · %- + Va2 + B2 sin 2(α - β)+
+l-\/a2 + b2^—-VA2 + B2sm2(a-P) = 2-V2y/a2 + b2>0,
ι
and a2 + b2 < 2,
and so (2) is proved.
230

1977/5
1977.
Remark. The converse of the statement is not true, (2) does not imply
f(x)>0; e.g. in case a = y/2, b = A = B = 0 a2 + b2<2, A2 + B2<\, but
/(0) = 1 - >/2 < 0.
9 9
1977/5. Let a and b be positive integers. When (a +b ) is divided by (a +
b), the quotient is q, the remainder r.
Find all pairs (a, b), such that
(1) g2 + r = 1977.
Solution. By the conditions
(2) a2 + b2 = (a + b)q + r, where 0<r<a + b,
and so
r _a2 + b2
a+b a+b
r
r<a + b implies < 1, and using the A.M.-G.M. inequality:
a + b
a2 + b2 a + b
a + b 2
giving
(3) 2(g+l)>a + 6.
(1) implies
By(3)
g2<1977, thus g<44.
r<a + b<2-45 = 90.
On the other hand, using (1) we get:
q2 = 1977 - r > 1977 - 90= 1887,
that is
1887 <q2< 1977.
As only 1936 = 442 is a square, g = 44 and so r = 1977- 1936 = 41. Substituting
to (2), we get
a2 + b2 = 44(a + b) + 41,
(a-22)2 + (b -22)2 = 1009.
None of the squares on the l.h.s can be smaller than the half of 1009 so they are
at least 505, hence the possible values are the squares between 505 and 1009:
529, 576, 625, 676, 729, 784, 841, 900, 961.
231

Solutions
\
1977/5
Considering the last digit, we get that only 784 gives a square difference
with 1009, with 152 = 225, so
{|a-22|,|b-22|} = {28,15},
hence the possible (a, b) pairs are:
(7,50), (37,50), (50,7), (50,37),
and these values satisfy the conditions of the problem.
1977/6. The function f is defined on the set of positive integers and its
values are positive integers. Let us assume that
(1) f(n + l)>f(f(n))
for every positive n.
Prove that for every positive η
f(n) = n.
First solution. We prove the statement by induction. First, we show that
/(1)=1. There has to be a ко such that f(ko)=l. If not, let us consider the
following sequence
&1=/(&о), &2 = /(&1-1), ·■·, h = f(ki_i-l), ...
As by our assumptions /(A^-i — l) = h ^1, h>l, hence k{ — \ is a positive
integer, and so this sequence exists. By (1) we get f(ki) > f(f(h - 1)) = f(ki+\),
hence there exists an infinite monotone decreasing sequence of integers
/(/со) > №) >■·> Kh) > /(fci+i) > · · ·
which is impossible.
Thus there is a positive integer kQ such that f(ko) = 1. But since if fcg > 1,
ko~l>l the inequality in (1) would imply
/(/(fy-i))</(fco) = i
which is impossible, as / cannot attain a value smaller than 1. Therefore
Now, assume that the statement is true for some positive integer к for every
function / satisfying the conditions of the problem, i.e. f(k) = k. We have to
prove that
f(k+l) = k + l.
Consider the function
<p(n) = f(n+l)-l.
232

1977/6
1977.
We show that φ(ή) satisfies the conditions of the problem. (1) implies f(n+ 1) >
1, and so f(n+ 1) — 1 is a positive integer. Hence — because of (1):
φ(φ(η))=φ(Κη+1)-1) = /(Κη+1))-1<Κη + 2)-1=φ(η+11
that is, φ(ή) satisfies (1) and by the assumption of the induction
ip(k) = k.
So
φ(]ζ) = /(к + 1) - 1 = к, f(k + 1) = к + 1,
And this is what we wanted to prove.
Second solution. We prove that / is strictly increasing. Let us divide the
set of integers into two disjoint subsets К and M.
Let К be the set of positive integers к such that
(2) f(k) < k;
and Μ be the set of positive integers η such that
(3) f(n) > n.
First we show that К is the empty set. Indeed, if К had an element, choose
m in К such that f(m) is the smallest. By the definition of К
(4) f(m) < m, that is f(m) < m — 1.
m — 1 is positive as it is not smaller than a value of the function, and so, f(m — 1)
exists. (1) and (4) imply that
(5) m>/(m)>/(/(m-l)).
(5) implies that f(f(m — 1)) is smaller than f(m). But / cannot attain values
smaller the f(m) in K, thus f(m — 1) G M. Hence, applying (3) and (5) we get:
(6) m > f(m) > f(f(m - 1)) > f(m - 1).
Similarly, m — 1 is not in К either, because f(m — 1) < f(m) and the values of
/ at the elements of К cannot be less than f(m), hence m — 1 G Μ and by (3)
f(m— 1) >m— 1.
Combining with (4) we get
Km-I) > f(m).
This contradicts (6), som-1 cannot be the element of K, so (3) holds for every
positive integer n. Hence applying (1) and (3)
/(п+1)>/(Дп))>Дп),
233

Solutions
1977/6
that is
f(n+l)>f(n)
follows, hence / is strictly increasing.
Now, it follows that in (3) equality holds because if for some η f(n) > n, it
means f(n) > n+ 1, but by the monotonity
f(f(n))>f(n+l)
holds, contradicting (1). Thus for every η
f(n) = n,
and this is what we wanted to prove.
1978.
1978/1. For m and η positive integers, η > m > 1, the last three decimal
digits of 1978m is the same as the last three decimal digits of 1978n. Find m
and η such that m + n has the least possible value.
Solution. 1978n and 1978m has the same last three digits if and only if
1000 divides their difference:
1978n - 1978m = 1978m(1978n~m - 1)
As 1000 = 8 -125, we have to examine divisibility by 8 and 125.
The second factor is odd and only the first power of 2 divides 1978. Hence
m has to be at least 3, m > 3.
Only the factor 1978n_m-l can be divisible by 125. As the last digits
of the powers of 1978 are 8, 4, 2, 6, 8, 4, ..., their period is 4. Obviously, 5
divides 1978n_m — 1 only in case the last digit is 6, that is if η — m = 4k. Since
the remainder of 1978 mod 125 is —22 and the remainder of (—22) is 6, 125
divides 19784A: - 1 if and only if 125 divides 6^ — 1 = (1 +5)^ — 1.
Applying the binomial theorem:
(l+5)k-l=5k + 52^^- + ...
2
the other terms are divisible by 5 = 125 hence we need to examine
5/c + 25^-^ = ^(5/c-3).
2 2
5k — 3 is relatively prime to 125 and 5k is divisible by 125, if 25 divides k, and
so к is at least 25, that is n — m = 4k > 100. Since m>3, n + m = n — m + 2m >
106. Thus the minimal value of n + m is 106, and m = 3, n = 103.
234

1978/2
1978.
1978/2. Ρ is a point inside a sphere. Three mutually perpendicular rays
from Ρ intersect the sphere at points U, V and W. Let Q denote the vertex
diagonally opposite Ρ in the parallelepiped determined by PU, PV, PW. Find
the locus of Q for all possible sets of such rays from P.
Solution. Let R denote the radius, О the centre of the sphere, OP=p. We
show that the locus is a sphere with centre О and radius у 3R? — 2p2.
Let a, b, c, p, q denote the vectors pointing from О to А, В, С, Р, Q,
respectively. As q — ρ is the diagonal vector of the parallelepiped,
q-p = (a-p) + (b-p) + (c-p),
Since (a — p), (b — p) and (c — p) are mutually perpendicular vectors and a =
h2 = c2 = R2,
q2 = (p + (a-p) + (b-p) + (c-p))2 =
= р2 + (а-р)2 + (Ь-р)2 + (с-р)2 + 2р(а + Ь + с-Зр) = ЗД2-2/,
thus we showed that the points Q are on the designated sphere.
We have to show that any point of the sphere can be the Q vertex of an
appropriate parallelepiped. We can choose the parallelepiped in a special way.
Let the plane
spanned by P, Q and О
intersect the original
sphere in the circle k\, and
the sphere with
diameter PQ in the circle /c2-
One of the points of
intersection of k\ and hi
is A. Let S complete
the right triangle PAQ
to the PAQS rectangle
(see Figures 1978/2.la,
b). Now, let us build a
square base prism such
that this rectangle is a
diagonal section of it,
with base square PBSC, and on the cover the points A and Q are opposite
vertices. Let F denote the midpoint of PS. As the diagonal ВС is orthogonal to
the plane of the diagonal intersection, i.e. to the plane of k\, the triangle OF В
is a right triangle.
Figure 78/2. la
235

Solutions
1978/2
In order to prove our statement it is
enough to show that В and С are on the
sphere; Since О lies on the
perpendicular bisector plane of ВС, it is enough to
show that OB = R.
We shall use that for an arbitrary
point Μ in the space, the distances from
the vertices of the rectangle XYZU
satisfy MX2 + MZ2 = MY2 + MU2 (see
Remark 1.)· We shall denote the vectors
by bold face lower case letters
corresponding to the points labelled by capital
letters with О as the vector origin.
Applying the theorem to the rectangle APSQ
and the point O, we obtain
Figure 78/2. lb
2 2 2
s = ρ + q -
=p2 + 3R2
a2 =
2p<
R2 = 2R2-v2.
In the square PBSC we have FB =
the right triangle OFB
PS |s-p|
, but \θΡ\ =
s + p
, hence from
2 ) ' V 2 / 2 2
В and С are on the given sphere and the square base prism satisfies the
conditions of the problem.
Remarks. 1. The rectangle theorem we
used above can be easily proved by vector
methods; let x, x + a, x + a + b, x + b represent
the vectors corresponding to the vertices of
the rectangle XYZU, choosing Μ as a
vector origin, where a, b denote the side vectors
of the rectangle (see Figure 1978/2.2). From
ab = 0 it follows
x2 + (x + a + b)2 = 2x2 + a2 + b2 + 2ax + 2bx,
(x + a)2 + (x + b)2 = 2x2 + a2 + b2 + 2ax + 2bx,
Μ
Figure 78/2.2
and so MX2 + MZ2 = MY2 + MU2.
2. In the second part of the solution we defined the points А, В, С to the
point Ρ such that the segments PA, PB, PC are pairwise orthogonal. In fact,
236

1978/3
1978.
on the circle determined by А, В, С there are infinitely many triples А', В', С'
having the same property.
If a cone has three pairwise perpendicular apothems, then it has infinitely
many. Moreover, every apothem is the member of such a triple.
3. The planar version of the problem is the following: Let Ρ be an inner
point of a circle and let А, В be points on the circle such that PA and PB are
perpendicular, then the vertices Q of the rectangles PAQB lie on a circle.
1978/3. The set of all positive integers is the union of two disjoint subsets:
F = {f(l)J(2),...J(n),...} and
G = {g(l),g(2)1...1g(n)i...}
where
/(1)</(2)< ... </(n)< ... and g(l)<g(2)<...<g(n)<...,
(1) g(n) = f(f(n)) + l for every n>\.
Determine /(240).
First solution. In principle, the values of / and g can be calculated, but it
would be long and tedious. We shorten our way with some observations:
Let us count the number of integers between 1 and g{ri). The values g(l),
g(2), ..., g(n — 1) precede g(ri) (this is η — 1 integers), moreover as g{ri) — 1 =
f(f(ri)), the f(n) values of /: /(1), /(2), ..., /(/(n)), thus we listed all numbers
between 1 and gin), hence
(2) g(n) = f(n) + n.
Combining with (1) this gives
(3) f(f(n)) = f(n) + n-l.
G cannot contain two consecutive numbers, because by (1) we have that
f(f(n)) precedes g(n). Thus the number following g(n) is /(/(n) + 1):
(4) f(№ + l) = f(f(n))+2.
The formulas (l)-(4) make the evaluation of the functions easier. 1 has to
be in F as (2) implies g(l) = f(l)+l, thus /(1) = 1 and so g(l) = 2. By (4)
/(2) = /(/(l)+l) = /(/(l)) + 2 = 3.
The following values are calculated by (3):
/(3) = /(/(2)) = /(2)+l=4,
/(4) = /(/(3)) = /(3) + 2 = 6,
f(6) = 9, /(9) = 14, /(14) = 22, /(22) = 35, /(35) = 56, /(56) = 90.
237

Solutions
1978/3
Now, applying (4) again:
/(57) = /(/(35) +1) = /(/(35)) + 2 = 92.
And from (3):
/(92) = /(/(57)) = /(57) + 56 =148,
/(148) = 239,/(239) = 386.
Finally, (4) gives
/(240) = /(/(148) +1) = /(/(148)) + 2 = 388,
and we answered the question.
Second solution. Although this solution is much more complicated than
the previous one, it gives a better view to the background of the problem.
First, we show that the conditions uniquely determine the functions / and
g. It is enough to show that / is uniquely determined, because by (2) /
determines g.
We prove by induction. By the previous proof /(1) = 1. Now, assume that
the value of /(n) is uniquely determined for every η < к for some positive integer
k. Let s denote the smallest positive integer that is not among the numbers /(1),
/(2), ..., f(k - 1), g(l), g(2), ..., g(k — 1). f(k) > s, because all the numbers
smaller than s are among the numbers listed above; f(k) > s cannot hold, because
by the monotonity of / it cannot belong to F, and by (2) we conclude that g(k)
is greater than s, so it cannot belong to G, either, contradicting the conditions of
the problem. Thus f(k) = s, and we proved that / and g are uniquely determined.
So if we produce a pair of functions satisfying the conditions of the problem,
that would be the unique pair of this kind.
We shall utilize the following theorem: if a and β are positive irrational
numbers such that
(5) i + I-l
α β
holds, then the set of integers is the disjoint union of
(6) {[α],[2α],...,[ηα],...} and {[/3], [2/3],..., [n/3],... }.
We shall use the theorem when Ι, α, β are three consecutive numbers of a
geometric progression, i.e. β = a , and by (5) in this case a is the positive root
of the equation:
(7) a2 -a -1=0.
a = —-— = 1,6180..., β = α = —-— =2,6180... Since a and β are greater
than 1 the sequences in (6) are strictly increasing.
238

1978/3
1978.
So our choice for
f(ri) = [na], g(ri) = [na2].
looks convenient.
To prove that our choice works, we have to show that (1) holds, i.e.
(8) [ηα2] = [[ηα]α] + 1.
[na] < na, so [ηα]α < па , implying that
[[ηα]α] < [па ],
Equality cannot hold, because by (6) the l.h.s. and the r.h.s. belong to different
sets, hence
(9) [[na]a]<[na2].
Reordering the inequality
na — [na] < 1 < a,
na<a + [na] follows and as (7) implies a — a = 1, we get
na<a + (a — a)[na].
Dividing this inequality by a, after reordering:
η + [na] < a[na] + 1,
[n(l + a)] < a[na] + 1
holds, or substituting l + a = a :
[na ] < [ηα]α + 1,
hence
[na ] < [[na]a] + 1.
Considering (9) we have:
[[ηα]α] < [na ] < [[na]a] + 1.
Since there is no integer between two consecutive integers, there is an equality
at the second place. Hence (8) holds.
Our result implies
/(240) = [240a] = [388,3281...] = 388.
Remark. There are several ways to determine the values of the functions
step by step. We mention the following:
239

Solutions
1978/3
There is a connection between the Fibonacci sequence defined by a\-\,
a2 = l, on+i = on + an_\ (n > 1) and the function f(m):
(10) Доп + 1) = оп+1+1. (п>1)
Since /(1)=1, /(2) = 3, /(3) = 4, (10) holds for the first few values. Using the
recurrence formulas and (3):
an+2 = f(an+l + 1) - 1 = /(/(α.η+1)) ~ ! =
=/(on + 1) - 1 + on = on+i + on,
so the connection is really true.
1978/4. In the triangle ABC we have AB-AC. A circle is tangent
internally to the circumcircle of the triangle and also to sides AB, AC at P, Q
respectively. Prove that the midpoint of PQ is the centre of the incircle of the
triangle.
First solution. Let к denote the circle tangent to the circumcircle through
Ρ and Q. The point of tangency of к and the circumcircle is denoted by Т. Т is
on the axis of symmetry of the triangle thus AT is a diameter of the circumcircle.
Let Μ denote the point of intersection of PQ and AT. We have to show
that Μ is the centre of the incircle, as by the symmetry Μ is the midpoint of
PQ (see Figure 1978/4.1).
A A
Τ Τ
Figure 78/4.1 Figure 78/4.2
Thales' theorem implies that ВТ and CT are orthogonal to AB and AC,
respectively. Let К denote the centre of the circle, and so Ρ Κ and QK are
orthogonal to the sides AB and AC, respectively. This implies that the
deltoids ABTC and APKQ are homothetic with centre A, hence if F denotes the
240

1978/4
1978.
midpoint of ВС, then
AT AK
AF AM
AT
Now, apply the enlargement with ratio —— and centre A to the circle k;
AF
this maps the circle к touching the lines AB and AC to a circle touching them.
As the image of Τ is F at the enlargement, the image of the circle contains F
hence the image is k\, the incircle. Thus К is mapped to M, hence the centre
of k\ is M, and this is what we wanted to show.
Second solution. We use the notations of the first solution. In order to
prove our statement it is enough to show that BM bisects /ABC, because in
this case Μ is the point of intersection of two bisectors, hence the centre of
the incircle (see Figure 1978/4.2). In the circle к we have LPTQ=LAPQ-
β because they subtend the same arc. Moreover, PBTM can be inscribed in
a circle as its two opposite angles are right angles, hence LPBMr~= LPTM,
β β
because they subtend the same arc, and as LPTM-—, LPBM equals —, as
well; now, IPBC = P, hence LMBC=^-. Thus BM is a bisector.
Third solution. We show that the statement holds for an arbitrary triangle,
we do not need the condition AB - AC.
Let к denote the
circle touching AB, AC
and g, the circumcirc-
le of ABC, moreover
denote by Τ the
point of tangency of the
two circles. Τ is the
centre of enlargement of
the two circles, hence
Q' and Ρ', the images
of Q and Ρ at the
enlargement, are the
midpoints of the arcs AC
and AB, because the
tangents AC and AB
are parallel to the
corresponding tangents (see
Figure 1978/4.3). Q,
Fugre 78/4.3
8
241

Solutions
1978/4
As 7 subtends the
7
arc AB, — subtends its
2
half, the arc BP', hence LBTP' = ^ Similarly, /Q'TC = ^. The point of inter-
section of the bisector BQ' and the segment PQ is a single point M, as in the
triangle P'TQ' the side P'Q' is parallel to PQ, and the points Q' and R ( the
point of intersection of the lines BQ' and TP'), are on different sides of the PQ
line.
OL
In the triangle Ρ MB the exterior angle at Ρ is 90° — —, hence IPMB =
90°-^-^=Д and so LPTB and LPMB are equal. PMTB is a cyclic
quadrangle, hence LPTM = LPBM-^-. LCTB is the external angle of a, thus
lCTB = P + ^ = lBTP' + lP'TM+lMTQ' + lQ'TC = ^ + ^ + lMTQ + ^,
and /MTQ = ^. This implies that LMTC= ^ + ^ = 90° - ^ = /.MQA, thus
MTCQ is an inscribed quadrangle, implying ZMTQ = ZMCQ = ^. Hence MC
is the bisector of 7, Μ is the point of intersection of the two bisectors, and so
the centre of the inscribed circle of ABC. Thus Μ lies on the bisector of the
isosceles triangle PAQ at A, and so Μ bisects the segment PQ.
Fourth solution.
Using well known trigonometric formulas we can give
a simple proof to the generalized
version of the problem.
Let Τ, Ρ, and Q denote the
points of tangency of the circle
к and the circumcircle, and the
segments AB and AC
respectively, О and R the centre and
the radius of the circumcircle, r
the radius of the incircle, and S
and ρ the centre and the
radius of the circle к (see Figure
1978/4.4). IQAO = 90°-P hen-
a
Fugre 78/4.4
ce \/OAS\=LU =
/З-7
(90° -β)
(Here, β > 7.)
Since O, S, Τ are colline-
ar, OS = R — ρ; the right triangle
242

1978/4
1978.
AQS satisfies AS = —. Applying the Law of Cosine for the triangle A OS
sinf
we get:
OS2 = О A2 + AS2 - 20 A ■ AS cos ω,
(R-ρ) =R +—-y- -^-^cos-——.
sin2 f sin f 2
After rearranging we obtain:
η α Λπ . a ( β — Ί . cx\
ρ cos — = 2R sin — cos sin — =
2 2 V 2 2)
Λπ . a ( /3-7 /3 + 7\ л„ . « . ^ . 7
= 2it sin — cos cos = 2R sin — sin — sin —.
2 V 2 2 ) 222
α β 7
Using the formula 2R sin — sin — sin — = r:
5 2 2 2
r
(1) e =
COS2 j
Let Μ denote the point of intersection of the segments AS and PQ (this is the
ex.
midpoint of PQ). In the right triangle PMS the angle ISPM=- implying
ex ex.
MS = Qsm—, MP = ρ cos— The geometric mean theorem applied to the right
triangle SPA gives MP2 = AM ■ MS, and using (1) we get:
AM =
MP2 ρ2 cos2 |
2 a
Г
MS ρ sinf sinf
As the distance of the vertex A from the centre of the incircle is .' n , Μ is the
centre of the incircle and this is what we wanted to prove.
r
sinf
Remarks. 1. The third solution is essentially the same as the second solution
of the problem 1969/'4. If there, we identify the points D and В we get this third
solution.
2. A careful investigation of Figure 1978/4.3 suggests the obvious solution
of the problem: we have to show that the points P, Q and Μ are collinear. This
is the straightforward consequence of Pascal's theorem:
Let 1, 2, 3, 4, 5, 6 be six arbitrary points of a conic (e.g. a circle), then the
points of intersection of the lines 12 and 45, 23 and 56 moreover 34 and 61 are
collinear.
On Figure 1978/4.3ЫА = 1,В = 2, Q'= 3, Τ = 4, P' = 5, С = 6. By Pascal's
theorem P, the point of intersection of AB and TP', M, the point of intersection
of BQ' and P'C and Q, the point of intersection of Q'T and С A are collinear.
243

Solutions
1978/5
1978/5. Let {a^} be a sequence of distinct positive integers (k=l, 2, ...,
n, ...). Prove that for every positive integer η
η _ η -ι
ω Σ^Σ^·
k=l κ к=\ л
Solution. We start with a lemma: let x\, x>i, . ■., xn\ У\, У2-> · · ·■> Уп be real
numbers, where x\ > x^ > ... > xn, y\ < У2 < · · · < Уп, moreover z\, Z2, · · ·, zn
a permutation of the ^-s, then
η к
(2) Σ^ί-Σ^*·
ν
=1 i=l
In (2) equality holds if the order of the Zi-s agrees with the order of the ^-s.
If to the contrary, we assume that they first differ at the к-th place, i.e. y\-z\,
y2 = z2> ■ · ·> Uk-l=zk-h but ук¥*к- Let zk = yr and yk = zs, where r and s are
greater than k, r> к and s > k. Using this notation
xkzk + xszs = хкУг + #sj//c·
Now, change the left hand side by substituting XkZk + xszs = xkyr + xsyk by
хкУк + хзУг- This way we increased (did not decrease) the sum on the left,
because
{ХкУк + хзУг) ~ (хкУг + хзУк) = (хк ~ хз){Ук ~ Уг) > 0.
This implies that the sum Y^Xiyi, where the Xi-s are in decreasing order
is the smallest if the ^-s are in increasing order. There is a minimal among the
above sums as there are finitely many of them and that can only occur when
the yi-s are in increasing order, otherwise we could decrease the sum using our
arguments.
We utilize the lemma in the following way: for a given η let
aix<ai2<---< αίη
be the first к elements in increasing order. But
1 1 J_ J_
I2>22>32>'">^2'
and hence by (2)
П г, П Π ■
(3) Σ^Σ^-
k=\ K k=\ K
^ > 1, cbi2 — ^' · · ·' aik — ^' · · ·' ain — n-> an<^ so (3) implies that
nn, n h n 1
h2 - 2^ l2 ~ 2^ и'
k=\ K k=l K k=\ л
and this is what we wanted to prove.
244

1978/6
1978.
Remarks. The full version of the lemma that we in the solution of problem
1975/1 is the following:
If x\, X2, · · ·, xn and z\, Z2, . ■., zn are real η-tuples, then the sum .
η
X%Z{
i=l
is maximal if the tuples are ordered on the same way and minimal if they are in
reversed order.
oo
2. A consequence of the statement of the problem is that the series У^ —r-
k=\ k
OO 1
is divergent (goes to infinity), because the same holds for the series У^ -. This
k=l k
is not true if there are only finitely many distinct a^-s. If о is the largest one,
J\ ak ( 1 1 1 \ / 1 1 1 \
n(n— 1)/
/,1111 1 1\ Л П ,
= al+ --- + --- + ...+ )=α 2-- <2α,
\ 1 2 2 3 η—1 η/ \ η/
οο
hence every partial sum of the series У^ —т· is smaller than 2a, thus the series is
k=\ k
convergent and its sum is smaller than 2a. If every a*, is 1, then we get the well
00 1 π2
known sum: У~^ —z = — = \ ,6449 ....
γ/c2 6
1978/6. An international society has its members from six different
countries. The list of members has 1978 names, numbered 1, 2, ..., 1978. Prove that
there is at least one member whose number is the sum of the numbers of two
members from his own country, or twice the number of a member from his own
country.
First solution. The statement of the problem is equivalent to the following:
there are two scientists from the same country such that the difference of their
number belongs to a scientist of their country.
Assume to the contrary that the statement is false. Let A, B, C, D, Ε and
F denote the countries. There is a country with at least 330 scientists, e.g. A,
because 6 · 329 = 1974 < 1978. Their numbers are:
a\ <a2< .. .<<2330·
By our assumption no scientist with number
(1) 02-01,03-αϊ,..., 0330-01
is from A, instead they are distributed among the other 5 countries.
245

Solutions
1978/6
As 5 · 65 = 325 < 329, there are at least 66 scientists from the same country
different from A, e.g. from B, from the list (1). Let their numbers be
b\ <62<...<666.
Now, none of the scientists with number
(2) 62-6b 63-61, ..., 666-6!
can be from B, but they are not from A either, because e.g. 62 — b\ = (0^ — a\) —
(aj — a\) = ak — dj were in A, contradicting our assumptions.
Now, since 4· 16 = 64 < 65, there are at least 17 scientists from the same
country different from A, and В e.g. from С from the list (2) with numbers
c\ <c2<...<ci7.
None of the scientist with number
(3) с1-с\,съ-сх,...,с\1-сх
can be from C, but not from В as e.g. c^ — c\ =(bs — b\) — (br — b\) = bs — br
would hold contradicting our assumptions about B. Similarly, it cannot belong
to A, either.
As3-5 = 15<16, there are at least 6 scientists from the list in (3) belonging
to the same country, let us say from D. They are
d\ <d,2< .. .<d(,.
Using our previous arguments, it is clear that none of the five scientists
(4) d2-di,d3-di,...,d6-di
belong to А, В, С or D. So there are at least 3 scientists from E, because
2·2 = 4<6:
e\ <e2<e3.
The scientists e2 — e\ and ез — e\ are not from the countries E, А, В, С, D,
hence they are from F:
h=e2~el and /2 = 63-61-
But now, the scientist with number f\ — /2 cannot be from any of the countries,
providing a contradiction, hence the statement is true.
Second solution. We prove a graph theoretical generalisation of the
problem.
Let
(5) „i = fe!(1 + i_ + i_ + ... + r) + 1.
246

1978/6
1978.
If we colour the edges of the complete graph with nk vertices by к colours, then
there are three edges in the graph that form a triangle. Of course, we do not have
to use all к colours...
We prove our statement by induction on k. Suppose first that k = 1, n\ =3.
Here, the existence of the monochromatic triangle is obvious. Now, assume that
the statement holds for some к and we have to prove that it is true for к + 1.
We shall use the following consequence of (5):
(6) (k + l)(nk-l) = nk+i -2<nk+i -1.
Choose an arbitrary vertex Ρ form the graph with nk+\ vertices and coloured
by k + l colours. Ρ is the endpoint of nk+\ —1 edges, coloured by the k + \
colours. There must be nk many edges of the same colour, as by (6) (nk — l)(k +
1) is smaller then the edges starting from P. Let these edges be coloured red.
The other vertices form a complete graph G with nk vertices. If G contains a
red edge, then we have a red triangle.
If G does not contain a red edge, then it is a complete graph with nk vertices
coloured by к colours, thus by the assumptions G contains a monochromatic
triangle. Thus we proved our theorem.
Now, we can solve the original problem: Let us assign a vertex of a
complete graph to every participant, and label it by the number of the scientists.
Assign a colour to all six countries. We colour the edge between i and j by the
colour of the country where the scientist with number \i — j\ belongs to.
Since щ = 1958 < 1978, the graph contains a monochromatic triangle with vertices
i<j<k. In this case the scientists with number j — i, k — j, k — i are from the
same country, and
(j-i) + (k-j) = k-i,
as we wanted. If j — i = к — j, then к — г is twice as j — i, so it fits the conditions
of the problem.
Remarks. 1. The graph theoretic theorem is a Ramsey-type theorem.
2. In the literature, the r.h.s. of (5) is denoted by [efc!] + l, where e is the
base of the natural logarithm. The notation is relevant, as:
(7) ekl + l-nk = ekl~(nk-l)<l.
his can be easily verified, since
111
β'!='!(1+π4+···+ί)+^ττ+(^ιί+2)+···
λ | ι ι
Пк + k + l + (k+l)(k + 2) + '"
247

Solutions
1978/6
If k>\,
1 ι ι
= r <1,
k+1 1-1 к
1 k+\
and as (7) is obvious for к = 1, (7) holds for every n.
1979.
1979/1. Let ρ and q be positive integers such that
pill 1 1
- = ! + - + ... _^^ +
q 2 3 4' 1318 1319
Prove that 1979 divides p.
Solution. Apply the following transformation to the sum:
Ρ л 1 1 1 1 1,11
- = 1 — - + - — -+.. · - -Г-7ТГ + ТТГТТГ = 1 + Г + ~ + . . .+
2 3 4 1318 1319 2 3 1319
/111 1 \ 1 1 1
2 w + - + - + . . . + ттгтт: =1 + - + - + ...+
.2 4 6 '"■ 1318/ 2 3 ■'" 1319
'11 14 1 1 1
1+ - + - + . · . + -T=- =^тт + ^т + . . .+
2 3 "" 659/ 660 661 1319
\660 1319/ V661 1318/ \989 990,
/ 1 1 1 \
= 1979 _„ + +... +
,660-1319 661-1318 989-990/
The sum in the parentheses is -, where α is a positive integer and b = 660 · 661 ·
...-1319. As 1979 is a prime and every factor of b is smaller than 1979, b
ρ 1979α
and 1979 are coprime. Thus - = —-— implies that pb= 1979aq and so 1979
q b
divides p.
Remark. The problem can be generalized for primes of the form 3k+ 2: if
ρ 1 1 1 1
- = 1 -- + - -...- — +
q 2 3" 2k 2/c+l'
then 3k+ 2 divides p. '
1979/2. A prism with pentagons Α\Α2Α2Α4.Αζ, and ΒχΒ^ΒιΒ^Β·^ as
top and bottom faces is given. Each side of the two pentagons and each of the
248

1979/2
1979.
25 segments A{Bj, (г, j = 1, 2, ..., 5) is coloured red or green. Every triangle
whose vertices are vertices of the prism and whose sides have all been coloured
has two sides of a different colour. Prove that all 10 edges of the top and bottom
faces have the same colour.
Solution. We call the coloured segments in the problem edges. First we
show that all five edges of the top and bottom faces have the same colour.
Suppose to the contrary that the bottom face has a red and a green edge.
Let Ai be the common vertex of the red edge A\Ai, and the green edge
AiAi (see Figure 1979/2.1). There are three edges of the same colour (e.g.
red) among the edges between A>i and the 5 vertices of the top face. As any
>A3
Figure 79/2.1
3 edges of a pentagon contain two adjacent vertices, there are two neighbour
vertices of the top face, B{ and Bk that are connected by a red edge with A^. As
every triangle has two distinct colours, the edges A\Bk and A\Bi are green, for
otherwise we would have A\AiBi or ΑχΑ^Β^ as red triangles. But then in the
triangle В{В^А\ the edge B{Bk has to be red, a contradiction, because then the
triangle AqBiBk is red. So the edges of the bottom face are of the same colour.
By symmetry the same holds for the top face.
Now, it only remained to show that the top and bottom faces have the same
colour. Let the bottom face be red. One of the end points of the red edge A\ A^ is
connected to the B{ vertex with green. So one of them (e.g. A\) is the endpoint
of 3 green edges. Two of these green edges end in adjacent vertices, so the edge
connecting them (on the top face) is red. But then, all edges of the top face are
red as we proved above. ^
Remark. We only used the fact that the top and bottom faces have odd many
vertices, we have not even used that they are of the same size. In fact we proved
the following graph theoretical statement:
249

Solutions
1979/2
Let ρ and q be two odd cycles in the graph G such that there is no edges
between the non-adjacent edges of ρ and q, but every vertex of q is connected
with every vertex of p. Colour the vertices of G by two colours such that there
is no monochromatic triangle. Then the edges of ρ and q have the same colour.
1979/3. Let k\ and fc2 be two circles on the plane and let A denote one of
their points of intersection. Starting simultaneously from A, two points, P\ and
P2 move with constant speed, each traveling along its own circle in the same
sense. The two points return to A simultaneously after one revolution. Prove
that there is a fixed point Ρ in the plane such that, at any time, the distance
from Ρ to the moving points are equal.
First solution. Let 0\ and O2 denote the centres of k\, and fc2, r\ and
r2 their radii, respectively. After examining some special cases (see Figure
1979/3.1), our guess is that Ρ is the reflected image of A to the perpendicular
bisector of the 0\Oi segment.
PI
Figure 79/3.1
Let Pi and P2 be a location of the moving points. Then LAO\P\ =
LAO2P2· The triangle AO\P is the symmetric image of the triangle PO^A,
LPO\A-LPOiA, hence LPO\P\ = LPO2P2· Moreover, the symmetry
implies that AO\ =P02 = r\ and АО^ = РО\ = r2, too. It follows that two sides and
their angle agree in the triangles PO\P\ and PO2P2, hence they are congruent
and so PP\=PP2.
Note that the length of the two segments equal even if the points P, 0\, P\,
and the points P, 02, P2 are collinear. The length of the segments in this case
are r\ + r2 or \r\ — r21.
Second solution. We start with the following observation: if the similar
triangles OA\B\, ОЛ2Б2, ОА^В?, are of the same orientation and the points
250

1979/3
1979.
Figure 79/3.2
A\, Αι, Αι are collinear, then the points B\, B2, Βί, are collinear, as well. (See
Figure 1979/3.2). This is simply true, because the rotation stretching with centre
О and ratio OB\/OA\ and angle ΑχΟΒχ map the points Αχ, Α<ι, A-$ to B\,
B2, B^, respectively. Since the rotation stretching maps line to line, the line
containing A\, A2, A3 is mapped to the line through B\, B2 and Б3. This holds
even if the triples Ο, Αγ, Β\\ О, А^ Вч\ О, А$, Б3 are collinear.
Now, let us assume that the points P\ and P2 both travelled an arbitrary
angle a, (0 < a < 2π) around the centres of the circles (see Figure 1979/3.3).
Thus the triangles AO\P\, AO2P2 (possibly degenerated) are similar isosceles
Figure 79/3.3
triangles of the same orientation. As the perpendicular bisector of the segment
AB is the line 0\02 (here, В denotes the other point of intersection of the two
251

Solutions 1979/3
circles), there is a point С on 0\0>i such that the triangle AC В is similar to the
triangle AO\P\ and they are oriented in the same way.
As the triangles AO\P\, AO2P2 and AC В are similar triangles of the
same orientation, and the points 0\, О 2, С are collinear, our introductory remark
implies that P\, P2 and В are always collinear. So we only have to show that
at any stage the perpendicular bisector of P\ P2 contains a fix point. The second
point of intersection of the perpendicular to AB at A with the circle k\ is X and
with the circle &2 is Υ (see Figure 1979/3.4). Let Ρ denote the midpoint of XY,
this is independent of the points.
Figure 79/3.4
The theorem of Thales implies that BX is a diameter in k\, and BY is a
diameter in fc2, hence the quadrangle ΧΡγΡ2Υ is a right trapezium. The
perpendicular bisector of Pi P2 is the median of the trapezium, hence it contains the
midpoint of XY.
Remark. The basic observation of the second solution is that the points P\,
P2 arid В are collinear. We can verify this by calculating the angles, but using
the rotation stretching needs less discussion. Of course, analytic or trigonometric
method may also lead to the solution.
1979/4. Given a plane π, a point Ρ in the plane and a point Q not in the
plane. Find all points R of the plane π such that the ratio
QP + PR
{ } QR
is maximal. ,
Solution. Let Q' denote the orthogonal projection of Q to the plane π, and
m denote the PQ' line; if P = Qf, then let m be an arbitrary line of the plane
252

1979/4
1979.
through Q'. We show that if R is not on m, then m has a point R' such that
QP + PR QP + PR'
QR < QR'
So for every point R not on m there is a point iu' on m that gives a greater value
at (1).
Indeed, if m does not contain R let R' denote the point on m such that
Q'R! -Q'R and R! is not on the same ray as Ρ (see Figure 1979/4.1). As the
triangles QQ'R and QQ'R! are congruent,
Figure 79/4.7
Figure 79/4.2
QR = QR', and from the triangle inequality PR< PQ'+ Q'R = PQ'+
Q'R' = PR'. Hence
QP + PR_QP + PR QP + PR'
^R ~ 'QR~~< Q£y
Thus the points providing the maximum are on m.
From now, we assume that m contains R (see Figure 1979/4.2). Let
/QPR-a and lQRP = x, and apply the Law of Sines for the triangle QPR:
QP + PR QP PR sin χ sin(a + z)
— = —— + = + =
QR QR QR sin α sin α
1 . 2x + ol a 1 ./ a\
= ——2 sin —-— cos — = ——- sin [x + — .
sin α 2 2 sin§ V 2
Our expression is maximal if sin I x + —1 = 1, that is if
OL
^ + -=90°,
2
ж = 90° - |. Hence IPQR= 180° -(x + a) = 90° - | = ж, and the triangle PQR
is an isosceles triangle, QP = RP.
Note that there are two points on m of distance QP from P. The
denominator of (1) is the same in both cases, but if P^Q', QR is smaller (i.e. (1) is
253

Solutions
1979/4
greater), where the angle QPR is acute. The unique solution, R is the endpoint
of the segment PR of length PQ on the PQ' line on the ray not containing Q'.
If Ρ = Q', our arguments show that the points R are the points of the circle
in π with centre Ρ and radius PQ.
1979/5. Find all real numbers b for which there exist nonnegative real
numbers x\, x2, Χ3, £4, £5 satisfying
5 5 5
3™ i2 V^,.5„ ib
(1)
^2kxk = b, ^k xk = b , ^k xk = b:
k=\ k=\ k=\
that is
(2)
First solution. By (1)
5 5 5
b2 Σ kxk -2bJ^k3xk + Y^ k5xk = b3- 2b3 + b3 = 0,
k=\ k=\ k=\
5 5
0 = Σ Xk(b2k - 2bk3 + k5) = Y^ k(b - k2)2xk.
k=l k=\
This latter equation is equivalent to
(b - l)2xx + 2(6 - 4)2x2 + 3(6 - 9)2x3 +4(6 - 16)2z4 + 5(6 - 25)2z5 = 0.
The 5 terms on the l.h.s. are all non negative. This can only hold if
Xl=X2 = X2=X4 = X5=0, 6 = 0,
or if xk 7^0, then in the k-th term 6 — к =0, i.e. b = k2. In this case in the other
terms b — k ^0 holds, and so xk = 0. The table of the solutions:
X]
0
1
0
0
0
0
x2
0
0
2
0
0
0
£3
0
0
0
3
0
0
X4
0
0
0
0
4
0
x5
0
0
0
0
0
5
6
0
1
4
9
16
25
Easy to verify that these values clearly satisfy (1).
Second solution. The Cauchy-inequality for arbitrary real numbers, a\, a2,
..., an; b\, 62, .. ·, bn states that
(3) ΣαΙΣ^(Σα^
254

1979/6
1979.
where equality holds if and only if there is a t such that ak = tbk (k = 1, 2,
n) or all bk = 0.
Set
Now (3) reads as:
&k
= y/bck, bk = ^k5xk.
5 5 / 5 \ 2
^2kxk-^2k5xk>l^2k3xk\ ,
by (1) we have b > b , thus equality holds in (3), hence there is а у such that
(4) . ^kx~k = ty[^x~k, xk(t2k4-l) = 0 (fc = l, 2, ..., 5)
or every 6^ = 0, so every xk = 0, and necessarily 6 = 0.
If in (4) Xi^O for some i, then t = ^j, but then by (4) for every other k,
xk = 0 (k^i). Substituting to (1) we obtain
. _, ·3 ._τ2 ·5 ._t3
implying b = i2 and а^=г; Thus the possible values of b are: 0, 1, 4, 9, 16 and
25; and these values satisfy the equation.
1979/6. Let A and Ε be opposite vertices of an octagon. A frog starts at
vertex A. From any vertex except Ε it jumps to one of the two adjacent vertices.
When it reaches Ε it stops. Let an be the number of distinct paths of exactly η
jumps ending at E. Prove that
fl2n-l=0, a2n = ^(xn-l-yn-1) (n=l, 2, 3, ...)
where χ = 2 + v2 and у = 2 — Vz.
Remark. A path of exactly η jumps is a sequence of the vertices Pq, P\,
..., Pn that satisfy the following conditions:
l.P0 = A,Pn = E·
II. for every i, 0 < г < η — 1 Pi is distinct from E.
III. for every i, 0<i<n—l, Pi and Pi+\ are adjacent vertices.
Solution. The conditions in the remark are the complicated formulation of
the following: the frog, starting from A reaches Ε with the n-th jump, but not
sooner, and it can jump only to adjacent vertices.
Let A, B, C, D, E, F, G, Η denote the vertices of the octagon (see Figure
1979/6.1). The vertices В and Η can be reached from A only with odd many
255

Solutions
1979/6
jumps, С and G with even many jumps, and so D and F by odd many jumps,
finally Ε by even many jumps. Thus an = 0, if η is odd.
From now we assume that η is even. We try to express an with the number
of paths shorter than n. Obviously 0,2 = 0 and 04 = 2, because the frog needs at
least four jumps from A to Ε and it can do it in two ways. From now we assume
that η > 4.
Let bn denote the number of the paths of η jumps from С to E. By
symmetry, there are bn many paths from G to E.
After two jumps there are two cases. First case: the frog is in A. Second
case: the frog is in С or G. In the first case there are 2an_2 paths from A to Ε
as there are two ways to get to A from A with two jumps. In the second case
there are 6n_2 paths to Ε starting from С or G. Combining the cases gives:
(1) an = 2an_2 + 2bn_2.
Let us find a similar formula for bn. Starting from С after two steps the
frog (on a continuable path) is in С or in A. From A to Ε there are an_i many,
from С to Ε there are bn_2 many η — 2 long paths, but there are two ways to
get from С to C, hence
(2) bn = an_2 + 2bn_2.
Combining (1) and (2), after reordering we get
256

1980/1
1980.
implying bn_i = an_2 — αη_4· Substituting to (1) we get a recurrence formula for
a
n·
(3) αη = 4αη_2-2αη_4.
Finally, we need to show that
(4) a2n = ^={xn-l-yn-1)
satisfy (3) for every positive integer n. We do it by induction. For n= 1, a2 = 0,
for n = 2 04 = 2 that coincides with our results. Now, assume that (4) holds for
η — 1 and we have to show that it is true for n, as well.
Observe that x + y = 4 and xy = 2, thus χ and у are the roots of the equation
(5) z2-4z + 2 = 0,
and this implies that χ and у satisfy the equation:
zn-l=4zn-2-2zn-3.
Using this:
A 0
02n =4fl2(n-l) - 2^2(n-2) = -Д (zn~2 - УП~2) ~ -i= (хП~3 - уП~Ъ) =
= -L ((4xn-2 - 2xn~3) - (4yn~2 - 2yn-3Yj = ±= (xn~l - yn~l) ,
and this is what we wanted to prove.
Remark. Before the competition the jury considered a difficult version of the
problem. Instead the "Prove that..." part they thought about the "Determine the
value of an", that is, the problem would be to find the value of (4). From (3) it
is easy to produce (4); some students first observed from (4) that (3) holds and
then they started to solve the problem.
1980.
In 1980 there was no International Mathematical Olympics. The
Organisation of the Teachers of Mathematics, Physics and Chemistry of Finland (MAOL)
invited the teams of England, Hungary and Sweden to compare their abilities.
Here we present the problems of that competition.
1980/1. Let α, β, and 7 denote the angles of the triangle ABC. The
perpendicular bisector of AB intersects ВС at the point X, the perpendicular
bisector of AC intersects it at Y. Prove that tan β -tan 7 = 3 implies BC = XY'.
Show that this condition is not necessary for ВС = XY, and give a sufficient
and necessary condition.
257

Solutions
1980/1
A(2a,2b)
1 N. '
Γ'/
/β Υ/
Β(-2, 0) /
, У
\Я'
y\\i
C(2,0) ^
First solution. The
most convenient way of
solving the problem is the
analytic method, as we
can avoid to distinguish
the several cases.
First, observe that
for right triangles the
problem does not make
sense, because the
perpendicular bisectors do
not intersect the ВС line
or they intersect it in the
same point. From now,
we assume that ABC is not a right triangle.
Insert the ABC triangle into the coordinate system such that the
coordinates of the vertices are: A(2a,2b), £(-2,0), C(2,0) (see Figure 1980/1.1); we
may assume that a < 0. The coordinates of the midpoints of the sides AB and
AC are: C'(a —1,6), B'(a+ 1,6); the side vectors: (o+ 1,6), (a —1,6), hence the
equations of the perpendicular bisectors are:
1, and (a— \)x + by = a +6 — 1.
They intersect the χ axis {ВС line) at the points
Figure 80/1.1
(a+l)x + by = a2 + b2
X
'a2 + b2-l
a + l
0
Υ
'a2 + b2-l
a
1
0
(a — 1 7^0, since a < 0 and a+ 1 =0 means that the coordinates of A are (—2,26)
and then the triangle ABC is a right triangle). Thus, the XY distance is:
(1)
XY =
a2 + b2-l a2 + b2-l
2V
(Γ
1
a— I a + l
And as ВС = 4, the sufficient and necessary condition required by the problem
is
262
(2)
a'
1
= 4, that is
cr
1
1
= 2.
Using our notations
tan/3 =
hence tan β · tan 7 =
a+l
62
tan7 = -tan(180°-7) =
α
1
a
2-l
(3)
and the condition in (2) rewrites as
I tan β tan 7 — 11 = 2.
258

1980/1
1980.
This splits into two parts, depending on the sign of the expression in the absolute
value:
(4) tan/3 tan 7 = 3,
and
(5) tan/3tan7 = -l.
Thus both (4) and (5) are sufficient conditions, and together they are
necessary and sufficient for XY = BC.
Second solution. Let us assume that ABC is an acute triangle. Use the
notations of Figure 1980/1.1. The perpendicular bisectors at B' and C' intersect
in O, the centre of the circumcircle, which is inside the triangle. The triangles
OB'С and OXY are similar and О separates the points B' and Y, and C' and
RC
X. If XY = ВС, as B'C = ——, the ratio of the similarity of the two triangles is
2:1. Hence — as the sum of the lengths of the altitudes of the triangles through
О is —^, the distance of О from ВС is —. This is the same as the distance of
3 3
the point of gravity of ABC from ВС, hence the line through О and the point
of gravity (that is the the Euler line of the triangle) is parallel to ВС.
It is a well known theorem that the sufficient and necessary condition for
the Euler line and ВС being parallel is (see the remark): in case β^-η:
(6) tan β tan 7 = 3
In case an acute triangle (6) is a sufficient and necessary condition of ВС = XY.
If β = 7, ВС = XY holds only for an equilateral triangle. The Euler line is
not defined, but tan β = tan 7 = tan 60° = л/3.
In our previous solution we pointed out that for right triangles the problem
is not defined. Assume that ABC is obtuse. The obtuse angle cannot be at A as
in this case X and Υ are inner points of ВС and XY < ВС. Let β be obtuse.
(See Figure 1980/1.2); O, the point of intersection of the perpendicular bisectors
is separated from В by AC. and so LBAC-LYOX = a, because the
corresponding lines are parallel. If, again, we assume that XY = ВС, then the altitudes
in ABC and OXY at ВС and XY have the same length. Indeed, B' and С
are midpoints of the sides of both triangles, ВС and XY are on the same line,
hence the altitudes of AB'C and OC'B' at A and О have the same length. As
two triangles are congruent if they have a side in common, the opposite angle
is the same and the appropriate two altitudes have the same length, moreover
they are the reflections of each other to the common perpendicular bisector, /,
the triangles AB'C' and OC'B' are the reflections of one another and the same
holds for ABC and OXY.
259

Solutions
1980/1
Figure 80/1.2
Since ΙΟΧΥ = β, in the triangle XB'C we have /3 = 90°+ 7 so
7 = -(90°-/3),
and tan 7 = - tan(90° -β) = - cot β, thus
(7) tan β tan 7 = -1.
Our arguments can be reversed, thus for an obtuse triangle BC = XY if and
only if (7) holds.
Remark. In the second solution we used the following theorem: The Euler
line of the triangle ABC is parallel to ВС if and only if
(8) tan/3tan7 = 3 (/V7).
Indeed, let the radius of the circumcircle R = 1. If the Euler line is parallel to
ВС, then the centre of gravity, the centre of the circumcircle and the orthocentre
lie on the same side of ВС. Therefore β and 7 must be acute angles and (8) can
only be satisfied by acute angles. So we may assume that β and 7 are acute. The
distance of the centre of the circumcircle and ВС is R cos a = cos a, the distance
of the centre of gravity from ВС is equal to the third of the altitude ha, hence
the Euler line is parallel to ВС if and only if:
(9) 3cosa = /ia.
Since It = aha = be sin 7 and the sides b = 2 sin β, с = 2 sin 7 and a = 2 sin a,
equation (9) implies that 3a cos a = 2t = be sin a, i.e.
6 sin a cos a = 4 sin a sin β sin 7,
3 cos a = — 3 cos(/3 + 7) = 2 sin β sin 7,
260

1980/2
1980.
—3 cos β cos 7 + 3 sin β sin 7 = 2 sin β sin 7,
tan β tan 7 = 3,
Thus (8) holds. Now, as all arguments are reversible, (8) implies that the centre
of gravity and the centre of the circumcircle are on the same side of ВС and
they are of the same distance from it, hence the Euler line is parallel to ВС.
1980/2. Define the numbers clq, a\, ..., an in the following way:
1 al
a0 = ^, ak+i=ak + — (n>l, fc = 0, 1, ..., n-l).
2 η
Prove that
(1) 1--<οη<1.
η
Solution. The definition of the a^-s implies that
(2) ад <ai < ... < an.
Examine the difference of the reciprocal of two consecutive terms:
,3ч J L_ = Qfe+1 ~ ak _ al = !
&k CLk+l ak^k+l ( ,ήΛ n + ak
nak \ak+ n I
Since ak<an, if к < η, we obtain from (2) and (3) that
1111
< <
Adding these inequalities for к = 0,1,..., η — 1 we get:
(4) _J!_<±_±<1.
n + an clq an
The right side of our inequality is equivalent to 2 < 1, i.e. to the inequality
On
an < 1 from the proof. The left side of (4) implies that:
1 η
— <2
an n + an
Using an < 1 this means that
1 ~ n ■, 1 , 1 η
— <2 7 = 1 + r<l +
a
η
n + 1 n+1 n—1 n—1'
or
n-l 1 l
an> = 1 .
η η
and this is the left side of the problem.
261

