volume 28 number 2 2015

M AT H E M AT I C S
C O MP E T I T I O N S

journal of the

WORLD FEDERATION OF NATIONAL

MATHEMATICS COMPETITIONS

AMT P u b l i s h i n g
VOLUME 28 NUMBER 2 2015

M AT H E M AT I C S
COM PE T I T I O NS

J O U RNA L O F T H E

WORLD FEDERATION OF NATIONAL

MATHEMATICS COMPETITIONS

AMT P U B L I S H I N G
MATHEMATICS COMPETITIONS
J ournal of the W orld F ederation of N ational M athematics C ompetitions

(ISSN 1031 – 7503)

Published biannually by
AMT P u b l i s h i n g
A u s t r a l i a n M at h e mat i c s T r u s t
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Australia

With significant support from the UK Mathematics Trust.

Articles (in English) are welcome.

Please send articles to:

The Editor
Mathematics Competitions
World Federation of National Mathematics Competitions
University of Canberra Locked Bag 1
Canberra GPO ACT 2601
Australia
Fax:+61-2-6201-5052
or
Dr Jaroslav Švrček
Dept. of Algebra and Geometry
Faculty of Science
Palacký University
Tr. 17. Listopadu 12
Olomouc
772 02
Czech Republic
Email: svrcek@inf.upol.cz

©2015 AMT Publishing, AMTT Limited ACN 083 950 341

MATHEMATICS COMPETITIONS VOLUME 28 NUMBER 2 2015

CONTENTS PAGE

WFNMC Committee 1
From the President 4
From the Editor 5
WFNMC Mini-Conference, Hamburg, July 23, 2016
Call for Papers 7
Call for Nominations 8
DIAGONAL: Etude in Six Movements 9
Alexander Soifer (USA)
Regular polygons and polynomial curves 16
Robert Bosch (USA)
Domain and range of functional equations and inequalities 25
Pavel Calábek and Jaroslav Švrc̆ek (Czech Republic)
The Powers of Two 37
Dennis Situ and Steven Xia (Canada)
The 36th Tournament of Towns, Selected Problems 42
Andy Liu (Canada)
The 56th International Mathematical Olympiad, Chiang Mai, Thailand, 2015 48
 Mathematics Competitions Vol 28 No 2 2015

World Federation of National Mathematics

Competitions

Executive

President: Professor Alexander Soifer

University of Colorado
College of Visual Arts and Sciences
P.O. Box 7150 Colorado Springs
CO 80933-7150
USA

Senior Vice President: Professor Kiril Bankov

Sofia University St. Kliment Ohridski
Sofia
BULGARIA

Vice Presidents: Dr. Robert Geretschläger

BRG Kepler
Keplerstrasse 1
8020 Graz
AUSTRIA

Professor Ali Rejali

Isfahan University of Technology
8415683111 Isfahan
IRAN

Publications Officer: Dr. Jaroslav Švrček

Dept. of Algebra and Geometry
Palacký University, Olomouc
CZECH REPUBLIC

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 Mathematics Competitions Vol 28 No 2 2015

Secretary: Sergey Dorichenko

School 179
Moscow
RUSSIA
Immediate Professor Marı́a Falk de Losada
Past President & Universidad Antonio Narino
Chairman, Carrera 55 # 45-45
Awards Committee: Bogotá
COLOMBIA
Treasurer: Emeritus Professor Peter Taylor
PO Box 6165
O’Connor ACT 2601
AUSTRALIA

Regional Representatives
Africa: Professor John Webb
Department of Mathematics
University of Cape Town
Rondebosch 7700
SOUTH AFRICA

Asia: vacant

Europe: Professor Nikolay Konstantinov

PO Box 68
Moscow 121108
RUSSIA
Professor Francisco Bellot-Rosado
Royal Spanish Mathematical Society
Dos De Mayo 16-8#DCHA
E-47004 Valladolid
SPAIN

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 Mathematics Competitions Vol 28 No 2 2015

North America: Professor Ian VanderBurgh

University of Waterloo
Waterloo
CANADA
Oceania: Professor Derek Holton
605/228 The Avenue
Parkville Vic 3052
AUSTRALIA
South America: Professor Radmila Bulajich
Facultad de Ciencias, UAEM
Av. Universidad 1001, Col. Chamilpa
Cuernavaca, Morelos 62209
MEXICO

The aims of the Federation are:

1. to promote excellence in, and research associated with,
mathematics education through the use of school math-
ematics competitions;
2. to promote meetings and conferences where persons inter-
ested in mathematics contests can exchange and develop
ideas for use in their countries;
3. to provide opportunities for the exchanging of information
for mathematics education through published material, no-
tably through the Journal of the Federation;
4. to recognize through the WFNMC Awards system persons
who have made notable contributions to mathematics edu-
cation through mathematical challenge around the world;
5. to organize assistance provided by countries with devel-
oped systems for competitions for countries attempting to
develop competitions;
6. to promote mathematics and to encourage young mathe-
maticians.

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 Mathematics Competitions Vol 28 No 2 2015

From the President

Dear Fellow Federalists!

As you know, in July 2016, Hamburg University, Germany, will host
what I would call the “Summer Olympic Games of Mathematics”: 13th
International Congress on Mathematical Education (ICME-13), http:
//icme13.org/. There will be many occasions to enjoy at the Congress,
including the Topic Study Group TSG 30: Mathematics Competitions,
co-chaired by our Past President Maria de Losada and I.

During the Congress, our World Federation of National Mathematics

Competitions (WFNMC) will have its General Meeting on Saturday,
July 30, 2016, where we will present the Paul Erdős Awards. Send
me issues of major importance to all members which you would like to
discuss during that meeting. The allotted time is limited, but I will see
what I can do.

The day before ICME-13, on July 23, 2016, WFNMC will hold its own
mini-conference. The logistics of this conference will be handled by Maria
de Losada—thank you Maria! The program of the conference will be
crafted by Kiril Bankov and I—see the Call for Papers in this issue.

Please, remember, our Federation is what we—together—make of it, no

more, no less. And so I urge you to be active, contribute papers, par-
ticipate in discussions, attend our meetings, and to write for our jour-
nal Mathematics Competitions, edited for very many years by Jaroslav
Švrček—thank you, Jarek!

I wish each of you a happy, healthy, and successful new 2016 year!

Alexander Soifer
President of WFNMC
December 2015

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 Mathematics Competitions Vol 28 No 2 2015

From the Editor

Welcome to Mathematics Competitions Vol. 28, No. 2.

First of all I would like to thank again the Australian Mathematics Trust
for continued support, without which each issue of the journal could not
be published, and in particular Heather Sommariva, Bernadette Webster
and Pavel Calábek for their assistance in the preparation of this issue.

Submission of articles:

The journal Mathematics Competitions is interested in receiving articles

dealing with mathematics competitions, not only at national and inter-
national level, but also at regional and primary school level. There are
many readers in different countries interested in these different levels of
competitions.

• The journal traditionally contains many different kinds of arti-

cles, including reports, analyses of competition problems and the
presentation of interesting mathematics arising from competition
problems. Potential authors are encouraged to submit articles of
all kinds.

• To maintain and improve the quality of the journal and its use-
fulness to those involved in mathematics competitions, all articles
are subject to review and comment by one or more competent ref-
erees. The precise criteria used will depend on the type of article,
but can be summarised by saying that an article accepted must
be correct and appropriate, the content accurate and interesting,
and, where the focus is mathematical, the mathematics fresh and
well presented. This editorial and refereeing process is designed to
help improve those articles which deserve to be published.

At the outset, the most important thing is that if you have anything
to contribute on any aspect of mathematics competitions at any level,
local, regional or national, we would welcome your contribution.

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 Mathematics Competitions Vol 28 No 2 2015

Articles should be submitted in English, with a black and white photo-

graph and a short profile of the author. Alternatively, the article can
be submitted on an IBM PC compatible disk or a Macintosh disk. We
prefer LATEX or TEX format of contributions, but any text file will be
helpful.

Articles, and correspondence, can also be forwarded to the editor by mail

to

The Editor, Mathematics Competitions

Australian Mathematics Trust
University of Canberra Locked Bag 1
Canberra GPO ACT 2601
AUSTRALIA

or to

Dr Jaroslav Švrček
Dept. of Algebra and Geometry
Palacký University of Olomouc
17. listopadu 1192/12
771 46 OLOMOUC
CZECH REPUBLIC

jaroslav.svrcek@upol.cz

Jaroslav Švrček
November 2015

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 Mathematics Competitions Vol 28 No 2 2015

WFNMC Mini-Conference,
Hamburg, July 23, 2016

Call for Papers

The Mini-Conference of the World Federation of National Mathematics

Competitions (WFNMC) will be held at the University of Hamburg,
Germany, on July 23, 2016. This is the day before the registration for
ICME-13 that will also take place at the University of Hamburg.