Solutions
1980/3
1980/3. Prove that the equation
(1) xn + l=yn+\
where η is a positive integer not smaller then 2, has no positive integer solution
in χ and у for which χ and η +1 are relatively prime.
Solution. Let us assume to the contrary that (1) has a positive integer
solution x, y, where χ and n +1 are relatively prime. Rewrite (1) in the following
form:
(2) xn = yn+l-l={y-l){yn + yn-l+... + y+l).
We show that у—I and yn + yn +... + y+l are coprime; if they are not, then
there is a prime ρ and integers A and В such that
(3) y-\=Ap, y = Ap + U
(4) yn + yn-1+... + y + l=Bp
holds. Substituting (3) to (4) the terms on the l.h.s give residue 1 mod p, hence
there is an integer С such that the l.h.s. of (4) is of the form Cp + (n + l), that is
(B-C)p = n + 1.
Then ρ divides η +1 and by (2) it divides x, contradicting the assumption that χ
and η + 1 are relatively prime.
Thus у — 1 and yn + yn +...+y+l are relatively prime. As their product
is a full n-th power, both of them are full n-th powers. But, this is impossible as
yn<yn + yn-l+... + y + l<(y + ir
and there is no full n-th power between the n-th powers of two consecutive
integers. This is a contradiction, so we proved our statement.
1980/4. Determine all positive integers η such that the following statement
holds: In the inscribed convex polygon A\A<i... A<in if the pairs of opposite sides
(AlA2,An+lAn+2), (А2А3,Ап+2Ап+з), ···> (Αι-ίΛι, Α2η-\Α2η) ■
are parallel, then the sides
are parallel as well.
First solution. First, we prove that the statement holds for even n-s, but
is false in general for odd n-s.
We precede our proof with two remarks. First: If Αχ, Α<ι, A-$ and B\, B2,
Βί, are consecutive points of a circle such that the two triples have the same
orientation, moreover A1A2 \\ B\B<i and A2A2 \\ B2B2, then the arc А\А^
containing A2 is equal to the arc B\B^ containing B2 (see Figure 1980/4.1).
262

1980/4
1980.
The angles ίΑχΑ^Αι - ίΒχΒ^Βι, because their sides are parallel and so
the angles ΑχΑ^ and ΒχΒ3 are measured by the same arc hence they are equal.
Second: if the arcs AB and CD of a circle are of the same length and
orientation, then the chords AD and ВС are parallel (see Figure 1980/4.2). The
Figure 80/4.1
Figure 80/4.1
condition implies that LACB- /CAD, because they are measured by he same
arc, hence AD || ВС.
Now, assume that in the convex polygon -41^2 II ^η+1^η+2>
A2A3 || An+2An+3, ..., Αη_χΑη \\ Aln_xAln. The convexity implies that
the triples of points ΑχΑιΑ3 and Αη+χΑη+2Αη+3, the triples A2A4A5 and
A„ _ 1 An. A
Ln-
n
>A
2n-2-£j2n-1^2n
A,
Αι+3Αι+4·Αι+5' ··■' moreover the triples An_
are oriented triers ame _wayJ_J^nceby__our first remark ΑχΑ2=Αη+χΑη+2,
Α2Αζ=Αη+2Αη+ζ, .. ■J_An-^An = Α2η-2Αΐη· Since η is odd, we listed all su-
barcs of ΑχΑη and Αη+χΑ2η, hence ΑχΑη = Αη+χΑ2η· Now, our second
remark implies that the sides ΑχΑ2η and ΑηΑη+χ are parallel as well (see Figure
1980/4.3).
We show a counterexample for
even n-s. We show that the parallelity
of η — 1 pairs of sides does not imply
that the remaining two sides are
parallel.
A
Choose the vertices of the polygon
such a way that for the consecutive
pairs of sides the following holds:
A2~An-l
n+2
Figure 80/4.3
ΑχΑ2=Α3Α4=.. .=An.
- 3π
■ΐΛι=;^=Αι+2Αι+3=Αι+4Αι+5=· ■ -=^2n-2^2n-b
A2A3=AAA5=.. =A
n-
2n
-2Αη-\=7Γ=Αη+1Αη+2=Αη+3Αη+4=· · ·=Α2η-\Α2η\
π
Αη,Αγ,Λ.λ — — ΑοηΑ
η
2ηΆ\·
263

Solutions
1980/4
ЗтТ 7Г
Visually, we have arcs of length — (type I.) and — (type II.) alternating
In LTi
between the vertices A\ and An and the vertices An+\ and A2n (see Figure
1980/4.4, where η = 4), and so our vertices are among the vertices of a regular 4n-
gon. Now, AkAk+i and Ап+кАп+к+1 are parallel for к ^n, as by the arrangement
Ak+\An+k = An+k+\Ak holds. On the other hand the parallelity of ΑηΑη+\ and
• ■ —' · ■ П
AinA\ would imply A\An = ΑηΛ.\Αιη. This is false, since A\An consists of —
edges of type I. and
η
edges of type II., hence
--——- η 3π η —2 ..
ΑλΑη = -· — + —— ■ — =
In An+\A2r we have
η
2 2n
π
2n
In-I
2n
-π.
in
-2 η
-— edges of type I. and — edges of type II., thus
η
An+\ A2n = —^— ' — + ~ · — =
3π η
2n 2
π
2n
2n — 3
-7Γ,
2 2n 2 2n 2n
so their length is different and the two sides are not parallel as we stated.
Second solution. Here we provide a simple proof for the first part using
complex numbers. Let a, b, c, d be points on a circle about the origin. It is well
known that the lines through a and b, and through с and d are parallel if and only
if ab = cd.
Consider the polygon Α\Αι·. ·Αιη on the complex plane such that the
centre of the circumcircle is the point O. Denoting the numbers of the vertices
by the corresponding lower case letters, for odd η the conditions reread as:
α1α2 = αη+ιαη+2,
an+2an+3 - α2α3 5
аЪаА~ an+3an+Ai
0'2n-la2n-an-la
n·
Figure 80/4.5
Multiplying the equations and
simplifying the products we obtain:
ala2n = anO'n+li
which means that the sides A\Ain and
AnAn+i are parallel.
This proof did not use the convexity of the polygon, so the statement holds
for self-intersecting polygons, too. For such an example see Figure 1980/4.5.
Remark. The counterexample of the first solution can be formulated with
complex numbers as well, but it requires some extra discussion.
264

1980/5
1980.
1980/5. In a rectangular coordinate system we call a line parallel to the χ
axis triangular if it intersects the curve with equation
4 3 2
y = x +px + qx +rx + s
in the points А, В, С and D (from left to right) such that the segments AB, AC
and AD are the sides of a triangle.
Prove that the lines parallel to the χ axis intersecting the curve in four
distinct points are all triangular or none of them is triangular.
First solution. Let the line e parallel to χ intersect the curve at the points
A, В, С, D, moreover let the coordinates of A be A(t, d). Translate A to the
origin. The equation of the curve in the new system is given by the substitutions
x-^x + t, y-^y + d:
y + d=(x + t)^+ p(x +t)3 + q(x +t)2 + r(x + t) + s,
hence
у = xA + {At + p)x3 + (6t2 + 3pt + q)x2+
+(4t3 + 3pt2 + 2tq + r)x + (i4 +pt3 + qt2 + rt + s)-d.
Let denote the new coefficients by P, Q, R and W, the equation rereads as:
у = xA + Px3 + Qx2 + Rx + S.
As the origin is on the curve, 5 = 0. The χ values corresponding to the points A,
B, С, D are the roots of the equation
x4 + Px3 + Qx2 + Rx = 0
and A belongs to x = 0. Let denote the χ coordinates of B, C, D by t\, t>i, Ц
(t\ <^2 <^з) respectively. These satisfy the equation
x3 + Px2 + Qx + R = 0.
A triangle can be constructed from the segments of length t\, t>i, Ц if and only
if the inequalities —1\ +ti + ts > 0, t\ — ti + ti > 0, t\ +t2 — £3 > hold. The first
two are obviously true, hence the condition is equivalent to the inequality:
(8) (-tl+t2 + t3)(tl-t2+t3)(tl+t2-t3)>0.
The connection between the roots and the coefficients of a polynomial say that
ti+t2 + t3 = —P, tit2 + t2t3+t3t\=Q and t\t2t3 = —R. It is easy to verify by
calculations that the l.h.s of (8) rereads as Ρ — APQ + &R, thus the sufficient
and necessary condition for e to be triangular is
6 = P3-4PQ + SR>0.
Easy calculation shows that
δ = P3 - 4PQ + SR = p3 - 4pq + Sr,
265

Solutions
1980/5
hence δ does not depend on the choice of e. Thus if for a line e the number
δ > 0,then it holds for every line; and if δ < 0, then this holds for all of them.
Hence all lines are triangular at the same time or none of them is triangular.
Second solution. A fairly simple solution can be given using visual
arguments. Observe on Figure 1980/5.1 that as AB <AC <AD, the condition for
the points А, В, С, D being triangular is
Figure 80/5.1
AD<AB + AC, i.e. AB + BC + CD <AB + AB +ВС
which implies CD < AB. The 4-tuple of points is not triangular if CD > AB.
Now, assume that there is a line e such that CD < AB, and another line e' such
that for the points of intersection C'D' >А!В'.
Translating e towards e' there will be a line e* such that for the points of
intersection А*, В*, С*, D* the equation C*D* = A*B* holds. Now, choose the
coordinate system such that χ coincides with e* and the midpoint of ВС is the
origin. Denote by —a, —b, b, a the χ coordinates of the points А*, В*, С*, D*
respectively. The polynomial (with leading coefficient 1) is uniquely determined
by its roots, the equation rewrites as:
y = (x- a)(x + a)(x - b)(x + b) = (x2 - a2 J (x2 - b2j ,
hence the curve is symmetric to the у axis.
But then the intersections of a line parallel to χ are symmetric, too, hence
for every intersecting line e AB-CD contradicting our assumptions.
266

1980/6
1980.
1980/6. Find the digits left and right of the decimal point in the decimal
form of the number
f ^1980
(V2 + V3J
First solution. The examined number changes only a "little bit" if we add
, ν 1980
its reciprocal, (V3 — ν 2 j . Set
, ^4 1980 / г ^\1980
A=[V3 + Vrj +(\/3-\/2j
Now,
(у/З + у/ΐ) =5 + 2\/6 and (y/b-Vij = 5-2л/б,
hence
f ^\990 / ^\990
А=(5 + 2л/б) +(5-2л/б) .
Using the binomial theorem we obtain:
A = 5990 + /990\5989 2v/^+ + /990\5 (2v^}989 +5990_
'9J°W89 · 2л/б +... - ^5 · (2л/б)989 + (2л/б)990 =
= 2-5990 + ... + 2^990V2-2988-6494 + 2-2990.6495.
\988/
Here, every term is an integer, hence A is an integer and all terms except
the last one is divisible by 10, hence the last digit of A equals the last digit of
the last term. The last term is 2991 · 6 . We know that the powers of 6 end by
6 and the powers of 2 end by 2, 4, 8, 6, ..., periodically. As 991 = 4 · 247 + 3,the
last decimal digit of 2991 is 8 and so the last digit of A is the same as the last
digit of 8 · 6, therefore it equals 8.
On the other hand, 5-2\/б<0,2, hence (5-2л/б) <(0,2)990 =
/ \990
0.4495 <0.1495, hence in the decimal form of (5 — 2\/б) there are at least
495 digits that are 0, thus
/ ^ч 1980 / ^ч 990 / г ^ч 1980
(V3 + V2J =(5 + 2л/б) =A-(V3-Vr)
= XXX ... 8 - 0,00... 0ХХ = XXX ... 7,999 ...XX ...
and the digits before and behind the decimal point are ... 7.9 —
Second solution. Along the lines of the previous solution let us examine
the sequence
An = (5 + 2л/б)П + (5 - 2л/б)П .
267

Solutions
1980/6
It is easy to observe that the elements of this series satisfy the equation
An + An+2 = lOAn+i.
Thus the sum of the second consecutive terms of the sequence, is divisible by
10. Since A\ = 10, A<i = 98, the last decimal digits of the terms are
0, 8, 0, 2, 0, 8, 0, 2, ...
The sequence has period 4. As 990 = 247-4 + 2, the number Л990 ends by8. As
seen in the first solution, in (5 — 2v6j the digit 0 stands after the decimal
point. Now
1980
990
990
(л/2 + л/з) =(5 + 2л/б) = А990-(5-2у/б)
/ г ^1980
implies that the digits by the decimal point in ί V2 + V3J are: ... 7,9
1981.
1981/1. Let Ρ he a point inside the triangle ABC and D, E, F are the
feet of the perpendiculars from Ρ to the lines ВС, С А, АВ, respectively. Find
all Ρ which minimise:
(1)
ВС CA AB
First solution. Use the notations of Figure 1981/1.1, let A denote the area
of the triangle and S the sum in (1).
П\
D
a
1 ^""-^
/ '''P
Ж У
I *кУ
Is' OCj
π βΓ^\
в
268

1981/1
1981.
A is the sum of the areas of the triangles APB, BPC and CPA, hence
2T = ax + by + cz,
„ a b с
S = - + - + -.
χ у ζ
S is minimal if ISA is minimal, because A does not depend on the choice of P.
Calculating 1ST and using that the sum of a real number and its reciprocal is at
least 2, we get:
2ST=(- + - + -)(ax + by + cz) = a2+ b2 + c2 + ab (- + -) +
\x у ζ) \y χ)
+ bc(- + -)+ac(- + -)>a2 + b2 + c2 + 2ab + 2bc + 2ca =
\z y) \z x)
= (a + b + c)2,
and equation holds if and only if χ = у = ζ, thus Ρ is the centre of the incircle.
Thus the minimal value of S is:
. (a + b + c)2
mm b = — ,
2Γ
and S attains it when ρ is the centre of the incircle.
Second solution. Use the notations of the first solution. The perpendiculars
and the lines through Ρ together with the vertices of the triangle determine six
right triangles. From the triangles PAF, and PBF we have AF = ζ cot ai, and
FB = ζ cot βι, hence as AF + FB = c,
с
- = cot «2 + cot β\
ζ
follows. Similarly,
a
- = cot βι + cot 7i,
X
b
- = cot 72 + cot a\.
У
From the sum of the latter three inequalities we get:
(2) S = (cot a\ + cot α2) + (cot βχ + cot /%) + (cot71 + cot 72).
Now, let us examine the value of the expression in the first parenthesis. As
αϊ = a — a\
.„. cos oq cos(a —αϊ) sin α
(3) cot αϊ +cota2 = — +
sin αϊ sin(a —αϊ) sin αϊ sin(a — a\)
Since sina\ and sin(a — a\) are positive, Axy <(x + y) implies that
9
. , ч /sinai+sm(a — αι)\ . ο a 9 a — 2a\ . 9 a
sin ai sin(a — a\)<\ I = sin ■=■ cos < sin —
269

Solutions
1981/1
and equality holds if and only if a = 2cq, that is, when PA is the bisector.
Applying it to (3) we get
sin a 2 sin Ц cos Ц a
cot a\ + cot «2 = —~— = ^ = 2 cot —.
sin | siir | 2
We get similar results for the other two parentheses in (2), thus
5>2(cot?+cot^+cot? ) ,
~ V 2 2 2) '
and equality holds if and only if PA, PB, PC are bisectors, hence Ρ is the
centre of the incircle.
Remark. In our second solution the equation с = ζ cot αϊ + ζ cot β\ holds
even if the angle β\ is obtuse, and the length of с is the difference of the length
of two sides, as cot/3i <0.
1981/2. Consider all subsets of size r of the set Hn = {1, 2, ... , n}, where
1 <r <n. Each subset has a minimal element, let F(n, r) denote the arithmetic
mean of these elements. Prove that
F(n,r) = -.
r + 1
First solution. Let S(n, r) denote the sum of the smallest elements from
the r element subsets of Hn. We prove that for η > 1 and r > 1
(1) 5(n+l,r) = 5(n,r-l) + 5(n,r).
Let us order the elements of every subset in increasing order. We can split
the r element subsets of Hn+\ into two disjoint groups as it contains the number
η +1 or it does not. If we omit η +1 from the subsets of the first group, we
get the set of r — 1 element subsets of Hn, and the second group is the set of r
element subsets of Hn. Thus (1) holds.
It is easy to observe that this formula is the same that holds for the binomial
coefficients. We prove by induction that
(2) S^>=("l!
If r = 1, the subsets of Hn are the sets {1}, {2}, ..., {n} hence
n(n+l) /n+l\ (η+Ϋ
5(^1) = -^-=^ 2 ) = {r + 1
and so (2) holds in case of r = 1 for every η and so, in case n= 1, too. It holds
in case of η = 2, r = 2, as #2 = {!> 2} is the unique two element subset of itself,
270

1981/2
1981.
thus 5(2,2) = 1, as in (2). Now, assume the (2) holds for until some η in case
1 <r <n. Combining (1) with the formula for the binomial coefficients gives
that
(3)
or 1 ч In+1\ ln+1
thus (2) holds under our assumptions.
As there are I J r-element subsets of Hn,
(4)
F(n,r) =
(n + 1)!
n!
n+1
(r + l)!(n —r)! r\(n — r)\ r + 1'
so we proved the statement.
Second solution. Let 5(n, r) denote the sum of the smallest elements from
the r element subsets of Hn. Assume that the elements of the subsets are in
increasing order. The possible values of the first elements are: 1,2, ..., η — r + 1.
The positive integer к is the smallest element of a subset, if the elements
were chosen from the elements k, k +1, ..., η but we chose к along with r — 1
other elements from the n — k elements left. The sum of the first elements for
η — к
these subsets is k\
r
1
and so
n—r+1
k=l
This sum is
S(n,r) =
η
r
Now, adding up the sum by the rows, using
n—k
(5) Σ
1=1
+
r
Τ
1+
+
271

Solutions
1981/2
gives
s(n,r)=i™]+r ί+γ.")+·■■+
Since the number of the subsets is [ J, the arithmetic mean, (as in the first
solution) is:
F(n,r) =
n+1
r + 1
that we wanted to prove.
Third solution. Let Rr denote the r element subsets of Hn and R'r+i the
r + 1 element subsets of H'n = {0, 1, 2, ..., n}.
Let us assume that к is the smallest number in an element of Rr (fc = l,
2, .. .)■ If we add one of the elements 0,1, 2, ..., к — 1 to the set, then we
get an element of R'r+\, and we can get exactly к elements of Rr+\ doing this
way. Obviously, every element of R'r+^ can be obtained in a unique way from
an element of Rr (by omitting the smallest element of it). This means that the
number of the elements R'r+i equals the sum of the smallest elements of the
elements of Rr.
So we determine the required arithmetic value on the following way:
the number of elements of R' +1
' the number of elements of Rr
Hence by (1)
F(n,r) = -.
r + 1
Remark. In our first solution we used the formulas (3) and (5) of the
binomial coefficients.
2. With a slight modification of our arguments we can prove that the
arithmetic mean of the largest elements is
n+ 1
r
r + 1
0 0
1981/3. Determine the maximum value of m + η where m and η are
integers satisfying m, η ε {1, 2, ..., 1981} and
9 9 9
(1) (n —nm — m) =1.
272

1981/4
1981.
9 9 9 9
Solution. First, m>n cannot be the case as it implies m >n ,n —m <
9 9
— 1, —nm< — 1, η —nm — m < —2, contradicting (1).
If n = m, the only solution of (1) is m = n= 1.
In case m<n, n — m is a positive integer. We show that if (n, m) is a
solution of (1), then (m,n — m) is a solution of (1), too:
im — m(n — m) — (n — m) j = (— (n — nm — m )j =1.
This means that if (n,m) is a solution, then m<n and the pair (n',mf) is a
solution, too, where
П =777-
(2)
m =n — m.
Reversing (2) we get that if (n', rri) is a solution, where m < n', then (n, m)
is a solution, too, where
n = m' + n'
(3)
m = n .
(3) implies that starting with the pair (1,1) we get infinitely many solutions:
(4) (1,1), (2,1), (3,2), (5,3), ..., (1597,987), (2584,1597), ...
We show that in (4) we listed all solutions of (1). Indeed, let (n, m) be a
solution; if m<n, then by (2) we can find a "smaller" solution (η',τττ,')· The
procedure terminates at a pair, where m' = n; but this has to be the pair n'=
m' = l. As a pair determines all the solutions given by (2) and (3), (4) provides
all the solutions. As in (4) the pair (1597,987) is the largest not exceeding 1981,
the maximum of η +m is:
15972 + 9872 = 3 524 578.
Remark. (3) implies that the numbers occurring in the solutions are the
Fibonacci numbers defined by oo = l, οι = 1, αη = αη_\ +αη_2· Thus the pairs
are
(ai,a0), (a2,ai), (a3,a2), ·.·, (an,an_i), ...
1981/4. a) For which η > 2 is there a set of η consecutive positive integers
such that the largest number in the set is a divisor of the least common multiple
of the remaining n— 1 numbers?
b) For which η > 2 is there exactly one set having this property?
First solution. Let first n = 3 and the first three consecutive integers be:
s — 2, s — 1, s. Since s and s— 1 are coprime, s can divide (s —2)(s— 1) only
273

Solutions
1981/4
if it divides s — 2, hence s = 1 or 2, but then the 3 numbers are not all positive,
thus for η = 3 the requested numbers do not exist.
Now, let η = 4, and the four consecutive numbers: s — 3, s — 2, s — 1, s. If
s divides the least common multiple of the other three numbers, then it divides
their product. It is coprime to s — 1 so it divides (s — 3)(s — 2). As
(s-3)(s-2) = s(s-5) + 6,
s divides 6. s^l, 2, 3 because then 0 were among the 4 numbers. Hence s = 6.
Then the 4-tuple is: 3, 4, 5, 6, and 6 divides 60, the least common multiple of
the other three numbers.
g
Finally, let η > 5. Choose s to be s = (n - l)(n - 2) and let к = -. We shall
provide two η-tuples of consecutive integers:
a) s —n+1, s — η+ 2, ..., s— 1, s;
β) к —n + 1, к —η+ 2, ..., к —I, к.
These numbers are positive: In case a) s — n+l = (n— l)(n — 2) — η + 1 =
(η — 3)(n — 1) > 8 since η > 5.
In case /3)
(n-l)(n-2) ^fo-5) ,^o
k — n+ 1 = n + 1 = +2 >2.
2 2 _
As a) and β) are η consecutive numbers, they contain numbers divisible by
η — 1 and η — 2. As η — 1 and η — 2 are coprime, their least common multiple is
divisible by (n — l)(n — 2), hence by s and k, as well.
In conclusion there is no solution for n = 3, there is a unique solution for
η = 4, and for η > 5 there are at least two solutions.
Second solution. Let s denote the largest of the η consecutive numbers.
Let n = 3; Now, our numbers are: s — 2, s — 1, s. If ρ is a prime divisor of
s, then (since s and s — 1 are coprime), ρ divides s — 2, and hence ρ divides 2,
too. Thus ρ = 2. The power of ρ in s is at most 1, because in other case 4 does
not divide s - 2, thus the only possibility is s = 2, but then, s - 2 = 0 would hold
and so there is no solution for n = 3.
Let η = 4; then s > 3, and our numbers are: s — 3, s — 2, s — 1, s. A prime
divisor ρ of s can only divide s — 2 or s — 3 hence ρ is 2 or 3 and they can only
occur on at most the first power in s. The only possible choice for s is 2 · 3 = 6.
And, indeed, in the 4-tuple 3, 4, 5, 6 we have that 6 divides 3-4-5.
Now, let η > 5. Let us put η between two powers of 2, thus let r be a
positive integer such that
2r<n<2r+1.
274

1981/5
1981.
Choose s = 3 · 21 or s = 5 · 2r, the η consecutive integers are:
(1) s+l — n, s + 2 — n, ..., s —1, s.
The lists in (1) consist of positive integers as, e.g., in the first case η < 2 · 2r <
3 ■ 2r = s, s — η > 0, and hence s +1 — η is a positive integer.
As η > 2r and η > 3, the integers s — 2r and s — 3 are in (1) if s = 3 · 2r. As
2r and 3 are coprime, the least common multiple of the first η — 1 numbers in
the first η-tuple is divisible by 3 ■ 2r = s. Similarly, if s = 5 · 2r, then for η > 5 the
least common multiple is divisible by 5 · 21 = s.
Summarizing: if n = 3 then there is no solution, in case n = 4 there is a
unique solution, and for η > 5 there are at least two solutions.
1981/5. Three circles of equal radii have a common point О and lie inside
a given triangle. Each circle touches a pair of sides of the triangle. Prove that
the incentre and the circumcentre of the triangle are collinear with the point O.
Solution. Let А, В, С denote the vertices of the triangle and А', В', С' the
centre of the circles touching the pair of sides originating from the corresponding
vertex. As these centres are of the same distance from 2-2 sides, they lie on
the appropriate bisectors and the sides of the triangles ABC and A'B'C' are
parallel. (See Figure 1981/5.1). Consequently, the triangles ABC and A'B'C'
Figure 81/5.1
275

Solutions
1981/5
are homothetic and the centre of the similitude is K, the incentre. О is of the
same distance from the vertices А', В', C' hence it is the circumcentre of the
triangle A'B'C'.
The enlargement with centre К mapping the triangle A'B'C' to the triangle
ABC maps О to the circumcentre O* of the triangle, ABC hence К, О, and O*
are collinear.
1981/6. The function f(x,y) satisfies
(1) /(0,j/) = j/ + 1,
(2) /(a + 1,0) =/0c, 1),
(3) /(rc + l,J/+l) = /(ic,/(ic + l,J/))
for every integer χ and y. Find /(4,1981).
Solution. Let у be an arbitrary positive integer. We want to find the values
/(1,2/), /(2,2/), /(3,2/); for simplicity we write over the equation sign the label
of the identity we use.
(4) f(l,0)(=)№l)®2.
/(l,2/+l)(=)/(0,/(l,2/))(=/(l,2/) + l(=
(=/(0,/(l,2/-l)) + l(=/(l,2/-l) + 2,
thus
/(l,2/) = /(l,J/-l) + l.
Repeating our arguments we obtain:
(5) /(l,2/) = /(l,0) + 2/=)/(0,l) + 2/=)2/ + 2.
Now, we determine the value of /(2, y):
/(2,2/)(=/(l,/(2,2/-l))(=/(2,2/-l) + 2.
Iterating the procedure we get:
(6) /(2,2/) = /(2,0) + 22/(=)/(l,l) + 22/(=)22/ + 3.
We continue with /(3,2/):
/(3,2/)(=)/(2,/(3,2/-D)=)2/(3,2/-l) + 3.
Repeating the arguments gives
/(3,2/) = 2(2/(3,2/-2) + 3) + 3 = 22/(3,2/-2) + 3 + 2·3 = ...=
= 3 + 2-3 + 22-3 + ... + 2y"1-3 + 2y/(3,0)(=)
(=)3(2y-l) + 2y/(2,l)(=3(2y-l) + 2y-5,
276

1982/1
1982.
(7) Ю,у) = 2У+3-3.
Finally, we determine the value of 7(4, y):
КА^Ю^^у-1))^2^У-^ -^
(7) 2(2/(4'^-2)+3-3)+3 _ 3 = 22/(4-2/-2)+3 _ 3
Repeating the arguments we get
2/(4,0)+3
№y) = 222' -3,
and there are у many 2-s on the r.h.s. As
/(4,0) (=)ДЗ,1)=)24-3 = 222-3,
2
Д4,т/) = 222' -3 (with у + 3 = 1984 2-s).
Remark. The function /(ж, y) in the problem is a so called doubly
recurrence function. The recurrence functions are functions defined on the nonnegative
integers attaining nonnegative integer values that are built up from some basic
functions with finitely many substitutions.
Our f(x, y) is a well known function introduced by W. Ackermann in 1928.
This form of the function is due to R. Peter (1934).
1982.
1982/1. The function f(ri) is defined on the positive integers and takes on
non-negative integer values. For all n, m
(1) f(m + ri) — f(m) — /(n) = 0 or 1,
(2) /(2) = 0, /(3)>0,
(3) /(9999) = 3333.
Determine 7(1982).
First solution. Let m = n= 1, then from (1) it follows that
7(2)-27(1) = 0 or 1,
(2) implies that 2/(l) = 0 or -1, hence /(1) = 0.
Similarly, if m = 2, η = 1
7(3)-7(2)-7(D=7(3)=o or l,
277

Solutions
1982/1
and by (2)we have /(3) = 1. Using induction we show that for every positive
integer к
(4) f(3k)>k.
Assume that (4) holds until some k. Let m = 3k, n = 3, then (1) implies
Ю(к +1)) = f(3k+3) > f(3k) + /(3) > к +1,
hence (4) holds for every k. If in the last step f(3k) > к then f(3(k +1)) > к +1
holds, that is, if for some ко we have f(3ko) > ко in (4), then for every k>ko
f(3k) > к holds, as well. This observation implies that in case к < 3333, f(3k) >
к cannot hold as then /(3 ■ 3333) > 3333 follows, contradicting (3); therefore
/(3 · 1982) = 1982. Now, (1) implies
1982 = /(3 · 1982) = /(2 ■ 1982 +1982) > /(2 ■ 1982) + /(1982),
and
Л1982+1982) > 2/( 1982).
Combining the last two results we get:
1982
(5) 1982>ЗД1982), Д1982)<——<661.
Using (1) again
Л1982) = Д1980 + 2)> /(1980) + /(2) = /(3 -660) + 0 = 660,
thus from (5)
660 </(1982) < 661,
and so
/(1982) = 660.
Second solution. We show that the value of /(1982) can be determined
without condition (2). (1) can be rewritten in the following form:
(6) /(m) + f{n) < f(m + n) < f(m) + /(n) + 1.
We prove by induction that
(7) xf(y)<f(xy)<xf(y) + x-h
where χ and у are positive integers.
For x= 1 and у arbitrary the statement is obvious. Assume that (7) holds
for x— 1, that is
(8) (x - \)f{y) < f((x - Dj,) < (x - \)f{y) + x-2.
From (6) with substitutions m = (x — \)y and n = y
(9) /((a:-l)j/) + /(j/)</(a;j/)</((a;-l)j/) + /(j/) + l.
278

1982/2
1982.
The left inequality of (8) implies:
№ -1)?/) + Ky) >(χ- i)/(j/) + f(y) = xf(y).
The right inequality of (8) implies:
f({x-\)y) + f(y) + l<(x-l)f(y) + f(y) + l + x-2 = xf(y)±x-l.
The two latter results combined with (9) gives the required (7).
Now, apply (7) for χ = 1982, у = 9999 and let Ρ denote f(xy):
(10) 1982 · 3333 < Ρ < 1982 · 3333 +1982
Now, we substitute χ = 9999 and у = 1982 into (7):
(11) 9999 ■ Л1982) < Ρ < 9999 · /(1982) + 9998
Comparing the l.h.s of (10) with the r.h.s of (11) and the r.h.s of (10) with the
l.h.s of (11) we get that
1982-3333-9998
/(1982) >
/(1982) <
9999
1982-3333 + 1982
9999
= 659,66...
= 660,86...
and since /(1982) is an integer, we obtain
/(1982) = 660.
1982/2. A non-isosceles triangle A\A2Ai, has sides щ opposite to A{. Mi
is the midpoint of side сц and Ti is the point where the incircle touches side щ.
Denote by Si the reflection of Ti in the interior bisector of angle Ai. Prove
that the lines M\S\, M2S2 and Mi,Si, are concurrent.
Solution. Under the assumptions of the problem, all upcoming points are
distinct; Indeed, if e.g. S\ coincides with S^, then the central angle belonging to
the arc T\T2 containing S\ is twice the angle of the bisectors of the angles A\
and A2, 90° + -, what is impossible (see Figure 1982/2.1).
Figure 82/2.1
279

Solutions
1982/2
Since S\ and T\, moreover Γ3 and T2 are symmetric to the bisector of A\,
we have TXT2 = SxTy, Similarly, TXT2 = S2Ty, hence SXT3 = S2T3.
The tangent of the circumcircle of the isosceles triangle Si S2T3 (that is the
incircle of the triangle A\A2A3) at the point Γ3 and the side S\S2 are parallel,
so A\A2 is parallel to S\S2. But the median M\M2 and A\A2 are parallel too,
hence S\S2 is parallel to M\M2. Similar arguments show that the segments
52 £3 and M2M3, and the segments S3S\ and M3M\ are parallel. Hence the
sides of the triangles S\S2S2 and MiM2M3 are pairwise parallel, so the two
triangles are similar.
As in this case the circumcircle of S\ S2S3 is smaller than the circle
through the midpoints of the sides, the triangle S\S2Si, is smaller than the triangle
МХМ2МЪ, hence the lines MXS\, M2S2, M3S3 concur in K, the centre of the
enlargement, and this is what we wanted to prove.
Remark. The enlargement with centre К maps the two circumcircles, that is,
the incircle and the Feuerbach circle to one another. By the theorem of Feuerbach
the incircle and the Feuerbach circle are touching, the touching point is at the
same time the centre of the enlargement, so К lies on both circles.
This observation simplifies the construction of the touching point of the two
circles. Note that the observations can be completed to a proof of Feuerbach's
theorem.
1982/3. Consider the infinite поп increasing sequence {xi} of positive reals
such that xq = 1.
a) Prove that for every such sequence there is an η > 1, such that
9 9 2
xf\ xi x„ 1
(1) Sn = -^ + ^- + ... + -^> 3,999.
b) Find such a sequence for which
5n = i0 + ii... + fLi<4.
Ju 1 »//0 *^7Ъ
for all n.
First solution. As (xi — 2xi+\)2 > 0, xf > 4xiXi+\ — 4xf+l follows, and so
,2
>4(xi-xi+i).
xi
xi+l
Applying this to the sum in (1):'
2 2 2
Sn = — + — + ... + -?-!>4(xo-^i + ^i -x2 + ... + xn-\ -xn) = 4(l-xn).
X\ X2 Xn
280

1982/3
1982.
If the limit of the sequence xn is 0, then there exists an n, such that xn <
and for this η
4000'
Sn>A 1
1
"n
4000.
= 3,999.
Now, let the limit с be different from 0, с > b > 0, so every element of the
4
sequence is greater than b. Let n>-. Then
π 2 η η
-0 Жг+1 ^=0 Ж*+1 г=0
г=и ■" * г
and the statement is true.
For b) let Xi = 2~l (i = 0,1,...). Now
uL-j5_ = 2i-<
and so
si+i 2-(*+D
Sn = 2+1 + - + ... +
2n-l
Extending it to an infinite geometric sequence the sum is 4, hence the finite sum
S never reaches 4.
Second solution. Introduce the notation
xk-l
ak = (& = 1, 2, ...)
here α/c > 1 and
a2 аз ^ ^
a
η
CL\ CL\CL2 а\&2 ··· an—l
1 / 1
1
= aj Η Q-2 "^ I аз + ... Η
a\ \ (*2 \ an-2
В
αη_ι +
a
η
an-\
Apply now the inequality a^ + — >2vB that is a consequence of the A.M.-
G.M. inequality if В > 0:
Sn>2
1
1
а
а2 + — а3 + ... + αη_ι +
η
>...>
\| α2
атг-2
αη-\
>2
Ν Ν
.. .2 /оэт_1 +
α
η
>
αη-\
l+i+i+...+- J
>2A/2J...2v^>2V2v/...\/2 = 2 2 4 ' 2»-1 .
281

Solutions
1982/3
The limit of the sum in the exponent is 2, so Sn can get arbitrarily close to 4,
hence we proved a). (Note that in case n = 14 Su > 3,999 is satisfied.)
See the first solution for the answer for b).
1982/4. Prove that if η is a positive integer such that the equation
3 9 3
x — 3xy+y =n
has a solution in integers x, y, then it has at least three such solutions. Show
that the equation has no solutions in integers for η = 2891.
Solution. As
χ — 3xy +y =(y — x) — 3(y — x)x — χ ,
if (ж,у) is a solution of our equation, then (x\,y\) is a solution, too, where
x\ =—x + y,
(1)
y\ = -x.
Similarly, with the pair (x\,yi), the pair (ж2>2/2) *s a solution, as well, where,
by(l)
x2 =-x\+y\ =-y,
У2 =-si =х-У-
The equality of (x,y) and (x\,y\) means that y = 2x and y = —x, implying x =
y = 0; hence the transformation in (1) gives a new pair, different from (x,y).
(1) gives (Х2,У2) fr°m (х1^Уг) an<i (Х)У) from (ж2>2/2)> tnus tne tnree Pa^rs °^
solutions are distinct, hence we have at least three solutions.
If η = 2891, the equation has no solution. We shall prove it by examining
the divisibility of χ and у by 3; 2891 gives remainder 2 mod 3 and mod 9. We
show that this cannot be the case on the l.h.s.
a) If χ and у are of the form 3k, then the l.h.s. is divisible by 3.
b) If x = 3k, y = 3k± 1, the remainder of the l.h.s mod 3 is ±1, the same
holds if χ = Зк ± 1 and y = 3k.
c) If χ = 3k +1, y = 3k — l (or in the reverse order), 3 divides x + y and the
9 9 9
l.h.s too, as it can be rewritten as (x + y)(x —xy + y ) — 3xy .
d) If гс = 3& + 1, y = 3k+l, the remainder of the l.h.s is 8 mod 9.
e) If χ = 3k — 1, y = 3k — l, the remainder of the l.h.s is 1 mod 9.
Thus we proved our statement.
1982/5. The diagonals AC and С Ε of the regular hexagon ABCDEF
are divided by inner points Μ and N respectively, so that:
AM CN
(1) ^с = сё=г-
Determine r, if Β, Μ and N are collinear.
282

1982/5
1982.
First solution. Choose the length of the diagonals AC = С Е as unit. In
this case (1) means that
AM = CN = r.
With this choice the length of the side of the hexagon and the radius of the
circumcircle: £ = —7= (see Figure 1982/5.1).
V3
Figure 82/5.1
The rotation around the centre of the hexagon by 120° maps the
triangle ВАС to DCE, the segment BM to DN, so LBND = 120°. Thus BD is
subtended by an angle of 120° from N and also from the centre О of the
circumcircle of the hexagon. As LBCD- 120°, the points Β, Ν and D lie on the
circle with centre С and radius BC = CD = q\ thus
Second solution. Let the length of the diagonals AC=CE be 1, then the
length of the side of the hexagon is: £> = —, and let G denote the intersection
v3
of the lines EC and AB. With this choice (1) can be rewritten as (see Figure
1982/5.2):
AM = CN = r.
In the triangle ACG we have IA = 30°; LC = 120° that implies ZG = 30°. Hence
it is an isosceles triangle: AC = CG=1, AG = v3 = 3ρ.
283

Solutions
1982/5
Figure 82/5.2
Apply Menelaus' theorem to the triangle ACG and secant BM:
AB GN CM ρ 1+r 1-r
1 =
BG NC ΜΑ 2ρ r
r
hence
3r2 = l
r =
Vs
= Q-
Third solution. Let В be the origin of the coordinate system and denote
the vectors by boldface lower case letters corresponding to the points labelled by
capital letters (see Figure 1982/5.2).
The vector pointing to О is a + c, hence e = 2(a + c). By the conditions:
m = (l — r)a + rc, n = (l — r)c + re = 2ra + (l + r)c.
As m and η are parallel, there is a real λ such that η = Am = λ(1 — r)a + Arc.
As the decomposition of η into components parallel to a and с is unique, we
have
A(l-r) = 2r, Ar = l+r,
1 — r 2r
r
1+r
3r2 = l.
1
r -
vT
Remark. 1. The third solution shows that we did not use the regularity of
the hexagon in full. Indeed, the statement holds for every affine-regular hexagon,
the projective image of a regular hexagon.
2. The structure of the problem came from the diagonals of a regular
dodecahedron. Its vertices are the midpoints of the arcs over the sides of the hexagon.
This shows the background of the problem and suggests other ways to solve it
(see Figure 1982/5.3).
284

1982/6
1982.
1982/6. Let S be a square with sides length 100. Let L be a path within S
which does not meet itself and which is composed of line segments AqA\, A\A<i,
..., An_xAn, where A0^An.
Suppose that for every point Ρ on the boundary of S there is a point of L
of distance from Ρ no greater than 1/2. Prove that there are two points X and
Υ of L such that the distance between X and Υ is not greater than 1 and the
length of the part of L which lies between X and Υ is not smaller than 198.
Solution. In our
solution we shall use that
the points of distance mo-
re than - from the seg-
2 &
ment RS lie outside of
a "stadium-shaped" region
(see figure at the solution of
problem 1973/4), therefore
the points of a line g, or e
that are of distance at most Figure 82/6.1
- from RS are the intersec-
2
tion of the line and the
region, that is a close segment or a single point (see Figure 1982/6.1).
285

Solutions
1982/6
Assume that starting from Aq the path L approaches first A from among
the vertices of S at a distance not greater than -. Then, at the point Μ the path
L approaches D at a distance of at least 1/2, DM < -, before it approaches B.
Μ divides L into two parts, to the polygonal paths AqM and MAn (see Figure
1982/6.2).
Figure 82/6.2
Now, colour red those vertices of the sides of the square that are of
distance at most 1/2 from AqM. According to our remark, this is the union of (not
necessarily disjoint) closed segments and individual points.
Similarly, colour green the points that are not farther from MAn than 1/2.
A is red and В is green because by the definition of M, we have that AqM
approaches A but AqM does not approach B, hence MAn approaches it.
By our colouring rules and the conditions of the problem, every point of AB
is coloured by red or green, and there may be points coloured by both colours. As
the monochromatic points consist of closed segments, there is a bicolour point
of AB.
Since Q is red, there is a point X on the polygonal path AqM such that
QX < -, and as Q is green, too, there is a point Υ on the polygonal path MAn
such that QY < -. Hence by the triangle inequality:
XY<QX + QY<1.
286

1983/1
1983.
The distances of the points Χ, Υ from AB, and the distance Μ from CD
is at most 1/2, hence XM > 99 and MY > 99, thus
XM < XM polygonal path ; MY < MY polygonal path,
and so XMY polygonal path =XM polygonal path +MY polygonal path >
ΧΜ + ΥΜ > 198, and this is what we wanted to prove.
1983.
1983/1. Find all functions f defined on the set of positive reals which take
positive real values and satisfy:
(i) f(xf(y)) = yf(x)
for every positive χ and y. Show that
(2) /O)^0, if z^oo.
Solution. Let the positive number a call a fix point of / if f(a) = a. f has
a fixpoint: since by (1) xf(x) = f(xf(x)), the number xf(x) is a fixpoint. Let b
be a fixpoint:
(3) f(b) = b.
We prove that bn (n positive integer) is a fixpoint, too. For n = 1 this is (3).
Now, assume that /(6n_1) = 6n_1; (1) implies
f(bn) = f(bf(bn-1)) = bn-l-f(b) = bn-1-b = bn;
thus bn is a fixpoint, too. Applying (1), again:
Ь =/(b) = /(1·δ) = /(1·/(b)) = 6/(1),
and by b > 0
(4) /(1) = 1
follows, thus 1 is a fixpoint. Hence
-'(1|-'(H"'(i'*>)-"©'
'GK
1 1
Thus - is a fixpoint, and by the preceding arguments — is a fixpoint, too.
This means that at the points
(5) b,b2,b3,...,bn,...
f attains the values
287

Solutions
1983/1
and at the points
(6)
the function / attains
b b2 b3 bn
_1 1 _1_ 1
Ъ' Ψ ψ "" W
\_ _1_ J_ 1
Ъ' 62' 63'···'^'·'·
If 6 > 1, (5) contradicts (2); if b < 1, (6) contradicts (2), thus b= 1 is the unique
fixpoint and so for every positive χ
xf(x) = l, that is f(x) = —.
χ
f(x) = — is the only possible solution, and it satisfies the conditions of the prob-
x
lem:
f(xf(y)) = f(-) = - and 2//(2θ = 2/·± = ^
\y J χ xx
and if
1983/2. Lei Л be one of the two distinct points of intersection of two
unequal coplanar circles C\ and Ci with centres 0\ and О2, respectively. One of
the common tangents to the circles touches C\ at P\, Ci at Pi while the other
touches C\ at Q\ and Ci at Qi· Let M\ be the midpoint of P\Q\ and Mi be
the midpoint of P2Q2- Prove that Ю\А02 = ίΜχΑΜ2.
First solution. We shall prove that
(1) ЮхАМх = ШгАМъ
which is equivalent to the statement. Let К denote the point of intersection of
the two tangents, this is the centre of similitude of the two circles (see Figure
1983/2.1).
The enlargement with centre К maps C\ to C2, 0\ to 02, P\ and Q\
to P2 and Q2, respectively. Hence the image of M\ is M2; let A' denote the
image of A.
The enlargement preserves the angles, so LO\AM\ = L02A2M2, hence it
is enough to prove that
(2) Ю2АМ2 = L02A2M2.
Observe that the power of К with respect to the circle C2 is KP2 = К А ■
KA2. Applying it to the right triangle KP202 we get KP2 = KM2 ■ K02, hence
КА-КА2 = КМ2-КОъ
288

1983/2
1983.
Figure 83/2.1
therefore the points ΛΑ^Ο^Μι lie on the same circle, thus A and A^ are the
angles of circumference of the chord M^O^, consequently (2) and so (1) holds.
We proved the statement.
Figure 83/2.2
Second solution. As in the first solution we shall prove that the equality
(1) holds. Let F denote the point of intersection of P\P2 and h, the common
289

Solutions
1983/2
chord of C\ and C2. F is the midpoint of P\P2, because its power with respect
to the two circles is the same, hence h is the median of the symmetric trapezium
F\Q\Q2P2, and s0 ^ is ше perpendicular bisector of the segment M\M2 (see
Figure 1983/2.2).
Let 0\ denote the reflection of 0\ to h, then LO\AM\ = LO\AM2. In order
to prove (1) we have to show that LO\AM2 = L02AM2, that is AM2 is a bisector
of the triangle 0[A02. For this, it is enough to prove that
A02 M202
(3)
AO[ 0[M2
The ratio of the enlargement mapping C\ to C2 is the ratio of their radii
-——·. Because of the reflection, АО] =АО'и hence the ratio is —r-г. As the
АО ι ι ΑΟ[
enlargement maps M\0\ to M202 and by the symmetry ΜχΟχ =0[M2,
A02 _ M202 _ M2Q2
~Щ~ MxOx ~ 0\M2
hence we proved (3), and so the statement of the theorem.
1983/3. Let a, b, с denote pairwise coprime positive integers. Prove that
2abc — ab — bc — ca
is the largest integer which cannot be expressed as
(1) xbc + yca + zab,
where x, y, ζ are поп negative integers.
First solution. Let S = 2abc — ab — bc — ca. First we show that S cannot
be written in the desired form. Indeed,
2abc — ab — bc — ca = xbc + yea + zab,
implies
(2) 2abc = (x + l)bc + (y + l)ca + (z + l)ab.
As a divides the l.h.s, it divides the r.h.s. so it has to divide x + 1, hence there
is a positive integer a such that χ +1 = aa holds. Similarly, there are positive
integers b' and с such that
y+l = bb', z+l = cc.
Substituting these values to (2) and dividing by abc we get that
2 = a' + 6' + c'.
That is impossible as 2 is not the sum of three positive integers.
290

1983/3
1983.
Now, we have to show that for an arbitrary positive integer к there are
positive integers x, y, ζ such that:
labc + k = xbc + yea + zab.
Consider the triples (x,y, z), where
(3) \<x<a, l<y<b, \<z<c.
In the set Η of triples under (1) with these coefficients there are abc many
elements. The remainder of any two elements of Η mod abc is different. Indeed, if
the remainder using the triples (x\,y\, z\) and (x2,y2,z2) as coefficients agree,
consider the difference of the two numbers:
(xx -x2)bc+(yx -y2)ca + (zx - z2)ab.
As a, b, с are coprime, the difference is divisible by a, b and с Now, by (3) it
can only hold if xx = x2, y\ = y2, z2 = z2.
Thus all mod abc remainders occur in H. Now, let (x,y,z) be a triple
satisfying conditions (3), such that the expression in (1) gives the same residue as
k. abc divides к— (xbc + yea + zab) and so k —(xbc + yea + zba) +labc is
divisible by abc, too, thus there is a positive integer η such that
к — (xbc + yea + zab) + labc = nabc,
that is
(4) labc + k = nabc + xbc + yca + zab.
Using values x, y, ζ satisfying (3) we get:
xbc + yca + zab <3abc,
hence (4) implies
к < nabc — labc + 3abc = (n + l)abc,
as к > 0 and η > 0. So (4) can be rewritten as:
labc +k = (na + x)bc + yea + zab,
hence with the positive integers na + x = x', y, ζ we presented labc+k in the
desired form.
Second solution. S = labc -ab — bc — ca is not of the form xbc + yea +
zab. Indeed, in case
xbc + yea + zab = labc — ab — bc — ca,
(5) (x + l)bc + (y + l)ca + (z + l)ab = labc
291

Solutions
1983/3
holds. As, for example, be and a are coprime, a divides χ + 1 hence a < χ + 1;
similarly, b<у + 1 and c<z+l, implying
labc = (x + l)bc + (y + 1 )ca + (z + l)ab> 3abc
that is impossible.
Now, we prove that for an arbitrary positive integer k' there are x, y, ζ
positive integers such that
(6) xbc + yca + zab = 2abc+k'
holds. Or, equivalently, if A; is a positive integer such that к > labc, then it is of
the form
(7) xbc + yca-l· zab = k
where x, y, ζ are positive integers.
Observe that the numbers
be, 26c, 36c, ..., (a — l)6c, абс
give distinct residues mod a since there are no two numbers among them with
their difference divisible by a. Hence they form a complete residue system, and
one of them, e.g. x\bc gives the same residue mod a as k. Thus
x\bc=k (mod a), \<x\<a.
Similarly, we get the positive integers y\, z\ such that
y\ca = k (mod b) l<y\<b,
z\ab = k (mode) 1 < z\ < b.
This implies that e.g.
{x\bc— k) + y\ca + z\ab = x\bc + y\ca+ z\ab — к
is divisible by a, and similarly, by b and c, too; moreover, as these numbers are
coprime, it is divisible by abc, thus
s = x\bc + y\ca + z\ab = k (mod abc).
This also means that s and к (thus s— 1 and к — 1) give the same residue
mod abc, that is
(8) к — 1 =q -abc + r,
(9) s-l = q' -abc + r (0<r <abc),
ι
where, к > labc implies q > 1, and the assumptions for x\,y\, z\ imply s < 3abc,
and so q' < 2.
292

1983/4
1983.
Considering the difference of (8) and (9) we obtain
k — s = (q — q')abc, (<7 — <?'> 0)
k = s + (q — q )abc = {x\ +(q — q')a)bc + y\ca + z\ab,
hence choosing x = x\+(q — q')a, y-y\, z-z\ we get (7) and this is what we
wanted.
Remark. Similarly, it can be proved that if the positive integers a\, a>i, ...,
an are pairwise coprime and s = a\a2 ... an, then the largest number that is not
of the form
s s
x\ — + X2— +
+ xn
s
with x\, X2, ..., Xk integers is
/ 14 / S S S
(n-l)s- [ — + — +... + —
\a\ 0*2 a
η,
1983/4. Let ABC be an equilateral triangle and Ε the set of all points
contained in the three segments AB, ВС and С A (including А, В and C).
Determine whether, for every partition of Ε into two disjoint subsets, at least one
of the two subsets contains the vertices of a right-angled triangle.
Solution. We show that
the answer to the statement is
affirmative. Let P\, P2, P3, P4.,
P$, Pg denote the points
dividing the sides AB, ВС, С A
into three equal parts, respectively.
These points are the vertices of
a regular hexagon. We shall use
only these vertices along with
the vertices of the triangle. We
show that these nine points
cannot be divided into two parts
avoiding a monochromatic right A
triangle, (see Figure 1983/4.1).
For simplicity, let us colour
the points of the two sets by
green and red. As two of the points А, В, С have the same colour, we may assume
that A and В are red. Considering all possible colourings of the points P\ and
P2 on the side AB we show that the assumption that there is no right triangle
leads to a contradiction.
В
rl r2
Figure 83/4.1
293

Solutions
1983/4
1. P\ and P2 are red· ^3 and ?6 nas t0 be green, as the triangles PiP2P
or PiP(yA were red. P5 is green, because of the triangle AP\P$, but then the
triangle P3P5P6 is green, contradiction.
2. Pi and P2 are green. P4 and P5 has to be red because of the triangles
P2P1P5, or РхР2Ра\ Рб and P3 are green for the triangles PP4P6 and AP5Py,
but, now, P3P5P6 is red, contradiction.
3. Pi red, P2 green (or vice verse). Now, P3 and P5 are green because of
the triangles P1P3B and AP1P5. Now, P2P3P5 is green, contradiction again.
Thus we proved that there is a monochromatic triangle.
1983/5. Is it possible to choose 1983 distinct positive integers, all less than
mal
ression?
or equal to 10 , no three of which are consecutive terms of an arithmetic prog
First solution. The answer is affirmative. Let us choose the set consisting
of all positive integers whose base 3 representations contain only the digits 0
and 1 (i.e. no 2-s). Since 310<10<3, these numbers have at most 11 digits;
the largest one among them is 11 111 111 111з = 88 573ю. The number of these
numbers is 2 — 1 =2047 (0 is not among them).
We show that these 2047 numbers do not contain an arithmetic progression.
If x, у and ζ are three elements of an arithmetic progression, then x + z = 2y.
The number 2y contains only digits 0 and 2. Hence χ and ζ must match digit by
digit thus χ = z, which is impossible. Hence the chosen 2047 numbers contain
no arithmetic progression of length 3, therefore the selection is possible.
Second solution. Let us define a sequence of sets on the following way:
Let Ηι = {1, 2} and let #n+i be defined such that first construct the set Sn, in
a way that we add 3n to every element of Hn (e.g. 52 = {4, 5}), and let
(1) #n+l=#nu5n.
Thus:
#2 = {1, 2; 4, 5}, #3 = {1, 2, 4, 5; 10, 11, 13, 14}.
Then the size of Hn is 2n. We prove by induction that the largest element of
3n+l 3 + 1
Hn is —л—· The largest element of Hi is —— = 2; suppose that the largest
3η_1 + 1
element of Hn_\ is , hence the largest element of Hn is
3^ + 1 ,3.-1^ + 1^
294

1983/6
1983.
Thus the elements of Hn are in the interval
3n+l + j
1 3n + l
and the elements of 5.
η
in the interval
3n + l.
2
We prove by induction that the sets Hi contain no arithmetic progressions of
length three. For H\ and #2 tnis *s obviously true, and suppose that it holds for
Hn, too. Consider the set #n+i =HnU Sn; we show that if in this set χ < у < ζ,
then χ + ζ = 2y is impossible. This is the condition of being three consecutive
members of an arithmetic progression.
If χ and ζ are from Hn, then the statement holds by the assumption. If χ
and ζ are from Sn, then there are some x' and z' in Hn such that
x = x' + 3n< z = z' + 3n
and so
This is not in Hn as it is greater than the largest element of Hn, but it is not in
χ1 + ζ1
Sn, either, because then —-— were in Hn, contradicting our assumption, thus
у is not in Ηη+γ.
Finally, if χ e Hn and ζ G Sn, then χ > 1, ζ > 3η + 1, and so
x + z 3n + l + l 3n + l
thus it is greater than the largest element of Hn, and
x + z 1 /3n + l 3n+1 + l\ 1 „ nn 1S OT? ,
^ —^(—+ ^) = 2(2·3 +l)<3 +l'
thus it is smaller than the smallest element of Sn, hence it is not in Hn+\.
Therefore Hn+i contains no arithmetic progression of length 3.
3n + l
We proved in general that there are 2n numbers not greater than —-— such
that they contain no arithmetic progression of length 3.
For η = 11 the result says that there are 2048 numbers among the first 88 574
integers not containing an arithmetic progression of length 3, so the answer to
the question is positive.
1983/6. Let a, b, с be the length of the sides of a triangle. Prove that
(1) a2b(a - b) + b2c(b - c) + c2a(c-a) > 0.
First solution. Introduce the following notations:
x = — a + b + c, y = a — b + c, z = a + b — c,
295

Solutions
1983/6
thus, χ, у, ζ are twice the length of the segments between the vertices and the
touching points of the incircle. So
y+z z+x Х+У
a= л , 6=——, c=——-.
2 2 2
Substitute them to (1) and multiply the inequality by 16:
(y + z)2(z + x)(y -x) + (z + x)2(x + y)(z -y) + (x + y)2(y + z){x -z)>0.
After reordering it reduces to:
(2) χ z + y x +ζ у >x yz + y zx +ζ xy,
and so
3 3 3 2 2 2
x z + y x + z y — x yz — y zx — z xy =
9 9 9
= zx(x — y) + xy(y — z) + yz(z — x) > 0.
The validity of the inequality is clear, as all the summands are nonnegative.
As χ > 0, у > 0, ζ > 0, equality holds if and only if χ = у = ζ, that is if a =
b = c, thus the triangle is equilateral.
Second solution. Transforming both sides to polynomials it is easy to see
that:
a2b(a-b) + b2c(b-c) + c2a(c-a) = a(b + c- a)(b - c)2+ b(a + b- c)(a - b)(a - c).
Since the l.h.s. of (1) remains the same at the permutation a-^b, 6—> с, с—»a,
we may assume that a>b, c; now, on the r.h.s. of (1) every term is positive thus
we proved the statement.
Remarks. 1. (1) holds in some cases when the numbers are not the sides of
a triangle, e.g. if a- 1, 6 = 3, с = 5.
2. (2) can be shown in different ways. e.g. dividing both sides of (2) by xyz
we obtain:
2 2 2
x У z ^
(3) — + — + — >x + y + z.
у ζ χ
ι Χ V
Now, the application of the Cauchy inequality for the triples
and (y/χ, y/y, yf£) gives:
о / X IS Ζ \ I X XI Ζ \
(x + y + z)2= —-_у/у+—^+—у/х~) <[— + — + —)(x + y + z),
\\/y Vz yfx J \y ζ χ J
and dividing by (x + y + z) we arrive to (3).
296

1984/1
1984.
1984.
1984/1. Prove that
7
(1) 0<xy + yz + zx-2xyz<—,
where x, у and ζ are nonnegative real numbers for which
(2) x + y + z=l.
First solution. To prove the left side of the inequality, observe that (2)
implies 0 < x, y, ζ < 1 and so
xy + yz + zx — 2xyz = xy{\ — z) + yz{\ — x) + zx>0,
since all three terms of the l.h.s. are nonnegative.
In order to show the right side of the inequality, introduce the notations
, 1 1
y = o+-, z = c+ -
y 3' 3
now, (2) gives
(3)
1
x — a + -,
3'
a + b + c-
1 , 2
--<a,6,c<-.
After substituting b + c=—a, we obtain
(4)
Τ 1 Τ 1
xy + yz + zx — 2xyz = — + -(ab + bc + ca — 6abc) = — + - (be— a — 6abc) .
Our original expression is symmetric in a, b, c, hence we may assume that
a<b<c. According to (3), a, b, с cannot be all positive or negative at the same
time, so there are two possibilities:
a) a<b<0<c, or /3) -~<a<0<b<c.
In order to prove (2) we have to show that the expression in the parenthesis
in (4) is not positive. In case a) every expression in the parenthesis is positive,
thus (2) holds.
In case β), we rearrange the expression for our purposes:
be-a2- 6abc = bc-(b + c)2 - 6abc =-(b-c)2- 36c(l + 2a).
Now, as 1 +2a > 0 and bc> 0, the expression cannot be positive, thus (2) holds.
Second solution. (2) implies that at least one of x, y, z, let us say x, is
not greater than - thus 1 — 2x > 0, and so
xy + yz + zx — 2xyz = xy + yz{\ — 2x) + zx > 0,
hence we proved the left hand side of (1).
297

Solutions
1984/1
To prove the right hand side, we distinguish two cases:
a) one of x, y, z, let us say z, is at least -; z>-, β) 0<x, у, z<-.
In case a), using x + y = l — zwe get
xy + yz + zx — 2xyz = z(x + y) + xy( 1 — 2z) =
= z{\ -z) + xy{\ -2z)<z(l-z)<-<—.
In case β) apply the following substitutions:
a=l — 2x, b = l—2y, x = l—2z.
These values give
(5) a, 6, c>0, a + b + c=l.
Substituting and applying (5) we obtain:
,-. „ l + abc
(6) xy + yz + zx — 2xyz=—-—.
Now, the A.M.-G.M. inequality gives
/a+b+c\3 1
_V 3 / 27'
hence (6) implies
28 j
xy + yz + zx- 2xyz < -21 = — ?
and so we proved (1) in every case.
Third solution. The problem involves the so called elementary symmetric
polynomials of x, y, ζ that leads us to the following idea: consider the polynomial
f(t) of degree 3:
(7) f(t) = (t- x)(t - y)(t - z) = t3 - t2(x + y + z) + t(xy + yz + zx) — xyz.
From (2) we get
2^ (2) = 4 ~ 2 +(ху + Уг + zx>>~2xyz>
xy + yz + zx- 2xyz = 2f I - J + -.
The problem says that
~ / \2) 4 ~27'
that is
(8) ~\<ϊ(ϊ\< l
8 -■' V27 - 216'
298

1984/2
1984.
1 1 1
We may assume that χ > у > ζ > 0. Choose χ, у, ζ such that x>-,y<-,z<-
holds. Then, (7) implies that / ( - } < 0, hence
-'G)-H)G-')(H-
Applying the A.M.-G.M. inequality and using that χ < 1
-,(1),(^)'=(^)'4 ,({)*-}.
1 1 1
If we choose the values of the variables such that χ < - and so у < -, ζ < -
- 2 y~ 2 ~ 2
holds, the previous method gives:
3
3
ι\ ./ι λ/i ^\ /i \ Ji~(x+y+z)\ ι
o</U <U-* U-i/ U-* <
ч2/-\2 / \2 / V2 /~\ 3 у 216
Thus we proved (8) and the statement of the theorem.
1984/2. Find one pair of positive integers, a, b, such that:
(1) ab(a + b) is not divisible by 7;
7 7 7 7
(2) (a + b) — a —b is divisible by 1 .
First solution. Using the binomial theorem and observing that both (a + by
and a+b is divisible by (a + b), (2) reads as:
(3) (a + b)1 - a1 - b1 = lab{a + b)(a2 + ab + b2)2.
Since ab(a + b) is not divisible by 7, we must choose a and b such that 7 divides
(a2 + ab + b2\ or equivalently: a2 + ab + b2 is divisible by 73 = 343.
We try to choose a pair a, b such that 6=1, that is, for some integer к
(4) a2 + a+l = 343k, that is, a2 + a + {\ -343/c) = 0
holds. This happens exactly if the discriminant of this equation of degree 2 in α
is a perfect square, which means that
l_4(l-343/c) = 1372/c-3
is a perfect square. This is true for k- 1, as 1369 = 37 . Hence (4) becomes:
α2 + α-342 = 0, α=18, 6=1,
and the pair (a, 6) = (18,1) satisfies the conditions, because 7 does not divide
18-1-19.
Second solution (due to Geza Kos). We finish the previous solution in a
different way to determine all pairs a, 6 that satisfy the conditions of the problem;
299

Solutions
1984/2
9 9
as we saw before, we need to find those pairs, a, 6 where 343 divides a + ab + b ,
but a, b, a + b are not divisible by 7. We start with the identity
18 (a2 + a6 + 62) =343ab + (a- 186)(18a-6).
18 and 343 are coprime so only one of a — 186 and 18a — b can be divisible by 7.
Otherwise their difference, П(а + Ь) is divisible by 7; hence there are two cases:
A) a =186 + 343k,
where 7 does not divide 6 and к is an integer such that 186+ 343A; is positive;
B)6=18a + 343/c,
where 7 does not divide a and к is an integer such that 18a + 343fc is positive;
It is clear that in none of the two cases is a + b divisible by 7. Indeed,
a + 6=196 + 343/c, and a + b= 19a + 343/c.
The values given in A) and B) provide all solutions to the problem.
1984/3. In the plane two different points О and A are given. For each
point X^O on the plane denote by ω(Χ) the measure of the angle between
О A and OX in radians counterclockwise from Ο Α (Ο < ω(Χ) < 2π). Let C(X)
be the circle with centre О and radius OX +
ω(Χ)
OX
Each point of the plane is
coloured by one of a finite number of colours. Prove that there exists a point
Υ for which ω(Υ) > 0 such that its colour appears on the circumference of the
circle C(Y).
Solution. Consider
the set of all
concentric circles with centre
О whose radii are less
than 1. Assign to every
circle the set of colours
that appear on the
circle. As there are finitely
many colours, we have
two circles with the
same combination of
colours. Call them В and S
with radii 6 and s
respectively, and let 6 < s < 1
(see Figure 1984/3.1).
Figure 84/3.1
300

1984/4
1984.
Pick a point У on Б such that ω(Υ) = b(s — b) holds. Such a point, Y, exists,
because 0 < b(s — b) < 1 < 2π. Hence
the circle S is the same as C(Y), and as S and В bear the same set of colours,
C(Y) = S contains the colour of Y, and this is what we wanted to prove.
1984/4. Let ABCD be a convex quadrilateral with the line CD tangent
to the circle on diameter AB. Prove that the line AB is tangent to the circle on
diameter CD if and only if ВС and AD are parallel.
Solution. Let F and G denote the midpoints of the sides AB and CD; F'
D
F
A
/ / ! / >л*1
/ / ι / .-"X
/ / ι ι - .'
/ / ι / -" .'
/ / \ J--" '
/ / 1 ~* '
1 / ' - / '
/ / """ ί / /'
I / -- / '
Ж ^' ι / f
/ \ ι S
/ <■' ''
/·-' U'
F
G
L "^
\
\
\
\
\
\
I
G'
В
Figure 84/4.1
denotes the foot of the perpendicular from F to CD, G' the one from G to AB
(see Figure 1984/4.1). From the conditions AF = FB = FF', and so
AF
FF'
= 1.
The radius of the circle over CD is CG-GD that touches the AB side if and
only if GG' = GD, that is
GD
~GG'
= 1.
It can be restated as: the circle with diameter CD touches AB if and only if:
AF GD . . AF-GG' GD-FF'
(1)
that is,
FF' GG': ' 2 2
This means that the area of the triangles AFG and DFG agrees. As FG is a
common side of the two triangles, the equality of their areas and the convexity
of ABCD imply that A and D are of the same distance from FG (on the same
301

Solutions
1984/4
side of FG), thus AD and FG are parallel. FG is the median of the quadrangle
ABCD, and the median is parallel to a side if and only if the quadrangle is a
trapezium, that is if AD and ВС are parallel; thus we proved our statement.
1984/5. Let d be the sum of the lengths of all the diagonals of a plane
convex polygon with η > 3 vertices. Let ρ be its perimeter. Prove that:
(1)
η
2d
3< —<
Ρ
η
2
n+1
Solution. First we prove the lower bound that is equivalent to the
inequality:
2d>(n- Ъ)р.
Choose two non-adjacent sides of the polygon XY and ZU such that the
quadrangle XY ZU is convex (see Figure 1984/5.1). The triangle inequality imp-
Figure 84/5.1
lies that the sum of the two diagonals XZ and YU is greater than the sum of
the two opposite sides XY and ZU.
(2) XZ + YU>XY + ZU.
A diagonal XZ is the diagonal of two quadrangles, because a pair of opposite
sides of a quadrangle can be chosen in two ways on the two sides of XZ. So,
if we choose all possible quadrangles on the way described above, and sum the
inequalities of type (2), the l.h.s. is 2d. For a side XY we can choose an opposite
side on η — 3 many ways, so for the XY side we can choose η — 3 quadrangles.
But we choose a quadrangle η times when we make all possible choices. Hence
the number of distinct quadrangles is n —3 and the perimeter of the polygon
comes up twice among their sides. Hence we get
2d>(n — 3)p,
and we proved the left side of (1).
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1984/5
1984.
In order to prove the right side, let first choose n = 2k + l. The diagonal is the
shortest path between its endpoints, it is shorter than each polygonial path
connecting them. We shall use the polygonial path that uses less terms. For example
we estimate the A\A2 ... An part of the polygonial path with the diagonals from
Ai in the following way: (see Figure 1984/5.2) (An+\ =A\):
AAi+2<AiAi+l+Ai+lAi+2
AiAi+3 < AiAi+i + Ai+i Ai+2 + Л+2^+з
(3) i
АА+к-i < AiAi+\ + · · · + Ai+k_2Ai+k_i
AiAi+k < Ai-A-i+i + · · · + Ai+k_iAi+k
Ai
Figure 84/5.2
Proceed the same way for every diagonal and sum these inequalities. On
the l.h.s every diagonal occurs twice, hence the sum is d. On the r.h.s.
2 + 3 + ... + fc = -(fc-l)(fc + 2) = i(fc(fc + l)-2)
TL
many sides come up at a vertex, so there are —(k(k + l) — 2) sides in the sum.
None of the sides were distinguished, and since each side occurs equally, on the
r.h.s. we have
|(fe(fe+l)-2),
that is
(4) d<|(fe(fe + l)-2)).
303

Solutions
1984/5
As - = k + -,
2 2
η
2
= к, moreover —-— = к + 1 and so
rewrites as:
2d
— <
Ρ
η
2
2
n+1
n+1
= fc+l, (4)
2.
Thus we proved (1) for odd many vertices.
Αι
Ai+k
Figure 84/5.3
For n = 2k, in (3) we omit the last
row, and substitute it by the obvious
estimate for the diagonal AiAi+^ (see Figure
1984/53):
-A-i-A-i+k < ~Z (г= 1? 2, . . . , fc).
At the summation the r.h.s. becomes the
sum d of the diagonals. On the r.h.s a
vertex gives
2 + 3 + ... + fc-l = -(fc-2)(fc + l)
many sides and repeating the previous
arguments this is -(k — 2)(k + l) times the
perimeter of the polygon; considering the
— sum coming from the estimates of the diagonals AiAi+k we get
d<^(k + (k-2)(k + l)) = ^(k2-2).
For n = 2k, k =
becomes
η
2
2
n+1
2d
— <
Ρ
hence (1) holds for every η > 3.
and so k2 =
n+1
n+1
, therefore our inequality
η
2
1984/6. Lei a, b, c, d be odd numbers, such that:
(1) 0<a<b<c<d,
(2) ad = be,
(3) a + d = 2k, b + c = 2m
holds, where к and m are integers. Prove that a=\.
Solution. In our solution we construct the integers a, b, c, d as functions
of к and m. First we show that к > т. Indeed, as с = — we have
b
2k -2m = {a + d) -(b + c) = (b- a) № - l) > 0.
304

1985/1
1985.
Now substitute d and с expressed from (3) into (2):
а(2к-а)=Ъ(2т-Ъ),
2mb - 2ka = b2-a2 = (b + a)(b - a).
Hence 2m divides (b + a)(b — a), b + a and b — a are even but they cannot be
both divisible by 4, since their sum (26), is divisible only by the first power of
2. Hence one of a — b and a + b is divisible by 2 and the other one by 2m_1. As
b - a <b <— = 2m-1
2
and
(4) 6 + a<6 + c = 2m,
only the first power of 2 divides b — a and so 6 +α is divisible by 2m . 6 + a =
r · 2m (r is a positive integer), but (4) implies r < 2, thus r = 1, hence
(5) b + a = 2m-1.
Moreover:
(6) c-a = (b + c)-(b + a) = 2m-2m-1=2m-1.
a and b, a and с are coprime because they are odd numbers, and any
common divisor of them divides 2m by (5) and (6). On the other hand (2) implies
that a divides be, thus
(7) o=l.
With this we proved the statement of the problem. (5), (6), (2) and (7) gives
(8) 6 = 2m-1-l, c = 2m-1 + l, d = 6c = 22(m-1)-l,
combining with (3) we get к = 2(ra — 1). It is easy to see that the numbers in (8)
are all the numbers that satisfy the conditions for m>2.
1985.
1985/1. A circle has centre on the side AB of the cyclic quadrilateral
ABCD. The other three sides are tangents to the circle. Prove that
AD + BC = AB.
First solution. Let О denote the centre of the touching circle, r its radius
and P, Q, R the points of tangency of the sides ВС, CD and DA respectively
305

Solutions 1985/1
Ao В
Figure 85/1.la
The length of the segments are denoted by BP = p, PC = CQ = q, QD = DR = s,
RA = t (see Figure 1985/1. la). With our notations we have to show that
AB=p + q + t + s.
As ABCD is cyclic, IBAD = a equals the external angle at C.
Cut and change the right triangles AOR, BOP and fix the other parts of the
quadrangle. We get the quadrangle A'CDB' that touches the circle with radius
r and centre О at three sides.
A'CDB' is a trapezium, because the sum of the angles at A' and С is 180°
(see Figure 1985/1.lb).
B'
ι
Figure 85/1.lb
306

1985/1
1985.
PCQO is a deltoid, so ОС is a bisector, hence LOCA' = 90° - ^. This
implies that Z,4'OC = 180° - α - (э0° - ^\ = 90° - |, hence A'OC is an
isosceles triangle, thus О A' = t + q. Similarly, OB' =p + s, thus
AB = A'B' = A'0 + OB' = t + q+p + s,
and this is what we wanted to prove.
Second solution. Let О be the centre of the circle, к the circumcircle of
the triangle CDO that intersects the AB side in a second point P. Ρ is an
Ο Ρ
Figure 85/1.2
inner point of AB; if P = О then AB touches the circle CDO at О (see Figure
1985/1.2). As the circle with centre О touches the sides AD, DC, ВС the lines
OD and ОС bisect the angles at D and С Set /Л = а, LB-β.
Assume that Ρ lies on the ray OB. As the quadrangle О PCD
is cyclic, LBPC=LODC = 90°-^, hence in the triangle РБС we have
LPCB = 180° -/3-^90° -f)=90°-f> thus PCS is an isosceles triangle,
therefore РБ = ВС (if Ο = Ρ, LBPC = LODC = 90° - ^ holds, as well).
As they are measured by the same arc in k,
LAPD = LOPD = LOCD = 90°-^,
2'
hence LADP = 180° - a - (э0° - у J =90° - у, and so PAD is an isosceles
triangle, implying AP = AD. Combining these two results:
AB = AP + PB = AD + BC.
Third solution. Use the notations of Figure 1985/1. la. ОС and OD bisect
the angles of the quadrangle at С and D, so LQCP- 180° - a, hence LOCP =
307

Solutions
1985/1
90° — — and in the right triangle О PC q = r cot I 90° — — I =r tan —. From the
r
right triangle ARO, t = r cot a, AO = ——; hence
sin α
As
/ a
t + q = r (tan — + cot a
a sinf cos α 2 sin2 % +1 -2 sin2 % \
tan — + cot a = j-+ - -
cos^ sin α 2 sin ^ cos ^ sin α
t + q = -^— =AO.
sin α
Changing the roles of PC and RD, AR and BP, the previous arguments give
p + s = OB.
Adding these results we get what we wanted.
1985/2. Let η and к be relatively prime positive integers with к <n. Each
number in the set Μ = {1, 2,3,..., η — 1} is coloured either blue or white, such
that:
(a) for each г in M, both г and n — i have the same colour;
(h)for each i in Μ not equal to k, both г and \i — k\ have the same colour.
Prove that all numbers in Μ must have the same colour.
Solution. Let a ~ b denote if a and b have the same colour. We start with
two lemmas:
a) if a = b (mod k), then a ~ b;
β) there exists a complete residue system mod k, b\, bi, ■.., 6ь such that
bi~bi+i (г = 1, 2, ..., fc-1).
These imply our statement: /3) means that all the bi-s have the same colour,
but every element of Μ belongs to a residue class mod k, hence they have the
same colour.
We assume that k>\, because in case к = 1 (b) implies i~\k — i\=i — \,
thus any two consecutive members of Μ have the same colour, hence it holds
for every element of M.
a) Let a = sk + r (0 < r < k).
(b) implies that r ~r + k, as r + к~ \k — (r + k)\ = r;
implying that the elements in the residue class of r mod к have the same colour.
308

1985/3
1985.
β) Consider the numbers
η, 2n, 3n, ..., kn.
Their residues mod к are:
h, b2, b3, ..., bk,
thus 1 < b{ < к (i= 1, 2, ..., k). Here, 6^ is k. The 6j-s form a complete
residue system mod k. Indeed, if hi and bj (bi > bj) were in the same class, their
difference is
bi — bj = (i — j)n = 0 (mod A;),
which is impossible as к and η are coprime and i—j<k.
Set 1 < г < к — 1. (b) implies bi ~ \k — b{\, and as k — bi>0, b{~k — b{. (a)
gives
к — bi^n — (k — bi) = n — k + bi.
Now
n — k + bi = n + bi = n + in = n(i + 1) = bi+\,
that is bi ~ η — к + b{ ~ bi+\, 6г ~ 6^+1 > and we proved the statement of the
problem.
1985/3. For any polynomial P(x) = ciq + a\ χ + ... + a^x with integer
coefficients let ω(Ρ) denote the number of odd coefficients, and let Qi(x) = (l + х)г,
where i = 0, 1, 2, ....
Prove that ifi\, i2, · ■ ■, in are integers such that 0<i\ < i2 < · · · < in> then
(1) "(Q^+Q^ + .-. + Qin)^^)·
Solution. We start with two observations:
/2m\
Lemma 1. If 0 < к < 2m (m > 1 integer), then II is even.
The definition of the binomial coefficients implies:
/2m\
as к < 2m, it shows that I J is divisible by 2, hence it is even.
Lemma 2. If P(x) is a polynomial of degree n, where η < 2m (m > 1
integer), then
(2) u(P-Q2rn) = 2uj(P).
\
309

Solutions
1985/3
By the first Lemma
(3) Q2m(x) = (l + xr =l + xl + R(x),
where R(x) is a polynomial of degree 2m — 1 with even coefficients, hence
P(x)Q2rn(x) = P(x) + x2mP(x) + P(x)R(x).
In the sum, P(x) and χ Ρ(χ) have no terms of the same degree, but their
coefficients agree, and all coefficients of P(x)R(x) are even, hence (2) holds.
We prove our statement by induction on in. If in = 0 or in = 1, the statement
is obvious. Suppose that (1) holds if in<2m, where m is a positive integer.
Starting with this assumption, we show that the statement holds if 2m <in<
2m ; this is clearly enough for the proof.
We distinguish two cases. First, assume that the exponents i\, i2, ■ ■ ■, in
2т<ц<г2<...<гк< 2m+1, (m > 1).
are between 2m and 2m+1, that is
In this case
Qix + Qi2 + · · · + Qik = Q2m ( Qix -2m + · · ■ + Qin-2rn) .
in<2-2m = 2m+1, therefore in -2m <2m and so our assumption implies that
^(Qil-2 + --- + Qin-2m) >^{Qil-2r>-) ,
ω [Q2m \Qix-2m + ... + Qin-2rnJJ = 2ω (р^-2гп + ... + Qin_2mJ >
> 2ω [Qix_2mj =ω [Q2mQil-2mj =ω [Q^J .
Thus we proved the statement.
For the second case, assume that for the exponents i\, i2, ..., in
i ι < ... < v_! < 2m < ir < ... < in < 2m+1, (l<q<n)
holds. The assumption is, again, that (1) holds for every in, where in < 2m, thus:
(4) ω (Qh +Qi2 + Qir_l) >ω (Qij) .
Set
QiY +Qi2 + . ■ - + Qir_i =CL0 + CLl + . . . + CL2m_\X
and use that
Qir + Qir+\ +--- + Qin= Q2™ (Qir-2m + ■■■ + Qin-2m) =
= (l+ Χ2""1 + Д(Ж)) (Qir_2m + . . . + Qin-2m) ■
Let the polynomial in the last parenthesis be:
7 7 7 2m-l
0Q + 0\X+. . . + 02m_\X ,
310

1985/3
1985.
where щ and bi are integers. Now,
(5) Qix + ... + Qir_ j + Qir + ... + Qin =
/т\ 2 2m -1
(1) =aQ + a\X + a2X +.. . + d2m_\x +
(П)
о оттг ι
ч-Ьо + ^я + ^я + · ■ - + b2m_ix +
(III) +6o^2W + 6i^2W+1+62x2m+2 + ... + 62--i^2W+1"1+^(^),
where all coefficients of Д'(ж) are even. Now, it is easy to see that
(6) ω (ρίχ +... + Qir_ l+Qir + ... + Qin) > ω (Qh +... + Qir_}) =
/ 2m-l\
= ω laQ + a\x+ .. . + a2m_\x I,
Let us denote the polynomials in the three lines of (5) by I., II. and III. If сц and
bi are even for some i, then the number of the odd coefficients does not change
in (5) compared to L; if щ is odd and bi is even, then since ai + bi is odd and the
i =
0
1
2
3
22
5 .
6
7
23
1
0
1
0
1
0
1
0
1
1
0
0
1
1
0
1
0
0
0
1
0
1
1
1
1
0
1
0
1
0
1
1
0
1
0
1
Figure 85/3.1
number of odd coefficients remains the same; if сц is even, bi is odd, then the
number of the odd coefficients is increasing in (5); finally, if щ and bi are both
odd, then щ + bi is even, but in III. bi is odd, hence the number of odd coefficients
is the same. Thus we proved (6), and (6) with (4) implies (1).
Remark. We can visualize the problem using the Pascal triangle. In Figure
1985/3.1 we presented the mod 2 residues of the shifted Pascal triangle. Thus
311

Solutions
1985/3
in the г-th row the 1-s represent the odd and the 0-s the even coefficients of the
polynomial (1 + x)1. The addition of the polynomial corresponds to the addition
of the appropriate rows, coordinatewise. We can reformulate the problem in the
following way: If we add η rows of the triangle, the number of 1-s in the sum is
at least he number of 1-s in the first of the rows.
1985/4. Given a set Μ ο/1985 distinct positive integers, none of which has
a prime divisor greater than 26. Prove that Μ contains a subset of 4 elements
whose product is the 4th power of an integer.
Solution. The possible prime divisors of the elements of Μ are
2, 3, 5, 7, 11, 13, 17, 19, 23
that is 9 distinct primes. Write the elements of Μ in the form
2Pi ·3ρ2.5Ρ3.....23ρ9.
Let us assign to every element the 9-tuple (p\,P2, ■ ■ ■,^9), where pi equals 1 if
it is odd and 0 if it is even. The number of these 0-1 9-tuples is 2=512, hence
there are two 9-tuples among any 513 that are identical. Choose two numbers
from Μ, α\ γ and a\ 2, such that the appropriate 9-tuples are the same. Then
α1,1α1,2 is a perfect square, as all exponents are even. Thus д/αι,1^1,2 is an
integer.
Similarly, from the remaining 1983 numbers we can choose a^ \ and 0^2
such that -\/ci2,la2,2 *s an integer. In a similar fashion we get:
(1) \/01,101,2> л/а2,1а2,2> Va3,la3,2> ■ · ■ > Va513,la513,2·
(We choose the last pair from 1985 — 2 · 512 = 961 numbers.)
The 513 integers listed in (1) have 9 prime divisors. Hence the previous
arguments give us two of them, у/аца^ \/aj,laj,2 sucn tnat me appropriate
0-1 9-tuples are the same. The product of them is a perfect square, hence
V/W ■ V4laj,2 = b2 (b integer)
and so %\a>i,2aj,\aj,2 = b*>
and this is what we wanted to prove.
Remark. The selection can be made from 2-512 + 513 = 1537 integers.
1985/5. The circle k\ with centre О passes through the vertices A and С
of the triangle ABC and intersects the segments AB and ВС again at distinct
points К and N, respectively. The circumcircles к of ABC and k<i of KBN
intersect at exactly two distinct points В and M. Prove that LO\ MB = 90°.
Remark. In ABC we have AB^BC, because otherwise by the symmetry
B = M would hold, which is not the case.
312

1985/5
1985.
First solution. Denote by к the circumcircle of ABC and by О and R its
centre and its radius. For hi, the circumcircle of BKN denote by 0>i and r its
centre and radius (see Figure 1985/5.1). First we show that BO2 and AC are
perpendicular.
Figure 85/5.1
Let Ρ and Γ denote the second points of intersection of &2 and the lines
BO2 and AC, respectively. As ACNK is a cyclic quadrangle, /ΒΚΝ = η
and LBPN = 7, as they subtend the same arc. Hence the quadrangle TCNP
is cyclic, and as by Thales' theorem LBN Ρ is a right angle, the opposite
angle /PTC = IBTC is a right angle, too. (We labelled our figure such that 7 is
acute.)
As /.ΒΚΝ = ί, the included angle of the line e parallel to KN through В
and the line AB is 7, too. Hence this is the tangent of к and the radius OB
is orthogonal to e. As e and if TV are parallel, OB and if TV are perpendicular.
0\02 is perpendicular to KN, their chord in common, hence OB || 0\Οι· 00\
is orthogonal to their chord in common, ЛС (see Figure 1985/5.2).
313

Solutions
1985/5
Figure 85/5.2
But, B02 and AC are perpendicular, hence the opposite sides of the
quadrangle OB02Ox are parallel, so OB = Oi02 = R and ΟΟλ=Β02 = τ. As 02P
is parallel and equal to 00\, 00\P02 is a parallelogram.
Denote by M' the second point os intersection of 0\P and k2. In the
quadrangle Ox002M' the sides OxM' and 002 are parallel, OxM'>002,
00\ =02M' = r, hence 0\002M' is a symmetric trapezium and so the lengths
of its diagonals are the same: 0\02 = OM' = R. This implies that M' is on k,
the circumcircle of ABC thus M = M'. In k2 the interval РБ is a diameter,
hence /РМБ is a right angle, thus ΙΟλΜΒ = IPMB = 90°, and we proved
our statement.
Second solution. Let r\ denote the radius of k\. The radical axes of the
pairs of the circle, the lines AC, KN, BM of the common chords of the circles
concur in Q the so called radical centre. This point exists as AB ^BC;
we may assume that A separates Q and С (Figure 1985/5.3).
Since ACNK and BMKN are cyclic quadrangles, LKNB-a and
/QMK = a, and so AKMQ is cyclic, too, as the sum of the angles at A and Μ
is 180°. Let fc3 be the circumcircle of AKMQ.
By the power of the point theorem:
(1) QOJ-rf = QA-QC = QM-QB = QM(QM + MB).
314

1985/5
1985.
Figure 85/5.3
and
(2) B02-r2 = BC-BN = BM-BQ = MB(QM + MB).
subtracting (1) from (2) we get
QO2 - BO2 = (QM - MB)(QM + MB) = QM2 - MB2.
which implies that BQ is perpendicular to MOj-re, and this is what we wanted
to prove.
Remarks. 1. At the last step of the second solution we used that if Μ is a
point of the line AB and Ρ is an arbitrary point, then PM is orthogonal to AB
if and only if
(3) AM2-BM2 = AP2-BP2.
Indeed,
AP2-BP2 = A?2 -Bf2 = (Wl-WPj -(ME-WPj =
= W12-W^2+2MP(M^-W1^ = am2-bm2+2'mP-aS.
Hence (3) holds if and only if = 0, that is if Ρ Μ and AB are
perpendicular.
2. This problem has an interesting connection to the theory of conic sections.
If a parabola touches the sides of the cyclic quadrangle ACNK then its focus
315

Solutions
1985/5
Μ lies on the line connecting В and Q, the points of intersection of the opposite
sides of the quadrangle. Moreover, BQ is perpendicular to the line through Μ
and the centre of the quadrangle.
{l) Xn+l — xn I "^n "■"
1985/6. For every real number x\ construct the sequence x\, xi, ..., xn,
where
n/
for every η > 1. Prove that there exists exactly one value of χ γ which gives
0<xn< xn+\ < 1
for every positive n.
Solution. Fist we show that there is at most one x\ that satisfies the
conditions. By the assumptions {xi} is strictly increasing and bounded, hence it has
a limit, x. By (1) it satisfies the equation
χ = χ Ι χ + lim — )
у η—>οο η/
\Xj ~~ \Xj
thus x = l.
Now, assume that there are two appropriate starting values, x\ and y\, where
x\ <y\. We prove by induction that xn <yn, for every positive integer n. By
the assumptions xn-\ <yn-i-
2 , Уп-1 2 xn-\ _
Уп-Хп-у^ + ^^-х^--^^-
(2)
= (Уп-1 - xn-\) Un-l +Уп-1 + -—[ ) > °·
By the convergence of the sequences there is an index N such that for N <n
3
the elements xn and yn are greater than -, hence from (2) we get:
3
Уп-хп>^{Уп-\ -χη-ύ
3 /3\2
2/ra+l ~xn+\ >2^Уп~Хп^>\2) ^Уп~1 ~Χη-^
/3\k+l
Уп+к - Хп+к > [^ ) (Уп-\-хп-\)·
Since yn_\ —хп-\ is fix, the r.h.s. goes to infinity, giving a contradiction. Thus
there is a single value for x\.
316

1985/6
1985.
We show that such an x\ exists.
Construct the elements x\, x2, .. ■ as a function of x = x\.
о о fo(x) Л -X Ъх X
x2 = f2(x) = x2 + x, /з(я) = /2(я) +^ = я +2z3 + —+ -,
(3) /п(ж) = /п_1(ж)(/п_1(ж) + ^—jj .
As f2(x) is a strictly increasing function of x, the same holds for fi(x) the
other functions of the sequence. (3) implies /^(0) = 0 and lim fi(x) = oo.
Define the sequences
ab a2, ·.., an, ... and Ьь b2, .··, 6n, ···
as follows:
fn(an) = l , fn(bn) = l.
η
As fi(x) strictly increasing and continuous, it attains every value on [0, oo[ and
so an and bn are uniquely determined.
The sequence сц is strictly increasing. Indeed,
. fn+\ (an+i) - fn+i (an) = 1 — - fn(an) I fn(an) +-) =
= 1 ι _ΛΙ)Λ_Ι+Ι)= ι >0,
n+1 \ n/ V η η J n(n+l)
hence by the monotonicity of fn+\(x) we have an+\ > an.
The sequence hi is strictly decreasing:
fn+l(bn+i) - fn+l(K) = 1 - fn(bn) (fn(bn) + -) = — <°?
\ η/ η
thus 6n+i <6n.
The [ai, bi] intervals are nested, because by definition fn(an) < fn(bn) and
so by the monotonicity of fn(x) we have an < bn for every n. If there were
a,k, h, such that b{ < a^ and i < k, then it would imply a^>bi> bi+\ > ... > bk,
contradicting the inequality a^<b^. We get a similar contradiction starting with
the г > к assumption.
Thus the [аг·, 6г] intervals have a common point c, such that for every г the
inequality ai<c<bi holds, as сц < ai+\ <c<bi+\ <b{.
We show that choosing x\ =c the sequence satisfy the conditions.
fn(dn) < fn(c), because an<c and so
(4) l--</n(c) and so /n(c)_i^</2(c))
η η
(5) /n(c)</fc) + —= /n+i(c).
η
317

Solutions
1985/6
Moreover, с < bn+\ implies
(6) /n+l(c)</n+l(bn+l) = 1>
furthermore (4), (5) and (6) mean that
0</n(c)</n+i(c)<l,
hence for the sequence starting with с
0 < xn < xn+i < 1
holds, and we proved our statement.
Remark, с can be calculated: с = 0,4465349 ...
1986.
1986/1. Let d be any positive integer not equal to 2, 5 or 13. Show that
one can find distinct a, b in the set {2,5,13, d} such that ab — 1 is not a perfect
square.
Solution. If a and b are among the first three numbers of the set, then the
values of (ab— 1), 9, 25, 64 are perfect squares. We only have to check whether
we can make perfect squares with d. These numbers are 2d — 1, 5d — 1, 13d — 1.
We have to show that they are not all perfect squares.
Assume to the contrary that there are positive integers a, b, с such that
(1) 2d-l=a2,
(2) 5d-l=b2,
(3) 13d- l=c2.
(1) implies that a is odd, hence its square is of the form a - 8fc+ 1,
2d-l=8fc+l,
(4) d = 4fc + l,
hence d is odd. From (2) and (3) we see that b and с are even. Let b = 2b\, c-2c\;
Now, the difference of the appropriate sides of (3) and (2) reads
&d = c2-b2 = 4(cj-b2l) = 4(ci+bi)(ci-bi),
(5) 2d = (c1+b1)(ci-bi).
As the sum and the difference of two numbers have the same parity, both c\ + b\
and c\ — b\ are even, since their product is even. Thus the l.h.s. of (5) is the
product of two even numbers, hence divisible by 4; let us denote it by 4r:
2d = 4r,
d = 2r
318

1986/2
1986.
But this implies that d is even contradicting that it was odd. Hence the equations
(l)-(3) have no solution.
1986/2. Given a point Ρ in the plane of the ΑχΑ2Α3 triangle. Define As =
As_3 for s>4.
Construct a series of points Pq, P\, Pi, · · ■ such that P^+χ is the image of
Pk under a rotation with centre Ak+\ through an angle —120° (k = 0, 1, 2, ...).
Prove that if Ρχ9&β-Ρθ> tnen the triangle ΑχΑ2Α3 is equilateral.
First solution. The problem can be rewritten as: compose the rotations
with centres A\, A2, A3 through an angle —120° and repeat it 662 times (662 · 3 =
1986). We have to prove that if Pq is a fix point of this transformation, then the
triangle ΑχΑ2Α3 is equilateral.
The composition of three rotations with an angle of —120° is a translation
since it maps every vector to itself. Let ν denote the vector of the translation.
The composition of 662 translations is a translation by 662v that has the fixpoint
Pq. Thus this transformation is the identity:
662v = 0, v = 0.
Apply these three rotations to Αχ. The first rotation fixes Αχ, the second
maps it to a point A[ and the third takes it back to Αχ. (see Figure 1986/2.1).
Thus ΑχΑ2Α\A3 is a quadrangle, where ΙΑχΑ2Α\ = ΙΑ[Α3Αχ = 120°, ΑχΑ2 =
Α2Α'χ, Α'χΑ2 = Αι)Αχ, thus it is a deltoid, hence the diagonal A2A3 bisects the
angle. But then ίΑχΑ2Α3 = ίΑ2Α3Αχ =60°, therefore the triangle is equilateral.
Ax A2
Figure 86/2.1
Second solution. Let us consider our configuration on the complex plane
with the centre of gravity as 0. Let a, b, с denote the numbers of the vertices. By
the choice of the origin
(1) a + b + c = 0.
319

Solutions
1986/2
Let e = cos(-120o) + zsin(-120°) = cosl20o-zsinl20°; now, e3 = l and
e3 - 1 = (e - l)(e2 + e + 1) = 0, which implies e2 + e + 1 = 0 and so
(2) e2 = -e-l.
The geometric meaning of the multiplication by e is a rotation by —120°
about the origin. Let ρ denote the complex number belonging to Pq, and pi the
one to Pi (i= 1, 2, ...)
p\ =a + (p — a)e = a+pe — ae.
Similarly, we can calculate the result after the rotation about the points A<i and
Ay.
9 9
P2 = b + (p\ —b)e = b + (a — b+pe — ae)e -pe — ae —be + ae + b.
P3 = с + (p2 — c)e=p + (a — b)e +(b — c)e + c — a.
Using (2) and (1) we get
(3) p3=p + 3(be-a).
So the composition of the rotations about the points A\, A>i, A3 is expressed
with (3). Define be — a = v.
P2=p + 3v, P6=p + 6v, p<)=p+9v, ..., P1986 =P+ 1986-г; = р.
Hence 1986υ = 0, υ = 0, be = a, (1) and (2) imply
be = a = —b — c, c = b(—e—l) = be.
This means that rotating A<i through 120° about 0 we get A\, and through 240°
we get A3, thus the triangle A1A2A3 is equilateral.
Remarks. 1. In the solution the only thing we used about 1986 is that it is
divisible by 3.
320

1986/3
1986.
2. Our solution was based on Napoleon's theorem: The centres of the
equilateral triangles constructed over the sides of an arbitrary triangle (outside) form
an equilateral triangle. The centres of the equilateral triangles constructed over
the sides of the Ρ0^Ί^2 triangle are A\, A<i, A3. (see remark after 1975/3).
1986/3. To each vertex of a regular pentagon an integer is assigned, so
that the sum of all five numbers is positive. If three consecutive vertices are
assigned the numbers x, y, ζ respectively, and у < 0, then the following operation
is allowed:
x, y, ζ are replaced by x + y, —y, z + y respectively. Such an operation is
performed repeatedly as long as at least one of the five numbers is negative.
Determine whether this procedure necessarily comes to an end after a finite number
of steps.
Solution. We shall base our proof on the observation that during an
allowed operation the sum of the numbers remains the same.
Denote the five numbers by: x\, X2, x3, £4 and χ ζ, their sum by S. Consider
the following function of the variables:
f(xi,x2,X3,X4,X5) =
= (xi - X4)2 + (x2 - Χ5Ϋ + (x3 - x02 + (XA ~ χΐΫ + (x5 ~ x3)2·
We show that after performing an allowed operation for the {xi}-s, the value
of the function is decreasing. As the range of / is the set of the non-negative
integers, and at every operation its value is decreased by at least 1, the procedure
terminates in finitely many steps. Hence it is enough to prove that / is strictly
increasing.
Let X{xi,X2,X2>ixAixs) be an appropriate 5-tuple such that one of the
numbers is negative. As / is symmetric, we may assume that £3 < 0.
After an allowed operation X transforms to У(х\,Х2 + хз> — ^3,^3 +
ж4, x5). We have to show that f(X) > f(Y), that is, f(Y) - f(X) < 0.
f{Y)-f{X) = {xx - x3 - x4)2 + (х2 + хз ~ χ5)2 + (~χ3 ~ χ\)2 + (χ4~ χ2)2+
+(χ5 + χ3)2 - (χι - χ4)2 - (Χ2 ~ χ5)2 - (χ3 ~ χύ2 ~ (χΑ ~ χ2)2 ~ (χ5 ~ хъ)2 =
= 2χι^3 + 2^2^3 + 2χ3 + 2χ2>χ^ + 2χ2)Χ5 = 2x3S <0,
hence we proved our statement.
Remarks. 1. You can replace / by any function that decreases after an
allowed operation.
Geza Kos produced a fancy function:
д{х) = 2{х\Х2 + Х2хЪ + Ж3Ж4 + Ж4Ж5 + ^5^1)+
+ 3(^1^3 +Χ2Χ^ + Χ?>Χ5 + Χ^Χ\ +Χ$Χ2)·
321

Solutions
1986/3
g is strictly increasing, but its value is bounded from above: g{X) < S . This is
a consequence of
2
The following c(X) works well, too:
c(X) = χ \ + x\ + x\ + x\ + x\ + (x\ + X2 + хз)2 + (χ2 + x3 + xa)2+
+(X2 + X4. + X5) + (x4 + x5 + xl) + (x5 + x\ + ^2) ■
c = f + 2S shows its monotonity.
2. The following function produced by an American contestant won a
special prize:
a(x\, X2, £3,24, x5)~ \х\\ + \х2\ + \хз\ + \ха\ + \х5\+
+ \x\ + X2\ + \X2 + хз\ + \x3 + xa\ + \x4 + x5\ + \x5 + xl\+
+ \Х1+Х2 + Хз\ + \Х2 + Х3+Х4\ + \Х3 + ХА + Х5\ + \Х4 + Х5 + Х\\ + \Х5 + Х1+Х2\ +
+ \x\ + X2 + %3 + X^\ + \x2 + x3 + x4 + x5\ + \x3 + x4 + x5 + xl\ +
+ \Χ4 + Χζ + Χ\ +X2\ + \x5 + x\ +х2 + хз\-
We may assume that X2 < 0. After the allowed operation:
a(X) — a(Y)= \x2+X4 + X5 + X\\ — \x2 + 2%2 + x3 + x4 + x5\
= \S-X2\ - \S + X2\ >0,
thus a(X) is strictly increasing. The advantage of this function is that it is easy
to handle.
3. The statement remains true for an n-gon instead of a pentagon.
1986/4. Let А, В be adjacent vertices of a regular n-gon (n > 5) with
centre O. A triangle XYZ, which is congruent to and initially coincides with
OAB, moves in the plane in such a way that Υ and Ζ each trace out the whole
boundary of the polygon, with X remaining inside the polygon. Find the locus
ofX.
Solution. Introduce the following notations: Let 1 be the radius of the
incircle, LAOB = 2тг/п = la, so a = тг/п; LOAB = ΙΟ Β Α = β=^--α.
The rotation with centre О and by angle 2a maps the polygon to itself, so
it is enough to examine the movement by the AB and ВС sides. Rotating the
points X obtained by angles 2a, 4a, ..., (n — l)2a we get the locus (see Figure
1986/4.1).
Let А, В, С be three consecutive vertices of the polygon, (in
counterclockwise order). If Υ is on the side AB, then Ζ lies on ВС, because the inner points
of the other edges are farther from AB.
322

1986/4
1986.
—А С
Figure 86/4.1
If Υ = A, then Ζ = В and so X coincides with O; The same holds for Υ = В.
Assume that Υ is an inner point of AB. Translate the triangle XYZ by XO, the
vertex X is mapped to O, moreover Υ to Y' and Ζ to Z'. As the distance of Y'
and Z' from О equal 1, they lie on the circumcircle of the polygon and
) )
YY = Ζ Ζ . Now, rotate the points Α, Β, Υ, Υ about О (counterclockwise)
through angle 2a their images are B, C, Y", Z', where Y" is an inner point of
ВС.
Since YY' = ZZ' = Y"Z', the triangle ZY"Zr is isosceles and its interior
angle is 2a, because Y"Z' was rotated by 2a. Now, the isosceles triangles
ZY"Z' and BCO are similar and they lie on the different sides of he line ВС.
Hence their sides are parallel, and so OB is parallel to ZZ', but then ZZ' = хб
is parallel to OB, too. This means that X is on the line BO, on the ray from О
not containing B.
Our result shows that while Υ moves
from A to Β, Χ passes along a segment,
starting at O, to its farthest location, Q, and
returns to O. We determine the exact
location of Q. Observe that the distance from
X to О equals ZZ' and this distance is the
largest if the ZY" Z' triangle is the largest.
This latter is the largest if the altitude of the
triangle is the largest. This altitude is the
distance of Z' from the line ВС, which is
the largest if Z1 is the midpoint F of the arc
ВС. Now it is easy to calculate the length
of OQ = ZF (see Figure 1986/4.2).
Figure 86/4.2
323