The Program Committee of the Mini-Conference hereby issues a call for

papers in the format of communications up to 15–20 minutes long.

The detailed proposals should be submitted in English. Please use Times

New Roman, 12 pts, 1.5 spaces. The proposals should not exceed 4 pages.

The deadline for submissions is January 31, 2016, and the applicants will
be notified by March 15, 2016.

Please, send your submission to both Alexander Soifer (asoifer@uccs.edu)

and Kiril Bankov (kbankov@fmi.uni-sofia.bg).

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 Mathematics Competitions Vol 28 No 2 2015

Call for nominations

Paul Erdös Award

The Paul Erdös Award was established to recognize the contributions of
mathematicians who have played a significant role in the development of
mathematical challenges at the national or international level and whose
contributions have been a stimulus for the enrichment of mathematics
learning.

Each recipient of the award is selected by the Executive and Advisory

Committee of the World Federation of National Mathematics Competi-
tions on the WFNMC Awards Subcommittee.

Paul Erdös Award 2016

The WFNMC Awards Committee hereby issues a call for nominations
for the Erdös Award 2016.

Send nominations to Awards Committee Chair, Marı́a de Losada

(mariadelosada@gmail.com) by January 15, 2016.

Awards will be given at ICME-13 in Hamburg, July, 2016.

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 Mathematics Competitions Vol 28 No 2 2015

DI
A
GO
NA
L:
Etude in Six Movements
Alexander Soifer

Born and educated in Moscow, Alexan-

der Soifer has for over 36 years been
a Professor at the University of Col-
orado, teaching math, and art and film
history. He has published over 300 ar-
ticles, and a good number of books.
In the past several years, 7 of his
books have appeared in Springer: The
Scholar and the State: In the search of
Van der Waerden; The Mathematical
Coloring Book: Mathematics of Color-
ing and the Colorful Life of Its Creators; Mathematics as Problem Solving;
How Does One Cut a Triangle?; Geometric Etudes in Combinatorial Math-
ematics; Ramsey Theory Yesterday, Today, and Tomorrow; and Colorado
Mathematical Olympiad and Further Explorations: From the Mountains of
Colorado to the Peaks of Mathematics. He has founded and for 32 years ran
the Colorado Mathematical Olympiad. Soifer has also served on the Soviet
Union Math Olympiad (1970–1973) and USA Math Olympiad (1996–2005).
He has been Secretary of WFNMC (1996–2008), and Senior Vice President
of the World Federation of National Mathematics Competitions (2008–2012);
from 2012 he has been the president of the WFNMC. He is a recipient of the
Federation’s Paul Erdős Award (2006). Soifer’s Erdős number is 1.

I will present 6 of the Colorado Mathematical Olympiad’s old problems,

to give you a flavor of our Olympiad. Number X.Y stands for the Y -th
problem of the X-th Olympiad.

19.1 What’s Hiding in the Corner? (A. Soifer, 2002).

Each square of a 2002 × 2002 chessboard contains a non-zero number
such that no matter where on the board we put 2002 rooks that do not

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 Mathematics Competitions Vol 28 No 2 2015

attack each other, the product of the 2002 numbers covered by the rooks
stays the same. The upper left corner contains 1, the upper right corner
contains 14, and the lower left corner contains 143. What number is
contained in the lower right corner? (We say that two rooks attack each
other if they are in the same row or column of the board.)
Solution. Place 2002 rooks on the main diagonal of the chessboard,
which goes from the square containing the number 1 to the square with
the number we want to determine, call it x (Figure 1). The 2002 rooks
do not attack each other.

1 14

a2

a3

..
.

a2001

143 x

Figure 1

We can now move the two rooks covering 1 and x to cover 14 and 143
while keeping the remaining 2000 rooks in their places on the main
diagonal. The 2002 rooks in a new position do not attack each other.
Therefore, the product of the numbers under the rooks in first position
and the second position are equal

1 · a2 · a3 · ... · a2001 · x = 14 · a2 · a3 · ... · a2001 · 143.

Therefore, 1 · x = 14 · 143, and x = 2002.

10.2 Four Knights (S. L. Tabachnikov and A. L. Toom, [TT]).

Four knights are placed on a 3 × 3 chessboard: two white knights in the

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 Mathematics Competitions Vol 28 No 2 2015

upper corners, and two black ones in the lower corners. In one step we
are allowed to move any knight in accordance with the chess rules to
any empty square. (One knight’s move is a result of first taking it two
squares in horizontal or vertical direction, and then moving it one square
in the direction perpendicular to the first direction.)
Is there a series of steps that ends up with the white knights in diagonally
opposite corners, and the black knights in another pair of the opposite
corners?

Solution. Number the squares of the 3 × 3 board and draw the diagram
showing how the knights can move (see Figure 2).

1
1 2 3 8 6

4 5 6 3 5 7

7 8 9 4 2
9
Figure 2

The moves diagram shows (to the surprise of many) that knights can
only move along the cycle (and never get into square 5). In the initial
position, the two white knights are in squares 1 and 3, and the two black
knights are in squares 9 and 7, i.e., on the moves diagram the two white
knights are followed by the two black knights.

In the desired final position, the white and the black knights would
alternate on the moves diagram. To get from the initial position to the
desired final position, the order of knights on the moves diagram has to
change, which is impossible because the moves diagram is a cycle.

2.3 (Russian Mathematical Folklore).

Each of the 49 entries of a square 7 × 7 table is filled by an integer
between 1 and 7, so that each column contains all of the integers 1, 2,
3, 4, 5, 6, 7 and the table is symmetric with respect to its diagonal D
going from its upper left corner to its lower right corner. Prove that the
diagonal D has all of the integers 1, 2, 3, 4, 5, 6, 7 on it.

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 Mathematics Competitions Vol 28 No 2 2015

Solution. Due to the given symmetry, for every integer 1 off the diagonal,
there is another integer 1, symmetric to the first, which also lies off the
diagonal (Figure 3).

Figure 3

Therefore, the number of integers 1 off the diagonal is even. The total
number of integers 1 in the table is odd (one per each of the seven
columns). Thus, at least one integer 1 must be on the diagonal.

Similarly, the diagonal must contain integers 2, 3, 4, 5, 6, and 7.

26.3 Red Square (A. Soifer, 2009).

Is it possible to color red some of the unit squares of a 2009 × 2009 grid
so that every unit square shares a side with exactly one red square?

Solution. Observe: a red square, neighboring a diagonal square, is a

neighbor to one more diagonal square! (Figure 4).

Therefore, pairs of diagonal squares share a red neighbor. But the total
number of squares on a diagonal (2009) is odd, therefore the required
coloring does not exist.

1.5 Forty-One Rooks (A. Soifer and S. Slobodnik, 1972).

Forty-one rooks are placed on a 10 × 10 chessboard. Prove that you can
choose five of them that do not attack each other. (We say that two

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 Mathematics Competitions Vol 28 No 2 2015

Figure 4

rooks “attack” each other if they are in the same row or column of the
chessboard.)

Second Solution, first found during the 1984 Olympiad (!) by Russel
Shaffer; the idea of using symmetry of all colors by gluing a cylinder,
belongs to my university student Bob Wood.

Let us make a cylinder out of the chessboard by gluing together two

opposite sides of the board, and color the cylinder diagonally in 10 colors
(see Figure 5).

One out of the ten

one color diagonals
is shown

Figure 5

Now we have 41 = 4 · 10 + 1 pigeons (rooks) in 10 pigeonholes (one-color

diagonals), therefore, by the Pigeonhole Principle, there is at least one

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 Mathematics Competitions Vol 28 No 2 2015

hole that contains at least 5 pigeons. But the 5 rooks located on the
same one-color diagonal do not attack each other!

11.5 Two Rook Game for Two Players (A. Soifer, 1994).
A move in our game consists of placing two rooks on a 1994 × 1994
chessboard so that no rook placed in previous moves is attacked. Two
players take turns. The last player to make a move wins. Find a strategy
that allows one player to win regardless of what the moves of the other
player may be, if
a) the two rooks of each move must not attack each other;
b) the two rooks of each move must attack each other.

(Two rooks are said to attack each other if they are located in the same
row or column of the board.)