Solutions
1986/4
Let Τ denote the foot of the altitude from О in the OBC triangle. From the
similarity of the right triangles FZT and OCT we get
ZF 1
Hence
ZF =
FT
FT ОТ'
l-ОТ 1 - cos a
ОТ ОТ
cos a
= OQ.
It is remained to show that every point of the OQ segment is the vertex
of a moving triangle. Indeed, translate ВС by XO, the image of the segment
intersects the ВС arc in two points (or touches at the midpoint). Let Z' be one
of the points of intersection. The line through Z' parallel to OB intersects ВС
in the point Z. There is a single point Υ on the side AB such that its distance
from Ζ is AB = ВС. The distance of vertex X of the triangle with base Υ Ζ
congruent to OAB from О — as we saw before — equals to ZZ' hence the
point is X (see Figure 1986/4.3).
Figure 86/4.3
Thus the locus is the union of η segments starting at О such that their other
vertices form a regular n-gon.
Remarks. l.In our solution we used visual arguments. These can be
formulated more precisely.
2. The problem belongs to the subject of kinetic geometry that has several
technical applications. We present a famous theorem of kinetic geometry:
Let be given two covering planes. Fix one of them and move the other one
in the plane. Every point of the moving plane describes a curve. At every stage of
the movement the normal vectors constructed to the points of the plane intersect
324

1986/5
1986.
В
Figure 86/4.4
in a point or they are parallel. This point of intersection is called the momentary
centre of the movement (Figure 1986/4.4).
In our problem the plane of the polygon is the fix plane and the moving
plane contains the XYZ triangle. During the movement the points Υ and Ζ describe
a segment, the point of intersection of the normal vectors ny, n^ intersect in M,
the momentary centre. As the quadrangle MYBZ is a cyclic quadrangle, both
Μ and X are on the circumcircle of XYZ, where MB is a diameter, hence Υ
and Ζ subtend a right angle. Μ lies on the normal vector of the orbit of Χ, ηχ,
too. Thales' theorem implies that BX is perpendicular to MX, thus BX is the
tangent of the orbit of X; hence X describes a curve such that all tangents go
through a fixed point, B. It can be shown that only a line bears this property,
thus X moves on a line.
1986/5. Find all functions f defined on the non-negative reals and taking
non-negative real values such that:
(a) f(x · f(y)) · f(y) = fix + y) for every поп negative χ and y;
(b)/(2) = 0;
(c)/(a;)^0, ifQ)<x<2.
Solution. First assume that χ > 2 and let у = 2; by (a):
f((x - 2)/(2)) · /(2) = f(x - 2 + 2) = f(x),
(b) implies /(2) = 0 and so for every χ > 2
(1) № = 0,
and this holds only in case χ > 2 according to (c).
325

Solutions
1986/5
Now, let 0 < у < 2. The l.h.s of (a) is 0, if xf(y) > 2, that is
(2) **m
the r.h.s. equals 0 if χ + y > 2, thus
(3) x>2-y.
This implies, that under the given conditions the equality
(4) ТГл=2-У
f(y)
holds. Indeed, if it does not hold for some y, than there is a smaller number on
the l.h.s. of (4) than on the r.h.s. and hence there were an x, such that
2
(5) —-<x<2-y
f(y)
held. But now, (2) implies that there is 0 on the l.h.s. of (a), moreover (c) implies
that the r.h.s. is not 0, that gives a contradiction. Similarly, it can be argued
against the opposite inequality in (5). Thus for every 0 < у < 2 the equation (4)
holds:
(6) /(2/) = ^-, (0<j/<2).
2-2/
Thus the only function satisfying the conditions (a), (b), (c) is the following:
2
№ =
■, if0<z<2,
2-х
0, if χ > 2.
It remains to show that this function satisfies the conditions, (b) and (c) are
obviously true, if у > 2, both sides of (a) is 0. Thus it is enough to consider the
case 0 < у < 2.
First, assume that x + y>2. Hence the r.h.s. is 0. Then x>2 — y, and
2x 2(2 — у) л
2-2/ 2-y
thus f(xf(y)) = 0, and so the l.h.s. is 0, too.
If χ + у < 2, then χ < 2 — y, and
2x 2(2 -у) л
2-у 2-у
and so
2 2 2(2 - у) ■ 2
fix ■ f(y)) ■ f(y) =
2-(x + y)
so f(x) satisfies (a).
2_ 2*l 2-2/ (4-2y-2a;)(2-y)
,2-y
= /(ж + 2/),
326

1986/6
1986.
1986/6. Given a finite set of points in the plane, each with integer
coordinates. Is it always possible to colour the points red or white so that for any
straight line L parallel to one of the coordinate axes the difference (in absolute
value) between the numbers of white and red points on L is not greater than 1?
First solution. Let us call the lines parallel to the axes containing lattice
points lattice lines and denote by Μ the set of the given lattice points. There are
two lattice lines through every point of M. We prove that there is a colouring
of Μ such that for any lattice line the difference between the numbers of white
and red points is not greater than 1.
For simplicity we may assume that both coordinates of the points in Μ are
positive.
If there is a point Ρ in M, such that neither of the two lattice lines through
Ρ contains a point of Μ different from P, then we colour Ρ white. From now,
we omit these points.
У
-О
-о
<-ϊ
-i,
-О
о-
о
-о
о о
ю-
Q-
-Q
I
6-^3-
ф
<^Аг
-о
о-
о-
о-
о—
о—
О
Figure 86/6.1
If an arbitrary lattice line contains more than 1 point of M, number the
points from left to right (bottom to top) by the integers 1, 2, .... Connect on
each line the pairs of points:
(1,2), (3,4), (5,6), ..., (2/c-l,2/c), .
327

Solutions
1986/6
The only points that are not connected to any other points are the "last" points
from the axes in case the appropriate lines contain odd number of points. Now
the points of Μ form a graph such that the degrees of the points are 0, 1 or 2
(see Figure 1986/6.1).
This graph is the disjoint union of open or closed polygonal paths that are
the concatenations of the line segments defined above.
Choose a point A\ of degree 1; it is connected to a unique lattice point
Ai- If Ai is of degree 1, the path terminated; if not, it is uniquely continued to
A3. We can proceed until we arrive to An, a point of degree 1. If these points
Αχ, Α2, ..., An do not cover M, choose B\, a point of degree 1 and construct
a polygonal path starting at B\. We continue the procedure until there are no
points of degree 1 left.
Now, omit these polygonal paths from M, and if there is a point of degree 2,
e.g. C\, then we can construct a closed polygonal path returning to C\. Thus we
connected the points of Μ by open and closed polygonal paths. Two paths cannot
share a point in common, because that was of degree at least 3, contradiction.
We colour the points path by path in the following way: choose a path and
an orientation of the path. Colour the points one after another on the path red
and white, alternating. If the polygonal path is closed, it has even many vertices
(see our remark), hence the endpoints of every segment are of different colours.
We prove that this colouring satisfies the conditions of the problem. The
endpoints of every segment are coloured with different colours, and on every
lattice line there is at most one point that is not the endpoint of a segment on the
line, hence the difference of the numbers of the red and white points is at most 1.
Second solution. We prove the statement by induction. For small numbers
(1, 2, 3, 4 points) the statement is obvious. Assume that for the numbers smaller
than η the colouring exists. We prove that there is an appropriate colouring for
η points.
If there is a point Ρ such that no lattice line through Ρ contains another
point, then the remaining η — 1 points can be coloured by the conditions and
colour Ρ arbitrarily.
If there is no point with the above property, choose two points of M, A and
В from the same line. We distinguish two cases:
1. There is no other point of Μ on the perpendiculars through A and B;
2. At least one of the perpendiculars contain another lattice point.
In case 1. the n —2 points distinct to A and В can be coloured by the
assumption; moreover let A be white and В red. Now, the number of red and
328

1987/1
1987.
white points remained the same in the lines distinct from AB, and on the AB
line the difference is the same.
A
В
λ
Χ
с
In case 2. there is a point С of Μ on the
perpendicular through B. Let X be the point
completing the triple А, В, С to a rectangle:
X is in Μ or not (see Figure 1986/6.2).
If X is among the given points, omit A,
В, С and X. The remaining η — 4 points can
be coloured by the assumptions and colour A
and С red, В and X white. Thus the
difference of the number of red and white points
remained the same.
If X is not in M, omit А, В, С and
include X to the set. The set of size n — 2 can
be coloured. Now, colour A and С the same
colour as X and В the opposite colour, then omit X. The colouring given this
way does not change the difference of the number of red and white points. Thus
we proved that the colouring exists.
Remark. In the first solution we used that if every consecutive pairs of sides
of a closed polygonal path is perpendicular then it has even many vertices.
Let us direct our vectors in some orientation. The angle between consecutive
vector pairs is 90° or 3 · 90°. The sum of these angles is a multiple of 360°. This
is an even multiple of 90°; only even many odd numbers can add up to be even,
so the number of the vectors is even.
Figure 86/6.2
1987.
1987/1. Let pn(k) be the number of permutations of the set 5 = {1, 2, ...
..., n} (n > 1) which have exactly к fixed points. Prove that the sum from k = 0
to η ofkpn(k) is n\.
First solution. Enumerate the altogether n! permutations of the η numbers
and mark the fixed points on the list i.e., an occurrence of 5, for example, is
marked whenever it is in the fifth column:
©0000-0
2 1 © 7 © ... 9
0© 5 © 7 ...©
329

Solutions
1987/1
There are к numbers marked this way in each row where there are к fixed
points. For the pn(k) permutations of the problem there are, altogether, pn(k)
numbers marked in the rows containing к fixed points for every k. Hence the
sum
η
is the total number of marked elements in the n! rows.
Let us also count the marked elements column wise. Every number, from
1 to n, occurs, by symmetry, the same number of times in each column, namely
Til
— = (n— 1)! times. Hence i occurs this many times in the ith column, in parti-
n
cular. Therefore, there are (n— 1)! marked elements in each column. Being so
there are n(n— 1)! =n\ marked numbers in the array altogether, which, when
compared to the previous result yields the claim:
η
Y^kPn(k) = nl
k=0
Second solution. Let's try to find a relation between pn(k), the
number of fc-fixed η-permutations, for short, and that of the (k— l)-fixed (n — 1)-
permutations, pn-\(k — 1).
Keeping any one of the η numbers fixed at its position, the (k— l)-fixed
permutations of the remaining n— 1 elements — there are pn_\{k — 1) of them
— form, with the fixed element, a fc-fixed η-permutation. There are η ways to
choose the fixed element so we obtain n-pn_\{k — 1) fc-fixed n-permutations
and every one of them has been checked this way. Each fc-fixed η permutation,
on the other hand, has been listed exactly к times: indeed, consider the 3-fixed
permutation 1 4 3 2 5 of the elements 1 2 3 4 5, for example. According to the
tally above this very permutation can be obtained by fixing 1 at the first position
and preparing the 2-fixed permutations of the remaining 4 elements. Since we
can also start by fixing 3 or 5 there are 3 · P^{2) of them, indeed. In general
(1) kpn(k) = npn_i(k-l).
Having respectively plugged 1,2, ... ,n in (1) for к and summing up the arising
equalities yields
η η η
(2) J2kPn(k) = J^npn_1(fe-l) = nJ^pn_1(fe-l).
k=\ k=\ k=\
Observe now that 0-£>n(0) = 0 and, as the sum of the numbers of 0-fixed,
η
1-fixed,... ,(n — l)-fixed (n — l)-permutations, respectively, ^pn_i(fc — 1) is
k=\
330

1987/1
1987.
equal to (η — 1)! Hence (2) can be written as
η
2_] kpn(k) = n(n— 1)! = n!
k=0
and the proof is complete.
Third solution. Denote, for brevity, the number of those n-permutations
where there are no fixed points at all by p(ri), that is let p(n) = pn(0).
Fixing к elements out of n, each at its own position and permuting, with
no fixed points this time, the remaining (n — k) ones yields /c-fixed permutations
and clearly p(n — k) of them. Since there are I J ways to fix к numbers the
total number of fc-fixed η-permutations is
η
к
(l) Pn(fc)=l?bfa-fc).
/ τι
Given that I
\K χ ,
(2) k-pn(k) = nl \p(n-k),
Since 0 · pn(0) = 0 this implies
Σ kpn(k) = J2n[k_i )P(n ~ ® = n Σ ( k _ ι )P(n ~ k^
According to (1) the sum left to be computed is
pn-\{n- 1) +pn-\(n -2) +... +pn_i(0)
and this is the sum of the number of (n — l)-fixed, (n —2)-fixed, ..., 0-fixed
(n— l)-permutations, respectively; the rest follows as before.
Remarks. 1. The origin of the problem is the infamous 'problem of the
confounded letters' from the first part of the 18th century. It goes like this: each
one of η letters is put into one of η addressed envelopes; what is the probability
that each letter goes to wrong destination. Labelling the letters by numbers the
number of favourable outcomes is exactly the number of 0-fixed permutations
of η elements, the number of so called derangements. N. Bernoulli (1687-1759)
has found the following formula:
/1 11 (-l)n\
(3) ρ(η)=Ρη(θ) = η!(--- + --... + ^).
(3), by the way, satisfies the recurrence p(ri) = (n — l)(p(n — 1) +p(n — 2)) and it
can also be used to complete the argument of the Third solution.
331

Solutions
1987/1
2. The original proposal also required the proof of
η
J2(k-lfPn(k) = n\;
k=0
this can be proved by any one of the outlined methods.
1987/2. In an acute-angled triangle ABC the interior bisector of angle A
meets ВС at L and meets the circumcircle of ABC again at N. From L
perpendiculars are drawn to AB and AC, with feet К and Μ respectively. Prove
that the quadrilateral AKNM and the triangle ABC have equal areas.
First solution. The idea of the proof is to show that the area left when
removing the triangle BNC from
the quadrilateral ABNC is equal to
what is left of the very same
quadrilateral if the two triangles, KBN
and MNC are removed. For this
it is enough to show that [BNC] =
[KBN] + [MNC] and this is going
to be done by dividing the triangle
BNC in such a way that the
respective parts are of the same area as the
triangles KBN and MNC Figure
87/2.1.
First of all, К and Μ are
interior to the respective sides AB and
AC because the triangle ABC is
acute. Denote the second
intersection of the line ВС and the circumcircle of the cyclic AKLM by P. (P and L
might coincide if the triangle is isosceles.) By the theorem of inscribed angles in
this circle and also in the circumcircle of AABC
N
Figure 87/2.1
LA
= LLAM = LLPM\
LA
= LB AN = LBCN = LPCN;
2 ' 2
hence the segments MP and NC separated by the line PC making equal angles
with the latter; therefore, they are parallel. Hence AMNC and APNC have
equal area: [MNC] = [PNC]. Replacing К and В with Μ and C, respectively,
we get, similarly, that [KNB] = [PNB]. This is the end since the triangles PNC
and PNB add up to the triangle BNC and thus [MNC] + [KNB] = [BNC],
indeed.
332

1987/3
1987.
Second solution. There are many ways to prove the claim using various
well known metric relations of triangle geometry. The following is an option.
We shall use the notations of the previous solution.
It is well known (c.f. the Note) that if h is the bisector of LA then
b + c
AN = h = - j.
2cos^
Being symmetric with respect to the bisector of LA [AKNM] = -AN · KM.
The angle bisector theorem, on the other hand, yields CL- . Hen-
b + c
ce in the right triangle CML ML = CLsinC; in AMKL we have
/π — A\ A
KM = 2ML sin I 1 = 2ML cos —. Putting all these together:
г.г^лп.^1 1 b + c ^^rT A 1 ч ab . _ absinC _ λ π^Ί
[AKNM] = -·- j-2MLcos- = -(b + c)-—smC=— = [ABC],
2 2cos| 22 b + c 2
the desired result.
Note. We prove now the result about the length of the chord bisecting I A.
In the isosceles triangle BNC BN = NC = j. Ptolemy's theorem in the
2 cos ^
cyclic ABNC yields
2cos^ 2cos^
= AN-a,
and hence
,„ b + c
AN =
2cos^'
1987/3. Let x\, xi> ■. ·, xn be real numbers satisfying
(1) x\ + x\ + ... + x\ = \.
Prove that for every integer k>2 there are integers щ (i = 1, 2, ..., n), not all
zero, such that \щ\<к — \ for all i, and
(2) \αιΧΐ+ωΐΧ2 + ... + αηχη\ <—-ζ—-—.
Solution. Applying the A.M-Q.M inequality for (1) yields
1 lx\ + X2 + .. . + x% \χ\\ + \χ>ι\ +.. . + |жп|
η V η — η
that is
|:ri| + |iC21 + ■ · · + \xn\ < y/n.
ЪЪЪ

Solutions
1987/3
Consider now the kn — 1 sequences of length n, none of them is constantly
zero, that are composed from the numbers (0, 1, 2, ..., к — 1); let one of them
be (a\, a2, ■ ■ ■, an). Set now the sign of щ in such a way that the product щх^
is not negative (i= 1, 2, ..., n). Then clearly
(3) a\X\ + .. . + anxn= \a\\\xi\ + la^ll^l + ■ ■ · + |αη||^η| <
<(k- 1)(|сс1| + |ж2| + · · - + |жп|) <(k~ l)y/n.
Split the interval [0; (k — 1)л/п\ into kn — 1 equal parts; the length of each part
(k — l)\/n
is — —. If any one of the sums above is inside the first interval, then (2)
kn — 1
clearly holds for this sum and we are done. If there is no such a sum then, by
the pigeonhole principle, there are at least two of them in some of the remaining
kn — 2 intervals. If these sums are
b\x\ +62^2 + · · ■ + bnxn and c\x\ +C2X2 + - · ■ + cnxn (\bi\, \ci\<k—l),
then, obviously, their difference cannot exceed the length of the interval:
\bixi+b2X2 + ·· . + bnxn-c\x\ -c2X2~ ...-cnxn\ =
(k — l)\/n
= |(bl -ci)xi+\(b2-C2)x2 + --- + (bn-cn)xn\ < n _ .
Since hi and Ci are of the same sign, \bi — Ci\<k — l. Now clearly ai = bi — c{ are
integers of the desired property:
(fc-l)v^
\a\x\ +a2x2 + · · · + αη^η| <
kn-l
Remark. (3) can be obtained immediately from (1) applying Cauchy's ine-
9 9
quality and then af < (k — 1) ; indeed
a\ x2 + a2x2 + ... + anxn < Ja^ + a^ + ■.. + a^Jx^ + x\ + ... + x\ <
< y/n(k - l)2\/T = (/c - l)v^.
1987/4. Prove that there is no function f from the set of non-negative
integers into itself such that
(1) f(f(n)) = n+1987
for all n.
Solution. Assume the contrary. Substituting f(n) for η in (1):
(2) /(/№))) = Дп) +1987,
and plugging also the respective sides of (1) for η
(3) ДДДп))) = Дп+1987).
334

1987/4
1987.
Combining (2) and (3) yields
(4) f(n+ 1987) = /(n)+ 1987.
Let t be an arbitrary positive integer. We now prove by mathematical
induction that
(5) f(n + 19870 = f(n) + 1987i.
Fix the value of n. For t = 1 the claim is but (4). By the induction hypothesis
/ (n + 1987(i - 1)) = f{n) + 1987(i - 1).
Substituting here n + 1987 for η (4) implies
/(n + 1987*) = /(n+1987)+ 1987(i-l) =
= /(η) + 1987 + 1987(ί-1) = /(η)+1987ί,
which is (5), the proof is complete.
Let s be now an arbitrary non negative integer less than 1987 and consider
the remainder r when f(s) is divided by 1987.
f(s) = 1987/c + r. (0 < к and 0 < r < 1986), /egno(6)
since, by condition, f(s) is not negative. Hence by (1)
fifis)) = s+ 1987,
and, from (6) and (5)
/ №)) = /(1987/c + r) = /(r) + 1987/c.
The last two results imply
(7) s +1987 = /(r) + 1987/c.
Since s< 1987
/(r) + 1987/c < 2 · 1987, /(r) < 1987(2 - /c).
Given that f(r)>0 there are two possible values of k: either k = \ or /c = 0. In
the first case (6) and (7) yield
(8) /(s)=1987 + r,
(9) /(r) = S,
and this overthrows r = s; indeed, plugging s for r in (8) and (9) forces 1987 = 0,
a contradiction.
In the other case к = 0. Combining again (6) and (7) implies
(10) fis) = r,
(11) /(r)= 1987 + 5;
335

Solutions
1987/4
and, like before, r = s leads to a contradiction.
Making the two ends meet (9) and (10) together imply that when acting on
the numbers
0, 1, 2, ..., 1986
/ is arranging them into pairs (a, b) in such a way that either
f(a) = b and /(b) = a+1987,
or f(b) = a and f(a) = b+ 1987,
and, of course, the numbers in each pair are different. Now this is a contradiction
since the number of elements of the set 0, 1, ..., 1986 is odd.
Remark. The claim holds, of course, for any odd positive number instead of
1987.
1987/5. Let η be an integer greater or equal to 3. Prove that there is a set of
η points in the plane such that the distance between any two points is irrational
and each set of3 points determines a non-degenerate triangle with rational area.
First solution. In the solutions we shall use a well known result about the
area of polygons formed by lattice points i.e., points of integer coordinates: the
area of such lattice polygons is a rational number. Consider now η independent
lattice points. (The points of a set are independent if the are no three collinear
one among them.) A simple induction argument shows that this can be done:
indeed, any finite set of points determines but a finite set of lines so these lines
cannot contain every lattice point; hence any finite set of independent points can
be increased.
As it was noted, the respective areas of the triangles formed by these points
is rational. Consider now the pairs from this set of points and denote the square
of their distances by d\, d>i, ..., dk, respectively. Being calculated in the usual
manner from integer coordinates these numbers are also whole numbers. Let ρ
be a prime which is not dividing any one of these di and magnify the lattice by
Vp·
Consider now the image of our set of η independent points. Multiplied by ρ
the area of each magnified triangle is still rational. Distances, on the other hand,
are scaled up by y/p and thus the ith. one magnifies into y/pdi. By the choice
of ρ its index in the factorisation of pdi is 1, pdi is not a square and thus its
square-root is irrational, indeed, which completes the proof.
Second solution. Pick η lattice points on the parabola y = x . The coordi-
nates of these points are of the form (a, a ) where a is integer.
336

1987/6
1987.
Since any straight line meets the curve at two points, at most, there are no 3
collinear among the selected points and the area of the triangles formed by them
is a rational number, as that of any lattice triangle,
The distance of the points A(a, a ), B(b, b ) is
AB = y/(a-b)2 + (a2-b2)2 = \a- b\y/l + (a + b)2.
о
Since there is no positive square whose neighbour is also a square, (a + b) +
1 is not a square, its square root is irrational; a — b is a whole number different
from zero and thus \a — 6|y (a + b)2 + 1 is also irrational: the pairwise distances
in our set are irrational numbers.
Note. The general result about the area of a lattice polygon clearly follows
from the special case of the triangle. Now any lattice triangle can be inscribed
into a rectangle whose sides are lattice lines and thus its vertices are lattice points.
Hence the area of the latter is an integer and the triangle itself can be obtained by
removing right triangles of integral legs from the rectangle decreasing its area by
the half of an integer each time: the area of our lattice triangle is indeed rational;
it is, in fact, the half of a whole number.
This result also follows from the triangle area formula: if the (integral)
coordinates of the vertices are (χγ, у ι), (x2, 2/2)> (ж3> 2/з)> respectively then the area
of the triangle computes as
1
*=2
X\
x2
ΧΊ,
y\
У2
УЗ
1
1
1
= 2 (ж1(2/2 —2/3> + ^2(2/3 — 2/l) + «3(2/i -2/2»
The celebrated result, known as Pick's theorem about the general lattice
polygon states that its area is equal to
A = i-l + \,
where г and b denote the number of lattice points inside and on the boundary of
the polygon, respectively. This also yields, as a byproduct, that the area of lattice
polygons is rational.
1987/6. Let η be an integer greater or equal to 2. Prove that if к + k + n
—, then к +k + n is prime for all
integers к such that 0 < к < η — 2.
Solution. Set f{k)-k +k + n. What is to be proved, in fact, is the
following: if /(0), /(1), ..., f
/(0), /(1), ..., Дп-2).
n/3
are all primes then so are the numbers
337

Solutions
1987/6
If the claim is false then there is a lowest non negative integer у such that
у < η — 2 and f(y) is composite. (Since η > 2 f(y) > 2 so if not a prime it has
to be composite.) By the choice of у for any к such that 0 < к < у — 1 f(k) is a
prime number.
Denote the smallest prime divisor of the hence composite f(y) by q. first
we show that q > 2y. Assume the contrary and consider the differences
(1) f(y)-f(k) = y2 + y + n-(k2 + k + n) = (y-k)(y + k + l)
as the integer к varies from 0 to у — 1. The array below displays the values of к
and the factors, y — k and y + k + l, of the respective differences:
k:
y-k:
y+k+l:
o,
У,
Ϊ/+Ϊ,
1,
y-h
У + 2,
2, ..
У -2, ..
2/ + 3, ■·
y-2 y-l;
2, 1;
., 22/-1, 2j/.
One can check that the union of the ranges of the respective factors, у —
к and y + k + l contains every integer from 1 to 2y and thus, in particular, it
contains q which, by assumption, is not exceeding 2y. Hence there is a value of
к such that 0 < к < у — 1 and (у — k)(y + к +1) = f(y) — f(k) is a multiple of q.
q, by its choice, divides f(y), therefore it also divides f(k). On the other hand,
both f(k) and q are prime numbers which leaves f(k) = q as the only option.
By the choice of y, however
y-k<n~2<n + k + k2 = f(k) = q,
y + k + l<n- l+k<n + k + k2 = f(k) = q.
These inequalities imply that neither y — k nor y + k+l is a multiple of q. Hence,
as a prime, q cannot divide their product, f(y) — f(k) = f(y) — q so it does not
divide f(y) contradicting its choice. The proof of q > 2y is complete and observe
also that since q is an integer, this slightly improves to q > 2y +1.
We are almost at the end now. Since q is the lowest prime divisor of the
composite f(y), they are different, moreover
f(y)>q2>(2y + l)2 = 4y2+4y+l,
y2 + y + n> 4y2 + 4y+l.
Rearranging we get 3y +3y + 1 — n<0. Satisfying this inequality, у cannot
exceed the higher root of the polynomial on the l.h.s.:
2/<"+л/-(-у2' and11™55 У <
According to the condition, however, if у satisfies this inequality then f(y) is a
prime and thus /(0), /(1), ..., f(n — 2) are all primes, a contradiction.
338

1988/1
1988.
Remark. The curious way that certain prime values force further values of a
polynomial to be prime makes this one a really intriguing problem. If η = 41, for
example, then having checked that к + к + 41 is prime for к = 1, 2, 3 the further
36 values, according to the problem, are also prime numbers: 61, 71,..., 1061.
(This perplexing property of χ +£ + 41 was, by the way, known to L. Euler in
the 18th century.
Being so, it was straightforward to ask for those values of η which, indeed,
have the property described in the problem. This has been undecided for a long
period, to the extent, that even false assertions have been announced; in an 1939
paper η = 72 491 was stated to have this property; this, however, has turned out
to be false. A result of Η. M. Stark from as late as 1967 implies that there are
seven such values of η altogether; they are
n=l, 2, 3, 5, 11, 17, 41.
There are certain algebraic number theoretical investigations of complex
numbers in the background of the problem. The reader can find further
information about the topic in E. Gyarmati: A note on my paper: "Unique prime
factorization in imaginary quadratic number fields" (Annales Univ. Sci. Budapest.,
Sectio Mathematica, 26 (1983), 195-196.)
1988.
1988/1. Consider two coplanar circles of radii R and r (R>r) with the
same centre. Let Ρ be a fixed point on the smaller circle and В a variable point
on the larger circle. The line BP meets the larger circle again at C. The
perpendicular I to Β Ρ at Ρ meets the smaller circle again at A (if I is tangent to
the circle at Ρ then A = Ρ).
(i) Find the set of values of ВС2 + С A2 + AB2.
(ii) Find the locus of the midpoint of AB.
Solution. Denote the common centre of the two circles by O. The smaller
circle divides the segment ВС into parts x, у and χ again. The chord of the line
I inside the smaller circle is z. Both у and ζ might reduce to zero if ВС, or I are
tangent to the smaller circle (Figure 1988/1.1).
Applying Pythagoras' Theorem on the perhaps degenerate triangles APC
and APB:
AB2 + В С2 + С A2 = x2 + z2 + (2x + y)2 + (x + y)2 + z2 = 6(x + y)x + 2(y2 + z2).
As perpendicular chords in the smaller circle, у and ζ are the legs of a right
9 9 9
triangle of hypotenuse 2r and thus у +z =4r ; furthermore, the power of Ρ
339

Solutions
1988/1
Figure 88/1.1
9 9
with respect to the larger circle is (x + y)x = R —r and hence
AB2 + ВС2 + С A2 = 6(R2 - r2) + 8r2 = 6R2 + 2r2
is constant; the first part of the problem is hence settled. We note here that В
and G can be swapped in the argument but this, of course, has no effect on the
result.
Moving on to the second part denote the mirror image of В through the
perpendicular bisector of AP by B' (if Ρ = A then, of course, В = В'). The
respective points clearly form a rectangle PAB'B whose centre, Q, is the midpoint
of AB. Thus the points Q in question are produced by reducing the larger circle
from Ρ by half. K, the centre of this circle к is the midpoint of PO and its
radius is —.
2
Each point of к belongs to the locus: indeed, if Q is on the diameter PO
then Ρ = A and thus В and С are the respective endpoints of the diameter PO. If
Q is not on PO then enlarging it from Ρ by a scale factor 2 yields a point B' on
the larger circle. Now clearly PB' > R — r and thus the Thales-circle of diameter
PB' meets the larger circle once more at some point B. For the corresponding A
the point Q is clearly the midpoint of AB that is Q belongs to the locus, indeed.
The locus of the midpoint of ВС is the circle of diameter R about the
midpoint of OP.
1988/2. Let η be a positive integer and let A\, A<i, ■ ■., ^2n+l be subsets
of a set B. Suppose that
340

1988/2
1988.
a) each A{ has exactly 2n elements,
b) each Ai nAj (1 < i < j < 2n + I) contains exactly one element, and
c) every element of В belongs to at least two of the Д.
For which values of η can one assign to every element of В one of the
numbers 0 and 1 in such a way that each Ai has 0 assigned to exactly η of its
elements?
First solution. We start by showing that each element of В belongs to
exactly two Ai.
The problem is worth rephrasing in graph terminology. Consider a graph
G of s + 2n+ 1 vertices. Assign each element b\, bi, . · ·, bs of В and also the
subsets Ai, i= 1, 2,..., 2n+ 1 to the vertices of G, labelling each vertex by
corresponding symbol. Vertex Ai is connected to vertex bj if and only if bj is in Ai\
these are the edges in the graph: G is a so called bipartite graph. The conditions
when also rephrased become:
a) the degree of each vertex Д is In,
b) for any two vertices Ai and Aj there is exactly one vertex b^ connected
to both of them,
c) the degree of any vertex bj is at least 2 (Figure 1988/2.1).
We are to prove that the degree of every element bj of В is exactly 2. Let
Ai Aj
Figure 88/2.1
be В arbitrary. Its degree, by c), is at least 2. Assume, to the contrary, that there
are three edges incident to b connecting it to the vertices A\, A2 and A$, for
example. From A\ there are further 2n — 1 edges, by a), connecting it to b\, b2,
..., &2n-l an<^' similarly, Ai and A-$ are adjacent to the vertices bin, ..., b^n_i
and 64^1, ..., b6n_3, respectively.
By b), none of these, altogether 6n — 3 vertices are connected to any one
of A\, Ai and A3. Indeed, if, for example, bin would be connected to A3, for
example, then the pair Ai and A3 would, in fact, be adjacent to two elements
341

Solutions
1988/2
of В, namely b and b2n violating b). The 6n —3 neighbours of A\, A2 and Αί,
listed above are thus all different.
Hence, by c), there are at least 6n — 3 edges connecting the above listed b\,
&2> · · · > ^6n-3 to ше remaining A-vertices, A4., A5, ..., A2n+\. There is,
therefore, some Ai (4 < i < 2n+ 1), adjacent to at least four of the 6-s listed above
since 3(2n — 2) < 6n — 3. Without loss of generality this Л-vertex can be
assumed to be Л4. There are four edges connecting Л4 to the three sets of vertices
{h, · ■ ·, b2n-i}, {b2n, · · ·, Ч1-2} a"d {hn-1> ■ · ■ > ъвп-ъ) so two of these edges
enter the same set: they arrive, for example, to the vertices h\ and b2, for
example. Now this is a contradiction, since A\ and Д4 are thus connected to two b-s
which is prohibited by b).
Hence the degree of each bj is exactly 2, indeed.
Assigning now 0 or 1 to some bj in the graph will be carried out by
colouring this bj red or green, respectively, and, at the same time, by colouring also
the edges starting from this bj with the very same colour. Hence, if the
assignment of the problem is feasible then the edges of G are all coloured and their
colour is the same as that of their 6-endpoint. Having thus coloured a proper
assignment there are η red and η green edges leaving each vertex Ai therefore
there are, altogether, (2n+ l)n red edges in G.
Any red edge is between some Ai and a (red, of course) b^\ the degree of
the latter is 2 and thus there is another (red, of course) edge connecting it to
some Aj. This simple observation implies that the red edges can be (uniquely)
divided into pairs, their total number, (2n+l)n is hence even and thus η itself
is also even. This is necessary for the required 0-1 assignment to exist.
We show now, representing the sets in question, that the condition for η to
be even is also sufficient. For arbitrary even value of η mark, on a circle, the
vertices A\, A2, ..., A2n+\ of a regular 2n+ 1-gon. Connect each vertex to the
Ti
neighbouring — vertices to its left and also to its right. Colour these segments
red and all the remaining diagonals green. Under this colouring there are clearly
η red and η green edges leaving each vertex. The respective sets, together with
the required assignment are now the following:
set В consists of the — coloured — edges, the red ones are labelled by 0
and the green ones by 1. The subsets Д are the edges leaving the vertex A^
a), b) and c) do obviously hold for this construction: there are 2n elements
of each Ai\ the edge connecting the corresponding vertices is the single common
element of Ai and Aj\ finally any element of В is contained by exactly two Ai,
these are the ones labelled by its endpoints.
Second solution. In what follows, a general approach to the first part of
the problem, is outlined.
342

1988/2
1988.
Αχ
Α2
Α3
A2n+l
h
1
1
0
0
ь2
0
1
0
1
ъъ
1
0
1
0
• . ·
. . .
•
. . .
bs
1
0
0
1
Let В have s elements, b\,bi, ■ ■ ■, bs.
The set В together with its 2n+l subsets
A\, A2, ..., A2n+\ can be displayed in a
(2n+l) by s array (matrix) whose entries
are 0 and 1, the so called incidence
matrix of the graph G of the first solution. The
columns — there are s of them — are to
represent the elements of B. Write 1 in the
respective entry if the very element b
corresponding to its column belongs the subset
corresponding to its row and 0 otherwise:
the rows are hence for the 2n+ 1 subsets.
The number of 1 entries, by the condition, is In in each row. Thus, if Μ is
the total number of 1-s in the array then Μ = 2n(2n+ 1). Denote, additionally, the
number of 1-s in the respective columns by c\, c2, ■ ■ ■, cs, respectively. Clearly
M = ci+c2 + . .. + cs.
The zth row of the array, for brevity, shall be considered an s-dimensional
vector щ whose components are the 0 and 1 entries of the respective rows. Thus,
for example
ai(l,0,l,...,l).
The conditions of the problem are then
а) а? = 2п(г=1, 2, ..., 2n+l);
c) c% > 2, (i = 1, 2, ..., s), and thus Μ >2s that is s < n(2n+ 1). Let a be
the sum of the vectors a^. Since its components are (c\, C2,..., cs), the square of
the vector a computes as
(1)
(a)2 = c\ + c\ +
+ c
s'
On the other hand, with a bit of algebra
2n+l /
(2) (a)2= Σ а2 + 2^]а^ = (2п + 1)2п + 2
1 i<3 \
2n+l
= 4n(2n+l).
The A.M.-Q.M. inequality provides a lower bound for the r.h.s. of (1) and
thus
(3)
Ml 1 о о
= -·4η2(2η+1)2.
s s
343

Solutions
1988/2
Comparing (2) and (3) yields
s > n(2n+1).
This when combined with the inequality c) forces s = n(2n + 1); there is equality
in (3) and hence the numbers q are all equal. Thus q = 2 for every i that is each
element b is contained in exactly two subsets, indeed.
Note. Set В with its subsets A{ as they are given forms a so called block-
design. Their research is an important task of combinatorics.
1988/3. A function f is defined on the positive integers by
(1) /d)=l, /(3) = 3,
(2) f(2n) = n,
(3) Д4п + 1) = 2Д2п+1)-Дп),
(4) Д4п + 3) = ЗД2п+1)-2Дп)
for all positive integers n.
Determine the number of positive integers n, less than or equal to 1988, for
which f(ri) = n.
Solution. Note, first of all, that formulas (l)-(4) determine function /
unambiguously, since the numbers of the form 2n,4n + 1 and 4n + 3 exhaust the
set of integers. Hence it is enough to provide a function / whatsoever satisfying
conditions (l)-(4).
Check a few values of /:
η 123456789
f(n) 113 15 3 7 19
Far from a pattern but as for the values satisfying f(n) = n they seem to
show up next to the powers of 2. This might suggest to rewrite the rows in base
2:
n10 12 3 4 5 6 7 8 9
n2 1 10 11 100 101 110 111 1000 1001
f(n)2 1 01 11 001 101 011 111 0001 1001.
Now it is hard to resist the suspicion that /, when in action, inverts the
order of digits in the binary representation of n. We shall, in fact, prove this by
induction on k, the number of binary digits of n.
Note, first of all, that the last binary digits of the numbers In, 2n+ 1, An,
An+1, An + 3 are respectively
In 2n+l An 4n+l 4n + 3
0 1 00 01 11
344

1988/3
1988.
In what follows binary strings are going to be overlined. If к = 1 or 2 then
the claim has already been verified above. Let к > 2 and assume that it holds
for any number whose binary form consists of less than к digits, i. е., if n =
ak_\ak_2 ... αχ {ak_\ = 1); then
(5) f(n) = aia2...ak_i.
1. Set now m = 2n = a\ai... ak_\0. Then clearly
f(jn) = f(2n) = /(n) = oA._1oA._2...oi= 0ak_iak_2 ... αϊ,
(5) holds for fc-digit even numbers. The odd case is checked in two parts.
2. Let m = 4n+l = aia2 .. -ak_20l, where η has now k — 2 binary digits
(k>2), n = a\a,2 ... ak-2· Hence 2n + 1 = a\a2 ... ak-21 > and thus, by (3)
f(m) = 2f(2n+l)-f(n) = f(2n+l) + (f(2n+l)-f(n)) =
= lak_2 ·. · 0,20-1 + lo/j._2 · · · 0201 — a>k—2 · ■ · 0201 =
= lak_2 . . . 0201 + 10 .. . 0= Юак_2 ... α>2α\·>
so (5) holds for fc-digit numbers of the form 4n +1 as well.
3. Let m = 4n + 3 = aia2.. .ak_2ll, where n, like before, is equal to
aia2...ak_2. Using (4) again
/(m) = 3/(2n+l)-2/(n) = /(2n+l) + 2(/(2n+l)-/(n)) =
= lak_2 ... a\+2 [\ak_2 ■. ■ αχ - ak_2 ... o2oij =
= lak_2 ... 01 + 10... 01 = llak_2 ... 01,
that is (5) holds for к -digit numbers of the form An + 3 and this one was left to
be checked for the induction to be complete.
Having understood the way / works it is obvious that f(ri) = n for those
numbers whose binary form is symmetric that is reversing the order of their digits
yields the very same string. These are the so called palindromes, like 10111101,
for example.
The leading digit of a 2n long binary palindrome is 1 and there are 2n~
ways to set the next η — 1 digits; hence there are this many 2n digit binary
palindromes, altogether. The number of 2n + 1 long binary palindromes is clearly
twice this much since each of them can be obtained from a unique 2n long one
by inserting a single 0 or 1 into the middle.
As 210 < 1988 <2n, the binary form of 1988 has 11 digits. The previous
observations hence imply that the number of at most 11 long binary palindromes
is the sum
20 + 2° + 21+21+22 + 22 + 23 + 23+24 + 24 + 25=94.
345

Solutions
1988/3
We are not finished yet. 1988 = 11111000100 and thus there are two 11-digit
binary palindromes above 1988: 11111011111 and 11111111111. Hence the
correct answer is 92.
1988/4. Show that the set of real numbers χ which satisfy the inequality
70 к 5
(1) T-^T>-
K J f-ix-к- A
is a union of disjoint intervals, the sum of whose lengths is 1988.
Solution. Denote the function on the l.h.s. of (1) by fix); its domain is
the set obtained by removing the numbers 1, 2, ..., 70 from the set of reals; it
is the union of the open intervals
]-oo,l[, ]1,2[, ]2,3[, ..., ]69,70[, ]70,+oo[.
The terms of the sum defining / are all continuous and decreasing inside these
intervals and so is their sum, /. In the open interval ] — oo, 1[ / is negative;
ι
at the values к (k = l, 2, ..., 70) the terms f(x) = (i^k) are continuous
x — i
and bounded; the limit of fk, as χ tends to к from the left and from the right
is —oo and +oo, respectively, so this holds for /, too. This means that as a
continuous function in the interval (к, k + 1) (1 < к <70) varying decreasingly
from +oo at the left endpoint to — oo at the right one, / admits every real value
in this interval. As for the last interval, ]70, +oo[ here / is positive, its limit at
the lower endpoint from the right is +oo, and at +oo it is 0. Being so / admits
every positive value in the interval ]70, +oo[. (To make life simpler on Figure
1988/4.1 we have sketched the 3 term sum / when к varies from 1 to 3.)
4 -
λ -
2 -
1 -
-ι ^! -i-
-4 -
- С
Ρ ι
χ
\
x:
1
\
4^
■^ ' ι
4 x35 6
I
Figure 88/4.1
346

1988/5
1988.
The diagram also shows that, apart from the first one, [—00,1[, / admits
5
the value - in every interval, each of which, hence, contains a subinterval where
5
/ > τ ( f is decreasing).
If / admits the value - at x\, x2, .. ·, £70 (k <Xk<k + l, if к < 69 and
5
£70 > 70) then / > - in the disjoint intervals
]l,x\], ]2,x2], ..·, ]69,z69], ]70,ж70].
Their total length is:
(2)
Я = Οι- 1) + (x2- 2) +... + O70- 70) = (χι + x2+ ■ ■.+ xjo) - (1 + 2 +... + 70).
The task remaining is to calculate the sum χ ι + x2 + ■.. + xjq. To make it simpler
switch to the function g(x) = f(x) . The zeros of this g are clearly the previous
Xi values. It is a first degree rational function, its denominator is (x — l)(x —
2)... (x — 70), while its numerator is
1 · (x - 2) ■ (x - 3)... (x - 70) + 2 · (x - 1)0 - 3)... (x - 70) +...
... + 70 · (x - l)(z - 2)... (x - 69) - - · (x - 1)0 - 2)... (x - 70).
The leading coefficient of this 70-degree polynomial is — - and thus the
coefficient of the 69th degree term can be computed by the corresponding Viete-
formula as
χΐ+χ2 + ... + χΊ0 = - Π+ 2+...+ 70 +-(1+ 2 + ...+ 70)) =
= ?(l+2 + ... + 70).
Hence, by (2), the total length of the intervals is
|(1 + 2+... + 70) - (1 + 2 +... + 70) = - · Ί— · 71 = 1988, indeed.
1988/5. ABC is a triangle right-angled at A, and D is the foot of the
altitude from A. The straight line joining the incentres of the triangles ABD, AC Ό
intersects the sides AB, AC at the points K, L respectively. S and Τ denote the
areas of the triangles ABC and AKL respectively. Show that S > XT.
First solution. AB < AC can clearly be assumed. Let the bisector f\ of
LB AD and BD intersect at X and the bisector f2 of LCAD and CD at Υ
(Figure 1988/5.1). Reflect AADX through fx obtaining AAD'X. This reflection
keeps the incircle of ABD fixed and thus this circle is also inscribed to the
kite AD'XD. A similar procedure — reflection through f2 — yields the cyclic
347

Solutions
1988/5
Figure 88/5.1
kite AD"YD whose incircle is the same as that of ACD. Denote, finally, the
intersection of D'X and D"Y by P.
The quadrilateral AD'PD" is a square since there are right angles at A,
D' and D" and, additionally, AD = AD' = AD" because of the reflection. The
diagonal D'D" bisects the angles of the square at D' and D" so it is passing
through the centres of the two incircles. Therefore D' = K and D" = L, the right
triangle AKL is isosceles, it is the half of the square AD'PD".
Reflect now В through f\ to R and С through /2 to Q\ these mirror images
are clearly incident to the altitude AD and, by AB < AC, we have AR < AQ.
Using the equality of areas implied by the respective reflections:
S = [ABD] + [ABC] + [ACD] = [ARD'] + [AQD"] > [AD'PD"] =
= 2[AD'D"] = 2[AKL] = 2Г,
indeed.
Second solution. Denote the incentres of ABD and ACD by Ο χ and O2,
respectively (Figure 1988/5.2). The similarity properties of right triangles imply
that the composition of the enlargement by ^ and the rotation through 90°, both
about Ρ is mapping ADC to BDA and O2 to 0\. Hence DO\ : DO2 = c:b and
thus ADO1O2 and A ABC are similar because the ratios of two corresponding
sides are equal and these sides now make right angles, respectively. This
similarity implies that IDO2O1 = IC. Hence the quadrilateral DO2LC is cyclic and
348

1988/5
1988.
Figure 88/5.2
its exterior angle at L is equal to Z02-DC = 45°. Since LALK = 45° in the right
triangle ALK, it is isosceles, AL = AK.
AALO2 and AADO2 are congruent because they have a common side,
their respective angles at A are equal, finally LADO2 = ZALO2 = 45°. Therefore
AL = AD. Hence
AD2
T = [ALK] = -—~.
be 9 9
Now a ■ AD = be yields AD = — and since Zr + с > 26c,
a
2 Л
T =
Vc
г л
bLc
6c_5
2^~2(b2 + c2)-J~2'
S>2T.
Third solution. Denote the incentres of AD В and ADC by 0\ and O2,
the inradii by r\ and Г2 respectively; also the touching points of the respective
incircles on AB and AD by Ε and I?' and on AC and A.D by F and F; (Figure
1988/5.3). The foot of the perpendicular from O2 to OjP is denoted by Q and
the altitude AD by m.
j.
/y
/ Q/
F / /c
A3
rl
4
Ρ
£'
\P
/2^
^2
r2
v^L
zN^
τ4^
β
D
a
Figure 88/5.3
349

Solutions
1988/5
We shall make use of the equality of the tangent segments to a circle from
an external point. Express, in terms of m, r\ and r2, the distances 0\Q and
02Q:
01Q = 01P-QP = AE-02F = AE' -r2 = m-ri-r2,
02Q = FP = AF-AP = AF' -r\=m-r2-rx.
This calculation shows that the right triangle 0\Q02 is isosceles and thus, having
their respective sides parallel, triangles KAL and 02FL are also isosceles right
triangles. Hence FL = r2 and thus
AL = AF + r2 = AF' + r2 = m.
Since m = c cos С = b sin С
_ m2 be sin С cos С be . ^^ S . n^
Τ = —- = = — sin 1С = - sin 1С,
2 2 4 2
so
Γ<|, 5>2Γ,
and we are done.
Remark. Every solution shows that equality S = 2T holds if and only if the
right triangle ABC is isosceles.
1988/6. Let a and b be positive integers such that ab+1 divides a + b .
Show that
(1) ^
K J ab + 1
is the square of an integer.
Solution. Denote the value of the ratio (1) by q. The claim can be rephrased
as follows: if there is a positive integer q such that the equation
(2) a2-qab + b2-q = 0
is satisfied by some pair (a, b) of positive integers then q is a square.
Suppose the contrary by assuming that (2) has a solution and q is not a
square. Consider the quadratic
9 9
(3) χ — qxy + y —q = 0 (q>0 integer).
Plugging numbers of opposite sign for χ and у clearly — qxy > q, the l.h.s. of (3)
is positive so equation (3) does not have roots — not even reals — of opposite
sign. Even if one of the solutions, say x, for example, is zero then this would
force q-y but exactly this was assumed to be false. Since any solution (x,y)
350

1988/6
1988.
implies also (—χ, —у) to be a solution of (3), from now on we can restrict
ourselves to the strictly positive integral solutions of (3). By condition this set is not
empty.
Consider now an element of the set of positive integral solutions of (3) for
which the sum of the square of the components is minimal; denote one of these
solutions of (3) by (A, B); here 0 < В < A can clearly be assumed. Hence
(4) A2-qAB + B2-q = 0
and there is no positive integral solution whose square sum is smaller than A +
B2.
Now (4) also says that Л is a positive integral solution of the quadratic
(5) x2-qBx + B2-q = 0.
Consider now the other solution A' of (5). By the previous observation A' is also
positive. Writing down the formulas of Viete
(6) A + A' = qB, (7) AA' = B2-q.
(6) implies the A' is also an integer, since A and qB are integers. (7) yields
AA! < B2 and this, together with A > B, shows that A' < В and thus
A,2 + B2<2B2<A2 + B2.
Now this is a contradiction. Indeed, the pair (A', B) is also a positive integral
solution of (3), assuming that q is not a square we have managed to produce a
solution of lower square sum than the certainly existing minimum. The proof is
thus finished.
Remarks. 1. It is natural to ask if there exist positive numbers (a, b) of the
given property at all. One can show that there are, in fact, infinitely many of
them and the following recurrence lists them all:
9 ....
£o = 0, x\=a, xn+\ —a xn — xn-\, where a is a positive integer, η > 1.
The solutions of (5) are the pairs (x\, x2), (x2, £3), · · ·, (^n-l > xn), · · ■', the value
of q is a .
2. The method of the solution is called infinite descent in arithmetic. It is
able to prove that a certain diophantine equation has no solution by choosing a
particular hypothetical solution of some minimal positive integral measure and
and then proceeding to another solution whose measure is definitely smaller. It
was invented by P. Fermat in the 17th century.
351