Solution. It is not obvious—is it?—to see which of the two problems a),
b) is easier. In fact, problem a) is very simple while problem b) requires
certain ingenuity.
a) Each placed pair of rooks reduces by 2 the number of rows and
the number of columns available for further placing. Since 1994/2
is odd, the first player wins regardless of his strategy.

Figure 6 Figure 7

b) Again we are watching two numbers, the number R of rows and

the number C of columns available for further placing respectively.
Each placed pair of rooks reduces R, C by 1 and 2, or by 2 and

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 Mathematics Competitions Vol 28 No 2 2015

1 respectively. This suggests the changes in coordinates, (1, 2) or

(2, 1), resulting due to a move of a knight (see Figure 6).
Observe also that one player can keep the knight on a certain
diagonal by replying to the reduction (1, 2) by (2, 1) and vice versa
(please see Figure 7). Let (m, n) be a position after a move in our
game, where m, n stand for the numbers of available rows and
columns respectively—after a player’s move. After the first move
of the game the position will be (1992,1993) or (1993,1992). Now
the winning strategy should be entering your mind.

Figure 8

Yes, the first player wins (Figure 8). After his first move, he leaves
the position (1992, 1993). If the second player reduces (R, C) by
(2, 1), the first player reduces (R, C) further by (1, 2) and vice
versa, thus always reducing (R, C) by (3, 3) after the first player’s
move. Since 1992 is divisible by 3, (1992, 1993) will eventually
become (0, 1) after the move of the first player.
Alexander Soifer
University of Colorado at Colorado Springs
1420 Austin Bluffs Parkway, Colorado Springs, CO 80918
USA
E-mail: asoifer@uccs.edu
http://www.uccs.edu/˜asoifer/

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 Mathematics Competitions Vol 28 No 2 2015

Regular polygons and polynomial curves

Robert Bosch

Robert Bosch is Coach and Coordi-

nator of Mathematics Competitions at
Archimedean Academy, USA. Involved
in problem solving since 2000, he won
a Bronze Medal in VIII Iberoamerican
Mathematical Olympiad for University
Students (2005).

In this note, we prove that only the square and the equilateral triangle
can have all their vertices on a cubic. From the study of the polynomial
p(x) = x8 + 3bx6 + 3b2 x4 + (b3 + b)x2 + (b2 + 1),
where b is a real parameter, we find that we can inscribe at most two
squares in a cubic, showing examples in both cases. Finally, we show
that every regular polygon can be inscribed in a polynomial curve.

1 Introduction.
√ √
The equilateral triangle T with vertices A(0, 0), B(− 3, 3) and C( 3, 3)
it is inscribed in the curve γ2 : y = x2 (Fig. 1).
Let Pn be a regular polygon with n sides and γn a curve with equation
y = q(x), where q is a polynomial of degree n with real coefficients. In
this paper, finding a regular polygon inscribed in a polynomial curve
means:
Find a pair (Pi , γj ) in the class C = {(Pn , γm )} of regular n-gons with
all their vertices on γm .

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 Mathematics Competitions Vol 28 No 2 2015

Figure 1

We will describe the set of regular polygons that can be inscribed in a

cubic.

2 γ3 : Cubic
Let y = x3 + ax2 + bx + c be a cubic curve, we are interested in finding
all regular polygons with all their vertices on the given curve. We shall
prove the following theorem.

Theorem 1. Let y = x3 + ax2 + bx + c be a cubic curve where a, b, c

are real numbers. The only regular polygons with all their vertices on
the given curve are the square and the equilateral triangle.

Proof. We can suppose, without loss of generality, that the regular

polygon has n sides and center at the O(0, 0). Since every regular
polygon is cyclic, let us study the intersection of the circumcircle and
the cubic, in other words, the system of equations

x2 + y 2 = R 2 , (1)
y = x3 + ax2 + bx + c. (2)

Squaring (2) and plugging in (1), we obtain a sixth-degree equation;

hence n ≤ 6. Let us analyze the different cases.

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 Mathematics Competitions Vol 28 No 2 2015

Triangle
√
Consider the equilateral √
triangle with√vertices A(0, 3), B(1, 0), C(−1, 0)
and the cubic y = x3 − 3x2 − x + 3 (Fig. 2).

Figure 2

Square
Assume that we only have one square with the given property. Denote
the curve y = x3 + ax2 + bx + c by γ and the vertices of the square by
A, B, C, D clockwisely beginning in the second quadrant. The symmetry
about O maps γ into the curve γ with equation y = −f (−x) = x3 −
ax2 + bx − c. Obviously, γ passes through A, B, C, D and, hence, has
four different intersection points with γ. Therefore 2ax2 + 2c = 0 has at
least four different solutions, which implies a = c = 0 or f (x) = x3 + bx.
In particular, γ passes by O and intersects all quadrants, so b < 0. Let γ 
be the curve obtained by rotating γ 90◦ around the origin. The curve γ 
has equation −x = f (y) and contains the points A, B, C, D, and O. The
intersection points (x, y) ∈ γ ∩ γ  are determined by −x = f (f (x)), and
hence the x-coordinates are roots of a polynomial p(x) = f (f (x)) + x
of degree 9. The number of times that two cubics intersect each other
is even. Since ABCD is the only square that satisfies the property on
γ ∩ γ  , the intersection points A, B, C, D have multiplicities 2. It follows
that
p(x) = x[(x − r)(x + r)(x − s)(x + s)]2 , (3)

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 Mathematics Competitions Vol 28 No 2 2015

where r and s are the x-coordinates of A and B respectively. On the

other hand, p(x) is defined by (x3 + bx)3 + b(x3 + bx) + x, with the same
coefficients of (3), so

3b =−2(r2 + s2 ),
3b2 =(r2 + s2 )2 + 2r2 s2 ,
b(b2 + 1) =−2r2 s2 (r2 + s2 ),
b2 + 1 =r 4 s4 .
√ √
Solving this system of equations yields b = − 8 and r2 + s2 = 18.
The point B(s, f (s)) is mapped to the point (−f (s), s) by the rotation
with center O and angle 90◦ . The image of B is A. We deduce that
f (s) = −r and f (r) = s. In fact, we can find the vertices of the square
  √ 
√ √ √
18 − 6 18 + 6
A − , ,
2 2
  √ 
√ √ √
18 + 6 18 − 6
B , ,
2 2
  √ 
√ √ √
18 − 6 18 + 6
C , − ,
2 2
  √ 
√ √ √
18 + 6 18 − 6
D − , − ,
2 2

√
and the equation of the cubic will be y = x3 − 8x (Fig. 3).

Pentagon

By the Pigeonhole Principle, there exist at least two vertices on the first
quadrant. Denote them by A, B (counterclockwise), the vertex C will
be on the second quadrant since the central angle of a regular pentagon
is 72◦ and 90◦ < ∠AOC = 144◦ < 180◦ . In such way D belongs to
the third quadrant and E to the fourth quadrant. Let xA , xB , xC ,
xD , xE be the x-coordinates of A, B, C, D, E respectively. The function

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 Mathematics Competitions Vol 28 No 2 2015

Figure 3

f (x) = x3 + ax2 + bx + c is continuous on R, in particular on the inter-

vals [xC , xD ], [xD , xB ], [xB , xE ], [xE , xA ]. Note that f (xC )f (xD ) < 0,
f (xD )f (xB ) < 0, f (xB )f (xE ) < 0, f (xE )f (xA ) < 0. Hence f (x) has
four different roots; but this contradicts the Fundamental Theorem of
Algebra.

In a similar fashion, we can arrive at the same conclusions in the case of

the hexagon.

3 Two squares inscribed in a cubic

Suppose there are two different inscribed squares. We fix one of these
squares and proceed as in the proof of Theorem 1 to obtain the cubic
equation y = x3 + bx, where b < 0 again. We conjecture that the
different squares must be close; this is because they produce changes of
monotonicity on the given curve. Considering again the rotation with
center O and angle 90◦ , the two resulting cubics will intersect each other
(apart from the origin) in 0, 4 or 8 points. In the first case, there are
no inscribed squares. In the second one, just one. In the last one we
have two inscribed squares. So a complete characterization of relation

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 Mathematics Competitions Vol 28 No 2 2015

square-cubic will be found by studying the polynomial

p(x) = x8 + 3bx6 + 3b2 x4 + (b3 + b)x2 + (b2 + 1) = 0.

Note that b ≥ 0 implies p(x) > 0. Hence there√ are no inscribed squares.
It just remains to consider b < 0 with b = − 8.

We have the following theorem.

Theorem 2. Consider the polynomial

p(x) = x8 + 3bx6 + 3b2 x4 + (b3 + b)x2 + (b2 + 1).

√
When b < − 8 we have two inscribed squares in the cubic y = x3 + bx.
The vertices are
   

 8b b 3
  √ b2 + √ −√ +4 
  
  b2 − 8 b2 − 8 b2 − 8 3b 3 
x1 = − − √ − , x1 + bx1  ,
 4 2 2 4 
 

   

 8b b3
  √ b2 + √ −√ +4 
  
  b2 − 8 b2 − 8 b2 − 8 3b 3 
x2 = − − − √ − , x2 + bx2  ,
 4 2 2 4 
 

   

 8b b3
  √ b2 + √ −√ +4 
  
  b2 − 8 b2 − 8 b2 − 8 3b 3 
x3 = − + √ − , x3 + bx3  ,
 4 2 2 4 
 

   

 8b b3
  √ b2 + √ −√ +4 
  
  b2 − 8 b2 − 8 b2 − 8 3b 3 
x4 = − − + √ − , x4 + bx4  ,
 4 2 2 4 
 

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 Mathematics Competitions Vol 28 No 2 2015

   

 8b b 3
 √ b2 + √ −√ +4 
  2 
  b − 8 b2 − 8 b2 − 8 3b 3 
x5 = − √ − , x5 + bx5  ,
 4 2 2 4 
 

   

 8b b3
 √ b2 + √ −√ +4 
  2 
  b − 8 b2 − 8 b2 − 8 3b 3 
x6 = − − √ − , x6 + bx6  ,
 4 2 2 4 
 

   

 8b b 3
 √ b2 + √ −√ +4 
  2 
  b − 8 b2 − 8 b2 − 8 3b 3 
x7 = + √ − , x7 + bx7  ,
 4 2 2 4 
 

   

 8b b3
 √ b2 + √ −√ +4 
  2 
  b − 8 b2 − 8 b2 − 8 3b 3 
x8 = − + √ − , x8 + bx8  .
 4 2 2 4 
 

in some order.
Proof. Using the substitution, y = x2 , the polynomial p(x) transforms
into q(y) = y 4 + 3by 3 + 3b2 y 2 + (b3 + b)y + (b2 + 1). We have that the
discriminant is
∆ = 4b6 − 60b4 + 192b2 + 256 = 4(b2 − 8)2 (b2 + 1) > 0.
Now we find
P = −3b2 < 0,
D = −(b2 − 8)(3b2 + 8) < 0.
Hence q(y) has four different real roots. Information concerning ∆, P, D
can be found in [1]. Now with the aid of WolframAlpha we find the four
roots.

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 Mathematics Competitions Vol 28 No 2 2015

√
In particular, when b = − 8 − 1 we have the following (Figure 4).

Figure 4

4 Regular polygons and polynomial curves

What regular polygons can be inscribed in the curve given by equation

y = am xm + am−1 xm−1 + · · · + a1 x + a0

if such a case is possible? (m ≥ 4).

We can fix the vertices of the n-gon; this is because we are interested
only in the existence of a certain polygon. Identify the vertices of the
regular n-gon with the n-th roots of unity and apply a central rotation
of angle α. So the problem has an interpolation flavor. If m = n the
result is a system of equations
    
2kπ 2kπ
p cos α + = sin α + , k = 0, ..., n − 1
n n
Using the function Interpolating Polynomial from Mathematica we
find that the answer to this question is the polynomial

Pn (x) = (Tn−1 (x) − cos(nα)x) csc(nα),

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 Mathematics Competitions Vol 28 No 2 2015

Figure 5

where Tn−1 (x) is the Chebyshev’s polynomial of the first kind. Let us see
the figures for n = 4, 8, 12 and certain values α (Fig. 5). We have proved
that every regular polygon can be inscribed in a polynomial curve. Note
that for the cubic we proved a little more; we get all possible inscribed
regular polygons.

5 Open Problem
Given the curve γ with equation

y = an xn + an−1 xn−1 + · · · + a1 x + a0 .

What regular polygons with m sides (m = n) could be inscribed in γ?

Acknowledgment. The author wishes to thank Professor Francisco Javier

Garcı́a Capitán for the figures and calculations in Mathematica.

References
[1] E. L. Rees, Graphical Discussion of the Roots of a Quartic Equation,
Amer. Math. Monthly 29 (1922) 51–55.

Robert Bosch
USA
email: bobbydrg@googlemail.com

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 Mathematics Competitions Vol 28 No 2 2015

Domain and range of functional equations

and inequalities

Pavel Calábek & Jaroslav Švrček

Pavel Calábek is a Senior Lecturer,

Department of Algebra and Geometry,
Palacký University of Olomouc. He
deals with a problematic of educating
mathematically gifted students. He is a
member of Czech Mathematical Olym-
piad committee and its problem com-
mittee. He has been deputy leader of
the Czech team in international com-
petitions (IMO, MEMO, etc.).

Jaroslav Švrček is a Senior Lecturer,

Department of Algebra and Geometry,
Palacký University of Olomouc. He is
Vice-Chair of the Czech Mathematical
Olympiad since 2000 and a member of
its problems committee. He works as
the editor of our journals Mathematics
Competitions and of the Czech journal
Matematika–Fyzika–Informatika. He
has been many times leader or deputy
leader of the Czech IMO and MEMO
teams. He is one of the founders and
organizers of the local international
mathematical competition DUEL.

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 Mathematics Competitions Vol 28 No 2 2015

This article contains the most important facts about using the funda-
mental properties of the domain and range in solutions of functional
equations and inequalities of the olympic type. We want to show here
the most significant role of both of these in the solutions of presented
problems (of both types—equations and inequalities), i.e. how important
the form of the domain is in the solutions of concrete problems. Fur-
ther we will present here some suitable methods for solving problems,
especially the substitution method and the use of symmetry.

Example 1 Find all functions f : 0; +∞) → R such that

f (xy) = f (x) + f (y)

holds for all non-negative x and y.

Solution. Let us assume that such a function exists. Since the given
equality must be fulfilled for all non-negative y, it holds also for y = 0
and so
f (0) = f (x · 0) = f (x) + f (0).
It follows f (x) = 0 for all non-negative real x. On the other hand we
see that such function satisfy the functional equation. Therefore this
equation has the unique solution f (x) = 0.

Example 1’ Find all functions f : (0; +∞) → R such that

f (xy) = f (x) + f (y)

holds for all non-negative numbers x and y.

Example 1’ differs from the previous Example 1 only in the domain,

which doesn’t include 0. But now we can easily check that every loga-
rithmic function f (x) = logc x for c > 0 and c = 1 is the solution. Only
such functions are the solutions in the class of continuous functions; in
the class of real functions there are a further rich set of noncontinuous
solutions (you can see the problematic of Cauchy’s functional equations).

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 Mathematics Competitions Vol 28 No 2 2015

From the method point of view we see, that in a solution of Example

1 we used deductive reasoning, so the check is a necessary part of the
solution.

Example 2 Find all functions f : R → R such that the equation

f 2 (x) = 1
holds for all reals x and y.
We can easily check that the solution of this problem is every real
function f with the range {−1, 1}. For example functions f (x) = 1,
f (x) = −1,
 
1 for x < 0 1 for rational x
f (x) = , f (x) =
−1 for x ≥ 0 −1 for irrational x

and many others.

Example 2’ Find all functions f : R → (0; +∞) such that the equation
f 2 (x) = 1
holds for all reals x and y.
Such an equation has unique solution f (x) = 1.
From the method point of view we note that the restriction of domain
(generally) extends the set of solutions and the restriction of range
restricts the set of solutions. Further we want in this example to present
a common mistake in solutions of functional equations, when solvers
claim, that the problem has only two solutions f (x) = 1 and f (x) = −1.

1 Substitutions
Now we describe the substitution method which is basic for many meth-
ods of solutions. We assume that there exists a solution of the given
functional equation or inequality. Now we specify variables and restrict
the set of possible solutions.

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 Mathematics Competitions Vol 28 No 2 2015

Example 3 Find all functions f : R → R such that

f (x)f (y) − f (xy) = x + y.

for all real x and y.

Solution. Initially, let us put x = y = 0. After an easy calculation we

obtain f (0)(f (0) − 1) = 0. It follows either f (0) = 0 or f (0) = 1.

Further let f (0) = 0. Substituting y = 0 yields

f (x)f (0) − f (x · 0) = x + 0.

This follows 0 = x for all real x. Therefore there does not exist any
solution for f (0) = 0.