Solutions
1989/1
1989.
1989/1. Prove that the set {1,2,..., 1989} can be expressed as the
disjoint union of subsets A\, Αι, · · ·, Α\\η in such a way that each A{ contains 17
elements and the sum of the elements in each A{ is the same.
First solution. The subsets A{ will be formed of 7 pairs and one triple each,
in such a way that the sum of the pairs and also that of the respective triples is
the same in every subset. In the first step the altogether 117-7 = 819 pairs will
be prepared and the 117 triples afterwards.
The first 351 integers are set aside for the triples and the pairs will be formed
from the integers 352, 353, ..., 1989 in a straightforward manner:
(352,1989)
(353,1988)
(1) .
(1170,1171).
These 819 pairs contain every number from 352 to 1989 and the sum of
each pair is 2341, the same.
With a bit more care the remaining numbers can also be arranged into 117
triples, with the same sum each:
(1,176,351) (60,118,350)
(2,177,349) (61,119,348)
(3,178,347) (62,120,346)
(2)
(58,233,237) (116,174,238)
(59,234,235) (117,175,236).
You can check that every integer, from 1 to 351 appears but once and the sum
of each triple is 528.
The subsets Ai now can be assembled one by one as the union of 7 pairs
from list (1) and a triple from list (2).
Second solution. As a more structural approach arrange the first 1989
integers in a 117 by 17 array in their natural order:
1 2 3 ... 17
18 19 20 ... 34
(1) :
1973 1974 1975 ... 1989
352

1989/2
1989.
The task is now to rearrange the entries in such a way that the sum of the
elements is the same in each row. If we succeed then the 117 rows will form
the required subsets. The numbers in the jth column are clearly j, j + l· 17,
j + 2 · 17, ..., j + 116 · 117. Denoting the г-th one by j + a{j · 17 (0 < α# < 116)
the sum of the zth row is
17 17 17
si = Y/(j + aij-n) = Y^j+n-J2 aij.
i=l .7=1 i=l
Since ^2,3 is constant it is enough to calibrate the numbers ац in such a way
that the sums \\ ац are equal in the respective rows.
This can obviously be done with any even number of columns by simply
reverting the order of numbers in every other column. We have to be careful,
however, since the number of columns is now odd: the last extra column ruins
balance. We need the last three of them to maintain equilibrium, as it is shown
below.
г— \
1
2
58
59
60
116
117
1
0
1
57
58
59
115
116
2
116
115
59
58
57
1
0
3
0
1
57
58
59
115
116
4 ...
116
115
59
58
57
1
0
... 14
116
115
59
58
57
1
0
15
0
1
57
58
59
115
116
16
59
60
116
0
1
57
58
17
115
113
1
116
114
2
0
Array (2) clearly shows that preparing the subsets Ai as it is described above
each number, from 1 to 1989 appears but once. In fact, the columns of array (1)
are permuted according to the numbers in array (2).
1989/2. In an acute-angled triangle ABC, the internal bisector of angle A
meets the circumcircle again at A\. Points B\ and C\ are defined similarly. Let
353

Solutions
1989/2
Aq be the point of intersection of the line AA\ with the external bisectors of
angles В and C. Points Bq and Cq are defined similarly. Prove that the area
of the triangle AqBqCq is twice the area of the hexagon AC\BA\CB\ and at
least four times the area of the triangle ABC.
Solution. The points Aq , Bq and Cq are clearly the excentres of the
triangle. The bisectors of the respective interior and the exterior angles are
perpendicular, AAq, BBq and CCq are the altitudes in AAqBqCq, the triangle ABC is
the pedal triangle of AAqBqCq. Therefore k, its circumcircle is the nine points
circle of the same AAqBqCq. Moreover, the angle bisectors of ABC meet at
K, the orthocentre of AAqBqCq (Figure 1989/2.1).
Figure 89/2.1
Since the nine points circle bisects the respective distances between
the orthocentre and the vertices, A\, B\ and C\ are the midpoints of
KAq, KBq and KCq, respectively. According to KA\=A\Aq, for
example, [KBA\] = [AiBAq] because they have an equal side and the
corresponding altitudes from В are identical. Similarly, [KCAi\ = [A\CAq\ but then
also [ΚΒΑ^] = [ΒΑ^Α0]. Repeating once more:[KCBiA] = [CBlAB0] and
[KAC\B] = [AC\BCq\. Adding the respective sides of these last three
equalities the l.h.s. is the area of the hexagon AC\BA\CB\ while the r.h.s. is the
354

1989/2
1989.
area of the region making the hexagon up to AAqBqCq which proves the first
proposition.
For the second one it is enough to show that the area of the hexagon is at
least twice of [ABC] since, as we have just proved, [AqBqCq] is the double of
the hexagon's area.
For the proof let us note first that among the triangles inscribed the minor
arc AB of the circle к it is AABC\ that has the greatest area because this one
has the greatest altitude to AB (Figure 1989/2.2). Denote the orthocentre of
ABC by Μ and its mirror image through AB by Mq. It is well known that Mc
Figure 89/2.2
is lying on the circle fc. As it was noted above
[AMCB]<[ACXB]
and equality holds if and only if AC = BC. Similarly, [BMAC] < [BAXC] and
[CMBA]<[CB\A]. Being acute, AABC contains Μ in its interior and thus
the following decomposition is valid:
[ABC] = [AMCB] + [BMAC] + [CMBA],
[ACXBAXCBX] > [AMCBMACMB] = 2[ABC],
the proof is complete.
Note. There are several ways to finish the proof, the second proposition, for
example, can be rephrased as
the area of an acute triangle is at least four times the area of its pedal triangle.
To prove this denote the sides of the triangle by a, b and c, respectively.
If its circumradius is R then, as it is well known, the sides are a = 2R-sinA,
b = 2R-sinB. The sides of the pedal triangle, on the other hand, are acos Д
355

Solutions
1989/2
Ъ cos Б and с cos С and its angles are 180° -2 A, 180° -2Б and 180° -2C,
respectively. Its area is
ab sin С л о . . . „ . _,
t = = 2Br sin A ■ sin В · sin C;
2
and that of the pedal triangle is
, _ a cos A ■ b cos В ■ sin(180° - 1С) _
= AR2 sin A · sin Б · sin С · cos A ■ cos Б · cos C.
The pending inequality At' < t hence becomes
16R2 sin Л · sin В ■ sin С · cos A · cos Б · cos С < 2R2 sin A · sin Б ■ sin C.
This simplifies to the well known
cos A · cos Б · cos С < —.
8
This reformulation of the claim now can be smashed if invoking the
generalization of the so called Sims on line theorem. This big gun states that if the
area of a triangle is a, its circumradius is R and the distance of its incentre from
a given point Ρ of the plane is d then the feet of the perpendiculars from Ρ to
the respective sides form a triangle whose signed area, a is equal to
a(R2-d2) /l df
a —
4R2 \4 AR2
Set now Ρ as the orthocentre of the triangle. Then a = a' is the area of the
pedal triangle and thus
t'<-, At' <t, indeed.
~ 4
Equality holds here if d = 0, that is the orthocentre and the circumcentre do
coincide, i. е., the triangle is equilateral.
1989/3. Let η and к be positive integers and let S be a set of η points in
the plane such that no three points of S are collinear, and for any point Ρ of S
there are at least к points of S equidistant from P. Prove that
k<- + V2n.
First solution. Rearranging and squaring the given inequality we get
(γ \ 2 γ
к — - I =k — k + - <2n which, when divided by 2, becomes
k2-k
356

1989/3
1989.
Hence it is enough to show that I J < η — 1.
For each point Ρ of the set S consider к points, also in S and being
equidistant from P, by condition. There are ( ^ ] pairs formed from these points and
hence there are
(1) η
pairs listed, altogether.
If a certain pair (A, B) is checked at point Ρ then, of course, Ρ is on the
perpendicular bisector of AB. Since there are no three collinear points in S, pair
(A, B) has been counted at most twice in (1). Product (1), hence, cannot exceed
the double of the total number of pairs formed by the points of S:
(2) η
The r.h.s. is n(n — 1) and dividing by η yields the claim.
Second solution. Surprisingly enough the first condition about collinearity
can be ignored. Denote the set of pairs formed by those points in S which are
equidistant from Pi £ S by Hi. If its cardinality is the usual \Щ\ then the
condition is
(3) \Щ\>
The points of the pairs in Hi are on a circle of centre Pi and those of Hj are on
another one about Pj. These two circles have at most two common points that
makes a single pair. Thus
(4) |#*П#,-|<1.
Notice now that the cardinality of the union of the sets Щ when taken for every
point PiE S can be estimated from below as
η η
(5) \jHi >£|#;|-£|#;П#,|.
This is a simple corollary of the Principle of Inclusion and Exclusion but it is
straightforward anyway: the pairs occuring in more Hi at the r.h.s. have been
subtracted more than once.
357

Solutions
1989/3
By (3) the terms of the first sum are at least I |. In the subtracted sum in
(5) there are I J terms, at most 1 each. Hence
η
2=1
>n
On the l.h.s. of (5), on the other hand, certain pairs, formed by the η points of
S, have been counted, their total is hence at most I 1. Putting the estimates
together:
that is 2M>n(jf
which completes the proof.
1989/4. Let ABCD a convex quadrilateral such that the sides AB, ВС,
AD satisfy AB = AD + ВС. There exists a point Ρ inside the quadrilateral at
a distance h from the line CD such that AP=h+AD and BP = h + ВС. Show
that
(1)
1
>
1
+
1
Vh~ Vad Vbc'
Solution. Draw circles k\ and fc2 of radii AD = r\ and ВС = r2 about the
vertices A and D, respectively. According to the conditions these circles are
touching each other externally inside AB and circle к of radius h about Ρ is
touching both of them and also the side CD (Figure 1989/4.1).
Figure 89/4.1
Consider now the varying quadrilaterals while the vertices A, B and also
the two circles k\ and ki are kept fixed ( r2 > r\ can clearly be assumed). In
358

1989/5
1989.
these quadrilaterals CD is connecting two points of the respective circles and
k, the varying circle is touching this segment. It is obvious that к can increase
until it is touching the common exterior tangent to the circles k\ and k2 and this
limiting position yields the highest value of h. Denote this value by h\ clearly
h\>h for any possible h, that is Jh\ > Vh or
(2)
Hence it is enough to prove (1)
for this extremal value h\ when
С and D are on the common
tangent of k\ and k2.
Quadrilateral ABCD is now a right
angled trapezoid. Denote the
projections of Ρ on AD and ВС by
X and Υ respectively and that
of A on ВС by Z. By
Pythagoras in the right triangles PXA
and PYB {Figure 1989/4.2)
1 1
>
Vh
%
Figure 89/4.2
XP = y/(r1+h1)2-(r1-htf = ifxhx,
PY = ^(r2 + h1)2-(r2-h1)2 = 2,f2T1
and also in the right triangle AZB
AZ = yj(r2 + rι )2 - (r2 - rι )2 = 2x/r1r2.
Since AZ = XP + PY,
Vrlr2 = 2Vri hl + 2y/T2hl ■
Dividing by 2Jv\v2h\
which, by(2), implies
1 1
+
^ лАТ v^'
1 >> ! T
> +
Vh~Vad л/вс:
indeed, and it also follows that equality holds if and only if ABCD is a right
angled trapezoid.
1989/5. Prove that for each positive integer η there exist η consecutive
positive integers none of which is a prime or a prime power.
359

Solutions
1989/5
First solution. For a given η we are looking for that value of Μ for which
none of the η consecutive numbers
(1) 2 + M, 3 + M, ..., (n + l) + M
is a prime power. An integer к is not a prime or the power of a prime if it has
two coprime divisors both less than k.
Let г + Mbea term in (1). If г divides Μ then πΐ{ Μ = irrii for some positive
integer or
i + M = i(l+mi).
The crucial observation is that if, additionally, i also divides πΐ{ that is Μ is
a multiple of i then i and l+rrii have no common factor: i and 1 + гтг are
coprime factors of Μ + i, as required.
A number of the given property is clearly M = ((n+1)!) yielding the η
element list
2+((n+l)!)2, 3 + ((n + l)!)2, ..., (n+l) + ((n+l)!)2,
with no prime powers, indeed.
Second solution. The claim clearly holds for n = 1 and 2; for η >2 let p\,
P2, .. ·, Pk primes not exceeding η (that is p\ = 2, p^ = 3, рз = 5, ...) and with
N = p\P2 · ■ -Pk consider the list
(2) N\ + l, Nl + 2: ..., N\ + n
of η consecutive integers. We prove that none of these numbers is a prime power.
Assume the contrary, that is some N\+j, 1 < j <η is equal to the rth power of
a prime p,
(3) N\ + j=pr.
Since N >n>j, this j is among the factors AT! and thus the l.h.s. of (3) implies
that j is a divisor of pr. This j, hence, is equal to ps, where s < r is a positive
integer, simplifying by j in (3) yields
(4) l-2-3-...-(j-l)-(j + l)-...-N+l=pr-s.
j =ps, on the other hand, implies that ρ < j < n, p, as a prime not exceeding η is
one of those listed as p\, p2, ..., Pk- By construction, however, ρ then divides
N, a contradiction.
Afofe. In the solution we used the estimate N >n. This should, in fact, be
proved. Assuming the contrary, ΐ)ί = ΡιΡ2 · ■ -Ρη^η, none of the primes p\, P2,
..., pk can divide the one less N — 1. Since these are all the primes up to n,
N — 1 as a number less than η cannot have prime divisors at all, it must be
equal to 1. This means that к = 1 and n = 2 which was excluded at the beginning.
360

1989/6
1989.
1989/6. A permutation x\, X2, · · ·, %2n-l> x2n of the set 1, 2, ..., 2n — 1,
2n where η is a positive integer is said to have property Ρ if \x{ — xi+\\ =n
for at least one i in {1, 2, ..., 2n — 1}. Show that for each η there are more
permutations with property Ρ than without.
First solution. Call, for brevity, permutations with property Ρ simply P-
permutations and the remaining ones Q-permutations. For any 1 < i < In set j
as the pair of i if 1 < j < 2n and \i — j\=nt that is the numbers whose difference
is η are arranged into pairs.
We shall prove the claim by defining a simple one to one mapping from
the set of Q-permutations to a proper subset of the P-permutations. If x\, X2,
..., X2n is a Q-permutation and the pair of X2n is Xk then define the image
of this permutation to be the one whose last element has been removed and
inserted behind its pair. This image is clearly a P-permutation and, additionally,
it contains but a single neighbouring pair that is not at the end of the permutation.
It is clear that from any P-permutation of this property we can reconstruct
the corresponding Q-permutation, the mapping is indeed one to one. Going on,
if η > 1 then 1, 2, 3, ..., η — 1, n+ 1, ..., n, 2n is a P-permutation and not
like those in the range of our mapping; it is hence a proper subset of the P-
permutations, indeed. Finally, if η = 1 then there is just one permutation and it
has property P, the proof is hence complete.
Second solution. Denote the number of those permutations of the 2n
numbers that contain i and i + n next to each other by Vi. These sets consist of P-
permutations only, their union is clearly the set of P-permutations that is denoted
by V. With usual notations
(1) |P| =
η
η
г=1 г<к
This is again a corollary of the Principle of Inclusion and Exclusion already
used in the Second solution of problem 3.
Now \Vi\=2· (2n — 1)! because the elements of Vi can clearly be obtained
by treating the pairs (i,i + ri) and (i + n,i) as single elements and preparing the
(2n — 1)! permutations of the hence obtained 2n — 1 elements.
Proceed likewise when counting the intersection Vi П Р&: 'glue' the numbers
i and i + n and also к and k + n. These pairs form (2n —2)! permutations with
the remaining 2n — 4 single numbers. Since there are 4 possible orders of the
numbers in the two pairs
\VinVk\=4-(2n-2)\ .
361

Solutions
1989/6
Remembering that there are 11 pairwise formed intersections, (1) becomes
|Ρ|>η·2·(2η-1)!-ί^)·4·(2η-2)! =
ч, 2n(2n-l)(2n-2)!(2n-2)_/0 ч, (2n)!(2n-2)_
~{П)' 2(2п-1) ~{П)' 2(2n-l) "
= (2n)41"2'2^lj>-T-
This, in fact, is nothing else but the assertion of the problem.
Note. Estimate (1) in the Second solution can be improved and then we
get that the number of P-permutations when divided by the total number of
permutations tends to the ratio
^— = 0.63212055...
e
as η tends to infinity; for sufficiently large values of n, hence, roughly 63.21 %
of the permutations have property P.
1990.
1990/1. Chords AB and CD of a circle intersect at a point Ε inside the
circle. Let Μ be an interior point of the segment EB. The tangents to the circle
through D, Ε and Μ intersects the lines ВС and AC at F and G respectively.
r,. r EG . „AM
Find ——- in terms oft = —-=-.
EF J AB
Solution. A careful check of the diagram (Figure 1990/1.1) reveals quite
a number of equal angles. By the theorem of intercepted angles
IACD = IABD = 1U ADAB = IDCB = 42,
IDME = IDEG= ίFEC = φ.
Simple calculation shows that ICGF= IMDB = ψ — η\.
The equality of the respective angles now implies that ACEF and A AMD
and triangles CGE, BDM are pairwise similar. Being so
СF Α λ/ί
that is CE-DM = AM-EF,
that is CE-DM = GE-MB.
(1)
and also
(2)
EF DM'
CE MB
GE DM'
362

1990/2
1990.
Figure 90/1. J
Combining (1) and (2) yields AM -EF=GE- MB, or AM = t · AB. Using
MB = AB - AM = (1 - t)AB we arrive to the desired result:
GE _AM _ t-AB t
~EF ~ ~MB " (1 - t)AB " T^t'
Remark. In the argument we used that A separates G and C. This order of
the points clearly follows from ψ>^\.
1990/2. Take η > 3 and consider a set Ε of2n — \ distinct points on a
circle. Suppose that exactly к of these points are to be coloured black. Such a
colouring is "good" if there is at least one pair of black points such that the interior of
one of the arcs between them contains exactly η points from E. Find the smallest
value of к so that every such colouring of к points of Ε is good.
Solution. We may assume that our points, A\, A2, · · ·, ^2n-l are tne ver"
tices of a regular (2n — l)-gon. In what follows, distances between these vertices
will be measured by one of the connecting arcs specified in due course. Setting
the circular distance between two adjacent vertices as the unit, some of these
distances might turn out to be greater than 2n — 1, that is the total perimeter; the
actual distance then is clearly the residue when dividing by 2n — 1; subscripts
should be reduced likewise, if necessary.
An arc between two vertices contains η points if its length is η +1; there
are two points this far from An+\, for example: А2(п+1) = Аз and Αιη-\· Start
363

Solutions
1990/2
now from An+\ and mark, in one direction, the points being (n+ 1) far from the
previous one:
(1) Αι+b A2(n+l)> A3(n+1)> ·■■> Λ<η+1)>
where с is the highest integer for which the points above are all distinct, the
length of the cycle in (1). Thus the next point, Α^ΐχη+ΐ) appears on the list
already, what's more, it is the very starting point, An+\. Indeed, the list contains
both (n+ l)-neighbours of any other point. Being so
(c+ l)(n+ 1) - (n+ l) = t(2n- 1),
for some positive integer t. Rearranging we get
(2) c(n+l) = t(2n-l).
Consider first when the cycle is maximal that is c = 2n— 1. With t = n+l
they satisfy (2) and now (1) includes every point of E. If η of them are coloured
then (1) contains at least one adjacent black pair. Since the length of their arc
along the chosen direction is η +1 there are η points from Ε on this arc,
indeed. With fewer points painted, on the other hand, it is clearly possible to avoid
neighbouring black ones in (1). Thus if the cycle contains every point then the
smallest value of к is n.
The cycle certainly locks at the (2n— l)th step but this can already occur
for smaller values of c. Let d be the g.c.d. of n+ 1 and 2n — 1. Then d divides
their difference, 2(n+ 1) — (2n — 1) = 3 that is either d=\ or d = 3. If <i = 3 then
the subscript of every vertex in (1) is a multiple of 3 and thus the list itself does
not exhaust the set E\ if the cycle is maximal then d is forced to be 1. If d=3
then 2n —1 as an odd multiple of 3 is equal to 3(2s —l) = 6s —3, and hence
n + 1 =3s. Substituting into (2):
3sc = t(6s — 3), or sc = t(2s — l).
In—I
с = 2s — 1 = —-— satisfies this equation and thus the following 2s — 1 points
(3) Лг+Ь 42(η+1)> ·■■> ^2n-l(n|1)
are all distinct. Repeating the previous argument (then s was equal to n) we get
that s — 1 can still be painted black without having adjacent black points in (3)
but this cannot be done any more if there are s points to be coloured.
Rotating the set (3) by one and two units respectively we obtain the points
of Ε with subscripts 3r + 1 and 3r + 2 and these points, together with (3) exhaust
the set E. In both sequences as in (3) one can colour s — 1 but no more points
without having adjacent pairs. Hence the maximal number of black points where
there are no two points of the given property is 3(s — 1) = η — 2, that is no matter
how we colour η — 1 points black there will always be two points as required.
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1990/3
1990.
Coming to the end: if the g.c.d. of In — 1 and η + 1 is equal to 1 then k = n;
if it is equal to 3 then k = n — l.
This can also be put as follows:
if 3 does not divide 2n — 1 then k = n;
if 3 divides 2n — 1 then k = n — l.
2n + \
1990/3. Determine all integers greater than 1 such that —~— is an integer.
nl
23 + l
Solution, η = 3 is a trivial solution as —=— = 1. We show that there is no
32
other solution. We shall proceed by proving a series of lemmas.
Lemma 1. If к is odd then s = 22k - 2k + 1 is divisible by 3 but not by 9.
For odd powers of 2
2!e2, 23ξ8, 25=5, 27ξ2, 29ξ8, ... (mod 9)
and the remainder, in general, of 2 r (r = 1, 3, 5) when divided by 9 depends
on r only, namely
Hence
>6fc+r = (64)fc-2r = 2r' (mod 9).
26we2, (26k+l)2 = 4, s = 4-2+l = 3,
26/c+3 = 8, (26/c+3)2 = 1, β=1-8+1 = 3, (mod 9)
26/c+5 = 5, (26/c+5)2 = 7, 5 = 7-5+1 = 3.
Thus, for every odd value of k, the corresponding power of 2 is of the form
3 + 9i, a multiple of 3 but not that of 9, indeed.
Lemma 2. If a, b and ρ are positive integers such that 2° =. 2b =. 1 (mod p)
and m = (a, b) then 2m = 1 (mod p).
Let a = bq + r (0 < r < 6). Then
(1) l=2a = 2^+r=(2b)g-2r = 2r (modp)
Thus if m is computed by Euclid's algorithm then congruence (1) holds for
the stepwise computed remainders and, in particular, for the last one different
from zero, the greatest common divisor of a and b\
(2) 2m = l (modp).
Lemma 3. If η is a positive integer then 2n + 1 is not divisible by 7.
This is true if n= 1, 2, 3. Assume that the lemma is false and there exists
some η > 3 for which 2n + 1 is a multiple of 7. Denote the smallest index of this
365

Solutions
1990/3
property by no- Then, for some positive integer N we have 7ΛΓ = 2ηο + 1. But
now
7(iY + l) = 23-2no-3 + i + 7 = 8(2no-3 + l)
2no +1 is also divisible by 7 contradicting to the choice of no- The proof of
the lemma is hence complete.
Let's turn now to the actual solution. 2n + 1 is odd and so is its divisor, n.
Let's write it as η = 3k · d where к > 0 integer and 3 does not divide the odd d. If
2n +1 -, ?
—ψ- is an integer then using the identity a +1 = (a + l)(a — a+ 1) it follows,
by induction, that
t k~l
3k
a
+ 1 = (α+1)Π (α2-3* - α3* + l)
г=0
Substituting 2 for α and using that n = 3kd we get
k-l . .
(1) 2n + l = (2d+l) fj (2гзч-2зч + 1).
г=0
By Lemma 1. the 3-term factors on the r.h.s. are divisible by exactly the first
power of 3 and hence the index of 3 in the product is k. This product, by
condition, is divisible by η and thus by 3 ; 2d + l, hence, is divisible by 3k. The
proof of Lemma 1. also implies that the index of 3 in 2d + 1 is at most 1. Thus
either к = 0 or к = 1 that is
(2) n = d or n = 3d.
It is enough to show that d=\. Assume the contrary, d> 1, and denote the
smallest prime divisor of d by p. By now we know that ρ is different from both
2 and 3 so it is at least 5. By its choice p—\ and d have no common factor,
(p—l,d) = l. We also know that ρ is a factor of η and hence it also divides
2n + 1 that is
(3) 2ηΞ-1 (mod p).
Squaring
22n ξ 1 (mod p),
on the other hand, by Fermat's theorem
2p~l = 1 (mod p).
Applying Lemma 2. to the latter two congruences yields
(4) 2m = \ (modp),
with m = (2n,p—l) and thus, by (2) m divides 6d. Since m divides ρ — 1 which
has no common factor with d we have succeeded to reduce the possible values
366

1990/4
1990.
of πι: it must divide 6. Hence, by (4), ρ is a prime factor, greater than 3, of one
of the numbers 1, 3, 7, 63. The only possibility is p = l and this, by (3), implies
that 7 is a proper divisor of 2n + 1. This, however, contradicts to Lemma 3 and
thus d, in (2), is indeed equal to 1. Since n> 1, the only solution is n = 3, the
proof is finished.
1990/4. Construct a function from the set of positive rational numbers into
itself such that
(i) №·№) = —
У
for all x, y.
Solution. First we note that if a function / satisfies the functional equations
(2) f(f(x)) = - and (3) f(x)f(y) = f(xy)
χ
then it also satisfies (1) and thus it is enough to find a functional solution of
equations (2) and (3). The solutions of (3), by the way, are the so called multiplicative
functions and the property given there is also called multiplicativity.
Denote, for the construction, the zth prime by pf. p\ =2, P2 = 3, .... Any
rational number can be written as
x-Pi P2 ■· -Vin-x Pin ··■
where c^ is positive or negative integer, maybe zero. The index of pi is zero after
a certain position anyway and the corresponding primes can be omitted from the
product above. Set now
(4) /(*) = Λ_αΐ-^"-.iC2"-1 - ■
This mapping works by grouping the prime numbers into pairs, swapping the
indices within each pair, finally, the index thus obtained is multiplied by -1 if
it is at even position. By the unique prime factorisation it is enough to verify
multiplicativity if the arguments are prime numbers and this is obvious for the
'prime-wise' acting function defined in (4). To prove that / satisfies (2) observe
that
which is but equation (2). Satisfying both (2) and (3) function (4) indeed has the
desired property.
Note. Those who have solved the problem usually realized that (1) also
implies (2) and (3). This is not necessary for the solution, however, we still
present a proof. Setting x = y=l in (1) yields
/(/(!)) = /(!), that is /(/(/(1))) =/(/(!)) = /(!).
367

Solutions
1990/4
Substitute now 1 for χ and /(1) for y:
f(l)
/(/(/(D)) = 77^ = 1-
Combining the results we obtain that /(1) = 1.
Using (1) again:
/(/(2/)) = /(l-/(2/)) = — = ",
2/ 2/
that is
(5) /(/(2/)) = -,
2/
which is (2), indeed. / when acting on both sides of (5) yields the following
equality:
(6) /(/(/(2/))) = /(-)-
Substituting /(i/) for j/ in (5) and comparing the result with (6) we get
1
f(f(f(y))) =
f(y)
/P l
.y) fiy)
Replacing у by f(y) again and applying (5):
f\l¥)) = f(f(y))=y'
Finally, evaluating (1) once more and using the last equality we get
1
Ж
Kxy) = f (xf (тгт)) = ^r = fMfM*
f(y)
we have arrived at (2) and this is the end.
1990/5. Given an initial integer щ > I, two players A and В choose
integers щ, 77-2, Щ, ... alternately according to the following rules: knowing ri2k>
A chooses any integer П2к+\ such that
П2к<п2к+1<П2к-
Knowing П2к+\, В chooses any integer П2к+2 such that
Щк+\ = r
> Щк+2
for some prime ρ and integer 1 < r. Player A wins the game by choosing the
number 1990; player В wins by choosing number 1. For which щ does
368

1990/5
1990.
a) A have a winning strategy?
b) В have a winning strategy?
c) Neither player have a winning strategy?
Solution. Suppose that the starting value no has just been announced and
it is now player A's turn.
If 45 < щ < 1990 then 1990 < 452 = 2025 and thus A can choose щ = 1990 and
win immediately.
If 27 < no < 44 then щ = 720 = 2-5 is a legal choice for A. To make sure that
ТЬл
— is a prime power В has to choose П2 between 45 and 360 and thus A can
™2
pick 1990 as щ because щ < 1990 < n^.
If 15 < no < 26 then n\ =210 = 2-3 ·5 -7 is available for A; player В is now
allowed to reply in the range [30,105] only and hence A can finish by choosing
1990 again.
If 11 < no < 14 then A should choose 77-1 = 105 = 3-5-7. The range for В is now
15 < П2 < 35 but these values have already been verified to be in favour of A.
If 8 < no < 10 then, similarly, if A starts with щ = 60 = 22 · 3 · 5 then 12 < П2 <
30 so, like before, A is to win.
If no < 8 then the triumphants of A are over: none of his options would force В
beyond 7.
The state of affairs so far is that A has a winning strategy as long as 8 < no <
1990.
If 2 < no < 5 then player В can win and there is no better way to prove this but
checking the starting values one by one.
If no = 2 then щ can be only 2, 3 or 4 and thus В can respond by saying 1 and
hence win.
If no = 3 then the legal values of щ are 3, 4, 5, 6, 7, 8, 9; if A chooses n\ =6
then В can reply with П2 = 2-t and can win as before; for any other choice of A
player В can prompt 1.
If no = 4 then if n\ is from the list {4, 5, 6, 7, 8, 9} then the players are back
in the previous case; the values remaining are 10, 11, 12, 13, 14, 15, 16 and the
winning replies of В are П2 = 2, 1, 3, 1, 2, 3, 1 respectively.
If no = 5 then the range n\ < 16 has already been checked. For the remaining
values
щ= 17, 18, 19, 20, 21, 22, 23, 24, 25,
В can choose
n2= 1, 2, 1, 4, 3, 2, 1, 3, 1
369

Solutions
1990/5
respectively and thus each case is reduced to some previously checked position
all in favour of B.
There are two cases left, no = 6 and щ = 1.
If no = 6 then A can reply from the range 6 < щ < 36. Apart from щ = 30 В can
always choose one of the numbers 1, 2, ..., 5 as П2 and, as we have already
seen, win afterwards. To avoid defeat A has only one choice, namely щ =30.
Now it is В in a similar dead end street: among the divisors of 30 he should
choose one which is less than 8 but there is only one such number left: П2 = 6.
Mutually forced to play against defeat they are back at beginning; from now on
the game goes on and on, indefinitely, none of the players has a winning strategy.
The story is similar if ng = 7. The numbers for A are now 7 < щ < 49. If щ is
different from both 30 and 42 then В can always choose some of his winning
numbers 1, 2, ..., 5. n\ =30 forces the infinite game above but the game is also
blocked if A chooses щ =42 because the only reasonable answer of В is the
vicious П2 = 6.Coming to the end: if no = 6 or 7 then none of the players can win
the game.
We show that A can also win if no > 1990. Consider the interval (no, 2no)
for no > 1990. It contains at least 1991 whole numbers and hence there are
certainly some of them of the form 9 · 2k, (k > 3). Let A choose one of these numbers
i.e., let щ = 9 ■ 2k. Now B's reply, щ, is at least 9 but not greater than 9 · 2 ;
В is forced to choose his number below no- Following this strategy A can push
his opponent below 1990 in finitely many steps keeping him above 9 all the time.
Since this range was proved to be in favour of A he can indeed win the game if
no is greater than 1990.
Summing up:
A has a winning strategy if no > 8;
В has a winning strategy if 2 < no < 5;
neither player has a winning strategy if no = 6 or 7.
Remark. The winning strategy of A is not unique; playing differently he can
also win the game. This, of course, has no effect on the conclusion.
1990/6. Prove that there exists a convex 1990-gon such that all its angles
7 7 7 7
are equal and the lengths of the sides are the numbers 1% 2% ..., 1989% 1990^
in some order.
Solution. The sides of the convex polygon to be constructed are parallel to
1990
those of a regular 1990-gon respectively. First we construct a 995 = —-— sided
polygon of suitable sides then each side will be divided into two parts and these
parts when rotated in proper order will form the required polygon.
370

1990/6
1990.
Working on the complex plane will make our life easier. Let e be the first
995th root of unity:
2π . 2π
e - C0S 7^77: + Ί- sln 77777: ·
995 995
This number obviously satisfies e = 1 that is e —1=0. Hence
(e-l)(l+e + e2 + ... + e993+e994)=0>
(1) 51 = l + e + e2 + ... + e993+e994 = 0>
/ c4199
on the other hand (e J —1 = 0 implies
(e5-l)(l + e5 + e10 + ... + e985 + e990)=0,
(2) S2 = l + e5 + e10 + ... + e985 + e990 = 0.
The sides of a regular 995-gon are parallel to the terms of (1) respectively.
Prepare now the following linear combination of these numbers/vectors:
198 /
(3) S = ^(b-e5b + (b+l99)e5b+l99 + (b + 2-l99)e5b+2l99+
b=0 ^
+ (6 + 3-199)e5b+3-199 + (6 + 4.199)e5b+4199Y
There are 5· 199 = 995 terms in the sum S. The coefficients, b + n· 199 (n = 0,
1, 2, 3, 4) are the integers from 0 to 994, each of them occurring once. We
show that the complex terms are but the roots of unity in (1), remembering that
e = 1. It is enough to show that the complex numbers — each of them is 995th
root of unity — in (3) are all distinct. The opposite would imply either
5b + n-199 = 56' + j-199, that is 5(b-b() = 199(j -n)
but then b — b' would be a multiple of 199 and the latter can happen only if b = b'
and, accordingly, j = n; or
56 + n· 199 = 564 j· 199 + 995
which is likewise impossible. The roots of unity in (3) are hence distinct and as
there are 995 of them they indeed exhaust the list in (1).
Next we prove that 5 = 0. Multiply for this by 1 — e ^ 0. Performing the
calculations we get
198
(l-e199)5=^e5b(l99e199(l+e199 + e2199 + e3199-4e4·199)).
Denote the 5-term sum in the brackets by E. Thus
198
(l-e199)s = ££e5» = i?(l+e5 + e10 + ... + e990))
and, by (2), this is zero, indeed.
371

Solutions
1990/6
Rearrange now the terms in (3) in the natural order of the coefficients b +
199j = k (j = 0, 1, 2, 3, 4) and denote the complex root of unity with coefficient
994
к by ek (eo = l). (3) then can be written as S^Y^/ce^. Remembering that the
A;=0
numbers e^ are the 995th roots of unity and thus their sum is zero, we now
consider the sum
/ 994 \ / 994 994 \
995· l997-J2^k + 2S)=995- (997-£efc + 2-£fcefc )=0;
this, by the way, can also be written as
994
(4) Σ 995 ■ (995 + 2(k +1)) ek = 0.
A;=0
Since 995(995 + 2(k + l)) = (995 + k + \)2-(k+1)2, (4) is the same as
994
(5) ^((995 + /c + l)2efc + (^ + l)2(-e^)) = 0.
k=o
The complex numbers ek together with their respective opposites — ek make
up the 1990th roots of unity. Formula (5) shows that the terms when multiplied
9 9 9
by 1 , 2 , ..., 1990 respectively and in the order as they appear in the sum,
form, in fact, a closed polygon whose sides are of the required length and
direction. Finally if the order of the terms is matched to the ascending order of
their respective arguments then the polygon obtained will be convex and thus
the construction is complete.
1991.
1991/1. Given a triangle ABC, let I be the incentre. The internal bisectors
of angles А, В and С meet the opposite sides at A', B', C' respectively. Prove
that
1 AI-BI-CI 8
(1) -< <—.
K } 4 AA'-BB'- CO ~ 27
First solution. Denote the area of the triangle by t, and that of the triangles
ABI, ВС I, CAI by tc, tA, ~Ьв respectively; clearly t = tA + tB + tc- Having ВС
as a common side the respective areas of the triangles ВС I and ВС A are in the
ratio of the corresponding altitudes, I A': AA' {Figure 1991/1.1).
tA_ IA' _AA'-AI AI
t " AA'~ AA' " AA'
372

1991/1
1991.
Figure 91/1.1
Thus
and similarly
AT tA
AA> t '
BI tB
BB' t ''
CI
~cc>
= 1
By the A.M.-G.M. inequality
AI-BI-CI
1
< —
AI BI
+
CI
tc
t
+
1 Λ tA + tB + tcy= 8
27 V t J 27
Л Л; -ВВ'-СС' ~ 27 V A A' ££' CC7
The second inequality is hence proved and observe that apart from being interior
to the triangle ABC nothing else was used about the point /.
Turning to the proof of the first inequality note that —— = 1 = ;
if r is the inradius then 2t = r(a + b + c), 2tA = ra, 2tB = rb, 2tc = rc and hence
AI b + c . ., ,
-— = —; similarly
AA1 a + b + c
BI
c + a
CI
a + b
BB' a + b + c
and thus the claim becomes
CC a + b + c
(2) 4(a + b)(b + c)(c + a)>(a + b + cy.
Performing the calculations and rearranging we get
-a3 -b3 -c3 + 2a2b + ab2 + a2c+ ac2 + bc + be2 + 2abc > 0.
which is the same as
(3) (a + b- c)(a + c-b)(b + c-a) + 4abc>0
and (3) obviously holds if a, b and с are the sides of a triangle.
373

Solutions
1991/1
Second solution. The bisector of the interior angle divides the opposite si-
CLC
de into the ratio of the neighbouring sides and hence В A1' = , for example;
b + c
applying this theorem for the bisector BI of AABA' yields
AT c-AA' (b + c)-AAl Δ u AI b + c
AI = — = - , and thus
С+Й1 a + b + c ' AA1 a + b + c
Similarly
ВI a + c CI a+b
BB' a + b + c' CC a + b + c
The double inequality (1) hence can be written as follows:
1 (a + b)(b + c)(c + a) _8_
{) 4 (a + b + c)3 'И'
By the A.M.-G.M. inequality:
1 , TWT w 1 /' a + b + b + c + c + a\3 8
(a + b)(b + c)(c + a)<
(a + b + c)3 (a + b + c)3 \ 3 /27'
which is the second inequality.
Using the identity
3(a + b)(b + c)(c + a) = (a + b + c)3 -a3 -b3 - c3
the first inequality becomes
1 (a + b + c)3 -a3 -b3 -c3
(5) -<- = .
4 3(a + b + c)3
To prove this we use the following Lemma. If x, y, x\, y\ are positive
numbers such that x + y-x\+y\ and x — y<x\—y\ (x>y, x\>y\) then χ +
3 3 3
у* <х\+у{.
When applying the lemma a>b>c can clearly be assumed. The cast will
be
7 a + b + c a + b —с
x = a, y = b, xi = —, yi = .
The conditions of the lemma now hold: x>y, x\>y\, x + y = x\+y\ and
x — y = a — b<c = x\—y\ so
-x 7-з /a + b + c\3 fa + b-c\3
tt · , · ι ! -r · ·.· (a + b — c\ 9 /a + b + c\3
Using this and also the easy to verify inequality I I + с < I I
(5) becomes
(a + b + c)3 -a3 -b3 -c3 ^(a + b + c)3 -^ψ)' - (^f - c3 ^
3(a + b + c)3 3(a + b + c)
374

1991/2
1991.
(а + 6 + с)3-(^^) -(3ψ£) ι
3(а + Ь + с)3 ~4'
indeed.
Remarks. 1. Both solutions show that equality holds in the second inequality
if our triangle is equilateral.
In the first inequality the lower bound - cannot be improved. If b = с are
fixed and a tends to zero then the limit of the ratio in (4) is - and thus the ratio
4
can be arbitrarily close to this number.
2. Let's prove now the lemma of the second solution.
2 2 2 2 2 2
From the conditions χ + 2xy + у -x\л-2х\у\Л-у\ and χ —2xy + y <
2 2
х\ — 2х\У\Л-у\. Their difference is —4xy<—4x\y\ that is xy>x\y\. Raising
the equality in the condition to the third power
x3+y3 + 3xy(x + y) = xl+yl + 3xiyi(xi+yi):
0 = x3l+y3-(x3+y3) + 3(x + y)(xiyi-xy),
x3l+y3l-(x3+y3)>0,
and this was to be proved.
1991/2. Let n>6 be an integer and let a\, a2, ..., ak be all the positive
integers less than η and relatively prime to n. If
a2-ai=a3-a2 = ... = ak-ak_\ >0,
prove that η must be either a prime number or a power of 2.
Solution. Rephrasing the problem: if the numbers less than η and prime to
it form an arithmetic progression then η itself is either a prime or it is a power
of 2.
The common difference of the A.P. is a positive integer, denote it by d.
Since 1 and η — 1 are both prime to n, a\ = l and ak = n — 1.
Let η be odd first. Then η — 2 is also prime to n. Indeed, their g.c.d. divides
2, their difference, but η as an odd number is not divisible by 2. Hence d=l, the
A.P. consists of the integers from 1 to η — 1, these numbers are all prime to η
which, with no proper divisors, is a prime number, indeed.
Let η be now even; then it can be written as n = 2r · s where r > 1 integer
and s > 1 is an odd integer.
If s = 1 then η is a power of 2, the numbers prime to η are the odd integers
and they form an A.P., indeed. It is left to show that s > 1 is impossible.
375

Solutions
1991/2
a) If s = 3 then n>6 and thus the sequence of the numbers prime to η
begins with as 1, 5, 7, ..., which is certainly not an arithmetic progression.
β) For s > 5 we show that s — 2 and s — 4 are both terms of the A. P. of the
problem. For the proof consider an arbitrary common divisor, c> 1, of n = 2r ■ s
and s - 2. Since 2r ■ s = 2r(s - 2) + 2r+l, it also divides 2r+1. On the other hand,
as a divisor of the odd s — 2, с is an odd divisor of a power of 2; thus с = 1. Hence
η and s - 2 are coprime, s - 2 belongs to the A.P., indeed. 2r(s - 4) + 2r+2 = 2r ■
s = n now implies that the g.c.d. of s — 4 and η is 1 and thus s — 4 is in the A.P.
as well. With s — 4 and s — 2 the number s is also in the arithmetic progression
of the condition; this, however, is a contradiction since s was a proper divisor of
n; the proof is complete.
1991/3. Let S' = {1, 2, 3, ..., 280}. Find the smallest integer η such that
each η-element subset of S contains five numbers which are pairwise relatively
prime.
Solution. First we show a 216-element subset of S which does not contain
five pairwise relatively prime numbers. Pick the multiples of 2, 3, 5 and 7 from
S and denote the subset thus obtained by N. Denote also by Ni the multiples of
г in S and its cardinality by the usual |iVi|.
Clearly
|i\r2| = 140, №| = 93, |iV5|=56, |iV7|=40.
\N2r\N3\, for example, is the number of the common multiples of 2 and 3, that
is the multiples of 6; for these sets
|JV2niV3|=46, |iV2niV5| = 28, |iV2niV7|=20,
1^3 niV5| = 18, |iV5niV7| = 8, |iV3niV7| = 13.
We also need
|Ν2ηΝ3ηΝ5| = 9, |Ж2ГШ3ПЖ7| = 6, |Ж2ГШ5ПЖ7|=4, |iV3niV5niV7|=2
\Ν2Γ\Ν3ηΝ5Γ\ΝΊ\ = 1.
By the principle of inclusion and exclusion
\N\ = (140 + 93 + 56 + 40) -(46 + 28 + 20 +18 + 8 + 13) + (9 + 6 + 4 + 2) -1=216.
Choosing now any five elements of N at least one of the four subsets ΑΓ2, ΛΓ3,
N5 and Νη will be represented twice and these two numbers are not coprime,
indeed.
Next we prove that no matter how do we select 217 numbers from S there
are 5 pairwise relatively prime ones among the selected numbers; the smallest
value of η we are looking for is hence 217.
376

1991/4
1991.
•19,
•23,
•29,
•31,
•37,
•41,
•43,
5-17,
5-19,
5-23,
5-29,
5-31,
5-37,
5-41,
7-13,
7-17,
7-19,
7-23,
7-29,
7-31,
7-37,
ll2},
11-13},
11-17},
11-19},
11-23},
13-17},
13-19},
52, 72, 132}.
There are 59 primes among the first 280 positive integers (cf. the remark
at the end of this solution). Adjoining also 1 to these 59 numbers we get a 60-
element set, denote it by P. Now there are 220 composite numbers in S; they
form the set T. We now list 8 disjoint subsets of T, each comprising five pairwise
relatively prime elements:
Μγ ={2 -23, 3
M2 = {2-29, 3
M3 = {2-31, 3
M4 = {2-37, 3
M5 = {2-41, 3
M6 = {2-43, 3
M7 = {2-47, 3
M8 = {22, 32,
Consider now an arbitrary 217-element subset Η of S. If Η and Ρ have at least
5 common elements then we are done, these 5 numbers are pairwise coprime.
We may hence assume that there are at most 4 elements of Ρ in Я and thus
its further 213 elements are all composite. This means that there are at most 7
missing from the altogether 220 composite numbers in S and thus one of the sets
Mi is contained in H. This M; comprises the 5 numbers required.
Remark. There are no calculators or tables available at the IMO. With such
aids around the number of primes below 280 can be found by either the sieve of
Eratosthenes or with bare hands by removing the primes 2, 3, 5, 7 from the set
N leaving 212 composite numbers; apart from these there are only 8 composite
elements in S: ll2, 132, 11 · 13, 11 · 17, 11 · 19, 11-23, 13 ■ 17 and 13-19.
1991/4. Suppose G is a connected graph with к edges. Prove that it is
possible to label the edges 1, 2, 3, ..., к in such a way that at each vertex which
belongs to two or more edges, the greatest common divisor of the integers
labelling those edges is 1.
[A graph is a set of points, called vertices, together with a set of edges
joining certain pairs of distinct vertices. Each pair of edges belongs to at most
one edge. The graph is connected if for each pair of distinct vertices x, у there
is some sequence of vertices x = vq, v\, ..., vm = y, such that each pair Vi, vi+\
(0<i<m) is joined by an edge.]
First solution. In the solution we shall try to do the labelling in such a way
that each vertex belongs to some pair of edges labelled by consecutive and hence
coprime numbers.
377

Solutions
1991/4
The degree of a vertex is the number of edges leaving from that vertex.
Denote the vertices by P\, P2, ■ ■., Pn. Since the graph is connected there is a
path between any two vertices. Start hence from P\ to P2, go on to P3 and so
on up to Pn, finally, from here go back to P\. Label the edges crossed along the
roundtrip as follows:
a) the first edge of the trip starting from P\ is labelled by 1;
b) travelling along an already labelled edge its label remains unaltered;
numbering is resumed only when arriving to an edge not labelled so far;
c) when arriving for the first time to a vertex whose degree is not 1 continue
the trip along any edge not yet labelled; this edge gets labelled now and
with the number one greater than the edge of arrival;
d) having entered the numbers 1,2, ..., s by the end of the trip if there are yet
unlabelled edges then these are numbered arbitrarily by the numbers s+ 1,
s + 2, ..., k.
This labelling has the desired property: indeed, apart from the first degree
vertices and maybe Ργ every edge belongs to at least one pair of edges labelled
by consecutive numbers written upon the first arrival to this vertex; the edge
labelled by 1 belongs to P\ which settles this vertex as well; finally, the first
degree vertices are clearly irrelevant.
Second solution. It is well known that in a graph the number of odd degree
vertices is even. Arrange the odd degree vertices into pairs and connect, in each
pair, the vertices by a red edge. Colour also the edges in G by black. Each vertex
of the graph G' prepared this way is of even degree. There might be multiple
edges (a black and a new red one) in G' but this does not matter. Note that there
is at most one red edge belonging to any vertex.
Graph G' is well known to be Eulerian. Starting from any vertex the edges
of such graphs can be traversed along a closed path without retracing any edge,
along a so called Euler-circuit.
Traverse now the edges of G' along an Euler-circuit. Label the first black
edge crossed by 1 and each subsequent black edge of the circuit is labelled by the
next integer. Red edges are not numbered at all. Since the circuit contains each
of them, every edge in G is get labelled. If a vertex belongs to black edges only
then its degree in G is even and thus there are consecutive ones among the labels
of these edges. If there is a red edge here and a single black one then its label has
certainly no effect. Finally, if there are more black edges then, by construction,
there are at least three of them, the circuit is passing through this vertex at least
twice and on one of these occasions both the arriving and the leaving edges are
black. These edges hence are labelled by consecutive numbers as required.
378

1991/5
1991.
1991/5. Let ABC be a triangle and X an interior point of ABC. Show that
at least one of the angles XAB, XBC, PC A is less than or equal to 30°.
First solution. We shall use the notations of Figure 1991/5.1.
Figure 91/5.1
If some angle of the triangle is at least 150° then there is another one strictly
less than 30° and thus the claim obviously holds; assume now, indirectly, that
30° <Αχ, Bi,C\< 150°. Hence, if the feet of the perpendiculars from Ρ to AB,
ВС, С A are С', A', B', respectively, then
. A PC . .„, 1
sin A\ -
> sin30° = -, that is IPC' > PA
and similarly
Thus
(1)
2PA'>PB,
PA' + PB' + PC'>
2PB' > PC.
PA+PB+PC
contradicting the assertion of the Erdos-Mordell inequality where, instead of '>',
(1) holds with '<'. This reference completes the proof.
Second solution. Assume again that A\, B\, C\ are greater than 30° which,
like in the first solution, implies that 30° <A\,B\,C\< 150°. Thus
1
(2)
sin A\ sin B\ sin C\ >
8
379

Solutions
1991/5
By the assumption A\ + Β γ + С γ > 90° and hence A2 + B2 + C2 < 90°. Applying
now the A.M.-G.M. inequality and also Jensen's inequality
. , . „ . ^ / sin An + sin Bo + sin Co \
sin A2 sin B2 sin C2 < ί — -1 <
<sin3^2 + B2 + C2<sin330^1
3 8
that is
ί >8,
sin Л2 sin B2 sin C2
which, when combined with (2), yields
sin A] sin5i sinCi
(3) ->1.
sin A2 sin B2 sin C2
Applying now the sine rale in the triangles APB, BPC, CPA respectively:
sin^! sin£i sinC!_P5 PC PA _
sinB2 ' sin C2 ' sin A2~~PA"pB ' ~PC~ '
contradicting to (3), the consequence of our assumption. There must be at least
one among A\, B\, C\, less than 30°, indeed.
Third solution. Let Q be one of the triangle's Brocard's points. Now with
the Brocard angle ω
QBCl = QCAl = QABl=u.
Since
cot ω = cot A + cot В + cot C,
the cotangent inequality implies
cot a; >
that is ω < 30°. Point Ρ of the problem is contained by one of the triangles
QBC, QCA, QAB, say QBC; hence LPBC < LQBC = ω <30°, and this was
to be proved.
Remarks. 1. If the respective distances from the vertices of a triangle of the
interior point Ρ are p, q and r and from the sides are x, у and ζ then
ρ + q + r > 2(x + y + z).
This is the Erdos-Mordell inequality. Equality holds if and only if the triangle
is equilateral and Ρ is its centre.
2. Points Q\ and Q2 are the Brocard's points of A ABC if
IBAQi = ICBQi=IACQi, and LABQ2 = IBCQ2 = LCAQ2.
380