Now let f (0) = 1. Similarly, we can substitute y = 0 and obtain

f (x)f (0) − f (x · 0) = x + 0,

or equivalently f (x) − 1 = x.

We can easily check that the function f (x) = x + 1 is the unique solution
of the given functional equation.

Example 4 Find all functions f : R → R such that

f (x + y) − 2f (x − y) + f (x) − 2f (y) = y − 2

for all real x and y.

Solution. Let x = 0 and y = 0. We obtain

−2f (0) = −2,

from which follows f (0) = 1.

Now let x = 0. Our functional equation is therefore in the form

f (y) − 2f (−y) + f (0) − 2f (y) = y − 2.

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 Mathematics Competitions Vol 28 No 2 2015

Simplifying with f (0) = 1 gives

f (y) + 2f (−y) = y − 3. (1)

Let t be an arbitrary real number. Since (1) is true for every real y, it
is true for y = t and y = −t.
f (t) + 2f (−t) = t − 3,
f (−t) + 2f (t) = −t − 3.

We can consider this two relations to be a system of two linear equations

with unknowns f (t) and f (−t). We solve it and obtain for all real t that
f (t) = t + 1.

Now we check that the function f (x) = x + 1 solves the functional

equation

f (x + y) − 2f (x − y) + f (x) − 2f (y) =
= (x + y + 1) − 2(x − y − 1) + (x + 1) − 2(y + 1) = y − 2.

Therefore f (x) = x + 1 is the unique solution of the given functional

equation.

2 Symmetry
In solving functional equations we often can use symmetry of some two
variables on some part of the functional equation. By this way we can
prove or assume that the unknown function is injective.

Example 4 Find all injective functions f : R → R such that

f (f (x) + y) = f (x + y) + 1.

for all real x and y.

Solution. We can see that the right side of the given functional equation
does not change if we interchange x and y (x is symmetric to y). So we
obtain
f (f (y) + x) = f (x + y) + 1 = f (f (x) + y).

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 Mathematics Competitions Vol 28 No 2 2015

Since f is the injective function then the arguments on the left and right
side are the same, it means f (y) + x = f (x) + y for all x and y.
Let us denote f (0) = c. Substituting y = 0 we obtain f (0)+x = f (x)+0,
so f (x) = x − c for every real x.
Now we check if the function f (x) = x − c is the solution of the given
functional equation. On the left side we have
f (f (x) + y) = f ((x + c) + y) = x + y + 2c,
while on the right side
f (x + y) + 1 = x + y + c + 1.
These terms are identical if and only if c = 1.
Since f (x) = x + 1 is an injective function it is the unique solution of
the given equation.

Example 5 Find all functions f : R → R such that

f (f (x) + f (y) + f (xy)) = x + f (y)
holds for all real x and y.
Solution. The left side of the given equality is a symmetrical function in
variables x and y so we have
y + f (x) = f (f (x) + f (y) + f (xy)) = x + f (y).
By the same way as in Example 4 we obtain f (x) = x + c, where c ∈ R.
But after easy checking we can see there is no solution of the given
problem.
Now we present some other easy problems without their solution.

Problem 1 Find all functions f : R → R such that

f 2 (x)f (y) = f (x − y)
holds for all real x and y.

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 Mathematics Competitions Vol 28 No 2 2015

Problem 2 Find all functions f : R → R such that

f (x) + y = f (f (x) + f (y)) − 1

holds for all real x and y.

Problem 3 Find all functions f : (0; +∞) → (0; +∞) such that

f (xf (y)) = f (xy) + x

holds for all positive x and y.

3 Functional inequalities
Now we apply the previous methods to functional inequalities.

Example 6 Find all functions f : R → R such that the inequality

f (x) + f (y) ≥ f (x)f (y) + 1

for all real x and y holds.

Solution. Let us put y = x. We obtain inequality

2f (x) ≥ f 2 (x) + 1.

We can rewrite this inequality to the equivalent form

(f (x) − 1)2 ≤ 0.

It follows if there is any solution of the given problem, then f (x) = 1.

We can easily check that the function f (x) = 1 is the unique solution of
the given problem.

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 Mathematics Competitions Vol 28 No 2 2015

Example 7 Find all functions f : R → R such that

2f (x2 ) + 2f (y) ≥ f (x)f (xy) + 4

holds for all real x and y.

Solution. We put x = y = 0 and obtain by rewriting

0 ≥ (f (0) − 2)2 ,

so f (0) = 2.

Now let y = 0. For all real x we have

2f (x2 ) + 2f (0) ≥ f (x)f (0) + 4,

and using f (0) = 2 it follows f (x2 ) ≥ f (x).

While the other way we put x = 0, we obtain for all real y

2f (0) + 2f (y) ≥ f 2 (0) + 4,

and using f (0) = 2 gives f (y) ≥ 2. So for all real x we obtain the
following estimate
f (x2 ) − 2 ≥ f (x) − 2 ≥ 0. (2)

Finally if we put x = y, by easy rewriting we obtain

(f (x) − 2)(f (x2 ) − 2) ≤ 0.

Using estimate (2) we can write

(f (x) − 2)2 ≤ (f (x) − 2)(f (x2 ) − 2) ≤ 0.

So f (x) = 2 for all real x. We can now easily check that this function is
the solution of the given problem.

Example 8 [Vietnam MO, 1991] Find all functions f : R → R such

that
1
f (xy) + f (xz) − 2f (x)f (yz) ≥
2
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 Mathematics Competitions Vol 28 No 2 2015

holds for all real x, y and z.

Solution. Initially we put x = y = z = 0 and we obtain

1
2f (0) − 2f 2 (0) ≥ .
2
This inequality we rewrite in the form (f (0) − 12 )2 ≤ 0. It follows

1
f (0) = .
2
Similarly by x = y = z = 1 we obtain
1
f (1) = .
2
Now let us put y = z = 1. For all real x we have
1
f (x) + f (x) − 2f (x)f (1) ≥ ,
2
or equivalently
1
f (x) ≥
. (3)
2
On the other hand, by y = z = 0 we obtain for all real x
1
2f (0) − 2f (x)f (0) ≥ .
2
It follows
1
f (x) ≤ . (4)
2

Joining (3) and (4) we see that for all real x necessarily f (x) = 12 . We
now easily check that this function satisfies the given problem.

Example 9 [8th MEMO, 2014] Find all functions f : R → R such that

xf (xy) + xyf (x) ≥ f (x2 )f (y) + x2 y

holds for all real x and y.

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 Mathematics Competitions Vol 28 No 2 2015

Solution. We put x = y = 1 and by easy manipulation obtain 0 ≥

(f (1) − 1)2 , so necessarily f (1) = 1. Now substituting y = 1 gives
2xf (x) ≥ f (x2 ) + x2 , which is equivalent to

2x(f (x) − x) ≥ f (x2 ) − x2 (5)

for all real x. For y = x we obtain by easy manipulation the inequality

0 ≥ (f (x2 ) − x2 )(f (x) − x). (6)

From (6) the substitution x = 0 follows f (0) = 0. If there exists real

x such that f (x2 ) − x2 > 0, then by (5) we have f (x) − x > 0. So
(f (x2 ) − x2 )(f (x) − x) > 0, which contradicts (6). For all x we have
f (x2 ) ≤ x2 . It follows that for all non-negative x the inequality f (x) ≤ x
holds. If furthermore there exists positive x such that f (x)−x < 0 holds,
then from (5) it follows f (x2 ) − x2 < 0, so (f (x2 ) − x2 )(f (x) − x) > 0.
This also contradicts (6). For all non-negative x we have

f (x) = x.

Now assume x < 0, y < 0. The given inequality gives us

x2 y + xyf (x) ≥ x2 f (y) + x2 y.

We obtain by rewriting and dividing by negative x2 y

f (x) f (y)
≤ .
x y
The inequality holds for all negative x and y. Interchanging x and y
gives
f (y) f (x)
≤ .
y x
The last two mentioned inequalities imply

f (y) f (x)
= .
y x

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 Mathematics Competitions Vol 28 No 2 2015

We then put y = −1 and denote −f (−1) = k ∈ R. Then for all negative

x the equality f (x) = kx holds, where k a real number. So, if there
exists any solution of the given problem then it is in the form

x for x ≥ 0,
f (x) =
kx for x < 0.

Now we check the function. Let us denote

L = xf (xy) + xyf (x), R = f (x2 )f (y) + x2 y.

a) Let x = 0 or y = 0. In both cases we have L = R = 0, so such

function satisfies the problem.
b) Let x > 0 and y > 0. Then P = 2x2 y and R = 2x2 y, so L ≥ R.
c) Now assume x > 0, y < 0. Then L = kx2 y + x2 y and R =
kx2 y + x2 y, so L ≥ R again for arbitrary real k.
d) Let x < 0, y > 0. Then L = 2kx2 y and R = 2x2 y and the
inequality L ≥ R is true if and only if k ≥ 1.
e) Finally let x < 0, y < 0. Then L = x2 y+kx2 y and R = kx2 y+x2 y,
so inequality L ≥ R holds for arbitrary real k.

We check that the function


x for x ≥ 0,
f (x) =
kx for x < 0.

is a solution of the given problem for arbitrary real k ≥ 1.

Now we state further unsolved problems.

Problem 4 Find all functions f : 1; +∞) → 1; +∞) such that

f (x) ≤ 2(x + 1) and xf (x + 1) = f 2 (x) − 1

hold for all real x ∈ 1; +∞).

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 Mathematics Competitions Vol 28 No 2 2015

Problem 5 Let function f : N → N satisfy for all positive integers n

the inequality f (n + 1) > f (f (n)). Prove that f (n) = n.

Pavel Calábek Jaroslav Švrček

Faculty of Science Faculty of Science
Palacký University Palacký University
Olomouc Olomouc
CZECH REPUBLIC CZECH REPUBLIC
pavel.calabek@upol.cz jaroslav.svrcek@upol.cz

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 Mathematics Competitions Vol 28 No 2 2015

The Powers of Two

Dennis Situ & Steven Xia

We consider some problems on expressing positive integers as sums of

powers of 2.

Problem 1. In how many ways can we express a positive integer n

as a sum of powers of 2, if distinct powers may not be used more than
once?

The number of ways is 1 for n = 1, namely, 1 itself. The number of

ways is 1 for n = 2, namely, 2. The number of ways is also 1 for n = 3,
namely, 2 + 1. Let the number of expressions for n be an . After some
experimentation, we notice a pattern. We have a1 = 1, a2 = 1, a3 = 1,
a4 = 1, a5 = 1, a6 = 1, a7 = 1, and so on. We can also throw in a0 = 1
for the empty sum. The pattern suggests that the general formula is
an = 1. We give a formal proof via two auxiliary results.

Lemma 1. a2k+1 = a2k .

Proof. Every expression for 2k + 1 must contain a 1. The exclusion of

this 1 yields an expression for 2k. On the other hand, every expression
for 2k contains no 1s. The inclusion of a 1 yields an expression for 2k +1.
This one-to-one correspondence establishes the desired result.

Lemma 2. a2k = ak .

Proof. Every expression for 2k contains no 1s. Dividing each term by

2 yields an expression for k. This process is clearly reversible, and the
one-to-one correspondence yields the desired result.

We now prove by mathematical induction that an = 1. Note that a0 = 1.

Suppose the result holds up to an−1 = 1. If n = 2k +1, then a2k+1 = a2k
by Lemma 1, and a2k = 1 since 2k ≤ n − 1. If n = 2k, then a2k = ak

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 Mathematics Competitions Vol 28 No 2 2015

by Lemma 2, and ak = 1 since k ≤ n − 1. This completes the inductive

argument.

Since the pattern for an is relatively simple, we seek an alternative

∞

approach to Problem 1. Let A(x) = an xn . Then A(x2 ) generates
n=0
the number of expressions with no 1s and xA(x2 ) generates the number
of those with one 1. It follows that

A(x) = (1 + x)A(x2 )
1 − x2
= A(x2 )
1−x
1 − x2 1 − x 4
= · A(x4 )
1 − x 1 − x2
= ···
1
= .
1−x
1
Now = 1 + x + x2 + x3 + · · · . Hence an = 1 for all n.
1−x

Problem 2. In how many ways can we express a positive integer n

as a sum of powers of 2, if distinct powers may not be used more than
twice?

Let the number of expressions for n be bn . The first 16 values are:

b1 =1 b2 =2 b3 =1 b4 =3
b5 =2 b6 =3 b7 =1 b8 =4
b9 =3 b10 =4 b11 =2 b12 =4
b13 =3 b14 =4 b15 =1 b16 =5

We can also throw in b0 = 1 for the empty sum.

The general pattern is unclear, but we make some observations. For

t ≥ 1, we have

b(2t − 1) = 1, (1)

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 Mathematics Competitions Vol 28 No 2 2015

b(2t ) = t + 1, (2)
t
b(2 + 1) = t, (3)
t−1
b(3(2 ) − 1) = 2. (4)

For a general approach, we first prove two auxiliary results.

Lemma 3. It holds that b2k+1 = bk .

Proof. The proof is exactly the same as that for Lemma 1.

Lemma 4. It holds that b2k = bk + bk−1 .

Proof. Every expression for 2k contains either no 1s or two 1s. In the

former case, dividing each term by 2 yields an expression for k. In the
latter case, dividing each term by 2 after the exclusion of the two 1s
yields an expresion for n − 1. These two processes are clearly reversible,
and the one-to-one correspondences yield the desired result.

A general formula for bn does not seem to be within reach. The method
of generating functions does not work here either. However, with Lemma
3 and Lemma 4, we can compute the value of bn for any positive integer
n. We can also prove that the observations are indeed correct.

We shall establish (1) and (2) by simultaneous induction. For t = 1, we

have b(2 − 1) = 1 and b(2) = 2. Suppose (1) and (2) hold for some t ≥ 1.
Then b(2t+1 − 1) = b(2(2t − 1) + 1) = b(2t − 1) = 1 by Lemma 3, and
b(2t+1 ) = b(2(2t )) = b(2t ) + b(2t − 1) = t + 1 by Lemma 4.

We can also establish (3) and (4) by mathematical induction. For t = 1,

we have b(2 + 1) = 1 and b(3(1) − 1) = 2. Suppose (3) and (4) hold for
some t ≥ 1. By Lemma 3, we have

b(2t+1 + 1) = b(2(2t ) + 1) = b(2t ) = t + 1

and

b(3(2t ) − 1) = b(2(3(2t−1 − 1) + 1) = b(3(2t−1 − 1) = 2.

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 Mathematics Competitions Vol 28 No 2 2015

Note also the blocks (1, 2, 1), (1, 3, 2, 3, 1), (1, 4, 3, 4, 2, 4, 3, 4, 1), . . . of
palindromes. They are centered at b(3(2t−1 ) − 1) and extend from
b(2t − 1) to b(2t+1 − 1) for t ≥ 1.

Problem 3. In how many ways can we express a positive integer n

as a sum of powers of 2, if distinct powers may not be used more than
thrice?

Let the number of expressions for n be cn . The first 7 values are 1, 2,

2, 3, 3, 4 and 4. We can also throw in c0 = 1 for the empty sum. The
pattern suggests that the general formula is c2n = c2n+1 = n + 1. A
slightly more compact way to express this result is cn =  n+2
2 

As before, we first prove two auxiliary results.

Lemma 5. It holds c2k+1 = c2k .

Proof. Every expression for 2k + 1 must contain at least one 1. The

exclusion of one 1 yields an expresion for 2k. On the other hand,
every expression for 2k can contain at most two 1s. The inclusion of
one 1 yields an expression for 2k + 1. This one-to-one correspondence
establishes the desired result.

Lemma 6. It holds c2k = ck + ck−1 .

Proof. The proof is exactly the same as that for Lemma 4.

We now prove by mathematical induction that c2n = c2n+1 = n+1. Note

that c0 = c1 = 1. Suppose the result holds up to c2n−2 = c2n−1 = n.
By the lemmas, we have c2n = c2n+1 = cn + cn−1 . If n = 2k, then
cn + cn−1 = c2k + c2k−1 = k + 1 + k = 2k + 1 = n + 1. If n = 2k + 1,
then cn + cn−1 = c2k+1 + c2k = k + 1 + k + 1 = 2k + 2 = n + 1. This
completes the inductive argument.
∞

Using the generating function approach, let C(x) = cn xn . Then
n=0
C(x2 ) generates the number of expressions with no 1s, xC(x2 ) generates

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 Mathematics Competitions Vol 28 No 2 2015

the number of those with one 1, x2 C(x2 ) generates the number of those
with two 1s and x3 f (x2 ) generates the number of those with three 1s.
It follows that

C(x) = (1 + x + x2 + x3 )C(x2 )
1 − x4
= C(x2 )
1−x
1 − x4 1 − x 8
= · C(x4 )
1 − x 1 − x2
1 − x4 1 − x8 1 − x16
= · · C(x8 )
1 − x 1 − x2 1 − x4
= ···
1
=
(1 − x)(1 − x2 )
α β γ
= + + .
1 − x (1 − x)2 1+x

Clearing fractions, we have 1 = α(1 − x)(1 + x) + β(1 + x) + γ(1 − x)2 .