1991/6
1991.
Apart from the circumcentre these are the only points with the property that their
perpendicular projections on the sides form a triangle similar to the original one.
This also implies that if the triangle is rotated about any one of it Brocard-points
then the triangles formed by the intersections of the corresponding pairs of sides
are similar.
3. The cotangent inequality states that in an arbitrary triangle
cot A + cot В + cot С > \/3.
Equality holds if and only if the triangle is equilateral.
1991/6. Given any real number a>\ construct a bounded infinite sequence
xq, x\, ... such that
(1) \xi-xj\-\i-j\a>l
for every pair of distinct i, j. [An infinite sequence xq, x\, ... of real numbers
is bounded if there is a constant С such that \xi\ <C for all i.J
First solution. Note first that if (1) holds with a = 1 for every pair of distinct
i, j then it clearly holds for a > 1. Being so it is enough to construct a sequence
satisfying (1) if a= 1. In due course we shall make use of the well known result
about the error when v2 is approximated by fractions of the form ^ (p and q
are positive integers):
(2)
Q
>
1
3g
2'
Denote now the fractional part of V2k by the usual {V2k} and let
(3) xk = 3{V2k} = 3(y/2k-[V2kfj , k = 0, 1, 2, ...
We show that this sequence satisfies (1) with a=l. Let i>j be non negative
integers. Since
xk <3,
the sequence is bounded. (1) also holds because if q = i — j and p= [iVl] — [jл/2]
then
|rc<-ici||i-j| = 3|(i-j)v^-([iv^]-b'V^])|(i-j) =
3]qV2-p]q = 3qx
Q
>1.
and the latter is true by (2). The sequence (3) hence satisfies the requirements
for every a > 1.
381

Solutions
1991/6
Second solution. The sequence will be defined in terms of the binary form
of natural numbers.
Let the binary form of г be:
i = b0 + bi-2 + b2-22 + ... + bk-2k,
where bs, the binary digits, are 0 or 1. With the given value of a > 1 prepare the
following real number hi from the digits of i:
hi = b0 + bl- 2~a + b2 · 2~2a + ... + bk- 2~ka.
As the sum of an infinite G.P.:
(4) 0 < hi < 1 + 2~a + 2~2a + ... + a~ka < —l- .
-г- \-2~а
Similarly, if j is different from i and j = cq + c\ · 2 + c2 ■ 2 +... + ck · 2 then
hj = c0 + ci · 2~a + c2 ■ 2~2a +... + ck ■ 2~ka.
Consider now the first binary position where г and j are different; if this is the
pair (bt,ct) then \i — j|>2*. The following chain of estimates is now clearly
valid:
\hi-hj\= (b0-c0) + (bl-cl)-2-a + ...+(bk-ck)2-ka
>
3\
> \bt - ct\2~ta - \bt+l - ct+l |2-(ί+1)β -...-\bk- ck\2~ka >
o-(i+l)o
y, <y—ta ο—(ί+1)α >y—ka ^ -л—ta
= 2~ta 1 =(2М > \i-j\ ■
Rearranging we arrive to
(5) |/4-/b-||i-jr>|^.
Define now the sequence as:
_ 2a - 1
%i — ~~ ~ ibi.
2° -2
We show that this sequence satisfies the requirements. First of all, by (4)
2a-2 a 2a-2
"i<2M'l-2-a"V-2 + 2-a <a'
the sequence is hence bounded. As for (1) we clearly have
m 2°-l·,, , ,, ,„ 2α-1 2α-2 Λ
\Xi — хА\г — э\= \hi - hj \\i - j Г > · = 1,
ι ' j и j \ 2° — 2 2° — 2 2° — 1
and the desired property is immediate.
382

1992/1
1992.
Remark. The method of the first solution when improved yields much better
estimations for the terms of the sequence.
The result about the error of the approximating fractions of v2 can be pro-
P
ved as follows. Once more, it states that if - is a positive rational number then
Q
V2-P
Q
>
1
з<г
If q = 1 then this is obvious so q > 2 can be assumed. Suppose now the assertion
is false and hence
1
V2-P-
Q
~3q2
Ρ
holds for some -. This means that
Q
1
<V2-P-< l
- ъЛ
q 3q
3q2~
which, when rearranged, becomes
-— +qV2<p< — +qV2
5q 3q
This implies
and by the assumption q > 2 we arrive to
,2 „ 2| 2V2 / 1 \2 2V2 1 24^2 + 1
\p -2q \<
3 +{tJ *
3 +36
36
<1.
О 01 0 0
Since \p — 2q | is a non negative integer, the only possibility is pz - 2qz = 0,
Q
contradicting the irrationality of v2. The proof is finished.
1992.
1992/1. Find all integers a, b, с satisfying 1 < a < b < с such that (a
(b — 1) · (c — 1) is a divisor of abc — 1.
1)
Solution. With
(1)
Q
abc— 1
(a-l)(b-l)(c-l)
383

Solutions
1992/1
we have to find those integral values of a, 6, с for which Q is an integer. Clearly
abc 1
(!') Q =
(a-l)(6-l)(c-l) (a-l)(6-l)(c-l)
a-l/V 6-1/V \c-lj (a-l)(6-l)(c-l)
therefore
a— 1/ \ b — l/\ c—1/
(1) implies that if any one of a, b, с is odd then so is the numerator; thus,
for Q to be an integer, the denominator has to be odd as well, that is a, b and с
are either all odd or all even. We also note that Q > 1 is immediate from (Г).
Next we show that a cannot exceed 3. Indeed, from what we know of these
numbers, a > 4 implies b > 6 and с > 8 and thus the r.h.s. of (2) is at most
(■*£)(·*£)(■♦£)-{§■
less than 2. But then Q as a whole number is strictly between 1 and 2, a
contradiction. The possible values of a are hence 2 and 3.
a) If a = 2, then both b and с are even, b > 4 and с > 6 and (2) implies
Q4<4
so either Q = 2 or Q = 3. The former is now impossible since, by (1), Q is odd;
therefore, if a = 2 then Q = 3 and (1) becomes
2bc-1 = 3(6-l)(c-l),
(6-3)(c-3) = 5 = l-5,
and thus 6 = 4 and с = 8. The triplet (2,4,8) is a solution, indeed.
β) If a = 3 then proceeding as above, 6>5, c>7; (2) now yields
^ 35 o
Q<^<3
that is Q = 2. Plugging these values in (1):
3c6-l
2 =
2(6-l)(c-l)'
(6-4)(c-4) = ll = l-ll,
and thus 6 = 5 and c= 15. The triplet (3,5,15) is the other solution of the problem.
1992/2. Find all functions f defined on the set of all real numbers with real
values, such that ι
(i) f(x2 + f(y))=y+(f(x))2
for all x, y.
384

1992/2
1992.
In what follows the notation f (x) will be used for (f(x)) , the square of f(x).
First solution. Guessing and checking f(x) = x turns out to be a solution;
we show that this is the only one.
Assume first that for some real number у
(2) Ку)<У: that is y-f(y)>0
and let χ be such that xz = y — f(y) that is у = χ + f(y). By (1), this implies
f(y) = f (χ2 + f(y)) = y + f2(x)>y,
contradicting (2); hence for every real number у
(3) f(y)>y.
Set now yQ-t to be smaller than — / (0) and denote /(г/о) by a. Combining (3)
and (1) we get
a < № = f(02 + f(y0)) =y0 + /2(0) < 0 that is
α < /(&) < 0, and thus
(4) a2 > f\a).
Let χ be now arbitrary. Using (3), (1) and (4) we arrive to
(5) x + a2<a2 + f(x)<f[a2 + f(x))=x + f2(a)<x + a2.
Since the lowest and the highest terms are equal, there is equality everywhere in
(5) and thus
f(x) = x,
for every real number x, indeed.
Second solution. Let a be an arbitrary real number. We show that /
assumes the value a and the corresponding argument is χ +f (a — f (x)j where χ
is arbitrary. For this set у = a — f (x) in (1):
(6) / (x2 + / (a- f\x))) = a- f2(x) + f\x) = a,
indeed. Next we show that / is one to one, that is y\ ^y2 implies f{y\) 7^/(2/2)·
Let χ be once more an arbitrary real number. Applying (1) twice:
f(x2 + f(yi)) = yi + f2(x):
1(х2 + КУ2))=У2 + 12(х)-
Thus f(y\) = f(V2) implies у ι =У2, f is one to one, indeed, it is a bijection
of the set of the real numbers.
385

Solutions
1992/2
Apply now (1) for each of the pairs (x: y) and (—x: y):
f(x2 + f(y))=y + f2(x):
f(x2 + f(y))=y+f\-x).
The difference of the respective sides can be factorized:
f2(x)-f2(-x) = 0,
(/(aO-/(-z))(/(s) + /(-s)) = 0.
Assume first that χ is not zero. Then f(x) = f(—x) is not possible because / is
one to one. Hence
f{x) = -f{~x\
so, apart from x = 0, our function is odd. Now we show that Д0) = 0 also holds.
If a = x = 0 then (6) becomes /(/(0)) = 0. Assume now that /(0)^0. Then, as
we have just shown, /(—/(0)) = —/(/(0)). But the latter is equal to zero, so if
/(0) ^0 then /(-/(0)) = 0. So /(/(0)) = /(-/(0)) and thus /(0) = -Д0) because
/ is one to one. The last equality means that /(0) = 0 that is /(0)^0 is indeed
impossible. Set now x = 0 in (1). Thus
(V) f(f(y))=y-
Finally we prove that / is increasing that is if x\ <X2 then f(x\) < f(x2)·
Let X2 = x\ + Xq (жо>0)· letting x = xq, y = f(x\) in (1) and using (7)
f(x2) = f(4 + xl) = f(xO + f(f^l))) = f(xi) + f2(xo)>f(xi)·
It is the monotonity that forces / to be equal to the identity. Indeed, if χ > f(x)
then, as an increasing function, f(x) > f(f(x)) which, by (7), is equal to x. Thus
χ > f{x) yields χ < f(x) and, similarly, from χ < f{x) we can deduce χ > f(x).
Therefore the only possibility is f(x) = χ and this is indeed a solution.
1992/3. Consider 9 points in space, no 4 coplanar. Each pair of points is
joined by a line segment which is coloured either blue or red or left uncoloured.
Find the smallest value of η such that whenever exactly η edges are coloured,
the set of coloured edges necessarily contains a triangle all of whose edges have
the same colour.
Solution. The problem when rephrased in usual graph-terminology is as
follows: at least how many edges of a complete graph with nine vertices have
to be coloured with red or blue if for any colouring there is a monochromatic
triangle.
We prove that the number we are looking for is 33. The complete nine-point
/9\
graph has 11= 36 edges. Delete 3 of them randomly and colour the remaining
386

1992/4
1992.
33 edges arbitrarily. Select 3 different ones among the endpoints of the deleted
edges and delete them from the graph, together with the coloured edges adjacent
to these erased vertices.
The remaining graph has 6 vertices and the edges connecting any two of
them are all coloured; this one is a complete 6-graph. According to a Ramsey-
type theorem, stated and proved in the solution of Problem 1964/4 one can always
find a monochromatic triangle in such a graph. Therefore, any 9-point graph with
33 edges contains a monochromatic triangle, indeed.
Figure 92/3.1
The graph on Figure 1992/3.1 shows a 9-point graph with 32 edges where
there is no monochromatic triangle. Hence the minimal value of η having the
given property is 33. There are the blue edges indicated on the figure, only, and
you can check that they do not form a triangle, indeed. Among any three vertices,
however, there are two connected by either a blue edge or an uncoloured one and
thus there is no red triangle, either. (The four deleted edges are indicated on the
figure by a dotted line).
Remark. According to a celebrated theorem of Paul Turan the problem can
be extended to graphs of arbitrary size.
1992/4. L is tangent to the circle С and Μ is a point on L. Find the locus
of all points Ρ such that there exist points Q and R on L equidistant from Μ
with С the incircle of the triangle PQR.
387

Solutions
1992/4
Solution. Consider a
triangle PQR satisfying the
conditions (Figure 1992/4.1). Circle С
is touching / at E. G is the
point diametric to Ε on the circle С
and the tangent to С at G is e.
Magnify now С from Ρ
into the excircle C' of APQR
touching the side QR. This
enlargement is mapping e into / and
thus C' and / are touching each
other at G'', the image of G under
this enlargement. As we know,
QE = RG' = s-PR where s is
the semiperimeter of the
triangle PQR. (PR > PQ) can be
assumed and hence the points Ε
and G' are symmetric through M.
This, of course, is including the
case when the two points, Ε and
G' happen to be identical.
It can be checked, on the
diagram, that the configuration E,
M, G1, G and also the position of the line GG' depends on the given data only;
the point Ρ hence is incident to the straight line G'G or, more precisely, it is on
the opened ray / starting at G and separated from the circle С by e.
We now prove that this ray is the locus itself. Consider an arbitrary point
Ρ on /. Being on the opposite sides of the line e, both tangents from Ρ to С
intersect e and also l; the latter at points Q and R respectively. By the
construction С is the incircle of APQR. As we have already seen, the point G' where
PG and / meet is the touching point of the excircle of the same triangle and thus
QE = G'R. О and Μ are the midpoints of the sides GE and EG1 respectively,
Μ is hence halving QR, indeed.
The locus of the points Ρ of the problem is the opened ray /.
1992/5. Let S be a finite set of points in three-dimensional space. Let Sx>
Sy, Sz be the sets consisting of the orthogonal projections of the points of S onto
the yz-plane, xz-plane, xy-plane respectively. Prove that
\s\2<\sx\\sy\\sz\,
where \A\ denotes the number of points in the set A.
Figure 92/4.1
388

1992/6
1992.
[The orthogonal projection of a point onto a plane is the foot of the
perpendicular from the point to the plane.]
Solution. Denote, for brevity, the cardinalities of the sets S, Sx, Sy, Sz by
n, a, b, с respectively. The task is now to prove
(1) n2<abc
and we shall proceed by induction on n.
(1) is clearly true if n = l = a = b = c. Let η > 1 and assume that (1) is true
for any set S for which \S\ <n and let S be an η-element set. Consider now
a plane, parallel to one of the three coordinate planes, which splits S into non
empty subsets: S\ and S2. К *s clear that one can always find such a plane, even
if the set S happens to be planar. Assume that this separating plane is parallel to
the xy-p\ane, for example. Let \S\\ = n\, \S2\ = τΐ2', then, of course, щ + n2 = n.
Denote the number of points in the projections of the two sets, S\ and 52 on the
yz-plane, z:c-plane, xy-plane by
ab bb cb and a2, b2, c2
respectively. We now clearly have
0-1+0-2 = a, b\+b2 = b, c\<c, C2 < с
Since both n\ and n2 are strictly less than η
П\<а\Ь\С\ and П2<«2^2С2
by the induction hypothesis. Hence
п2 = (щ +п2)2 < (Jaibici +у/а2Ь2С2) <
<U<4b\c + y]a2b2c\ =c\Jalbi+^a2b2J .
Cauchy's inequality now settles the issue since:
η < c(a\ + a2){b\ + b2) = abc:
and the proof is complete.
Remark. Equality might hold in (1) if, for example, the given points are all
on a line parallel to one of the axes.
1992/6. For each positive integer n, S(n) is defined as the greatest integer
such that for every positive integer к < S(n), η can be written as the sum of к
positive squares.
(a) Prove that S(n) < η — 14 for each η > 4.
(b) Find an integer η such that S(n) = n — 14.
(c) Prove that there are infinitely many integers η such that S(n) = n — 14.
389

Solutions
1992/6
Solution. To get an impression about the flavour of S(ri) let's check how
it works if η = 13. By definition 5(13) is the greatest positive integer for which
13 = 169 can be written as the sum of 1, 2, 3, ..., 5(13) squares. (By a
square now we mean a positive number, zero is excluded.)It is easy to check, for
example, that 5(13)> 12, because 13 = 169 can be written as the sum of 1, 2,
..., 12 squares as follows:
(1) 169=132 = 52 + 122 = 32 + 42 + 122 = 42 + 52 + 2·82 = 52+4·62 =
ΐ32 + 42+4·62 = 5·42 + 52 + 82 = 4·32 + 52 + 3·62 = 42 + 5·32 + 3·62 =
Ю
9.42 + 52 У 32 +10 ■ 42 l= 12 + 4 ■ 22 + 5 ■ 42 + 2 ■ 62;
The numbers upon the equality signs indicate the number of terms in the
subsequent sum.
Let's turn now, one by one, to the solution of the three parts.
9 9
(a) It is enough to show that η cannot be written as the sum of η —13
squares if η > 4. Assume the contrary: suppose that there exist η — 13 positive
integers such that
2 2 2 2
αι+α2 + ... + αη2_ι3 = η .
Rearranging we get
(2) (a2 - l) + (a2 - 1) + .. . + (a2_13 - 1) = 13.
9 9
The terms in the brackets are not negative and thus af — 1 < 13. Hence af — 1 is
either 0 or 3 or 8. There is at most one 8 in (2) because 2 · 8 is already greater
than 13. If there is no 8 at all then the l.hs. is a multiple of 3, a contradiction.
Similarly, if there is exactly one 8 then the l.h.s. is congruent to 2 modulo 3
while 13 is not. Thus (2) is impossible, indeed.
(b) We prove that n=13 will do by showing that 5(13) = 132 - 14 = 155.
Having already seen, in part (a), that 5(13) < 13 —14 = 155 it is left to show
that 169 can be written as the sum of 1, 2, 3, ..., 155 squares. This means that
for each к between 1 and 155 there exist positive integers щ such that
169 = af + a2 + ... + a2·
As in (2) this sum is written as
(3) (a2 - 1) + {a\ - 1) +... + (a2, - 1) = 169 - k.
9 9
Since af — 1 < 169 we can prepare the list of the possible values of af — 1 again:
(4) 0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168.
ι
We start by (a\ — 1); its value is adjusted by a kind of greedy algorithm; it is set
to be the highest element not exceeding 155 — к on the list above. The possible
values of a\ — 1 are arranged in the following array:
390

1992/6
1992.
a2-\
0
3
8
15
24
35
48
63
80
99
120
143
CL\
l2
22
32
42
52
62
72
82
92
102
ll2
122
169 —fc
14< 169-/c< 16
17<169-/c<21
22<169-/c<28
29 < 169 - к < 37
38<169-/c<48
49<169-/c<61
62<169-/c<76
77 < 169 - к < 93
94 < 169- к < 112
113 < 169- к < 133
134 < 169- к < 156
157 < 169- к < 168
к
153 < к < 155
148 </с < 152
141 < /с < 147
132 < к < 140
121 < А; < 131
108 < к < 120
93 < к < 107
76<к<92
57<к<15
36<к<56
13</с<35
1</с<12
Having set the value of a\ the further c^-s have to be chosen according to
(a2 ~ l) + (a3 ~ l) + ■ ■ ■+ (al - l) = 169 - A; - (a2 - l) .
By the choice of a\ we have 14 < 169 — к — (of — l) < 36 and now this number
has to be written as a sum whose terms are of the form a — 1. There are various
ways to do this. If, for example, 169 — к — ία2 — 1J is a multiple of 3 then it can
be written as the sum of at most twelve 3-s; if it is equal to 31 +1 then two 8-s
and at most six 3-s are needed; finally, if it is of the form 31 + 2 then one 8 and
at most nine 3-s will do the job.
Thus 169 — к — (of — 1 j is decomposed into the sum of at most 12 numbers,
each of the form a —I. If к> 13 then the remaining terms in (3) are simply
zero, therefore 169 can also be expressed as the sum of 13, 14, ..., 155 positive
squares; as for fewer square terms, well, this has already been verified in the
introductory part. Hence 13 is indeed a number requested in part (b).
(c) For this we prove that if n > 8 and S(n) = n2 - 14 then S(2n) = (2n)2 -
14. The claim now follows by induction: n= 13 · 2m has the desired property for
every m.
For m = 0 that is n = 13 we have already proved in part (b) that S(n) = n —
9 9
14. Assume now that S(n) = n —14. We show that An can be written as the
sum of к squares for every к between 1 and An — 14; this, when combined with
(a), already implies that
S(2n) = 4n2-14.
391

Solutions
1992/6
Consider first
(5) 1 <k<n2 -14.
By the induction hypothesis η now can be written as the sum of к positive
squares; multiplying each term by 4 yields a decomposition of An .
Let k\, &2, &з, к^ be now positive integers not exceeding η — 14. By the
induction hypothesis η can be expressed as the sum of k\, &2> кт,, к^ squares
respectively:
n2 = a2+ (% + ...+ a2k{, n2 = b\ + b% + ... + 6^,
п2 = с2 + 4 + ... + с2кз: n2 = d2 + 4 + ... + d2k4.
Adding these sums yields a decomposition of An into the sum of k\ + k^ + кт, +
k/\. squares.
An2 = a2 + ... + a2ki+b2 + ... + b2k2+c2 + ... + c2k3+d2 + . .. + d^.
As for k\ + ki + k% + /c4, the number of terms clearly
(6) 4 < к < A (n2 - 14) = An2 - 56.
Finally write η as the sum of 1, 2, ..., η — 14 squares respectively and
append 3n copies of 1 to each decomposition. This yields a square sum repre-
sentation of An where k, the number of terms satisfies
(7) 3n2 + l<k<An2-lA.
Observe additionally that if
and thus every к satisfying
7 7 9
Observe additionally that if η > 8 then 4 < η — 14 and also 3n + 1 < An — 56,
l</c<4n2-14,
fits to one of the conditions (5), (6) or (7) and thus the corresponding
decomposition works.
1993.
1993/1. Let f(x) = xn + 5xn~ +3, where n>\ is an integer. Prove that
f{x) cannot be expressed as the product of two non-constant polynomials with
integer coefficients.
First solution. Assume, to the contrary that f(x) can be factorized as f(x) =
g(x)h(x) where ι
g(x) = anxn + αη_\χη~ +... + a\x + aQ,
h(x) = bnxn + bn_ixn~l +.. . + b\x + bQ·,
392

1993/1
1993.
the coefficients are integers. Comparing the coefficients of f(x) and those of the
product, g(x)h(x) yields
(1) a0b0 = 3,
(2) aib0 + a0bi=0,
(3) a2bo + aibi+aob2 = 0:
a^bo + a2bi +a\b2 + aob2 = 0,
an_2bo + an_3bi+an_4b2 + ... + aibn_3+a0bn_2 = 0.
(1) implies that clq or 6q is equal to 3 or —3; we may clearly assume that
uq = 3 and thus bo = l. Plugging these values to (2) yields that 3 divides αχ; hence
(Z2 also and so on: the coefficients аз, сц, ..., an_2 are all divisible by 3.
Since h(x) is at least first degree, by condition, an has to be zero otherwise
the degree of the product g(x)h(x) would exceed that of f{x). We show now
that an_i cannot be zero. Indeed, the opposite implies that every coefficient of
g(x) is divisible by 3 and thus its integer values as well. Then, of course, this
also holds for the product f{x). But /(2) = 7 · 2n_1 +3 is not a multiple of 3, a
contradiction.
Since an_\ =^0 the degree of g(x) is (n — 1) and thus h(x) is a first degree,
it can be written as h(x) = b\x + 1. Comparing once more the coefficients yields
a,n_\b\ = 1 that is \b\\ = 1, h(l) = 0 or 2, an even number anyway. On the other
hand /(1) = 9 is odd, a contradiction and thus the proof is finished.
Second solution. Suppose again that there is a factorization f(x) =
g{x)h{x); g and h are at least first degree integer polynomials. We may
clearly assume that the leading coefficient of g(x) — and hence that of h(x) — is
equal to 1. Thus
g(x) = x + ak_\x + ak_2Xk +.. . + a\x + aQ.
Comparing again the coefficients of f(x) and g(x)h(x) the product of the
constant terms, g(0)h(0), is 3 that is |д(0)||/г(0)| = 1-3. Hence
(4) |ao| = W>)| = 1
can clearly be assumed, к = 1 is not possible otherwise g(x) would be equal to
x±l with a root —1 or 1 but none of these numbers fit f(x); hence
(5) fc>l.
Consider now the roots o>\, ct2, ..., α^ of g(x). Then g{x) can be factorized
as
(6) g(x) = (x-ai)(x-a2)...(x-ak).
393

Solutions
1993/1
In the general case these roots are complex and, of course, they also make f(x)
zero. (4) now implies that
(7) ^(О)!^!^... 0^1 = 1.
Since f(ai) is also zero (г = 1, 2, ..., к),
(8) α?-1(αί + 5) = -3.
Prepare now the product of the relations of the kind (8) for г = 1, 2, ..., k:
(ага2 ... ak)n~l (ax + 5)(a2 + 5)... (ak + 5) = (-3)k.
By (7), this implies
|(ai+5)(a2 + 5)...(afc+5)| = 3fc.
The product on the l.h.s., by (6), is equal to g(—5); thus
Ы-5)\ = зк.
On the other hand
^(-5Ж-5) = Л-5) = (-5Г + 5(-5)п-1+3 = (-5Г-1(-5 + 5) + 3 = 3.
Hence \3k -h(-5)\ = 3 that is \3k~l -h(-5)\ = 1. We have seen, however, that
к > 1 and thus the last equality cannot hold because h(—5) is also an integer; the
required factorization is hence impossible, indeed.
1993/2. Let D be a point inside the acute-angled triangle ABC such that
LABD = 90° +1 AC В and AC-BD = AD- ВС.
, ч ^ , , , AB-CD
(a) Calculate the ratio —————.
AC-BO
(b) Prove that the tangents at С to the circumcircles of ACD and BCD are
perpendicular.
First solution. It is worth noting that the two parts of the problem are just
vaguely related; the second condition, for example, is not necessary for the
second claim to hold.
Let 1С AD = a', LCBD = β'; besides the usual notations let AD = a', BD =
b' and CD = c' (Figure 1993/2.1). Consider part (b) first,
(b) Summing the angles in AABD:
180° = α-αί + β-β' + ^ + 90ο:
(3) that is α' + β' = 90°.
394

1993/2
1993.
Draw a ray from С which makes
an angle a' with DC opposite to A. By
the theorem of inscribed angles this ray
is touching the circumcircle of AACD
at C. Similarly, the ray making ίβ'
with DC at C, opposite to В is
touching the circumcircle of ABCD. The
angle of these two tangents is a + β1
which, by (3), is equal to 90°: the two
tangents are thus perpendicular indeed;
the second proposition is hence
proved.
(a) Rotate and enlarge A ABC
a
about A by a with scale factor
υ
mapping hence AC to AD. Denoting
the image of D under this
transformation by D1 clearly I ADD1= 7 and
ID AD'= a.
With our notations (2) becomes
(4) bb' = aa'.
D'
Figure 93/2.1
a
W
. = — = b'\ moreover LBDD'L =
b b
For the image DD' of С В clearly DD1 = a
90°; ABDD' is an isosceles right triangle and thus BD' = b'y/2. Since LDAD' =
a' IB AD' = a. Hence AB = cis mapped into AD' and thus for the latter we have
AD'
ca
Observe that ACAD and ABAD' are similar; indeed, they have an equal
angle both at A and the respective ratios of the sides forming this angle are also
equal. The scale factor of similarity is - and thus
cc
or
BD' = y-CD = ^ = b'V2:
b b
AB-CD
yielding
bb' '
= л/2,
AC-BD '
the answer for the first question of the problem.
Second solution. (For part (a) only.) Drop perpendiculars from D to the
sides AB, ВС, С A; the feet are interior to the sides, by condition. Denote them
395

Solutions
1993/2
Figure 93/2.2
by C, A', B' respectively. (See Figures 1993/2.1 and 1993/2.2) The
quadrilateral AC'DB1 is clearly cyclic and thus, by the theorem of inscribed angles,
lDC'B' = a' and, similarly, IDC'A'= β'. Hence, by (3)
B'C'A'l = a' +β'= 90°,
AB'C'A' is right angled.
The diameter of the circumcircle of AC'DB' is a', therefore
B'C = a'sin a.
If, for simplicity, the circumradius of A ABC is 1/2 then a = sin a and
B'C' = aa'.
Similarly, C'A' = bb' and A'B' = cc'. The second condition as it is written in (4)
now implies
B'C' = C'A',
the right angled AB'C'A' is isosceles. Its hypotenuse is A'B' = C'A'V2 and
thus, by the previous cc' = bb'V2,
cc' _ AB-CD _ r-
W~ AC-BD~ '
Third solution. We apply inversion; this approach reveals a closer relation
between the two parts and also the possible origin of the whole problem.
The following property of inversion will be used: if the images of the points
X and Υ under the inversion of pole О and radius R are X' and Y' respectively
396

1993/2
1993.
then — unless Χ, Υ and О are collinear — AOXY and ΑΟΥ'X' are similar
and thus
lOXY = /OY'X' and /OYX = lOX'Y',
moreover
(5)
x'y'-r2-xy
OXOY
Let's see the proof. By definition OX ■ OX' = OY ■ OY' = R2 therefore
OX OY'
OY OX'
(Figure 1993/2.3)
AOXY and ΑΟΥ X have one common angle at О and the ratios of the respec-
X
Figure 93/2.3
tive sides forming this angle are also equal; the two triangles are hence similar,
indeed, therefore
X'Y' OX' , . .„ OX'-XY
that is X'Y' =
XY OY '
OY
Substituting
OX' =
Rl
OX
yields (5). Observe that (5), of course, holds even if the points O, X and Υ are
collinear.
397

Solutions
1993/2
С
Figure 93/2.4
Apply now inversion of pole D and radius R = Va'b'c' to the points A, B,
С in Figure 1993/2.1. By our previous remark these points transform into the
triple А', В', C' such that
A'B' = cc', B'C = aa', C'A' = bb'. (Figure 1993/2.4)
With the notations of Figure 1993/2.1 the previous results become
a' = lDAC = lDC'A' and β'' = IDBC= LDC'B''.
Hence lA'C'B1 -ol + β' and in the previous solutions we have already
seen that these angles are complementary. AA'C'B' is hence right angled and
isosceles and thus
bb' C'A'
The circumcircles of AC Ό and BCD are mapped to the straight lines A'C'
and B'C' respectively. Since inversion is preserving angles and A'C' and B'C'
are perpendicular, the two circles above are also orthogonal; their respective
tangents at С make a right angle, indeed; proposition (b) is hence proved.
1993/3. On an infinite chessboard a game is played as follows. At the start
η pieces are arranged in an η χ η block of adjoining squares, one piece on
each square. A move in the game is a jump in a horizontal or vertical direction
over an adjacent occupied square to an unoccupied square immediately beyond.
The piece which has been jumped over is removed. Find those values of η for
which the game can end with only one piece remaining on the board.
Solution. We show that the object of the game can be achieved if and only
if 3 does not divide n. We shall proceed by induction and, getting started, two
particular cases are going to be checked.
398

1993/3
1993.
In case a) consider a board where there are two squares adjoined to the
extreme right field of a 3 χ 1 rectangle. There are four pieces altogether on the
board and one of the adjoined squares is unoccupied. Figure 1993/3.1 shows
how this board can be emptied. The pieces are shown as spots and the arrows
indicate the three moves. As a result, a 3 χ 1 rectangle can always be emptied if
there are two adjacent squares such that just one of them is occupied.
— t · · _
•1· · 1·|· ι | · | |
>
Figure 93/3.1
In case b) consider a saturated 2x2 square with a three long empty row
attached to it (Figure 1993/3.2). The three steps to empty this board are shown
in Figure 1993/3.2. Note that the single left piece ends up on one of the attached
squares.
>
A | _· |A ·_·__! ·
·_·_ _·_
Figure 93/3.2
After this preliminary solitaire let's turn to the induction proof; we show
that if n = 3k+l or n = 3k + 2 then one can arrive to a single piece on the board.
This is obvious if n = l and if η = 2 then this is case b). Let n>\ and
assume that the task can be done on any board whose dimension is not divisible
by 3 and less than n.
If η = 3k + I then consider a framing pattern of 3 χ 1 rectangles of the η χ η
board as it is shown in Figure 1993/3.3. Since the externally adjacent squares
are all empty, the frame can be cleaned if proceeding piecewise as in case a).
Having finished we are left with a 3k+1 —2 = 3(k — l) + 2 saturated board and
the rest is induction.
The story is essentially the same if n = 3k + 2. The frame now has two layers
(Figure 1993/3.4) and beginning from the outside both layers can be cleaned
up following again the procedure in a). The full part of the board is of side
ЗА; + 2 — 4 = 3(k — 1) +1 so, by the induction hypothesis, we are done, again.
399

Solutions
1993/3
Figure 93/3.3 Figure 93/3.4
Finally, we prove that the solitaire has no solution if the side of the board is
3k. Divide the infinite board into 3x3 squares and label the fields as it is shown
below.
I
0
1
2
1
2
0
2
0
1
0
1
2
2
0
0
1
0
Consider now a 3 k χ 3 k occupied block and mark each piece with the label
of the currently occupied square. There are two squares emptied during any move
and another one so far empty gets occupied. Jumping to the right, for example,
from the indicated square in the figure the two pieces removed were labelled by
1 and 2 respectively while the apparently new piece pops up with label 0. Thus,
as a byproduct of every move, the number of labelled pieces is changing by 1
for each label; there are two of them, like 1 and 2 in the example, whose total
is decreasing and for the third one it is increasing. Initially there are the same
number of pieces labelled by 0, 1 and 2 respectively (n is now a multiple of 3)
and thus, as long as we are playing, the parity of the number of pieces is the
same for the respective labels.
This is the heart of the issue; now it is clear that we cannot end up with a
single piece: the label of its current field would occur but once while the other
two labels would be represented zero times, thus breaking the parity invariance.
1993/4. For three points P, Q, R in the plane define m(PQR) as the
minimum length of the three altitudes of the triangle PQR (or zero if the points are
collinear). Prove that for any points A, B, C, X
m(ABC) < m{ABX) + m(AXC) + m(XBC).
400

1993/4
1993.
Solution. In what follows by 'the longest one' in a system of distances we
mean one of the longest distances in the system.
The heart of the issue is the
following Lemma. If S is an interior
point of the segment А В then m(ABC) >
m(ASC).
To prove this it is clearly enough to
show that m(ABC) is not shorter than
some altitude of AASC because then
it obviously cannot be shorter than the
smallest altitude of the triangle.
Assume first that AB is the longest
side of A ABC (Figure 1993/4.1). Since
there is the shortest height perpendicular
to the longest side now it is the height
CT; this, at the same time, is the height of AASC and thus the claim is now
true.
Let side ВС be now the longest one in AABC. The line of AR, the shortest
height of AABC meets CS at some point Q because A and ВС are separated
by CS (Figure 1993/4.2). Denote the foot of the height from A of AASC by Z.
From the right triangle AZQ we have AZ < AQ < AR which settles the claim
in this case.
Figure 93/4.1
Figure 93/4.2
Finally, if the longest side of AABC is AC then the smallest height is ВТ
and for the perpendicular SR from S to AC clearly SR < ВТ (Figure 1993/4.3).
Being so
m(ASC) = SR<BT = m(ABC).
401

Solutions
1993/4
Figure 93/4.3
Turning to the actual problem the argument depends on the position of the
point X relative to AABC. There are three possibilities:
1. X belongs to the interior or the boundary;
2. X belongs to the so called U-regions that contain the excircles of the
triangle;
or
3. X is in the so called V-regions formed by the angles vertical to the angles
of ABC.
1. X is now contained by AABC. Assume also that with standard
notations a>b>c (Figure 1993/4.4). Expressing the area of AABC in two different
ways:
(1) 2[ABC] = a ■ m(ABC) = ax ■ m(ABX) + a2 · m(AXC) + a3 ■ m(XBC),
where αϊ, a2 and аз denote the longest sides of the triangles ABX, AXC, ABC
С а В
Figure 93/4.4
402

1993/4
1993.
respectively. Since the longest distance contained in a triangle cannot exceed the
longest side of the same triangle we have a\ <a, 0,2 < а, а^<а and thus (1)
implies
a ■ m(ABC) < a · m(ABX) + a ■ m(AXC) + a ■ m(XBC).
Dividing through by a the claim follows.
2. X is now in one of the C/-regions, say the one opposite to A. {Figure
1993/4.5) Denote the intersection of AX and Б С by Η. By our lemma
(2) m(ABH)<m(ABX) and m(AHC)<m(AXC).
We reuse the results proved in part 1. For A ABC and the point H:
(3) m(ABC) < m(ABH) + m(AHC) + m(HBC).
Now m(HBC) = 0 and thus (2) and (3) yield
m(ABC) < m(ABX) + m(AXC) + 0.
Since m(XBC) > 0 the estimation can be extended and we get the claim again:
m(ABC) < m(ABX) + m(AXB) + m(XBC).
Figure 93/4.5 Figure 93/4.6
3. The last case to be checked is when X is in one of the V-regions, say
the one at A, the rays forming the V-shape included (Figure 1993/4.6) If BX
and AC meet at К then by the lemma
(4) m(ABC) < m(KBC) and m(KBC) < m(XBC).
Since m(ABX) and m(AXC) are not negative (4) yields the claim
m(ABC) < m(ABX) + m(AXC) + m(XBC);
each case has been checked, the proof is complete.
403

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1993/5
1993/5. Does there exist a function f from the positive integers to the
positive integers such that /(1) = 2, f(f(ri)) = f(ri) + nfor all n, and f(ri) < f(n + 1)
for all n?
First solution. Denote the positive root of the equation
(4) ^-^-1 = 0
by a; it is equal to -(1 + V5) = 1,16180....
Define now the sequences
g(n) = an, f(n) = g(n) + — = an +
1
We show that f(n) satisfies the conditions and hence the answer for the question
of the problem is affirmative. By (4)
(5) g(g(n))-g(n)-n = 0.
Since /(1) = [1,6180... + 0,5] = 2, (1) holds; moreover, a > 1 implies
f(n+l) =
1
an + - + a
2
>
1 i"
an+ — + l
2
an +
1
+ l = f(n)+l>f(n)
and thus (3) is also satisfied.
Writing it as n + f(n) — f(f(n)) = 0 we follow a tricky way to prove (2):
since the values of / are whole numbers it is enough to show that
\n + f(n) — f(f(n))\<l, for every η εΝ.
(6)
We note first that by the irrationality of g(n)
\f(n)-g(n)\<1-.
Now clearly д(п\) — g(n2) = а(п\ —П2) (щ gN, п^ Ε Ν) and (5) can be
rearranged as n = g(g(n)) — g(n); comparing these pieces:
η + fin) - f{f(n)) = g(g(n)) - g(f(n)) +g(f(n)) - f (f(n)) + f(n) - g(n) =
= a(g(n) - f(n)) - (g(ri) - f(n)) +g{f(n)) - f{f(n)) =
= (a- l){g(n) - f(n))+g{f(n)) - f{f(n)).
Hence
. \n + f(n)-f(f(n))\<(a-l)\g(n)-f(n)\ + \g(f(n))-f(f(n))\<
/ 1ч1 1 a л
<(β"1)2 + 2 = 2<1'
indeed, the proof is hence complete.
404

1993/5
1993.
Second solution. In the solution we shall use the Fibonacci-sequence
defined by the recurrence
«1 = 1, u2 = 2, un+i=un + un_i (n>2).
The initial terms of this increasing sequence are
щ = 1, U2 = 2: щ=3, щ = 5, щ = 8, щ = 13, ηη = 21,
the first few Fibonacci numbers. First we provide an algorithm which uniquely
splits every positive integer into the sum of Fibonacci numbers such that there
are no consecutive terms of the sequence in the representation. Single term sums
are also accepted here. The proof goes by induction. For the first few positive
integers this can be checked with bare hands. Let n be greater than 1 and assume
that the claim holds for every positive integer less than n.
If n = uk then this is already the desired representation. If n itself is not a
Fibonacci number then consider its Fibonacci neighbours, uk and uk+\, that is
(7) uk<n<uk+l.
Hence, by n < uk+\ = uk + uk_\ < 2uk
0<n — uk<uk<n.
By the induction hypothesis n — uk as a positive integer less than n can be written
as
n - uk = щ1 + щ2 +... + uis
and thus
(8) п = щ1+щ2 + ... + щ8+ик:
where щ8 and uk are not consecutive Fibonacci numbers; the opposite, by the
defining recurrence, would imply
п = щ1+щ2 + .. .+uk+i
contradicting to (7). The induction is finished, the required representation does
exist for every n.
Next we prove that this "Fibonacci representation" of positive numbers is
unique. The very same induction works similarly and all we need to show is that
uk, as defined above, has to be present in any Fibonacci sum for n. Assume the
contrary and consider a sum
n = щх + щ2 + ... + uis + Щ8+1
which does not include uk. Thus щ , <uk and this implies the stronger inequa-
Hty uis+l <uk_i.
405

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1993/5
Let к be even first; /c — 1 is then odd and since there are no consecutive
Fibonacci numbers in the sum, clearly
п = щ. +щ2 +.. . + щ , < (1 +щ) + щ +щ +... + ик_\ =
= (u2 + u3) + u5 + ...+uk_i=(u4 + u5) + ... + uk_i=uk<n:
a contradiction. Similarly, if к is odd that is к — 1 is even and thus
п = щ1 +Щ2 +.. . + щ , <(щ 4- U2) + Щ + щ +. . . + uk_\ =
= (щ + щ) + щ +... + uk_ \=ик<п,
a contradiction again; the decomposition in (8) is unique, indeed.
We can now define the function / in terms of this decomposition. If η is
written as a Fibonacci sum in (8) then let f(n) be
(9) f(n) = uil+l +uil+i +.. .+uis+i +uk+h
We have to show, of course, that this well defined function satisfies the conditions
(D-(3).
that's about (1).
f{f(n)) = f(uil+i +uil+i + .. . + uis+i +uk+ij =
= uil+2+ui2+2 + ... + uis+i+uk+2 =
= \Щх+\ +ui1+\ + ■ · - + Щ8+1 +uk+l) + (uii +Щ2 + ·· ■ + uis+uk) =
= f(n) + n,
so (2) is also settled.
We prove (3) by induction.
№) = f(ui) = 2<f(u2) = u3 = 3,
(3) thus holds for the initial value. Let η > 1 and assume that (3) holds for every
positive integer less than η and let uk<n< uk+i for some k.
We distinguish two cases:
A) n+1 <ик+г;
B) n + l=uk+i.
A) The induction hypothesis now also implies that if η <η" <η are
arbitrary positive integers then
(10) f(n)<f(n").
ι
Indeed, applying (3) for the numbers all less that η yields the following chain of
inequalities:
f(n/)<f(n/ + l)<f(n/ + 2)<...<f(n").
406

1993/6
1993.
We know that both η and n+ 1 can be written as Fibonacci sums:
(11) п = щ{ +щ2 + .. . + щ3 + uk = n +uk, (n'<n)
U+l=Ujl + Uj + . . . + Ujr + Uk = Tb" + Uk (n' <n).
We have to show that f(n) < f(n + l) that is
(12) uil+i +щ2+1 +.. . + uis+i +uk+i < ujl+i +uJ2+i +.. .+ujr+i +uk+i,
or
u4+i + ui2+i + ■■■ + щ8+\ <ujx +1 + uJ2+i + ... + ujr+i.
By the definition of / this is nothing else but
fin'XKn"),
and this follows, as we have seen, from the induction hypothesis.
B) With the notation in (11) f(n) < f(n+l) is now
(13) Uil + l+Ui2+l+---+Uis+l+Uk+l <Uk+2=uk+Uk+i
that is
By the definition of / this last inequality is nothing else but
/(n,)</K_1);
since n' <n and uk_\ < n, the induction hypothesis yields the claim again.
We note that if n-uk then the sum щ. +.. . + щ8 is replaced by zero and
thus (12) and (13) obviously hold.
1993/6. There are η> 1 lamps Lq, L\, ..., Ln_\ in a circle. We use Ln+k
to mean Lk. A lamp is at all times either on or off. Perform steps sq, s\,. ..
as follows: at step s{, if L^_\ is lit, then switch Li from on to off or vice versa,
otherwise do nothing. Show that:
(a) There is a positive integer M(ri) such that after Μ (ή) steps all the lamps
are on again;
(b) If η = 2 , then we can take Μ (ή) = η — 1.
(c) Ifn = 2k + l then we can take M{n) = n2 -n + 1.
Solution, (a) The following chart contains the instructions of the problem
sfore Sji
after Sj:
Lj_i Lj
ON ON
ON OFF
Lj_x Lj
ON OFF
ON ON
Lj_x Lj
OFF ON
OFF ON
Lj_i Lj
OFF OFF
OFF OFF
407

Solutions
1993/6
From the chart it is clear that the state after Sj uniquely determines the state
before Sj. On the other hand there are but finitely many states of this n-lamp
system and thus some state has to occur more than once. This repeating state, as
we noted, uniquely determines the previous states, one by one, and thus, after a
certain number of steps — this is the number M(n) of the problem — each lamp
will be ON, again.
It can be checked from the chart that if ak = 1 (mod 2) then the kth lamp
is ON after the (A; — n)th step and it is OFF if ak ξ 0 (mod 2). Remember that
the lamps are numbered mod n.
In what follows we shall make use of the sequence clq, a\, ... defined as
(1) α0 = αι=... = αη_ι = 1, ak = ak_i+ak-n (k>n).
The characteristic equation of recurrence (1) is
xk = xk-1+xk-n.
which can also be written as
(2) χη-χη~ι-1 = 0.
This equation has no multiple roots (see the Note) and thus, as it is well known
from the theory of linear recurrences, with given coefficients C{
(3) ak = ciak+C2a2 + ... + cnakl,
where a\, a2, ..., αη are the roots of equation (2).
(b) Turning to part (b) let η be some positive integral power of 2 and a a
root of (2) that is an = an~l + 1. Consider now an +k. Clearly
an2+k = ak . an2 = ак(ап)П = ak(an-l + 1)n =
= α*^(η-1)+Μα(η-1)(η-1) +>i> +
on the other hand
an2+k = ak . an2=ak . an2-n .an = ak. an2-n(an-l + l) = akfan2-l+an(n-l)
Subtracting these two equalities
Hence, by (3)
ι
n\ ( η \
χ I %_ i)2+fc + ··.+ Ι1Κ-1+* + α*- an2_i+k =
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1993/6
1993.
\ / ι=1 \ ι г=1 г=1 г=1
Ciaf-1+k
1 г
Σ г. ((П\ (гъ-1)2+к^ ,(П \„п-1+к,„к „п2-1+к\_ V^ П-П
Since η is a power of 2 the binomial coefficients in the above expressions are all
even and thus, by the previous result,
°=( J%-l)2+fc + ---+Ll!1jan-l+fc+afc-V-l+fcS
= a>k- ап2_\+к (mod 2) that is
a>k = an2_i+k (mod 2).
Since ak = 1 if k = 0, 1, ..., n— 1-re α& = 1
the desired result.
(c) Now it is (n — l)'s turn to be a power of 2; using the notations and also
the results of part (b) consider an _n+fc, again. First of all
an2-n+k = ak . an2-n = ak (αη}η-1 = ak ^an-l + ήη~1 =
On the other hand
an2-n+k = ak . an2-n = ak . a(n-l)2 .an-l=ak, a(n-l)2 (q,ti _ χ) =
Subtracting these equalities
0 = ak Uab-1? + Н1н(«-1) + ...+ Н|й-1 + 1_^1
Proceeding now as in part (b)
The binomial coefficients are even again and thus
o-k = an2_n+i+k (mod 2) for every k.
Since a\ = ... = an = 1
an2_n+2 = αη2_η+3 = · · · = an2+l = 1 (mod 2),
and this was to be proved.
409

Solutions
1993/6
Remark. We have used that the polynomial
χη_χη-1_1=0
has no multiple roots. Indeed, multiple roots also fit nxn — {n—\)xn =
xn~ (nx — (n — 1)), the derivative polynomial; the roots of the latter are ob-
71 — 1
viously 0 and and none of these do satisfy the original polynomial.
η
1994.
1994/1. Let m and η be positive integers. Let a\, a,2, ■.., am be distinct
elements of {1, 2, ..., n} such that whenever щ + a,j < η for some i, j (possibly
the same) we have ai + clj = a& for some k. Prove that
ai+a2 + ..- + am n+1
m ~ 2
Solution. Denote the set {a\, a2, ..., am} by A; we may assume that the
elements are in decreasing order that is a\ > а>2 > ■ ■. > am. Assign, to any щ
element of A, the element am_i+\ as its pair. This matching is symmetric, the
pair of am_i+\ is a{ and if m is odd then the median, am+\ is equal to its own
pair.
First we show that the sum of any pair is at least η +1. If, to the contrary,
щ + am_i+i < n+ 1 for some (1 < i < m) then the ordering of A yields
(1) ai<ai + am<ai + am_\ <...<щ + am_i+\ < n.
Hence, by the condition, the i sums in (1) produce but elements of A, moreover,
all of them are greater than щ, a contradiction since there are only i — 1 elements
in A beyond c^. Therefore
a2 + am-l >n+ 1;
am + a\ > n+ 1.
Summing these inequalities yields the claim:
2 (a\ + 0,2 +... + am) > m(n + 1).
1994/2. ABC is an isosceles triangle with AB = AC. Μ is the midpoint of
ВС and О is the point on the line; AM such that OB is perpendicular to AB. Q
is an arbitrary point on ВС different from В and C. Ε lies on the line AB and
F lies on the line AC such that E, Q and F are distinct and collinear. Prove
that OQ is perpendicular to EF if and only if QE = QF.
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1994/2
1994.
Figure 94/2.1
Solution. Everything is
obvious if M = Q; assume hence
that MiQ. AM is on the axis of
the triangle and thus О is
diametrically opposite to A on the cir-
cumcircle {Figure 1994/2.]);
besides 10BC = 10CB.
Assume first that OQ and
EF are perpendicular. The
quadrilaterals OEBQ and OCFQ
are then cyclic because 10BE =
/OQE = lOCF = lOQF = 90°.
Hence, as inscribed angles
/OBQ=lOBC=lOEQ,
LOCQ = LOCB = LOFQ.
The right triangles OEQ and
OFQ are thus congruent so
QE = QF, indeed.
The converse is proved by the method of reductio ad absurdum:
assume that Q bisects EF but OQ
and EF are still not
perpendicular (Figure 1994/2.2). Draw
perpendicular from О to EF;
this line cuts ВС at Q'. We
need also the perpendicular to
OQ' at Q'; it intersects the
lines AB and AC at E' and F'
respectively. By the first part
of the proof Q'E' = Q'F' and
now, by our indirect
assumption the points Q and Q' are
definitely different.
Consider now the median
through Q' of AAE'F'. It cuts
EF at the internal point К that
bisects EF. This, however, is
impossible, since Q is also the
midpoint of EF and it is
obviously different from К since the latter is not incident to ВС. The contradiction
proves that OQ is indeed perpendicular to EF.
Figure 94/2.2