Setting x = −1, 1 = 4γ so that γ = 14 . Setting x = 1, 1 = 2β so
that β = 12 . Setting x = 0, 1 = α + β + γ so that α = 14 . Now
1 2 3 1 2 3
1−x = 1 + x + x + x + · · · and 1+x = 1 − x + x − x + · · · . On the
other hand,
1
= (1 + x + x2 + x3 + · · · )2 = 1 + 2x + 3x2 + 4x3 + · · · .
(1 − x)2

Hence cn = 12 (n + 1) + 14 (1 + (−1)n ) =  n+2

2 .

Dennis Situ & Steven Xia

Old Scona Academic High School
Edmonton
CANADA

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 Mathematics Competitions Vol 28 No 2 2015

The 36th Tournament of Towns

Selected Problems

Andy Liu

1. K and L are points on the sides AB and BC of a square ABCD re-

spectively, such that KB = LC. Let P be the point of intersection
of AL and CK. Prove that DP and KL are perpendicular.
Solution by Charles Leytem.
Since triangles ABL and DAK are congruent, AL is perpendicular
to DK. Since triangles CBK and DCL are congruent, CK is
perpendicular to DL. Hence P is the orthocentre of triangle DKL,
so that DP is perpendicular to KL.
....A K B
..............................................................................................................................................................
... ...... .. ... .
....
... ....... ... ....
... ...... ... ........ ...
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.
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...... .
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.
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.
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...
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.
.....
.
. . . ... . .
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. L
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.
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...
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... . .
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....... ........................... ......
...........................................................................................................................................................................

D C
2. In the 40 tests Andrew had taken, he got 10 As, 10 Bs, 10 Cs
and 10 Ds. A score is said to be unexpected if this particular score
has appeared up to now fewer times than any of the other three
scores. Without knowing the order of these 40 scores, is it possible
to determine the number of unexpected ones?
Solution.
Consider the first A, the first B, the first C and the first D that
Andrew gets. The last one to come along must be unexpected,
and none of the other three can be unexpected. The same applies
to the second A, the second B, the second C and the second D

42
 Mathematics Competitions Vol 28 No 2 2015

that he gets, and so on. It follows that exactly 10 of the scores are
unexpected.
3. Peter writes down the sum of every subset of size 7 of a set of 15
distinct integers, and Betty writes down the sum of every subset
of size 8 of the same set. If they arrange their numbers in non-
decreasing order, can the two lists turn out to be identical?
Solution.
This is possible. Take the set

S = {−7, −6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6, 7}.

Then f (S) = 0, where f denotes the sum of all the elements in

the set. Let A = {a1 , a2 , . . . , a7 } be a subset of S. Let B =
{b1 , b2 , . . . , b7 } be such that bk = −ak for 1 ≤ k ≤ 7. Then
f (B) = −f (A). Now f (S − A) = f (S) − f (A) = −f (A) = f (B).
Similarly, f (S − B) = f (A). If A = B, then the two sums f (A)
and f (B) on Peter’s list correspond respectively to the two sums
f (S − B) and f (S − A) on Betty’s list. If A = B, then we must
have f (A) = 0 so that f (S − A) = 0 also. In this case, the sum
f (A) on Peter’s list corresponds to the sum f (S − A) on Betty’s
list. It follows that the two lists are identical.
4. At the beginning, there are some silver coins on a table. In each
move, we can either add a gold coin and record the number of
silver coins on a blackboard, or remove a silver coin and record the
number of gold coins on a whiteboard. At the end, only gold coins
remain on the table. Prove that the sum of the numbers on the
blackboard is equal to the sum of the numbers on the whiteboard.
Solution.
Suppose there are m silver coins at the beginning and n gold coins
at the end. In each of the m + n moves, we either remove a silver
coin or add a gold coin. Place the m + n coins in a row so that,
for 1 ≤ k ≤ m + n, the coin involved in the k-th move is in the
k-th position from the left. Replace each silver coin by a girl facing
left, and each gold coin by a boy facing right. Each girl counts the
number of boys in front of her, and each boy counts the number of
girls in front of him. For any girl and any boy, either both count
the other or neither does. Hence the total count by the girls must

43
 Mathematics Competitions Vol 28 No 2 2015

be equal to the total count by the boys. The numbers counted by

the girls are precisely those recorded on the blackboard and the
numbers counted by the boys are precisely those recorded on the
whiteboard. Hence the sum of the numbers on the blackboard is
equal to the sum of the numbers on the whiteboard.
5. On a circular road there are 25 equally spaced booths, each with a
patrolman numbered from 1 to 25 in some order. The patrolmen
switch booths by moving along the road, so that their numbers are
from 1 to 25 in clockwise order. If the total distance travelled by
the patrolmen is as low as possible, prove that one of them remains
in the same booth.
Solution by Po-Sheng Wu.
Consider the motion plan which accomplishes the desired result
in which the total distance covered by the patrolmen is minimum.
Suppose all of them move. Let m be the number of those who
move clockwise and n be the number of those who move counter-
clockwise. Then m = n since m + n = 25. By symmetry, we may
assume that m < n. Ask each patrolman who moves clockwise to
go one booth farther and each patrolman who moves counterclock-
wise to stop one booth earlier. Then the patrolmen’s numbers will
still be in order. However, the total distance they have covered
will be reduced by n − m times the distance between two booths.
This contradicts the minimality assumption on the original motion
plan.

6. In triangle ABC, ∠A = 90◦ . Two equal circles tangent to each

other are such that one is tangent to BC at M and to AB, and
the other is tangent to BC at N and to CA. Prove that the
midpoint of M N lies on the bisector of ∠A.
Solution by Po-Sheng Wu.
Let P be the centre circle tangent to BC at M and Q be the
centre of the circle tangent to BC at N . Let T be the point
of tangency of these two circles and let D be the midpoint of
M N . Drop perpendiculars from P to CA at E and from Q to
AB at F , intersecting each other at R. Then AERF is a square
whose side length is equal to the common radii of the two circles.
Hence ∠RAF = ∠ARF = 45◦ . Now DM P T and DN QT are also

44
 Mathematics Competitions Vol 28 No 2 2015

squares. Hence ∠P DQ = 90◦ = ∠QRP . Hence DP QR is a cyclic

quadrilateral and ∠DRQ = 12 ∠DT Q = 45◦ . It follows that A, R
and D are collinear, so that D lies on the bisector of ∠A.

A ....
..........
...... . ...
...... .. .....
F .
.........
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.....
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R .
. E ...
.
.....
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.
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.
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........ ........ ... .............. ....................... ................ ... ............. ...
.
.......................................................................................................................................................................................................................................
. . . . . . . .

B M D N C

7. Gregory writes down 100 numbers on a blackboard and calculates

their product. In each move, he increases each number by 1 and
calculates their product. What is the maximum number of moves
Gregory can make if the product after each move does not change?
Solution by Po-Sheng Wu.
Suppose a1 , a2 , . . . , a100 are real numbers for which there exists a
real number k such that (k+a1 )(k+a2 ) · · · (k+a100 ) = a1 a2 · · · a100 .
Treating this as an equation for k, there are at most 100 real roots,
one of which is 0. It follows that Gregory can make at most 99
moves. This maximum can be attained if Gregory starts with the
numbers from −99 to 0. The initial product is 0, and this value is
maintained for the next 99 moves, until he hits 100! on the 100th
move.

8. The incircle of triangle ABC is tangent to BC, CA and AB at

D, E and F respectively. It is given that AD, BE and CF are
concurrent at a point G, and that the circumcircles of triangles
GDE, GEF and GF D intersect the sides of ABC at six distinct
points other than D, E and F . Prove that these six points are
concyclic.
Solution by Po-Sheng Wu.
Let the circumcircle of triangle F GD intersect the line AB at U
and the line BC at P . Let the circumcircle of triangle DGE inter-
sect the line BC at Q and the line CA at R. Let the circumcircle

45
 Mathematics Competitions Vol 28 No 2 2015

of triangle EGF intersect the line CA at S and the line AB at T .