Solutions
1994/2
Remark. In terms of the theory of conic sections the problem can be
rephrased as follows:
If AB and AC are tangents to the parabola, moreover, they are symmetric
with respect to its axis then the tangent at the vertex bisects the part cut by AB
and AC of any other tangent.
The reason of this symmetry is that EF is touching another parabola whose
focus is О and it touches at its vertex the line ВС.
1994/3. For any positive integer k, let f(k) be the number of elements in
the set Ak = {k+ 1, к+ 2, ..., 2k} which have exactly three Is when written
in base 2. Prove that for each positive integer m, there is at least one к with
f(k) = m and determine all m for which there is exactly one k.
Solution. Call a positive integer 'good', for brevity, if its binary form
contains exactly three Is; that is, it is the sum of three different powers of 2. Observe
that к is good if and only if 2k is also good.
(a) Compare now the number of good elements in the sets Ak and Ak+\.
Ak = {k + 1, k + 2, k + 3, ..., 2k-\, 2k, },
Ak+i = { k + 2, k + 3, ..., 2k-], 2k, 2k+l, 2k + 2}.
These sets differ in the elements k + l, and 2k +1, 2k+ 2 only. Since k+ 1 and
2k + 2 are equally good or not, 2k +1 is responsible for any difference between
f(k) and f(k+l). Thus
f(k + l)-f(k) = 0, if 2k +1 is not good,
f(k + 1) - f(k) = 1, if 2k +1 is good.
Increasing к by 1 either leaves f(k) unchanged or it also goes up by 1 and the
latter happens if and only if 2k + 1 is good.
Let us calculate now the value of f(2n). Since 2 is not good there are the
same number of good elements in the intervals [1,2 ] and [1,2 — 1]. The
binary form of these good numbers contains η +1 digits (leading zeros are certa-
(n +1
) such numbers.
Similarly, f(2n)= (П\. Thus the set
A2n = {2n + 1, 2n + 2, ..., 2n+1} contains / (2n) =
good numbers. Since f(4) = f\2 ) = 1 and the counting above implies that /
assumes arbitrarily large values jumping at most 1 at a time, it follows that /
admits every positive integer.
412

1994/4
1994.
(b) Given the value of m the equation f(k) = m has a single solution if / is
strictly increasing at к that is
f(k+l)-f(k) = f(k)-f(k-l) = l.
For this to happen both 2k +1 and 2k— 1 have to be good as we have seen
above. If 2k — 1 is good then both its first and last binary digits are 1. We show
that the third unary digit is next to the last one. If not then adding 2 (Ю2) to
2k— 1, the result, 2k + 1 would contain four Is and thus it were not good. Thus
2/c-l=2a + 2 + l, (a>3)
therefore
k = 2a~l+2. With n = a-l this is k = 2n + 2 (n>2).
The converse is obvious, if к = 2n + 2 then 2k +1 and 2k —I are both good.
We are left to determine / (2n + 2), the number of good elements in the set
A2n+2 = {2n + 3, 2n+4, ..., 2n+1+4}.
Since 2 +3 is good while there is no good one among the numbers
2n, 2n + \, 2n + 2, 2n+l, 2n+1 + l, 2n+1+2, 2n+1+4
there is one more good number in the set ^2n+2 tnan m tne interval
f2n + l, 2n + 2, ..., 2n+
Here, as we have seen, there are { 1 good numbers, therefore
f(2n + 2)=(n\ + l-
the equation f(k) = m has one solution only if m = I 1 +1.
1994/4. Determine all ordered pairs (m,n) of positive integers for which
(1) ^Щ
mn — 1
is an integer.
Solution. First we show that in spite of the apparent asymmetry, if a pair
(m, n) makes (1) a whole number then so does the pair (n, m). Since m is prime
to mn — 1 the latter divides η +1, the numerator if and only if it also divides
rr?(r? +1). On the other hand
о о о о
πι (η +1) — (πι + l) = (mn) —1, a multiple of (mn—1)
413

Solutions
1994/4
therefore πι (η +1) and πι +1 are congruent modulo πιη— 1 and the claim
follows. Hence we may assume that m > η in (1).
Consider the case m = n first. Then (1) becomes
n3 + l 1
= n +
n2 — 1 η — 1
and this is a whole number only if η = 2; the pair (2,2) is hence a solution.
2
Let now m> n. If η = 1 then is an integer only if m = 2 or m = 3;
πι — 1
the pairs (1,2) and (1,3) are also solutions. If η > 2 then denoting the value of
(1) by e we obtain
3 n3 + 1 + e
η +l = emn — e, m = .
en
For m, to be an integer, 1 + e has to be divisible by η therefore 1 + e = kn for
some positive integer k. Hence e = kn—l so(l) yields
n3 + l n3 + l 1
kn—l = < —* = η +
πιη — 1 η2 — 1 η—1
that is
(fc-l)n<l+ *
η— 1
Since the r.h.s. is strictly less than 2, the only possibility is k= 1 and then
n3 + 1 = (n — l)(mn — 1).
Isolating m
n2 + l л 2
m = =n+l + -,
n—1 n—1
which is an integer only if n = 2 or n = 3; the corresponding value of m is 5 in
each case and thus the remaining solutions are the pairs (2,5) and (3,5).
Remembering our initial remark there are 9 solutions altogether, namely
(2,2), (1,2), (1,3), (2,5), (3,5),
(2,1), (3,1), (5,2), (5,3).
1994/5. Let S be the set of all real numbers greater than (—1). Find all
functions f from S to S such that f(x + f(y) + xf(y)) = y + f(x) + yf(x) for all χ
f(x)
and y, is strictly increasing on each of the intervals — 1 < χ < 0 and 0<x.
χ
Solution. Let x = y^S. The functional equation hence becomes
(1) f(x + (l+x)f(x))=x + (\ + x)f(x).
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1994/5
1994.
Consider now the solutions of
(2) f(u) = u,
the so called fixed points of /. By (1) u = x + (l + x)f(x) is a solution of (2) for
every χ G S. If uj^O then (2) can be written as
(3) £ϊ> = 1.
U
Since is strictly monotone, it admits the value 1 at most once in both in-
u
tervals ] — 1,0[ and ]0, +oo[.
Let f(u) = и for some ue]-1,0[. Substituting χ = и (1) becomes
f(u2 + 2u) = u2 + 2u.
Observe that the mapping ia->u + 2u keeps the two parts of S fixed that is
if и G] - 1,0[ then u2 + 2u G] - 1,0[ and, similarly, if и G]0, +oo[ then u2 + 2u G
]0, oo[. Thus there are no solutions щ, U2 both in S such that щ G] — 1,0[,
2 2
U2 G]0, oo[ and υ,γ + 2u\ = щ and n2 + ^u2 - Щ ·
Since there is at most one ue]-l,0[ satisfying (2), if и +2u£] — l,0[
then
η +2u = u.
But this quadratic has no solutions in ] — 1,0[, at all, and thus (2) cannot hold
if и G] — 1,0[. A similar argument shows that (2) cannot hold in ]0, +oo[ either
and thus f(u) = и forces и = 0.
Thus (1) holds only if
x + (l+x)f(x) = 0 x<eS
and hence the only possibility is
(4) №) = -TT-
1 +x
We have to check, of course, if this one is indeed a solution.
I. f(x) is defined in S and
/(aO = -l + J_
1 +x
shows that χ G S implies f(x) > — 1 that is f(x) G S, indeed.
II.
χ 1 +x
is strictly increasing if χ > — 1.
415

Solutions
1994/5
III.
у + /О)+yf(x)=y- τ- -r— = tt~
1 + x 1 + x 1 + x
x-y
x + f(y) + xf(y) =
l+y
f(X~y\= l+y = Х~У = У~Х
J\l+y) 1 + f^ 1+ж 1+ж'
ж
so f(x) = satisfies the given functional equation and thus it is the one and
1 +x
only solution.
1994/6. Show that there exists a set A of positive integers with the following
property: for any infinite set of primes, there exist two positive integers m in A
and η not in A, each of which is a product of к distinct elements of S for some
k>2.
Solution. Define the set A as follows: it consists of those positive integers
whose prime factorisation satisfies that
a) the index of any prime factor is 1, the elements of A are all squarefree
numbers;
b) the number of prime factors is not less than the smallest prime divisor of
the given number.
Let S = {pi, p2, . · ·} where p\ <p2 <рз < ... < is an arbitrary infinite set
of primes.
Set к to be equal to p\ (p\ > 2) and let
m = pip2...pPv η = ρ2Ρ3··-Ρρι+ι-
Both these numbers are p\-factor products and their factors are all from S. By
the definition of A, the first number, m is in A. On the other hand, nGiis
impossible otherwise n, having only p\ prime factors, should possess p2 °f them,
by definition, while p\ >P2- The set A as defined above thus satisfies the
requirements.
1995.
1995/1. Let А, В, С, D be four distinct points on a line, in that order. The
circles with diameter AC and BD intersect at X and Y. Let Ρ be a point on
the line XY other than Z. The line CP intersects the circle with diameter AC
at С and M, and the line Β Ρ intersects the circle with diameter BD at В and
N. Prove that the lines AM, DN, XY are concurrent.
416

1995/1
1995.
First solution. The orthocentre О of ABPC is clearly incident to the
straight line XY since the latter is perpendicular to ВС. (Figure 1995/1.1). Point
Z, on the other hand, is incident to the radical axis of the circles with diameters
AC and BD therefore the respective powers are equal:
Ζ A ZD
ZA-ZC = ZB-ZD, that is -—= —— = A.
ZB ZC
Figure 1995/1.1
AM and DN are parallel to BO and CO, respectively, since they are
perpendicular to the same lines. Enlarge ABOC from Ζ by A. This enlargement
is mapping В to А, С to D and, at the same time, the straight lines BO and
CO are mapped into AM and DN, respectively. Therefore O, the intersection
of BO and CO is mapped into the intersection O1 of the corresponding image
lines. Hence the points Ζ, Ο and O' are collinear and thus AM, DN and XY
are passing through the point Ο', they are concurrent, indeed.
Second solution. The circles with diameter AC and BD are denoted by
c\ and C2, respectively (Figures 1995/1.2 and 3.) intersections of AM and DN
with XY be Qi and Q^. Thus we have to show that these two points, Q\ and
Q2 are identical. By Thales' theorem LAMC = IBND = 90°. The angles at Μ
and Ζ subtended by the chord Q\C are both right angles therefore the points Z,
C, M, Qi are lying on a circle C3. A similar argument yields that the four points
Ζ, Β, Ν, Q2 are incident to a circle C4.
417

Solutions
1995/1
Figure 1995/1.2
c\/
a\
B\
M/
Qy
~-^x
Qi
π
Ζ
Υ
\N
Iе
\c2
}d
Figure 1995/1.3
Ρ is hence incident to the radical axes of the circle pairs (cj, οι), (c\, c-y)
and (pi 04) respectively. Thus its powers with respect to the circles C3 and C4 are
418

1995/2
1995.
also equal:
PQl-PZ = PQ2-PZ, yielding PQl=PQ2.
We are to finish now; since Q\ and Q2 are not separated by Ρ on the line
XY, the two points, Q\ and Q2 are identical, indeed.
Remark. Two different configurations due to the position of Ρ are shown
in the respective figures; the apparent difference, however, has no effect on the
proof.
1995/2. Let a, b, с be positive real numbers with abc= 1. Prove that
l/a3(b + c)+l/b3(c + a) + l/c3(a + b)>3/2.
First solution. Denote the sum on the l.h.s. by S and apply Cauchy's
inequality for the two awkward triples
(jl/a3(b + c), Jl/b3(c + a): Jl/c3(a + b?j
and
I Jb + c/bc, J с + a/ca, J a + b/ab I .
We thus obtain
2
1 /1 / 1 \ „fb + c c + a a + bs
that is
a3bc V ab3c V abc3 I \ be ca ab
1 1 1\2 /1 1 1\
a be \a о с
2\a b c,
Going on with the A.M.-G.M. inequality:
~2 3 -2]jabc 2'
By the last estimation equality holds only if a = b = с and then it clearly does.
Second solution. With the notations x = —, y = -r, z = - we have xyz = l.
a b с
Now the first term, for example, of the l.h.s. is
1 χ χ yz χ
a3(b + c) i + - У + z y + z
У ζ
419

Solutions
1995/2
Rewriting the remaining two terms similarly the claim becomes
y + z z + x x + y 2
we shall proceed by proving (1).
We need the extended A.M.-H.M. inequality for weighted means; the terms
are now , , with the respective weights x, y, z:
y+z z+x x+y
χ у ζ
c x +У + ζ
y+z z+x x+y x+y+z
х+У+z x+y+z χ^+Ζ ιyZ+X ιzX+^'
χ у ζ
that is
(x + y + z) x + y + z
s>-—-—— = —-—.
2(x + y + z) 2
The A.M.-G.M. inequality settles the problem again; since the product of
the new variables is 1, we get
Ъх + y + z 3 „. 3
~2 3 ~ 2У У 2
Equality holds if and only if χ = у = ζ that is a = b = c.
Third solution. There are various generalizations of the problem; we show
one of these proving hence the claim again.
Additionally to the given conditions let β>2; we are going to prove the
inequality
1 1 13
+ T7, + -T- — >
аР(Ъ + с) ЪР{с + а) cP{a + b)~2'
Introducing, as in the previous solution, the variables x = —, y=-r-> z = - the
a b с
claim becomes
χβ-1 β-l ζβ-1 з
+ + > -
y + z z + x x + y 2
To make it simpler let α = β — 1 (a > 1); what we have to prove now is
„a „.a ya о
Sa = J^- + ^— + -^—>5 (a>i).
y + z z + x x + y I
First we check the case a = 1 (we don't need xyz= 1 now).
χ у ζ x + y + z x + y + z x + y + z
Si = + —— + = —-— + —-— + 3 =
y + z z + x x + y y + z z + x x + y
11 1
+ +
ч У + Z Z + X X + y 0
= 3(x + y + z)·- --3.
420

1995/3
1995.
The A.M.-H.M. inequality now yields
S\ >3(x + y + z) 3 = — — 3 = -.
y + z + z + x + x + y 2{x + y + z) 2
Assume now that x>y>z; then clearly xa~l >ya~l >za~l and the
inequalities
χ у ζ
>—— >
y + z z + x x + y
also hold. Indeed, x>y, y+z<z+x and > , therefore
y + z z + x
χ у
y+z x+z
We use now Chebyshev's inequality. It states that if
ab a2, .. ·, an and Ьь 62, ■··, bn
are two, similarly ordered sequences of real numbers then
а\Ь\ +α2^2 + · · · + ΑτΑι α\+α2 + · ■ · + αη Ь\ + fr2 + · · · + Κ
η ~ η η
(If the sequences are arranged in opposite order then the inequality holds the
other way round; we have equality if and only if αϊ = α2 =... = an or b\ = b2 = ...
... = bn.)
Applying Chebyshev's inequality for the triples
a—l a—I a—I ■, % У %
x , у , ζ and
y+z z+x x+y
yields
о ^ о xa~l+ya~l+za-1 3xa-l+ya-l+za~l
Sa ■> ο ι > —
a~ l 3 ~ 2 3
The end is A.M.-G.M. again. Since xyz = 1:
33/—T^T_3
Sa>-^](xyzy
2yy у J 2,
and equality holds if and only if χ = у = ζ that is a = b = c.
1995/3. Determine all integers η > 3 for which there exist η points A\, .A2,
..., An in the plane, no three collinear and real numbers r\, r2, ..., rn such
that for any distinct i,j,k, the area of the triangle AiAjA^ is ri + Vj + r&.
Solution. We prove that the requirements can be satisfied only if η = 4.
If this is the case then consider a unit square with vertices A\, A2, -A3, -A4
and let r\ = г2 = гз =r^ = -. Now any triple of the given points forms a triangle
6
1 1
of area 3 ■ - = -. (Instead of a square any parallelogram of unit area will do.)
6 2
421

Solutions
1995/3
Next we prove that if there are five points given then there can be no
equal ones among the numbers r^. Assume, to the contrary, that, for example
Г4 = Г5- Then [.Ai.A2.A4] = ri +Г2 + Г4 and [A\A2A$\ = r\ +Г2 + Г4; the two areas
are equal, therefore A1A2 and A4A5 are parallel. A similar argument for the
triangles A2A3A4 and A2A3A5 yields that under the assumption г^ = г$ the lines
A2A3 and A4A5 are also parallel. Now this is already a contradiction since it
means that the points A\, A2, A3 are collinear. (Figure 1995/3.1).
Assume now that there are five points on the plane with the required
property. If their convex hull is a pentagon (Figure 1995/3.2) then, clearly,
[А^АзЛк! = [ΑχΑ2Α3] + [ΑχΑ4Α3] = [A2A3A4] + [ΑΑΑγΑ2] that is
rχ + Г2 + r3 + r\ + Г4. + r3 = V2 + τ3 + Γ4 + Г4 + r\ + V2, and hence
rl+r3-r2 + r4· A4
Figure 1995/3.1 Figure 1995/3.2 Figure 1995/3.3
The same argument when applied to the convex quadrilateral Α\Α2Α3Αζ,
yields
r\ +Г3 =Г2 + г$
and thus Г4 = Г5, which we have already seen to be impossible.
If the convex hull of the points is a quadrilateral, A1A2A2A4. for example,
then Αζ is an interior point of the convex hull (Figure 1995/3.3). The previous
argument can be repeated, without essential modification, to the quadrilaterals
A1A2A2A4. and .A1.A2.A3.A5 yielding once more the impossible r^ = r^.
Finally, if the convex hull is the triangle A\ A2A3 then clearly
[AlA2AA\ + [A2A3A4] + [A3A1A4] = [Ai A2A5] + [A2A3A5] + [A3 Ai A5]
that is
r\ + Г2 + r$ + Г2 + Г3 + Г5 + Г3 + r\ + Г5 = r\ + Г2 + Г4. + Г2 + Γ3 + Г4 + Г3 + r\ + Г4,
and hence
again. There are no five points (and more, of course) satisfying the conditions.
422

1995/4
1995.
1995/4. Find the maximum value of xq for which there exists a sequence
xq, x\, ..., £1995 of positive reals with ^0-^1995 such that for i = \, 2, ...,
1995
(i) xQ = Xi995\
2 1
(n) Xi-i + = 2xi + —.
X{—\ X{
Solution. With a bit of algebra, relation (ii) can be written in more tractable
form as a quadratic in xf.
*?-(^+ l
and hence
X\ 1
(1) Xi = -^- or (2) Xi = .
2 Χί_λ
This shows that starting from xq each term of the sequence can be obtained from
the previous one by either halving it or as its reciprocal. Therefore, each term of
the sequence is either of the form
2k
(3) Xi = 2kx0, or Xi = —
x0
for some integer k.
Hence, by (i), the 1995th term satisfies either
(4) #1995 = 2 xq = x0, or (5) ^1995 = — = x0: and thus £0 =
x0
Arriving from xq to #1995 the iteration steps (1) and (2) have been performed
1995 that is an odd number of times altogether. (4) can be reached only if (2)
has been applied even times that leaves odd steps of type (1). Observe that
applying (1) changes the index of 2 by 1 and, at the beginning we have xq-2 xq.
Therefore if #1995 = 2^0 then к has to be odd. Since xq >0, £1995 now cannot
be equal to xq-
To arrive to (5) step (2) has to be applied at least once and thus (1) can be
used at most 1994 times. Hence 2k, the power factor in (5) cannot exceed 2
and thus the maximum value of Xq is at most 2 . Hence
χ0<^2^ = 299Ί.
On the other hand xq = 299 is already possible. For this the first 1994
iteration steps are all of the first kind and the last step is (2); the sequence produced
from the maximal value of xq is hence
„ _o997 _. _9996 _. _o0_i „. 0-l _. _9-997
Xq-1 , X\-Z , ..., £997-2 =1, £998= 2 , ..·, Ж1994-2 ,
_ o997
^1995-2 ·
423

Solutions
1995/4
Remark. Calculating the first few terms we can verify that
™. _ 0&ίγ£ί
dj% — ^ JsQ ,
where \kj\ <i and ei = (—\)l+ki\ this can be proved by mathematical induction
but it also follows from the argument of the solution.
1995/5. Let ABCDEF be a convex hexagon with AB = BC = CD and
DE=EF=FA such that IBCD = IEFA = 60°. Suppose that G and Η are
points in the interior of the hexagon such that LAGB = ID ΗΕ =120°. Prove
that
(1) AG + GB + GH + DH + HE>CF.
Solution. Let AB = a and AE = b. By the conditions ABCD and AEFA
are equilateral triangles of sides a and b respectively. Draw equilateral triangles
AC'В and DF'E externally upon the sides AB and DE respectively {Figure
1995/5.1). Since ABDE is a kite of sides a and b the whole figure is symmetrical
to the axis BE = t of the kite. Hence CF = C'F'.
Figure 1995/5.1
G and Η are incident to the circumcircles of the equilateral triangles AC'В
and DF'E respectively and thus,by Ptolemy's theorem
GC' = AG + GB, HF' = DH + HE.
424

1995/6
1995.
Hence the sum on the l.h.s. of (1) is equal to the length of the broken line
C'GHF1. Since the latter is clearly at least C'F1, the l.h.s. of (1) is at least
C'F' = CF and the proof is hence finished.
It is clear that equality holds in (1) if and only if Η and G are incident to
C'F'.
1995/6. Let ρ be an odd prime number. How many p-element subsets A of
{1, 2, ..., 2p} are there, the sum of whose elements is divisible by p?
Solution. Split Я = {1, 2, ..., 2p} into two p-element subsets as follows
A = {\,2, ...,p}, Б = {р+1,р + 2, ...,2p}.
Any p-element subset С of Η different from both A and В contains elements
from both A and B; consider its intersection with A:
ab <22, · ··, an (1 <n<p-l).
Add now 1 to each element of А П С and replace every number thus modified in
С by its remainder when divided by p; if this remainder happens to be zero then
write ρ instead.
Denote by C\ the set obtained this way from C. Similarly: adding 2, 3 ...
..., ρ to the elements сц inside С respectively and reducing them bmod ρ in the
above sense yields the p-element subsets C2, C3, ..., Cp; the set Cp is clearly
identical to С itself. These subsets form a closed class of the p-element subsets
of H, any one of them generates the class via the above procedure and subsets
outside of this class generate different subsets.
For an arbitrary p-element subset С of Η denote the sum of its elements
by s(C). s(A), for example, is equal to ^-—-, s(B) = s(A) + p2. Both s(A)
and s(B) are divisible by ρ because it is odd. Since the respective sums of the
elements in C; and Ci+\ differ by η (mod p, of course) we have the following
congruence:
s(C{) — s(Cj) = (i — j)n (mod p).
\i — J\<P and n<p imply that (г — j)n is not a multiple of p; hence the numbers
s(Ci) belong to different residue classes mod p. Since there are ρ members of the
C-class there is exactly one C; such that s(Ci) = 0 (mod p), the sum of whose
elements is divisible by p.
Apart from A and В there are ( 1—2 p-element subsets of H. These
w
subsets, as we have seen, can be arranged into - ( ( ) — 2 ) classes and there
p\\p/ J
is a single subset in each class the sum of whose elements is divisible by p.
425

Solutions
1995/6
Therefore the number of p-element subsets (A and В included) of Η where the
sum of the elements is divisible by ρ is
+ 2.
Ρ
Note. An interesting arithmetical byproduct of the result is that ( -2a
w
multiple of ρ whenever ρ is an odd prime; indeed, by its meaning, the ratio in
the result is an integer.
1996.
1996/1. We are given a positive integer r and a rectangular board divided
into 20 χ 12 unit squares. The following moves are permitted on the board: one
can move from one square to another only if the distance between the centres of
the two squares is \fr. The task is to find a sequence of moves leading between
two adjacent corners of the board which lie on the long side.
(a) Show that the task cannot be done if r is divisible by 2 or 3.
(b) Prove that the task is possible for r = 73.
(c) Can the task be done for r = 91?
Solution. Label its corners and place the board ABCD in a Cartesian
system with the centre of the .Α-square at the origin, the centre of the 5-square
at (19, 0), finally, let the centre of the D-square be (11, 0). The moves can be
resolved into the sum of horizontal and vertical components and thus we may
assume that our token is moving parallel to the axes.
(a) Let r be even first and consider a vector \(x: y) for a legal move; then,
2 2
by condition, χ + y =r. Its value is now even and thus x = y (mod 2). Colour
the squares as on a usual chessboard with the field of A black. If χ and у are
both even then any move, horizontal or vertical, takes us to a field of the same
colour; if both components are odd then the colour of the field flips step by step
and thus, eventually, the destination and the start have the same colour again.
Since B, the target square is white, no way to reach it from the black A.
2 2
1 Let v-x +y be now divisible by 3. Since a square
mod 3 is equal to 0 or 1, the sum of two squares is divisible by
3 if and only if both χ and у are of the kind. Draw a 3 χ 3 grid
on the infinite board and colour the 9 fields of each square of
the grid by one of three colours according to the pattern:
2
3
1
3
1
2
1
2
3
426

1996/1
1996.
The length of any move, horizontal or vertical, is now divisible by 3 and thus we
are travelling along squares of the same colour, all the time. On the other hand,
if A gets colour 1 then В has colour 2, therefore it cannot be reached from A.
(b) There is just one way to split r = 73 into the sum of two squares: 73 =
2 2
8 +3 . Thus every legal move is by one of the following vectors:
(3, 8), (8, 3), (-8, 3), (3, -8), (-3,8), (8, -3), (-3, -8), (-8, -3)
The task is to produce (19, 0) as the sum of vectors from this list; the token,
during the trip, is not allowed to exit from the board, of course. One can find such
a trip with trial and error (Figure 1996/1.1); the list below displays the stepwise
positions of the moving token, the corresponding steps are indicated above the
arrows, respectively.
(3,8) (8,-3) (8,-3) (-3,8) (-8,-3)
(0, 0) —- (3, 8) > (11, 5) > (19, 2) > (16, 10) > (8, 7) -
(-8,-3) (8,-3) (3,8) (-8,-3) (8,-3) (8,-3)
' > (0, 4)' 1 (8, 1) K-4 (11,9) K—i (3, 6)' 1 (11, 3)' > (19, 0).
Figure 96/1.1
2 2
(c) 97 can be written as the sum of two squares only as 9 + 4 and thus
the length of any move parallel to the axes is 4 or 9. Colour the fields of the
board once more, according to the pattern on Figure 1996/1.2; note again that
the .Α-square is black and the target В is white.
The argument should be familiar by now: starting from a black field, a
horizontal step by 4 takes us to a black field again and so does a vertical step of 9;
a horizontal 9 is white but a vertical 4 is black again. The story is the same with
427

Solutions
1996/1
-—
.—-
—
_
_
—
—
r"
.1
\
1 S !
_L_J_JL
~r Τ
! i
τ ζ
ι ;
:
..„;
!
i
/zxzit:::
1 i
"~~
1
■
"irzir:
. j_ L J L
1 , ' 1
,..J._L..i_; 4_
, ! ; 1 1 !
ί Mill
ι'ΠΤΊΊ i ;
"'"'Π ■ ΠΊ"
1 : i !
„ 1 . J_
Mil
i
^L·
_
ι ι
_J__L_
—
—„
—
.^«.
™—
—
—
в
~~*™
__
™«™
—~
™™«™
—
«_™.
~~—
Figure 96/1.2
a vertical 4 start: ending up on a white square, a horizontal 9 returns to a black
field; finally, moving vertically by 9 and horizontally by 4 we are not leaving
the safe black area at all. Coming to the end: if we start from a black field then
every legal move will take us to black fields and thus the white В is inaccessible
from the black A.
Note. A reasonably simple argument shows that the 11 long sequence of
moves presented in part (b) is the shortest possible trip from A to B; the moves,
by the way, can be found by an algorithmic approach, although this was not
required in the competition. Most students have quicly found one of the several
solutions with bare hands, anyway.
1996/2. Let Ρ be a point inside the triangle ABC such that
(1) LAPB-LACB=LAPC-LABC.
Let D, Ε be the incentres of triangles APB, APC respectively. Show that AP,
BD and CE meet at a point.
First solution. Since D and Ε are incentres, BD and CE are bisecting
the angles LABP and LACP. By the angle bisector theorem these two lines
meet on the segment AP if and only if they divide it into parts of equal ratios,
respectively. This common ratio is hence equal to the ratio of the sides enclosing
AP (Figure 1996/2.1). This happens if
AB AC , . AB PB
(2) -rr^ = -^-^, that is
Ρ В PC
AC PC
428

1996/2
1996.
A
Figure 96/2.1.
We solve the problem by proving the latter equality.
A
X
Figure 96/2.2
The lines АР, Β Ρ and С Ρ meet the circumcircle of AABC for the second
time at Χ, Υ and Z, respectively (Figure 1996/2.2). The very same lines are
dividing the angles α, β, j of AABC into a\, α^; β\, βϊ, 7ь 72. respectively,
as it is shown in the figure. By the theorem of inscribed angles we can write
down the angles of AXYZ:
IAPB, on the other hand, is the sum of the respective exterior angles of
AACP and ABCP; thus
IAPB — LAC В = a.\ +72 +/З2 +7i — 7i -72-^1 +βΐ·
Similarly,
LAPC- LABC = a2 + β\ +/?2 + 7l -β\ -/#2 = ^2 + 71)
429

Solutions
1996/2
and hence, by (1)
(3) αι+β2 = α2 + ^\·
Thus, in AXYZ we have IZ= LY, it is isosceles,
(4) XY = ZX.
Observe now that the pairs AABP, AYXP and AACP, AZXP are
respectively similar because the corresponding angles are equal in each pair. By
these similarities
AB_ _ XY_ AC _ZX
¥b~Yp: a ¥c~Yp:
u· u u /„χ · ,j AB AC u
which, by (4), yields —— = ——, the claim.
Second solution. We use some pieces from the previous solution.
Reviewing Figure 1996/2.2 we can check that in APBX and APCX we
have ΙΡΧΒ=η and IPXC = β and also ΙΡΒΧ = αχ+β2 and LPCX = a2+7l ·
This, by (3), yields IPBX = IPCX. Apply the sine rule in APBX, APCX,
finally in AABC:
PB_PB PX _ sin 7 sin PCX I _ sin j _AB
~PC~Tx' ~PC~ sinPBXl ' ski/? ~sin/?~AC
which, by (2), is equivalent to the claim.
Third solution. An interesting relation was found by Peter Frenkel, a
Hungarian contestant. He used the following theorem: if Ρ is an interior point of the
triangle and PA is reflected through the bisector of LA and, similarly, PB and
PC are reflected through the bisectors of LB and 1С, respectively, then the
mirror image lines are concurrent.
The straight lines PA, PB, PC are dividing the respective angle of AABC
as it is shown in Figure 1996/2.2. Reflect now the lines PA, PB, PC as in the
above theorem, through the bisectors fa, fa and fc, respectively. The image lines,
by the assertion, meet at some point P' {Figure 1996/2.3). Because of reflection
clearly ΙΡ'ΒΑ = β2 and ίΡ'ΟΑ = ηχ. The straight line AP' divides IBP'С into
the parts ψ\ and ψ2\ hence, by the exterior angle relation
ψ\=αγ+β2, (£2 = ^2+71,
and this, by (3), yields ψ\ =φ2 that is AP1 is bisecting IBP'С
Denote the incentre of ABP'C by K. Apply the theorem again: the
mirror images of the lines К A, KB, КС through the respective angle bisectors of
AABC are thus concurrent. The lines К A and P'A are identical and their
common mirror image through fa is PA. Since reflection is mapping angle bisectors
430

1996/3
1996.
Figure 96/2.3
into angle bisectors, the mirror image of KB is DB and that of КС is EC; the
claim hence has been proved.
Remark. According to relation (2) the ratio of the respective distances of
the point Ρ of the problem, from the vertices В and С is equal to c/b, The locus
of such points is the so called Apollonius' circle; its diameter is the segment cut
from В С by the bisectors of the interior and the exterior angle at A.
1996/3. Let S be the set of non-negative integers. Find all functions f from
S to itself such that
(1) f(m + f(n)) = f(f(m)) + f(n)
for all m, n.
Solution. If m = n = 0 then (1) implies
/(/(0)) = /(/(0)) + /(0)
that is /(0) = 0. Set now m = 0 in (1):
f(f(n)) = f(f(0)) + f(n) = f(n).
Hence / keeps the elements of its range fixed,
(2) f(f(n)) = f(n).
Therefore, (1) can be rewritten as:
(3) f(m + f(n)) = f(m) + f(n);
in what follows we shall be using this property of /.
431

Solutions
1996/3
The constant zero function is clearly a solution so we assume that in what
follows / is not all zero. Denote the smallest positive fixed point of / by a.
This value, by (2) and our previous assumption, is well defined. We prove, by
induction on k, that ka is also a fixed point for every positive integer k. For к = 1
this is the definition of a. Let k>\ and assume that f((k — l)a) = (k — l)a.
Substituting m = a and n = (k — \)a in (3) yields
f(ka) = f(a + f(k — l)a) = f(a) + f((k - l)a) = a + (k-\)a = ka
so ka is also kept fixed by /.
Using the minimum property of a in the usual way we now show that every
fixed point of / is equal to some positive multiple of a. Indeed, if b = aq + r
(q > 0, 0 < r < a are integers) is an arbitrary fixed point then, by the previous
result, qa likewise and thus, by (3)
aq + r = b= f(b) = f(aq + r) = f(r + f(aq)) = f(r) + f(aq) = f(r) + aq,
yielding
f(r) = r.
Thus r is also a fixed point and being smaller than a, the smallest positive
fixed point, it is forced to be zero, b = aq, indeed. Making the two ends meet,
since the elements of the range are all kept fixed, f(ri) is equal to ka with some
к for every n.
If 0 < г < a then let f(i) = ща (щ is a positive integer, no = 0). If η = ka + г
is an arbitrary integer (0 < i < a) then, by (3)
f(n) = f(ka + i) = f(i + f(ka)) = f(i) + f(ka) = ща + ka,
and thus
(4) f(n) = f(ka + i) = (rii + k)a (0<i<a).
Now we are ready to characterize the solutions of (1). Set an arbitrary positive
integer a as the smallest fixed point of / and choose the a — 1 values /(1) = щ,
f(2) = 77-2, ..., f(a — 1) = na_i from S arbitrarily, apart from the restriction i ^щ.
Set, finally, /(0) = 0. By (4) these functions can uniquely be extended to S.
To finish properly we are left to show that these functions do satisfy (1).
For this let
m = kia + i, n = k2a + j (0<i,j<a).
Now clearly
f(m + f(n)) - f(k\ a + i + rija + k2a) = f((k\ + А% + nj )a + i) =
= (k\ +к2 + Щ + rij)a,
f{f{m)) + f{n) = f{{kl+ni)a) + f{k2a + j) = {kl+ni)a + {k2 + nj)a =
= (k\ +к2 + щ + щ)а,
432

1996/4
1996.
and thus (1) does hold, indeed.
Remark. Since for n = ka + i clearly к =
f(n) =
η
a
, (4) can also be written as
η
a
+ n,- a.
1996/4. The positive integers a and b are such that 15a +166 and 16a — 156
are both squares of positive integers. What is the least possible value that can
be taken by the smaller of these two squares?
Solution. If 15a+166 = r^ and 16a-156 = sz for some whole numbers r
and s then
Γ4 + 54 = 225(α2 + 62) + 256(α2 + 62) + 2·15·16α6-2·15·16α6 =
= 481(α2 + 62) = 13·37(α2 + 62).
(13, 37)= 1, therefore r +s is divisible by both 13 and 37.
Simple case checking yields the possible remainders of an arbitrary 4th
power when divided by 13; they are
0, 1, 3, 9.
The sum of two 4th powers is hence divisible by 13 if and only if both terms are
of the kind.
The story is similar if the remainders of 4th powers when divided by 37 are
checked; they are
0, 1, 7, 9, 10, 12, 16, 26, 33, 34.
Being a multiple of 13 and 37, r +s thus forces both r and s to be a
multiple of these numbers and hence that of 481, their product. Thus
r>481, s>481.
Observe, on the other hand, that 481, the smallest possible value of r and s
can, in fact, be realized because the system
15a+166 = 4812,
16a-156 = 4812
can be solved in the set of positive integers. Indeed
α = 481·31 = 14 911, 6 = 481
and thus the least positive value of the smaller one of the two squares is
4812 = 231361.
433

Solutions
1996/4
.4
Remark. Fermat's small theorem provides a uniform way to prove that χ +
у* is divisible by 13 and 37 respectively only if both χ and у are divisible by
these numbers.
It is obvious that if χ +y is divisible by 13 and any one of χ and у also
has this property then so does the other one. Assume hence that none of the two
numbers are divisible by 13 but the sum of their 4th powers is.
χ +y =0 (mod 13) implies χ = — у ; then also χ
.12
у . By the small
Fermat theorem, however,
.12 _
x
.12
1 (mod 13),
1 and у
and thus 1 = — 1 ογ2ξ0, a contradiction.
The point is that (13-l)/4 is an odd number and thus the signs of the opposite
terms in the congruence relation remain opposite after the index has reached 12.
The same proof works whenever ρ is a prime of the form 4k +1 with к odd, like
37.
1996/5. Let ABCDEF be a convex hexagon of perimeter ρ such that AB
is parallel to DE, ВС is parallel to EF and CD is parallel to FA. Let Ra, Rc,
and Re denote the circumradii of triangles FAB, BCD, DEF respectively.
Prove that
(1)
Ra + Rc + Re>~.
Solution. Since the opposite sides of the hexagon are given to be
parallel, lA = lD = a, ΙΒ = ΙΕ = β and IC=IF = j. Construct a rectangle XYZU
about the hexagon by drawing perpendiculars to the lines ВС and EF from the
vertices A and D, respectively (Figure 1996/5.1).
X В b С Υ
и
F e
Figure 96/5.1
With AB = a,BC = b, CD = c,DE = d, EF = e,FA = f clearly
XA = asinB, AU = f sin C, YD = csinC, DZ = dsinA.
434

1996/5
1996.
Observe that BF>XU = Υ Ζ and thus 2BF >XU + YZ; therefore
2BF>XA + AU + YD + DZ = (a + d)sinB + (c + f)smC.
The circumradius of AFAB is
Similarly
BF 2BF 1 / „sin Б , „4sinC\
2 sin A 4 sin A 4 \ sin Л sin A)
D 41/, ,4sin,4 sin£\
4 V sin С sine;/
„ 1 Λ, sin С , , sincA
RE>~;((b + e)—-+(a + d)—-).
4 V sin В sin Б/
Adding these inequalities
4(Ra + Rc + Re)>
T4 /sin Л sinIA , „ч /sin Л sin7\ ,t ч /sin С sin ΙΑ
>(a + d) -^б + ^—7 )+(с + Л Η + ^—7 +(Ь + е) -ν—+ - ).
Vsinl* sin A) \sin7 sin A) \%m.B sin 7/
Since the sum of a positive number and its reciprocal is at least 2,
4(RA + RC + Re) > 2(a + b + с + d + e + /) = 2p,
Ra + Rc + Re>%,
indeed. Equality holds if and only if the hexagon is regular.
Remark. The problem is closely related to the Erdos-Mordell inequality and
its extension.
Construct the parallelograms FABP, BCDQ, DEFS inside the hexagon
and draw perpendiculars from the vertices F, B, D to the lines PF, QB, SD,
respectively. These perpendiculars enclose a triangle XYZ. X is the intersection
of the lines perpendicular to PF and QB and so on, see Figure 1996/5.2.
XFPB is cyclic because a right angle is subtended by PX at both F and
B. The circumradius of this quadrilateral is equal to that of AFP В and, as
they are congruent, also of AFAB; the latter is but Д4. As the diameter of this
circle, PX = 2RA. Similarly, QY = 2Rq and SZ = 2Re- The parallelograms on
the diagram show that PF = a, SF = d, QB = c, PB = f, SD = e,QD = b.
Rewriting the claim as
2Ra + 2Rq + 2Re >a + b + c + d + e + f
and substituting the previous results yields
(2) PX + QY + SZ>PF + SF + QB + PB + SD + QD;
this, as it stands, is a generalization of the celebrated Erdos-Mordell inequality.
The actual theorem itself goes like this:
435

Solutions
1996/5
X F Ζ
Figure 96/5.2.
If F, B, D are interior points of the sides ZX, XY, Υ Ζ of AXYZ,
respectively, and the perpendiculars to the respective sides at these points meet at
P, Q and S (Figure 1996/5.2) then (2) holds.
If the hexagon happens to be centrally symmetric and thus the opposite
sides, apart from being parallel, are also equal then P = Q = S and (2) becomes
the original Erd6s-Mordell inequality.
The solution of the problem hence also proves this generalization; (2),
nevertheless, can be demonstrated directly thus yielding an alternative solution.
1996/6. Let p, q, η be positive integers withp + q < n. Let xq, x\, ..., xn be
integers such that XQ = xn = 0, and for each 1 < i < η, χι — Xi_\ -p or —q. Show
that there exist indices i<j with (г, j)^(0, n) such that Xi = Xj.
Solution. The g.c.d. of ρ and q divides Xi for every i. This is immediate
for x\ by the initial conditions x\ =x\ —xo=p or — q and then follows by
induction. Dividing through by (p, q) the conditions of the problem still hold and
thus (p, q) = 1 may be assumed.
Consider the 'telescopic' sum
On-£n-l) + On-l -xn-2)+- ■ - + {x2-x\) + {x\ -xq) = Q.
If there are к ones among these differences equal to ρ then the remaining (n — k)
ones are — q and thus pk — q(n — k) = 0. Since (p, q)=l, q divides к which thus
can be written as aq (a is a positive integer). Hence apq — q(n — aq) = 0,
(1) n = a(p + q);
observe that the condition n>p + q implies a>\.
436

1996/6
1996.
Consider now the differences
Vi = Xi+P+q -Xi (i = 0, 1, ..., η - (p + q));
it is enough to show that some of them are equal to zero; indeed, if yi = О then
•"i ~ ^i+p+q·
Write down the (p + g)-term telescopic sum for yf.
Уг — \Xi+p+q ~ %i+p+q—\) + \Xi+p+q—\ — X{+p+q—2) + · ■ ·
■•• + (Xi+2-Xi+l) + (Xi+l-xi)·
If there are r ones among these differences whose value is ρ then the remaining
p + q — r ones are all equal to —q, therefore
(2) yi = rp-(p + q-r)q = (p + q)(r-q).
Hence (p + q) divides every г/j. Consider now the difference yi+\ —yi\
Уг+l ~У% — \xi+p+q+l ~ xi+l) ~ \xi+p+q ~ xi) = \xi+p+q+l ~ xi+p+q) ~ \xi+l ~ xi)·
In the last form each difference is equal to ρ or —q and thus
0,
Уг+l ~Уг =
or p + q,
or -(p + q),
that is
(3) yi+l-yi = c(p + q), where c = 0, 1 or -1.
Now the sum
(Xp+q ~ X0) + (X2(p+q) - Xp+q) + (X3(p+q) ~ ХЦр+q)) + ■■·+ (xa(p+q) ~ x(a-l)(p+q)) =
remembering also that, by (1), xn -хп-(р+д) = ха(р+д) -X(a-l)(p+qy In terms of
the ^/-variables this means that
2/0 + Ур+q + УКр+q) + У3(р+Я) + ··■+ Уп-ip+q) = °·
Hence it is not possible that the y-s are all of the same sign, in the sequence
(4) 2/0, 2/1 > 2/2, ···, Уп-(р+д+1), Уп-ip+q)
either there is a positive term followed by a non positive one or vice versa.
Assume, for example, that y^ >0 and yk+i <0. Now, by (3), yk = c\(p + q) and
2/fc+l - C2(P + <?)» cl > 0, C2 < 0, therefore C2 — c\ < 0. On the other hand, by (2)
2/fc+l -yk = (c2~ci)(p + q) = -(p + q),
and thus C2 — c\ = — 1 that is c\ = C2 + 1. Then
0<c2 + l<l,
437

Solutions
1996/6
and, finally, we are at c2 = 0 that is yk+\ = 0, indeed.
The same argument works if there is a negative term in (4) preceding a non
negative one.
Remarks. 1. The following sequence, in fact, has the given property if тт. =
15, ρ = 3, q = 2 and a = 3:
0, 3, 6, 9, 7, 5, 8, 11, 9, 7, 10, 8, 6, 4, 2, 0.
2. Having checked the solution one can — and should — ask that where did
the idea of the sequence у ι come from. This seems to be quite a rabbit out of the
hat, however, if one attempts to prepare an actual sequence of the problem and
remembers also the hard to miss property (1) then the p + q neighbouring terms
might come to sight as potential candidates for being equal.
1997.
1997/1. In the plane the points with integer coordinates are the vertices
of unit squares. The squares are coloured alternately black and white as on a
chessboard. For any pair of positive integers m and n, consider a right-angled
triangle whose vertices have integer coordinates and whose legs, of lengths m
and n, lie along the edges of the squares. Let S\ be the total area of the black
part of the triangle, and S2 be the total area of the white part. Let
f(m, n) = \Si -S2\.
(a) Calculate f(m, n) for all positive integers which are either both even or
both odd.
(b) Prove that f(m, n)<- max(m, ή) for all m, n.
(c) Show that there is no constant С such that f(m, n) < С for all (m, n).
Solution, (a) Denote, for brevity, the black and white areas covered by a
polygon XYZU ... by Si(XYZU ...) and S2(XYZU ...), respectively. Let
AB = m, AC = η and, by reflecting it to the midpoint F of the hypotenuse,
complete the right triangle ABC to the rectangle ABDC. If both legs are even then
F itself is a lattice point and if the legs are both odd then F is the centre of a
lattice square. Therefore, the reflection through F preserves the colouration, any
square and its image have the same colour {Figure 1997/1.1).
Si (ABC) = Si (DCB) and S2(ABC) = S2(DCB).
Hence
Si(ABC) = ^-Si(ABDC) and S2(ABC)=^Si(ABDC).
438

1997/1
1997.
В
т
А
η
D
Ρ
с
в
D
т
А
4F
η
С
Figure 97/1.1
Thus
f(m, η) = \Si(ABC)-S2(ABC)\= hsi(ABDC)-S2(ABDC)\.
If both m and η are even then there are the same number of black and white
squares contained in the rectangle; if both of them are odd then the area of the
rectangle is also odd, therefore one colour is ahead of the other one by a single
unit. Thus
0, if m and η are both even and
f(m, n)= ■ 1
-, if m and η are both odd.
I 2
(b) If η ξ πι (mod 2) then the claim
follows from (a) because both πι and η are
at least 1.
Consider now, for example, when η is
even and the other leg, m is odd.
Choose point Ε on AB such that BE = 1
(Figure 1997/1.2). Now clearly EA = m-\
and, since both πι —I and η are even,
f(m — \, n) = 0. Besides
Si (ABC) = Si (AEC) + Si (EBC),
S2(ABC) = S2(AEC) + S2(EBC).
m— 1
Figure 97/1.2
Hence
/(m, n) = \SX(ABC) - S2(ABC)\ = \Si(EBC) - S2(EBC)\ <
η 1
< [EBC] = -t<tz max(m, n),
the desired result.
439

Solutions
1997/1
(с) What we are to prove, actually, is that f(m, n) admits arbitrarily large
values, it is unbounded. We show that, in fact, that is the case if the legs are
consecutive integers, m = 2k+l and η = 2k.
Consider, as in part (b), the point Ε on the leg AB for which BE = 1; then
AC = AE = 2k and
f(2k+l, 2k) = \Sl(EBC)-S2(EBC)\.
2k
В
■Ε
2k -1
\f \r
A
\
\
\1
\
S.
\
2k
С
Figure 97/1.3
Observe now the diagram; we clearly may assume that CE, the hypotenuse
of the isosceles right triangle ACE is passing through black fields. Calculate
now the total white area inside the scalene triangle EBC. The white ABEF
as well as all the smaller and smaller white triangles inside AEBC are similar
к 2к
to AABC. [BEF] = ——— because BE = 1 and EF = ——-. The subsequent
.Z/C + 1 Z,t\i 4" 1
white triangles can be obtained from ABEF by reduction by
2k-l 2k-2 2k-3
2k
£it\j Ait\j Ζ*Γυ
respectively and hence, for their areas, [BEF] should be scaled up by the squares
of these numbers:
2k-l
2k
'2k~T
2k .
'2к-У
2k
2k)
440

1997/2
1997.
Therefore, the total white area is
к
S2(EBC) =
χ1 + (^ζλ)\(^ζ1)\^+(λ
2k+l\ V 2k J V 2k J \2k
к
((2/c)2 + (2£;-l)2 + ... + l2) =
(2k + l)4k2
1 2k(2k+l)(4k+l) 4k +1
4k(2k + l) 6
Since [EBC] = k, its black part is
12
Si(EBC) = k
4k+l 8/c-l
Hence
/(2fc+l, 2k) =
12 12
8/c-l 4/c + l 2/c-l
12
12
which is unbounded, indeed.
1997/2. The angle at A is the smallest angle in the triangle ABC. The
points В and С divide the circumcircle of the triangle into two arcs. Let U
be an interior point of the arc between В and С which does not contain A.
The perpendicular bisectors of AB and AC meet the line AU at V and W,
respectively. The lines BV and CW meet at T. Show that
AU = TB + TC.
В
/ι
'Ji
a
U
Solution. Let the lines BV and
CW meet the circumcircle at B' and
С (Figure 1997/2.1) and denote the
perpendicular bisectors of AC and AB
by /i and /2, respectively.
The mirror image of CC'
through f\ is AU is mapped to BB'
when reflected on through /2· Hence
CC = AU = BB'. Two equal chords in
a circle are spanning a symmetrical
trapezium and thus ТС = ТВ' yielding
AU = BB' = TB + TB' = TB + TC:
the claim.
Remark. Since f\ is taken to /2 by a rotation of LA, the composition of
the two reflections in the solution is a single rotation by L2A. If LA is right
angle then the resulting rotation is by 180° and thus CC' and BB' are parallel
c\
^_\\^/
^^vt\
2^^
B'
Figure
97/2.1
~~~^0\У
/1
441

Solutions
1997/2
or identical; in the first case point Τ does not exist. The restriction about the
magnitude of L A is hence to guarantee the existence of T.
1997/3. Let x\, x2, · ■ ·, xn be real numbers satisfying
\x\ +X2 + ■ ■ · + χη\ - 1
and
77,+ 1
Ы<—г— (г = 1, 2, ..., η).
Show that there exists a permutation г/i of χι such that
ι ι П+1
\y\ + 2y2 + 3y3 + ... + nyn| < -y-.
Solution. We shall use the well known fact that every permutation can be
obtained from any other one by a sequence of transpositions that swaps two
consecutive elements. Denote the chain of permutations of x\, x2,... ,xn leading
from Pq = (xi,x2,.··, x-n) to the reversed Ρω = (xn, xn_\,...,x\) by Pq, Ργ, ...
..., Ρω; each permutation in the chain is obtained from the previous one by a
71+ 1
transposition. For brevity, denote —-— by r and for Рг = (у\,У2: ■ ■ ■:Уп) tet
S(Pi) = yi+2y2 +.. . + nyn.
Let's check now the change of S(Pi). Note that
S(Pq) + S(PU) = {x\ +2^2 + · · · + nxn) + (xn + 2xn_i + .. . + nx\) =
= (n + 1) · {x\ + X2 + · · · + xn) = 2r(x\ + X2 + · · · + xn)i
and thus
(1) \S{PQ) + S{Pu)\=2r.
If |5(Ро)| <r or |5(Ρω)| <r then we are done; assume, hence, that both
inequalities hold the other way round:
\S(P0)\>r and 3(Ρω)>ν.
This, by (1), means that S(Pq) and S(PU) are of opposite sign:
(2) S(P0)<-r and S(Pu)>r,
for example. Consider now Pi and Pj+i, two consecutive permutations in the
chain. Let
Pi = (2/b 2/2> ■ · · > Ук-Ъ Ук, Ук+1 > Ук+2, · ■ · ,Уп) and
Pi+l =(У\:У2:···: Ук-l : Ук+l: Ук:Ук+2 · · · 5 2/п)>
with у к and i/fc+i being swapped. Here clearly
\S(Pi+i)-S(Pi)\ = \k(yk+l-yk) + (k + l)(yk-yk+l)\ =
= \Ук ~ Ук+l I < \Ук\ + \Ук+11 < 2r-
442

1997/4
1997.
Thus in the sequence S(Pq), S(P\), ..., S(PU) the difference of consecutive
terms does not exceed 2r; S(Pq) and S(PU), on the other hand, are separated by
the interval [—r,r] by (2).
S(P0)
0
S(Pu)
Starting from below — r at S(Pq) we hence arrive above r to S(PU) by steps not
exceeding 2r; therefore, the sequence of permutations contains a term Pj such
that S(Pj) is in the interval [—r, r] that is
\S(Pj)\<r
which completes the proof.
1997/4. An nxn matrix whose entries come from the set 5 = {1, 2, ...
..., 2n — 1} is called silver matrix if, for each i = 1, 2, ..., n, Й£ ζί/ι row шг^
Й£ гй column together contain all elements of S. Show that:
(a) there is no silver matrix for n= 1997;
(b) silver matrices exist for infinitely many values of n.
Solution, (a) We show, in general, that there is no silver matrix for η odd,
n> 1.
The union of the kth row and the kth co-
7. Ί
lumn is going to be called the kth 'cross' (k=l,
2, ..., n); each number from S is contained in
every cross, by condition. We shall need now
an element χ of S which does not occur in the
main diagonal at all; such a number certainly
does exist since \S\ = 2n — 1 and there are only
η entries in the main diagonal (n > 1).
Consider now such an entry and the two
crosses, say the ith one and the jth one passing
through this occurrence of χ (Figure 1997/4.1).
The size of a cross is 2n — 1 so this one is the
only copy of χ in these two crosses, otherwise, some element of S would be
missing from one of these two crosses. Consider now any other cross: this, by
condition, also contains an ж-entry. By the choice of χ the other cross through
this entry is different from the first one. Coming to the point, there is a matching
among the crosses of the matrix, via the occurrences of χ and the members of
each pair are different. This is the end because, by this matching, the number of
crosses, η is even, indeed.
(b) We construct silver matrices of size n = 2k for every positive integer k.
In fact, we show how to prepare a 2n χ 2n silver matrix from an η χ η one.
χ
Figure 97/4.1
443

Solutions
1997/4
Consider A, a silver matrix of order n. Prepare matrix В by adding 2n to
each entry and matrix С by replacing the entries in the main diagonal in В by
2n. Matrix D is now assembled from these components according to the pattern
D =
A
С
В
A
We still need a 2 χ 2 silver matrix; A below will do. Then
A =
1 2
3 1
B =
5 6
7 5
C =
4 6
1 4
D =
12 5 6-
3 17 5
4 6 12
L7 4 3 1.
Any cross in D is contains the numbers 1, 2, ..., 2n — 1 in its A-minor and
the remaining 2n, 2n+ 1, ..., An — 1 values are in В and C; D itself is hence
indeed silver.
Remark. The attribute 'silver' is definitely unusual in terms of matrices;
the name follows the IMO tradition: whenever there is a chance to do so, the
formulation of problems indicate some characteristics of the host country. In
Finland or Poland, for example, some points in a problem have been coloured
white and blue or white and red, the colours of the national flags, respectively,
of these countries. Argentina, in latin languages, means silver or rich in silver
(the Latin itself for silver is argentum).
1997/5. Find all pairs (a, b) of positive integers that satisfy
(1)
a
= ba.
Solution. Let a and b be solutions of (1) and denote (a, b) by d. Then
a = du, b = dv, with positive integers u, ν such that (u, v) = 1. Rewriting (1) as
(du)d2v2 = (dv)du yields
(2). (du)dv2 = (dv)u
We shall proceed by distinguishing three cases, namely
(a) dv2 = u, (b) dv2 > u, (c) dv2 < u.
(a) (2) now becomes
(du)u = (dv)u, and thus u = v.
о
Since и and ν are coprime, u = v = \ and the arising dv =u implies d=\.
Therefore a = b = 1 and this is clearly a solution.
(b) If dv >и then dv —u>l and now (2) yields
„2
(3)
ddv2-umUdv-=vu
444

1997/5
1997.
Thus и divides ν; using (и, v) = l again u= 1 and (3) hence becomes
(4) ddv2~l=v.
If d = 1 then we get ν = 1 contradicting dv —u>\; if d>2 then
2 2-1
which is again impossible; indeed, 2 > υ contradicts (4). Hence we get no
solution at all, this time.
9 9
(c) The tough part is case (c). Now dv < и and thus и — dv > 1. This time
(2) becomes
udv2 = du-dv2-vu.
Now it is ν dividing и so, by (u, v)=l again, v = l and our equation is
(5) ud = du-d.
Here, of course, и > d but then d<u — d, therefore и > 2d.
If ρ is an arbitrary prime factor of и then, by (5), ρ also divides d. Let a
and <5 be the indices of ρ in η and d, respectively, that is the highest integers for
which pa | и and ρδ \ d. Thus the indices of ρ on the l.h.s and the r.h.s. of (5) are
ad and <5(ii — d), respectively. Remembering that и > 2d, we obtain
ad = 5(u — d) > Sd,
that is α > δ. Hence any prime power divisor of d divides u, as well, the former
is a proper divisor of the latter:
(6)
furt)
(7)
lermo
Plug
те, by и > 2d,
this into (5):
we
u-
have k>3.
(kd)d--
k =
■ kd,
-- dkd-d
= dk~2.
d=l is not possible since к >3. If k = 3 then, by (7), we get d = 3; и = 9 by
(6) and hence a = 27, finally, v = \ yields 6 = 3. Here there is another solution, the
pair (27, 3).
If k = 4 then proceeding similarly: (7) yields d-2, hence и = 8, by (6), then
a= 16 and 6 = 2, one more solution.
If fc>5 then dk~2 >2k~2, by d>2. It is easy to check, by induction, for
example that if к > 5 then 2 > к and thus (7) cannot hold any more.
There are three solutions of (1), altogether. They are
(1, 1), (27, 3), (16, 2).
445

Solutions
1997/6
1997/6. For each positive integer n, let f (n) denote the number of ways
of representing η as a sum of powers of 2 with non-negative integer exponents.
Representations which differ only in the ordering of their summands are
considered to be the same. For example, /(4) = 4, because 4 can be represented as
4; 2 + 2; 2 + 1 + 1; 1 + 1 + 1 + 1.
Prove that for any integer η > 3
2 2
η η
2~ <f{2n)<2T.
First solution. Getting started we verify two simple recurrences, both
satisfied by the function /.
If η is odd then 1 has to be present in its representations; removing it we
get a representation of η — 1. Conversely, adding a 1 to any representation of the
even η — 1 yields a representation of n. This one to one mapping shows that
(1) f(2k + l) = f(2k) (fc = l,2, ...)
If η = 2k is even then, depending on the presence of 1, its representations can
be divided into two disjoint groups. Removing 1 from the elements of the first
group yields representations of 2k — 1 and, clearly, all of them can be produced
this way. Hence there are f(2k — l) representations of n = 2k that contain 1.
The terms in any representation of the second kind, when halved, yield a unique
representation of n/2 = к so there are f(k) of them. Therefore
(2) f(2k) = f(2k-l) + f(k).
Comparing (1) and (2) yields
f(2k)-f(2k-2) = f(k):
f(2k -2) = f(2k - 1) = f(2k) - f(k) < f(2k) = f(2k + 1).
Function / is hence increasing.
Clearly /(1) = 1. To make life simpler set also the usual /(0) = 1. Rearrange
(2) as f(2k) — f(2k — 2) = f(k) and substitute here к = 1, 2, ..., η, respectively:
/(2)-/(0) = /(l),
f(4)-f(2) = f(2),
f(2n)-f(2n-2) = f(n).
The internal terms do cancel as these differences are added and we obtain
(3) f(2n) = f(0) + /(1) +... + f(n) (n is an arbitrary natural number).
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1997/6
1997.
Remember now that / is increasing:
f(2n) = 2 + (f(2) + /(3) +... + f(n)) < 2 + (n - l)/(n),
(4) f(2n)<nf(n).
Starting with n = 2n apply the above estimation to the decreasing powers:
r\n— 1 <in- 2 <л1.
ζ , ζ , . . ., ζ .
/ (2η) < 2n~lf (2n~l\ < 2η~ι ■ 2n-2f(2n-2) <...<
/ ι\ / ο\ .1 η(η—Ι) η —η+2
< 2(η-ΐ)+(η-2)+...+ΐ j(2) = 2~^~ ■ 2 = 2—2 .
2 2
Since η — η + 2 < η if η > 3,
2
η
/ (2η) < 2Τ,
indeed, the upper estimation of the problem is hence settled.
To prove the other one note first that if a = b (mod 2) and b > a > 0 then
(5) /(b+l)-/(b)>/(a+D-/(a).
This is obvious if both a and b are even because then, by (1), both sides are
equal to zero. If a and b are both odd then rearranging (2) as f(2k) — f(2k — 1) =
f(k) and plugging here b = 2k — 1 and a = 2k — l, respectively, yields
f(b+i)-f(b) = f(^y f(a+i)-f(a) = f(^±Ly
Since b+ 1 > a+ 1 the monotonity of / implies (5).
Let r be now even and apply (5) к times, to the pairs
(r, r), (r-l,r + l), ..., (r-k + l, r + k- 1).
Thus
/(r + l)-/(r)>/(r + l)-/(r),
/(r + 2)-/(r + l)>/(r)-/(r-l),
f(r + k)-f(r + k-\)>f(r-k + 2)-f(r-k+\).
When the resulting inequalities are added the internal terms cancel again yielding
/(r + AO-/(r)>/(r + l)-/(r-fc + l).
Remembering that r is odd and thus, by (1), f(r + 1) = /(r), the above inequality
can also be written as
f(r + k) + f(r-k + l)>2f(r) (k = l:2,...,r)
447

Solutions
1997/6
Writing this down for the possible values of к
/(r + l) + /(r)<2/(r),
/(r + 2) + /(r-l)<2/(r),
/(r + r) + /(l)<2/(r)
and summing once more yields
/(l) + /(2) + ... + /(2r)>2r/(r).
The l.h.s., by (3), is equal to f(4r) - /(0) = f(4r) - 1, therefore
/(4r)-l>2r/(r),
that is
(6) f(4r) > 2rf(r), if r > 2 is even.
For the powers of 2 this yields
(7) f(2m)>2m-lf(2m-2).
(We have set r - 2m~2 (m > 3) in (6)) It is obvious that (7) also holds if m = 2:
then/(4) = 4 > 2/(0) = 2.
After this long preparation we finally take off to prove the lower estimation.
The main weapon is (7) and it is applied to the values m = n, n— 1, ..., n —
2s+ 2 (n> 2s), respectively:
/ (2n) > 2n~lf (2n-2\ > 2n~l ■ 2n~3f {2n~A\ > ...
^ r\(n-\)+(n—2)+...+(n-2s+l) £ ijn—2s\ _>ys(n—s) ш £ {r\n—2s\
η
If η is even then set s = — -t:
2
Л
otherwise let s =
f(2n)>2T2f(2°j=2~
η— I
n-\ n+1 / ι\ η -1 n2+3 n2
f(2n)>2-^—~f(2l\=2~^-2 = 2~?T >2T;
the lower estimation is hence proved. (It is, by the way, obviously true for η = 1.)
Second solution. In what follows you can see a more direct approach to
the upper estimation. Consider a representation of 2n\
(8) 2η = α0·2° + αι-2ι + ... + αη·2η,
448

1997/6
1997.
where щ is non negative integer. A particular solution, for j = 0, 1, 2, ..., n, is
dj = 2n~J and a,i = 0 for i^j. Hence we obtain n+ 1 trivial solutions of (8). Note
that these solutions are all for the very same, single term representation of 2n.
We now write down an upper bound for the number of non trivial solutions of
(8) which, of course, provides an upper bound to the actual number of
representations. This multiple counting, by the way, yields a rather poor upper bound,
luckily enough it still works.
a\ can admit the values 0, 1, 2, ..., 2n~l — 1; 2 possibilities;
a2 can admit the values 0, 1, 2, ..., 2n~2 - 1; 2n~2 possibilities;
9 9
an-2 can admit the values 0, 1, 2, 3 = 2 — 1; 2 possibilities;
an_i can admit the values 0, 1 = 21 — 1; 21 possibilities.
Observe that once these values have been set, the value of ciq is already
determined and thus a 2n factor can be saved.
We don't care about overlaps, this brutal estimate for the number of
solutions — which does count each actual representation several times — will already
do for our purpose. Indeed
r) ι n(n-l)
/(η)<η+1+2·22·...·2η_1=η+1 + 2
2 =
2
η η
= П + 1+22 2.
If η > 3 then this can be weakened on:
η _ ι l l η
=2ч у ■ 2
-2 1 1 1 п2
/(η)<η+1+2Τ~1 = -(2η + 2) + -2Τ-.
2
— Ч 4 5'
For η = 3 clearly 2n + 2 < 2 2 because 2 is already less than 2 and from here
rt
onwards 2n + 2 is growing linearly versus the exponential growth of 2 2 . Being
so
1 n2 1 n2 n2
/(η)<-·2-2-+-.2-2-=2-,
indeed.
For the lower estimation we need some bits from the first solution. We start
by showing that / is weakly convex on the set of even numbers. This means that
(9) 2/(2£;) < f(2k - 21) + f(2k + 21) (I < k).
To prove this property it is enough to show, by (3), that
2(/(0) + /(l) + ... + /(A;-Z) + ... + /(A;))<(/(0) + /(l) + ... + /(A;-Z)) +
+ (f(0) + f(\) + ... + f(k-l) + ... + f(k) + ... + f(k + l)),
f(k-U\) + f(k-l + 2) + ... + f(k)<f(k+\) + f(k + 2) + ... + f(k + l):
449

Solutions
1997/6
which clearly holds, term by term, since / is increasing (as it has been proved
in the first solution).
The lower estimate now can be proved by induction on n. f [2 J = 2 > 24,
/ ί 2 J = 4 > 24 = 2, the claim hence holds for η = 1, 2. Let η > 2 and assume that
(10)
/ (2n-2) > 2—Γ-;
we show that it follows also for n.
Consider the recurrence (3) for / (2n) and replace the values admitted at
odd arguments according to (1):
f(2n) =
= f(2n-l) + f(2n-l-l)+f(2n-l-2) + f(2n-l-3) + ... + f(l) + f(0) =
= f(2n~l^+2f (2n~l - 2) +2/ (2n~l -4) +...
... + 2/ (2n~l - (2n~2 - 2)) +2f (2n~l - 2n~2) +...
... + 2/(2)+ 2/(0) =
~f(2n-l) + f(0)
+ 2
+ 2
/(2я"1-4)+ /(4)
/(2-1-2)+/(2)
.. . + 2 [/ (2"-1 - (2n~2 -2)) + / {2n~2 -2)1 +2/ {2n~2\ + /(0).
Apply now (9) for the 2n terms in the brackets:
/ (2n) > 2/ (2n~2) + 4/ (2n~2) +... + 4/ (2n-2) + 2/ (2n~2) + 1 >
>4·2η-3/(2η~2).
The r.h.s. can be estimated, by the induction hypothesis, from below:
+ ...
, (n-2)z
/(2η)>4·2η-3·2^Γ-=2
the proof is hence finished.
п-ПЦг-n+l ^Щ-
2
= 2~:
1998.
1998/1. In the convex quadrilateral ABCD, the diagonals AC and BD are
perpendicular and the opposite sides AB and CD are not parallel. The point
P, where the perpendicular bisectors of AB and CD meet, is inside ABCD.
Prove that ABCD is cyclic if and only if the triangles ABP and С DP have
equal areas.
Solution. Place the quadrilateral in the Cartesian system in such a way that
the diagonals AC and BD are lying along the χ and у axes, respectively; they
450

1998/1
1998.
hence meet at the origin O. {Figure 1998/1.1). The vertices are ДО, a), B(b, 0),
C(0, c) and D(d, 0); the perpendicular bisectors of the opposite sides AB and
CD meet at P(x, y), and, accordingly, the feet of the perpendiculars from Ρ to
the diagonals are T(x, 0) and 5(0, y), respectively. In what follows we shall be
working with signed areas.
D(d,0)
A(0, a)
Figure 98/1.1
Note first that
2[ABP] = 2[ABO] - 2[APO] - 2[BPO] = ab-ax-by and
2[CDP] = 2[CDO] - 2[CPO] - 2[DPO] = cd-cx- dy.
Observe that the two areas [ABP] and [CDP] have been computed with the
same sign. For their difference
(1) 2 ([ABP] - [CDP]) = (a- y)(b -x)-(c- y)(d - x).
Turning to the proof, assume first that ABCD is a cyclic quadrilateral; then
Ρ is the incentre and thus S and Τ are bisecting the respective diagonals:
b+d a+c
χ
У =
2 ' 2
Substituting these results to the r.h.s. of (1) yields
2 ([ABP] - [CDP]) = ^ [(a - c)(b -d)-(c- a)(d - b)] = 0,
[ABP] = [CDP], indeed.
Assume now, for the converse, that [ABP] = [CDP]. Hence, by (1),
(2) \a — y\\b — x\ = \c — y\\d — x\.
451

Solutions
1998/1
We show that Ρ is equidistant from the vertices of the quadrilateral. Since PA-
Ρ Β and PC = PD, it is enough to show that PA-PC. Assume the contrary,
for example PA > PC and, at the same time, Ρ Β > PD. Checking the (maybe
degenerate) right triangles CSP, ASP and BTP, DTP the indirect assumption
yields SA > SC and also ТВ > TO or, in terms of the respective coordinates:
\a — y\>\c — y\ and\b — x\>\d — x\,
contradicting to (2). A similar contradiction follows from the assumption PA <
PC. Therefore, PA and PC are indeed equal, ABCD is cyclic.
Remarks. 1. Here there is a
simple proof of the first part that
in a cyclic quadrilateral [ABP] =
[CDP] (Figure 1998/1.2). Ρ is the
centre of a circle with radius r
about O. With the notation LBCO=a
clearly /CBO = 90°-a,
furthermore, as central angles, LAP В = 2a
and LCPD = 180° - 2a. Thus
r2
[ABP] = — sm2a =
2
= ^-sin(180°- 2a) = [CDP].
2. There are various ways to
solve the problem; there is,
however, a technical difficulty: as Ρ varies in the triangular regions which the
quadrilateral is divided into by the diagonals, the area relations have to be updated and
verified over and over again. The tedious case checking can be bypassed via the
analytic approach.
In what follows an alternative, still analytic solution will be sketched and
thus we don't have to worry about the position of P.
The equations of the perpendicular bisectors of the sides AB and CD are
(with the notations of Figure 1998/1.1):
ι ι
2bx — 2ay = b —a ,
2dx-2cy = d2-c2.
Solving this simultaneous system yields the coordinates of the intersection
Figure 98/1.2
x0 =
c(b2 - a2) - a(d2
2(bc — ad)
c2)
and
2/0 =
d(b2 - a2) - b(d2 - c2)
2(bc — ad)
452

1998/2
1998.
Since Ρ is interior to the quadrilateral, AABP and ACDP are oriented
similarly. The well known analytic area formulas
2[ABP] = axQ + byo — ab and2[CDP] = cxq + dy$ - cd
hence compute the respective areas with the same sign. These areas are equal if
and only if
(a — c)xq + (b — d)yo + cd — ab = 0.
Substituting here xq and y$ as computed above yields
{ac - bd) Ua - c)2 + (b - d)2~\ = 0.
The term in the brackets is not zero because a and c, as well as b and d are of
opposite sign. Hence ac — bd = 0 and thus
|a||c| = |b||d|, that is OA-OC = OB-OD,
yielding that А, В, С and D are lying on a circle, indeed.
1998/2. In a competition there are a contestants and b judges, where b>3
is an odd integer. Each judge rates each contestant as either "pass" or "fail".
Suppose к is a number such that for any two judges their ratings coincide for at
most к contestants. Prove
к b-\
~> ■
a 2b
Solution The problem can be represented in a bipartite graph. Assign a
vertex to every contestant and also to every judge and denote these vertices by
v\,v2, ■ · ■, va and w\, u>2, · · ·, и>ь, respectively. Vertex ν is connected to vertex
it; by a green edge if contestant υ has passed in the rating of judge w and in
case of fail the connecting edge is red. (The corresponding line in the diagram
is continuous for green and dotted for red.) To make it simpler a person and a
corresponding vertex will not be distinguished.
νλ V2 V3 Vj
W\ u>2 u>2
Since every judge has rated all the contestants, the degree of the vertices
Wi is a and that of the vertices Vj is b. A pair of similarly coloured edges will
be called hook if it connects two w vertices via a v vertex. Any hook in the
graph means that contestant υ has been rated similarly by the corresponding two
453

Solutions
1998/2
judges. There are II pairs formed from the judges and since, by condition,
there are at most к hooks connecting any two judges, the maximal number of
Let us count now the hooks from the contestants' side. If there are gi green
edges and η red ones incident to the vertex Vi then clearly gi + n = b and there
are I г 1 + I г J pairs of the same colour through this vertex. Therefore, there
are
hooks at Vi. On the other hand
, 9i(9i ~ 1) , П(п - 1) 1/2,2 / ^ Л
hi = 2 + 2 = 2\9i { ~^9i + ri)) =
= sii+r} _b> (gj + гЛ2 _Ь = Ь^_Ь = (b-l)2 _ 1
2 2~\ 2/242 4 4'
Here hi is integer and since Ъ is odd, — is also an integer. Therefore we
have the stronger Si > — . The number of hooks from the contestants side
° a(b— l)2
is hence ^ hi > · Comparing this with the previous estimation yields
к b-l
a ~ 2b
indeed.
1998/3. For any positive integer n, let d(n) denote the number of positive
divisors of η (including 1 and n). Determine all positive integers к such that
d(n2)
d(n)
for some n.
= k
Solution. Note first the well known property of the function d(n): if the
prime factorisation of η is n = p^ lp22 ■ · -p^r then
d(ri) = (a\ + l)(o!2 + a)... (ar + 1);
454

1998/3
1998.
from here already follows that d(n) is multiplicative: if a is prime to b then
d(ab) = d{a) · d{b); this, of course, holds for any number of factors. Hence
d(n2) _2αχ + \ 2α2 + 1 2ar + l _
d(n) a\ +1 ct2 +1 ar + \
Observe that the numbers к we are looking for are the ones that can be
written in the form (1) because the corresponding numbers щ, a>2, ■. ·, ctr can
be then combined arbitrarily with distinct primes p\, P2, ■ ■ ■, Pr yielding к as the
value of d(n2)/d(n).
Note, first of all, that, by (1), к has to be odd. Checking a few small odd
values with bare hands is quite reassuring: they can indeed be written as required:
W*) 1 Φ4·32)2) ^5
d(l) ' d(24-32) 5-3 '
There seems to be no obvious obstacle and the opposite would be quite
awful to prove so let's set out to prove that, indeed, к can admit any odd value.
Induction seems to be the right strategy and the initial step has already been
completed. Let к > 3 and assume that for every odd /cq, less than k, there exists
no such that
d(nh
(2) -^ = *0>1.
We are to prove that к can also be written as desired.
Let's try to set the corresponding indices a\, ct^, ·. ·, ar in such a way that
in the fraction form of к on the r.h.s. of (1) we should be able to cancel as much
as possible.
As an even number, k+ 1 can be written as 2rkQ where /cq > 1 is an odd
number and r is a positive integer. Since /cq < k, it can be written in the form
(2) by the induction hypothesis. Let
a\ =(2r — l)A?o — 1 and a.i+\=2lo>i (i = 1, 2, ..., r)
Choose now the primes ργ, p2, ..., pr all coprime to no and set
With this choice
al a2 ar
n = pllp2l ...prrn0.
d(n2) _ 2щ + 1 2a2 + 1 2ar + 1 ά(τφ _
d(n) o>\ +1 OL2 +1 ar + \ d(riQ)
OL2 + 1 Οίτ> + 1 ar+l + 1 ar+l + 1
= . .. ~Tk0 = ТГк0 =
o>\ + 1 ct2 + 1 ar + 1 ai + 1
2ra1 + l _2rcq + l _2г(а! + 1)
~ (2r - \)k0 °~ 2r-\ ~ 2r-\
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Solutions
1998/3
2Γ(2Γ - \)kp
= ~ : 1=2 Acq - 1 = «,
2r -1
indeed, the proof is finished.
1998/4. Determine all pairs (a, 6) of positive integers such that (ab +6 + 7)
divides (a b + a + b).
Solution Let A = a2b + a + b and B = ab2 + 6 + 7'. Observe that if 5 divides
Л then it also divides bA — aB; we are hence looking for those pairs (a, b) for
which В divides
bA-aB = b2- la.
Since a> 1 we have 62 <ab2, therefore b2 — la<ab2 + b + 7 = B.
Let b — la > 0 first. Then В is dividing a non negative integer smaller than
9 2
Б itself and thus the latter, 6 — la is equal to zero. Then b =la and thus b,
9 9
as a multiple of 7, can be written as b = 1c; hence 49c — la = 0 and α = 7c (c
is a positive integer). These pairs (a, 6) = (7c , 1c) already satisfy the conditions
since
4 = 49c4-7c + 7c2 + 7c = 7c(49c4 + c+l),
Б = 7с2-49с2 + 7с + 7 = 7(49с4 + с+1)
that is A = cB.
Assume now that b2 — la< 0; its opposite, the positive integer la — b2< la
is divisible by B. Note that b now cannot exceed 2, otherwise B = ab +6+7 >
> 9a+ 10 and thus В cannot divide a positive number less than la. Hence 6=1
or 6 = 2. The rest is straightforward:
a) if 6 = 1 then la — b2 = la — 1. If this is a multiple of В = a + 8 then writing
ηα—\ = 7(α + 8) — 57 shows that В has to divide 57. The positive divisors of 57
are: 57, 19, 3, 1, and thus a + 8 is 19 or 57, a= 11 or a = 49. These pairs can be
now checked:
4=133 = 7-19, Б=19, ill. 4 = 2451=43-57, 5 = 57.
b) if 6 = 2 then la-b2 = la-4 and B = 4a + 9. The division 4(7a-4) =
7(4a + 9) — 79 yields that В is now dividing 79. Since this prime number has no
divisor of the form 4a + 9 we get no solution if 6 = 2.
Coming to the end: the solutions are the pairs (7c , 7c), с is arbitrary
positive integer and the two extras: (11, 1) and (49, 1).
1998/5. Let I be the incentre of the triangle ABC. Let the incircle of ABC
touch the sides ВС, С A, AB at K, L, M, respectively. The line through В
parallel to Μ Κ meets the lines LM and LK at R and S, respectively. Prove
that the angle RIS is acute.
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1998/5
1998.
Solution In the isosceles triangle LAM clearly IAML= IBM Rl =90° -
ή-\ similarly: ICKL= IBKS = 90°-^ and LBKM = LBMK = 90° - ?-.
<£j Aj £t
С
Besides we also have ILMК = IMRS = 90° — —. Applying the sine rule to
ABRM and ABSK:
(1)
cos
с
RB _ sin (9°° ~ ί) _ cos4 SB__
BM = nn(90o_cycosc> 81т11аг1У BK = cos л
Figure 98/5.1
Hence, by В К = BM
(2) RB-SB = BM2.
KM and RS are parallel and IB is perpendicular to KM; hence IB is also
perpendicular to RS. In the right triangles RBI and SBI
(3) IR2 = RB2 + IB2 and IS2 = SB2 + IB2.
The cosine rale is ready now in ARSI:
RS2 = IR2 + IS2 - 21R ■ IS cos Zi?JS.
Hence, by (3) and (2) we obtain
21R ■ IS cos IRIS = IR2 + IS2 - RS2 = RB2 + SB2 + 21B2 - (RB + SB)2 =
= 21B2 + RB2 + SB2 - RB2 - SB2 - 2RB · SB = 2(1 B2 - BM2).
457

Solutions
1998/5
Finally, in the right triangle BIM we have IB2 - BM2 = IM2 and hence
IM2
cosIRIS=7rJs>
a positive number and thus IRIS is acute, indeed.
1998/6. Consider all functions f from the set of all positive integers into
itself satisfying
(1) f(t2f(s)) = s(f(t))2
for all s and t. Determine the least possible value of f (1998).
Solution Denote the set of the solutions of (1) by Я and let /еЯ be an
arbitrary solution. Denote /(1) by a. In the first part of the argument we are
going to construct an element д of Η smaller than / in the usual sense that is its
values are smaller (not greater) than the respective values of /.
Setting t = Tin (1) and also s= 1 yields
(2) f(f(s)) = a2s; (3) f(at2) = (f(t))2.
In the computations below the labels above the equality signs are to indicate the
actual relation that has been used in that very step.
(f(e)f(t)f = (/(s))2 (/(t))2 ® (f(s))2 f(at2) ^
■ (i' / (*2/ (fiat2))) £> / (sV · at2) = / (a(aet)2) © (f(ast))2 .
Since the values are positive, here we can write
(4) f(s)f(t) = f(ast)
and hence, plugging t=\, we obtain f(as) = af(s). Perform here the substitution
s—> st:
f(ast) = af(st),
which, by (4), yields
(5) af(st) = f(s)f(t).
A straightforward induction on к shows that (f(t))k = ak~lf(tk). If k=\
then this is the meaningless f(t) = f(t). Let k> 1 and assume that (f(t))k~l =
ak~2f(tk~l). Multiplying by f(t) and applying (5) yields
(6) (f(t))k = ak-2f(tk-l)f(t) = ak~lf(tk),
that's it.
Next we prove that a divides f(t) (t e N). If ρ is an arbitrary prime and its
indices in a and f(t) are a and β, respectively, then β>α has to be checked.
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1998/6
1998.
Clearly (f(t))k is a multiple of pk(3 and ak~l is that of p^-lV for every k.
Equality (6) hence yields kp>(k — l)a and if this holds for every к then β>α,
indeed.
Introduce now function g as
g(t) =—.
a
The values of g, by the previous divisibility relation, are whole numbers, g is
mapping N to itself. Since, by (3), f(a) = a
(7) g(a) = a.
Divide now (5) by a :
(8) 9(st) = g(s)g(ty,
function g is hence totally multiplicative and this, of course, can be extended to
the product of more arguments. Furthermore (with the previous labelling
convention)
ag(g(s)) = g(a)g(g(s)) = g(ag(s)) = g(f(s)) = = — = as,
a a
that is
(9) g(g(s)) = s.
We are now able to prove that g also belongs to H. The properties (7)-(9)
can be used to verify that g indeed satisfies (1):
s(tVs))(=V)steW)(8^Wi))2.
Thus, for arbitrary /еЯ there is a g g Η such that for every η e N g(n) <
f(ri). This means that, as far as the problem is concerned, we are free to restrict
ourselves to functions satisfying (7)-(9).
Properties (8) and (9) imply that g is one to one and for any prime ρ the
value g(p) is also a prime number. Indeed, by (9)
s = g (g(s)) = g (g(t)) = t.
For the other property note first that
a a
Let ρ be now an arbitrary prime and assume that g(p) = uv (u and υ are positive
integers). By (8) and (9)
p=g (g(p)) = g(uv) = g(u)g(v),
459

Solutions
1998/6
and thus g(u) or g(v) is equal to 1. Since g{\) is also 1 and g is one to one, и or
υ also has to be equal to 1, that is g(p) is a prime, indeed.
Finally, we can turn to the actual question, the minimal possible value of
/(1998). As we have already seen, /(1998) > #(1998) and since 1998 = 2 · 33 · 37,
#(1998) = #(2)·(#(3))3#(37).
In this product every factor is a prime or the power of a prime. The first
primes are 2, 3 and 5, therefore
0(1998) >3·23 -5 = 120.
/(1998) is hence at least 120 for every /еЯ. We show that there is an element
of Η — in fact, it will be of the у-kind — such that #(1998) = 120. It can/should
be defined as follows:
а)я(1) = 1;
b) 0(2) = 3, g(3) = 2, g(5) = 37, #(37) = 5;
c) g(p)=P f°r every prime ρ different from 2, 3, 5 and 37;
d) if η = 2αΐ3α25α3 37°ν*5 .. .p*\ (ai = 0 is possible) then
#(n) = (#(2))«i (g(3)T2 (#(5))α3 (#(37))a4 fo(p5))«5 ... (#(Pr))^ .
This function satisfies the conditions (7)-(9) since #(1) = 1, (8) holds by the
definition, finally
9 (9(n)) = g (3αι · 2«2 .37^3.5α*·ρ°5·.... ρ?ή =
= 2αι .3α2·5α3 ·37α4·Ρ55 •...•р«г=щ
which is (9). Thus g is in H, indeed, and #(1998)= 120, as we have seen, is
smaller or equal to /(1998) for every /еЯ.
1999.
1999/1. Find all finite sets S of at least three points in the plane such that
for all distinct points А, В in S, the perpendicular bisector of AB is an axis of
symmetry for S.
Solution. If \S\ = n then the vertices of a regular n-gon clearly satisfy the
requirements. We show that there is no other solution, S is the set of vertices of
some regular n-gon. If С is the centroid of S then it is also the centroid of any
set which is produced by reflecting S in some perpendicular bisector. Therefore
С is kept fixed by any symmetry and thus the perpendicular bisectors are all
concurrent, they are passing through С Since reflection preserves distances, this
implies that the points of S are all lying on a circle с of centre C.
460

1999/2
1999.
Consider now A\, A2 and A3, three consecutive points of S, that is A2 is
between A\ and A3 and there is no other point of S on the arc A\A^ Figure
1999/1.1. Such a triple certainly exists since the set S is finite.
The heart of the issue is that
consecutive arcs are equal, A\A2 = A2A^.
Indeed, if, for example, A1A2 <A2A^ then
A'2, the mirror image of A2 through the
peipendicular_bisector of A1 A3 is interior
to the arc A2A^, a contradiction. The two
arcs are equal indeed and thus A2 is
incident to the perpendicular bisector of Αχ Α.3.
We have finished since if any point of S
on с is of the same distance from its two
neighbours then S forms a regular n-gon
inscribed с
Remarks. 1. The condition about the
finiteness of S was essential; apart from
the explicit reference it was also used when
the centroid was introduced. If this
restriction is released then, besides the points of a circle, there are other solutions, for
example, the points of a straight line or the whole plane itself.
2. The original proposal was the more demanding 3D version of the
problem: what can be said about the finite set S if it is symmetrical in the
perpendicular bisector plane of any two of its points? The answer in this case is that S
either forms a regular n-gon (if it is planar) or a platonic solid, a tetrahedron or
an octahedron.
1999/2. Let η > 2 be a fixed integer. Find the smallest constant С such that
for all non-negative reals x\, · · ·, xn:
(1) Σ XiXj(xj + xj)<c( J2 xi
l<i<j<n M<i<n
Determine when equality occurs.
Preliminary remarks. If there is at most one among the given numbers
different from zero then the l.h.s. of (1) is zero, it holds for any non-negative constant
C. In what follows it will be assumed that there are at least two positive ones
among the numbers X{.
Figure 99/1.1
4-
461

Solutions
1999/2
In the subsequent solutions the new variables
CLi
Σ
1<г<п
Xi
will be used. With these new variables clearly Σα; < 1 and 0 < a; < 1 (г = 1,2,...
4
, η). Dividing (1) by ί ^ χι \ yield
4<г<п '
(2) Σ aiaj(a^ + aj)<C.
l<i<j<n
Our task is to find the maximum of the l.h.s.; its actual value, C, is the
answer for the first question of the problem.
First solution. Observe that
7 9 3 3
CLiCLj{<H + CLj) - ai aj +CLiCLj,
and thus the l.h.s. of (2) can be written as
(a\ + 0,2 + ■ · · + αη){α\ + cl\ + ■ · ■ + 4) — (aj + a\ + ■ ■ ■ + a^) =
/33 3\ / 4 4 4\
= (aJ + a^ + · ■ · + an) — {α[ + θ2 + ···+ α„).
Hence the job is to find the maximum of the function
F(a\, a2,..., an) = (a] + a\ + ■· · + a3) - (aj + a\ + · ■ · + a\)
under the condition Y^ ai = 1. 1 > a\ > a2 > ... > an > 0 can clearly be assumed
here. Let ak+\ be the last positive one among the numbers a; in this order; then
clearly к > 1. (The case к = 1 will be checked in due course.) Consider now the
values of F if the arguments are
\(cii, a2, · · ·, ak+i, 0,..., 0) and \'(аг, α2,..., ак_г, ак + ак+г, 0,..., 0)
respectively. We show that switching from ν to v' the value of F is increasing;
the sum of the coordinates, at the same time, remains the same.
F(\')-F(\)=(al + --- + a3k_l+(ak + ak+i)3^-
(aj + ■ ■ ■ + aj_j + (ak + ak+x)4) - (a\ + · ■ ■ + a3k+l) + (aj + ■ ■ ■ + aj+l) =
/ \3 3 3 4 4 / \4
= {CLk + ak+i) -CLk-ak+l+ak + ak+l-(ak + ak+i) =
= ^CLk^k+l(ak + CLk+l) - ^ak^k+\(2ak + ^k^k+l + 2ak+l) =
= ^№+1 (αΛ + αΑ;+ι,)(3-4(αΛ + αΑ.+1)) + 2αΑ;αΑ.+1
To show that this difference is indeed positive it is enough to check that
ak + CLk+l <3/4.
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1999/2
1999.
This is straightforward since a\ > ak > ak+\ implies 2a\ > ak + ak+\ and thus
CLk + &к+\
l>ai+ak + ak+i > + ak + ak+\.
Hence
3 4
1 > -^iflk + ak+i)> -(ak + ak+i)
therefore
F(V)>F(y).
Starting from an arbitrary argument ν the value of F hence can be increased
stepwise until every coordinate, apart from the first two, a\ and a2, of the argument
is equal to zero. At this point
9 9
F(a\, a2: 0,... ,0) = a\a2(a^ + α^)= αΐα2(1 ~^a\al) =
*{ 1\4 1 1
—2[aia2--j + -<-.
1 1
Equality holds if a\a2 = - that is α^(1 — αγ) = - and the latter holds if and only
•f l
if ai=a2 = -.
The smallest constant С of the problem is hence -g- and equality holds if
and only if two of the given numbers Xi are equal and the remaining ones are all
zero.
Second solution. As in the previous argument we proceed by analyzing
2
'3
9 9
inequality (2). The idea of the proof is now to humbly estimate the terms af + a
by the total sum of the squares in the l.h.s. of (2). Since
9 9 9 9 9
Щ + CLj < CL\ + &2 + ' ' ' + an ~ Q>
(3) F= Σ aiCLj(cii+aj)<Q Σ a*ai"
l<i<j<n \<i<j<n
Since
1 = ( Σ a4 =Q + 2 X) а+а
(3) yields
l<i<n 1 <i<j <гг
f<Q^ = i[1-(2Q-1)2]4
The indicated maximum of F can be attained and this happens only if Q = - and
for every г, j in (3)
Π.-Π .Y/ТГ . Λ- tu
9 9
(4) α; aj (af + a ■) = ai aj Q.
463

Solutions
1999/2
Since there are at least two positive ones among the щ by assumption the
ordering of these numbers implies that αϊ and a^ are certainly positive. Thus, by
(4)
2 2 1
ai+a2 = -.
In case of equality, on the other hand, there can be no more positive ones
among the numbers щ otherwise (4) would imply
9 9 9 9 9
d\d2Q = ala2(a\ +СФ <d\d2(a\ + a2 + ai) — ala2Qi
a contradiction. Hence, if there is equality in (2) then щ = 0 if i > 2. From here
one can conclude as above, in the first solution.
1999/3. Given an η χ η square board with η even. Two distinct squares of
the board are said to be adjacent if they share a common side, but a square is
not adjacent to itself Find the minimum number of squares that can be marked
so that every square (marked or not) is adjacent to at least one marked square.
Solution. For simplicity any neighbour of a marked square will be called
to be covered by the very marked square. The task is then to find the minimum
number of marked squares if they cover all the η squares on the board. For this
we shall introduce a tricky, labelled colouring of the board.
Colour the fields of the board
black and white as if on a chessboard
Figure 1999/3.1, label the corners
and place the board ABCD to have
its diagonal BD horizontal. Starting
now from the fields adjacent to the
side AB enter 1 or · (dot) in the white
fields above the horizontal diagonal
in the following way: the 'rows'
parallel to BD are labelled by the same
symbol; every other field of each odd
row is labelled by 1 and every other
field of each even line is "dotted".
Flip now the dotted fields in BD but
watch out: the reflected fields below
are also labelled by 1.
Let's count now how many squares are labelled by 1, altogether. Their
number is clearly the same as the number of fields, 'oned' or dotted, above the
horizontal diagonal. Starting from A there are exactly к squares labelled in the k-th
С
Figure 99/3.1
464

1999/3
white row parallel to BD and thus there are
M = l+2+... + -= g
fields oned on the board.
Observe now that each oned square (they are all white) is covered by some
black squares but there are no two oned squares such that both of them would
be covered by the same black square. Thus we need at least Μ black squares to
cover the white part of the board. The oned squares, on the other hand, do cover
the black squares and thus at most Μ white squares cover the black part.
Now η is even and thus the whole arrangement is symmetric with respect
to the two colours. Rotate the board by 90° about its centre and label the images
of the oned fields by 2 respectively. The 'twoed' fields are all black and they are
displayed in Figure 1999/3.2.
С
Figure 99/3.2
Swapping the colours temporarily the previous argument now yields an
opposite estimate: to cover the black squares we need at least Μ white squares and
the white part can be settled by Μ marked black squares. Coming to the point
there are at least
ЛГ ~*,Г П(П+1)
N = 2M = —
4
squares have to be marked and the fields labelled by 1 and 2 when marked show
that this many is enough.
465

Solutions
1999/3
Remarks. 1. In a good
marking pattern each marked square is
adjacent to some other marked
square; this can be checked on Figure
1999/3.3. There are many other
patterns, of course, with the same,
minimal number of marked squares.
2. If η is odd then the
symmetry of the coloured board is broken;
a more tedious argument shows that
the required minimum is
n+1
and
(n+\y
С
Figure 99/3.3
\ 2 ) 4
if η = 4k +1 and η = Ak + 3
respectively.
1999/4. Find all pairs (n,p) of positive integers, such that ρ is a prime,
n<2p and (p — l)n + 1 is divisible by np .
Solution. It is worth checking a few particular cases first.
1 1
a) If n= 1 then (p— 1) =p is divisible by lp = 1 and thus the pair (l,p)
is a solution for any prime p.
b)lfp = 2 then np~l = η should divide ln + 1 = 2 and thus if η > 1 then η = 2;
the pair (2,2) is another solution.
c) If η > 2 and ρ > 3 then consider again a special case; let n = p. Now
(p-l)p + \=p'
-2
f~l-
+
V
\V
Ρ
Apart from the last one each term in the brackets is divisible by ρ and thus
the whole sum is not. The index of ρ hence is at most 2 in the factorisation of
(p — Yf +1 and thus ρ — 1 < 2 yielding ρ = 3 as the only possibility. The
corresponding pair (3,3) is another solution.
Note that n<2p holds for every solution so far.
We now show that beyond those already found there is no other solution.
Note first that for any further one η > 2 and ρ > 3. Now (p — l)n +1 is odd and so
is n, its divisor by condition. Hence equality cannot hold in the restriction, now
η < 2p. Consider the smallest prime, divisor of n, let it be q. Since q also divides
(p— l)n + l, the two numbers, q and p— 1 are coprime. Observe, additionally,
that η and q — 1 are also coprime since, by g's choice, any proper divisor of η
is at least q.
466

1999/5
1999.
We need a well known piece from the theory of first degree diophantine
equations: if (n, q — 1) = 1 then there exist whole numbers χ and у such that
nx + (q — l)y = 1.
Here g — 1 is even and thus χ is odd; hence
(1) (р-1)1=(р-1)пх+Ь-1)у = (р-1)пх -(р-1)(д-1)у.
The divisibility of (p — l)n +1 by q can also be written as
(ρ-1)ηΞ-1 (mod q).
On the other hand, since (q,p—l) = l also holds, Fermat's theorem can be
invoked as
(ρ-1)ΐ_1Ξΐ (mod q).
These two congruences and (1) imply
р-\=(-1)хЛу = -\ (mod q);
thus ρ = 0 (mod q) that is q > 1 divides the prime number p. The only possibility
is hence ρ = q so the smallest prime factor of η < 2p is ρ and thus η = ρ which
was already checked in case c).
The solutions of the problem are hence the pairs (2, 2), (3,3) and (1, p) with
any prime p.
Remark. Using the assertion of the third problem of the 31st IMO (1990)
one can show that there is no other solution even if the restriction n<2p is
released.
1999/5. The circles C\ and C2 lie inside the circle C, and are tangent to
it at Μ and N, respectively. C\ passes through the centre of C2. The common
chord of C\ and C^, "when extended, meets С at A and B. The lines MA and
MB meet C\ again at Ε and F. Prove that the line EF is tangent to C2.
Solution. We shall use the following
Lemma. If circle c\ is touching circle с internally at P, a chord AB of
с is touching q at Q then F, the point where PQ intersects с bisects the arc
AB (the one not containing P) and, additionally, FA2 = FB2 = FQ ■ FP (Figure
1999/5.1).
Putting it differently: given a circle c, a chord AB of it and F bisecting arc
AB, q touching с at Ρ and AB at Q, and AB separating c\ and F, the power
of F (FQ ■ FP) is constant with respect to any such circle c\. For the proof of
the lemma see the Remark.
Let the centres of c\ and C2 be 0\ and О2 respectively and let t\ and t2 be
the common tangents (Figure 1999/5.2).
467

Solutions
1999/5
Figure 99/5.1
Figure 99/5.2
Now, by the lemma, the powers, with respect to the two circles, c\ and C2,
of the midpoints of the respective arcs cut by these tangents are equal; these
midpoints are hence incident to the radical axis of the two circles. This line is
now the common chord of the circles and thus the respective midpoints are but
A and B. Our lemma also implies that t\ and t^ are touching c\ at С and D
respectively.
468

1999/6 1999.
Consider now the similarity of centre Μ which maps c\ to с The image of
CD, under this enlargement, is clearly AB and thus they are parallel and also
perpendicular to 0\02. Consequently, 02 bisects the arc CD of c\ (the one
which does not contain M) that is LCOx02 = LDOx02.
Denote the point of contact of t\ and c2 by T. By the theorem of inscribed
angles in c\ we get
lTCOl = l-lCOl02 = ^lDOl02 = LDC02.
Hence C02 bisects LTCD, 02 is equidistant from t\ and CD and thus CD is
touching circle c2, indeed.
Remark. We now prove the lemma; for the notations please check Figure
1999/5.1. Let's see! The similarity by PF/PQ from Ρ maps c\ to с and, at
the same time, AB into the tangent A'B' of c; the equality of the arcs is now
straightforward since the point of contact of a tangent parallel to a given chord
bisects the arc cut by the chord. Additionally, subtended by equal arcs, IF Ρ Β =
FB FP
LQBF and thus AQFB and ABFP are similar; this yields —— = —- which,
when rearranged, becomes FB = FQ · FP, indeed.
1999/6. Determine all functions f : R —> R such that
(1) fix- Ky)) = f (f(y)) + xf(y) + fix) -1
for all x, у in R. [R is the set of reals.]
First solution. Let /(0) = c. Substituting x = y = 0 (1) yields
/(-c) = /(c) + c-l.
This shows that c = 0 is not possible; the contrary would imply 0 = —1. Choose
now an arbitrary element χ from TZf, the range of /; if χ = f(y) then (1) becomes
c=f(x) + x2 + f(x)-l,
(2) c+l-x2
/(*> = —2 —
We show that / is surjective, IZf is the set of reals. Setting y = 0 in (1)
f(x-c) = f(c) + cx + f(x)-l,
f(x - c) - f(x) = cx + f(c) - 1.
Let xq be arbitrary. Since c=^0, χ can be isolated from
cx + f(c)-\=x0.
With the solution (3) becomes
£0 = f(x ~ c) - f(x)
469

Solutions
1999/6
and thus any real number can be written as the difference of two elements in IZf.
Consider now two arbitrary elements in the range, y\ and г/2 and let x$ = y\ —г/2·
Setting χ = yi and Kv) = V2 in (1) we get
/Oo) = /(2/1 ~ VI) = Д 2/2) + 2/12/2 + /(2/1) ~ 1 ·
Comparing this with (2) yields
tt С+1~У2 , , c+1~2/2 л 2с-(У1-у2)2 2c — rc§
/(^0 = 2— 2/1 ^2 2 = 2 = —2—'
Accordingly
2c-x2
/0*0 =—2—
for any real number x. Comparing this with (2) again
c+l-x2 2c —x1
2 2 '
that is-c=l. Hence
!, 4 2-χ2
and / can be checked as a solution.
Second solution. Let f(0) = c again and set x = f(y). Then (1) becomes
c = /(/(2/)) + (/(2/))2 + /(/(2/))-l,
and hence
(4) /(/(2/)) = ^(c+l-(/(2/))2)·
Substitute this to (1) and introduce function g as
р(ж) = /(ж) + у.
Then
/(s - /(2/)) = ^ (c - 1 - (/(2/))2) + xf(y) + /(ж),
/ jy чл f< f< \\i (x~f(y^2 c-l-(f(y))2
9(x- f(y)) = f(.x- /(2/)) + 2 = 2 ж/(2/)+
^ ^ f \ χ1 ■ (x~f(y»2 t ч^-1
The constant zero function is not a solution and thus f(y)^0 for some y. Plug
1 ■ m
now £ = —- in (1):
f(y)
(6) f(x-Ky)) = f(f(y))-Kx)·
470

1999/6
1999.
Introducing the notations и = χ — f(y), ν = f(y) and w = χ (6) becomes
(7) f(u) = f(v) + f(w).
Setting у = и in (5) we obtain
g(x-f(u))=g(x) + c-l
2.
Applying (5) twice and using also (7) yields
c-1
g(x-f(u)) = g[x-f(v)-f(w)]=g(x-f(v)) + -^-=g(x) + c-l.
c- 1
Comparing these last two equalities we get —-— = с — 1 that is с = 1. Hence,
by (5)
g(x- f(y)) = g(x)
which means that g is periodic and each value of / is a period. Since /(0) = с = 1,
(4) implies
/ (/(0)) = /(1) = |(1 + 1 - l)2 = i.
Hence - is in the range of / and thus it is a period of g. Set now y = 0 in (1):
f(x-l) = f(x-f(0)) = f(f(0)) + xf(0) + f(x)-l = ^+x + f(x)-l =
Being in the range of /, f(x) + x — - is also a period of g which makes altoget-
1 1
her three periods found so far: f(x), - and f(x) + x — -. Hence, as their linear
combination,
x = f(x) + x---f(x)+-
is also a period and thus any real number is a period of g. The better for us,
function g is constant then. By its definition
02
0(O) = /(O) + y = l
so g(x) = 1 for every real number x. This is the end; switching back to / we get
9 9 9
χ .. χ 2-х
f(x) = g(x)- — = \- —
2 2 2
and, as substitution shows, this one is satisfying (1) indeed.
471

ISBN 189-8855-48-Х
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