Since P DF U is cyclic, BP · BD = BU · BF . Since BD = BF ,
we have BP = BU . Since QEDR is cyclic, BQ · BD = BG · BE.
Since T F GE is cyclic, BG · BE = BF · BT . From BD = BF ,
we have BQ = BT . It follows that P QT U has a circumcircle.
Similarly, we can show that P RQU and P ST U have circumcircles
too. The three circles cannot be distinct as otherwise P Q, RS and
T U are their pairwise radical axes and will be concurrent at the
radical centre of the three circles. However, these three lines are
the sides of ABC, and cannot be concurrent. It follows that two
of these three circles must coincide. Then all three will coincide,
and P, Q. R, S, T and U all lie on this common circle.
S....................................................................... T
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9. Peter prepares a list of all possible words consisting of m letters

each of which is T , O, W or N , such that the numbers of T s and Os
is the same in each word. Betty prepares a list of words consisting
of 2m letters each of which is T or O, such that the numbers of T s
and Os is the same in each word. Whose list contains more words?
Solution by Po-Sheng Wu.
For each word Peter writes down, convert every T to T T , every O
to OO, every W to OT and every N to T O. Since Peter’s word
has an equal number of T s and Os, the new word has an equal
number of T T s and OOs. Regardless of the numbers of OT s and
T Os, the new word has an equal number of T s and Os overall, and

46
 Mathematics Competitions Vol 28 No 2 2015

is therefore on Betty’s list. By reversing this conversion process,

every word Betty’s writes down is on Peter’s list. It follows that
the two lists have equal length.

Andy Liu
University of Alberta
CANADA
email: acfliu@gmail.com

47
 Mathematics Competitions Vol 28 No 2 2015

The 56th International Mathematical

Olympiad,
Chiang Mai, Thailand, 2015

The 56th International Mathematical Olympiad (IMO) was held 4–16

July in Chiang Mai, Thailand.

This was the largest IMO in history with a record number of 577 high
school students from 104 countries participating. Of these, 52 were girls.

Each participating country may send a team of up to six students, a

Team Leader and a Deputy Team Leader. At the IMO the Team Leaders,
as an international collective, form what is called the Jury. This Jury
was chaired by Soontorn Oraintara.

The first major task facing the Jury is to set the two competition pa-
pers. During this period the Leaders and their observers are trusted to
keep all information about the contest problems completely confidential.
The local Problem Selection Committee had already shortlisted 29 prob-
lems from 155 problem proposals submitted by 53 of the participating
countries from around the world. During the Jury meetings one of the
shortlisted problems had to be discarded from consideration due to be-
ing too similar to material already in the public domain. Eventually, the
Jury finalised the exam questions and then made translations into all
the more than 50 languages required by the contestants. Unfortunately,
due to an accidental security breach, the second day’s paper had to be
changed on the night before that exam was to be taken. This probably
resulted in a harder than intended second day.

The six questions that ultimately appeared on the IMO contest are
described as follows.

1. A relatively easy two-part problem in combinatorial geometry pro-

posed by the Netherlands. It concerns finite sets of points in the
plane in which the perpendicular bisector of any pair of points in
such a set also contains another point of the set.

48
 Mathematics Competitions Vol 28 No 2 2015

2. A medium classical number theory problem proposed by Serbia.

3. A difficult classical geometry problem in which it is asked to prove

that a certain two circles are mutually tangent. It was proposed
by Ukraine.

4. A relatively easy classical geometry problem proposed by Greece.

5. A medium to difficult functional equation proposed by Albania.

6. A difficult problem in which one is asked to prove an inequality

about a sequence of integers. Although it does not seem so at first
sight, the problem is much more combinatorial than algebraic. It
was inspired by a notation used to describe juggling. The problem
was proposed by Australia.

These six questions were posed in two exam papers held on Friday 10 July
and Saturday 11 July. Each paper had three problems. The contestants
worked individually. They were allowed four and a half hours per paper
to write their attempted proofs. Each problem was scored out of a
maximum of seven points.

For many years now there has been an opening ceremony prior to the first
day of competition. HRH Crown Princess Sirindhorn presided over the
opening ceremony. Following the formal speeches there was the parade
of the teams and the 2015 IMO was declared open.

After the exams the Leaders and their Deputies spent about two days
assessing the work of the students from their own countries, guided by
marking schemes, which had been discussed earlier. A local team of
markers called Coordinators also assessed the papers. They too were
guided by the marking schemes but are allowed some flexibility if, for
example, a Leader brings something to their attention in a contestant’s
exam script that is not covered by the marking scheme. The Team
Leader and Coordinators have to agree on scores for each student of
the Leader’s country in order to finalise scores. Any disagreements that
cannot be resolved in this way are ultimately referred to the Jury.

The IMO paper turned out to be quite difficult. While the easier prob-
lems 1 and 4 were quite accessible, the other four problems 2, 3, 5 and 6

49
 Mathematics Competitions Vol 28 No 2 2015

were found to be the most difficult combination of medium and difficult

problems ever seen at the IMO. There were only around 30 complete
solutions to each of problems 2, 3 and 5. Problem 6 was very difficult,
averaging just 0.4 points. Only 11 students scored full marks on it.

The medal cuts were set at 26 for gold, 19 for silver and 14 for bronze.1
Consequently, there were 282 (=48.9%) medals awarded. The medal dis-
tributions2 were 39 (=6.8%) gold, 100 (=17.3%) silver and 143 (=24.8%)
bronze. These awards were presented at the closing ceremony. Of those
who did not get a medal, a further 126 contestants received an hon-
ourable mention for solving at least one question perfectly.

Alex Song of Canada was the sole contestant who achieved the most
excellent feat of a perfect score of 42. He now leads the IMO hall of
fame, being the most decorated contestant in IMO history. He is the
only person to have won five IMO gold medals.3 He was given a standing
ovation during the presentation of medals at the closing ceremony.

The 2015 IMO was organised by: The Institute for the Promotion of
Teaching Science and Technology; Chiang Mai University; The Mathe-
matical Association of Thailand under the Patronage of His Majesty the
King; and The Promotion of Academic Olympiad and Development of
Science Education Foundation.

Venues for future IMO’s have been secured up to 2019 as follows.

2017 Brazil
2018 Romania
2019 United Kingdom

The 2016 IMO is scheduled to be held 6–16 July in Hong Kong.

1 This was the lowest ever cut for gold, and the equal lowest ever cut for silver.

(This was indicative of the difficulty of the exam, not the standard of the contestants.)
2 The total number of medals must be approved by the Jury and should not

normally exceed half the total number of contestants. The numbers of gold, silver
and bronze medals must be approximately in the ratio 1:2:3.
3 In his six appearances at the IMO, Alex Song won a bronze medal in 2010, and

followed up with gold medals in 2011, 2012, 2013, 2014 and 2015.

50
 Mathematics Competitions Vol 28 No 2 2015

Much of the statistical information found in this report can also be found
at the official website of the IMO www.imo-official.org.

Angelo Di Pasquale
Department of Mathematics and Statistics
University of Melbourne
AUSTRALIA
email: pasqua@ms.unimelb.edu.au

51
 Mathematics Competitions Vol 28 No 2 2015

1 IMO Papers
First Day
Friday, July 10, 2015

Problem 1. We say that a finite set S of points in the plane is balanced

if, for any two different points A and B in S, there is a point C in S
such that AC = BC. We say that S is centre-free if for any three
different points A, B and C in S, there is no point P in S such that
P A = P B = P C.
a) Show that for all integers n ≥ 3, there exists a balanced set
consisting of n points.
b) Determine all integers n ≥ 3 for which there exists a balanced
centre-free set consisting of n points.

Problem 2. Determine all triples (a, b, c) of positive integers such that

each of the numbers
ab − c, bc − a, ca − b
is a power of 2.
(A power of 2 is an integer of the form 2n , where n is a non-negative
integer.)

Problem 3. Let ABC be an acute triangle with AB > AC. Let Γ be

its circumcircle, H its orthocentre, and F the foot of the altitude from
A. Let M be the midpoint of BC. Let Q be the point on Γ such that
∠HQA = 90◦ , and let K be the point on Γ such that ∠HKQ = 90◦ .
Assume that the points A, B, C, K and Q are all different, and lie on Γ
in this order.
Prove that the circumcircles of triangles KQH and F KM are tangent
to each other.

Language: English Time: 4 hours and 30 minutes

Each problem is worth 7 points

52
 Mathematics Competitions Vol 28 No 2 2015

Second Day
Saturday, July 11, 2015

Problem 4. Triangle ABC has circumcircle Ω and circumcentre O. A

circle Γ with centre A intersects the segment BC at points D and E,
such that B, D, E and C are all different and lie on line BC in this
order. Let F and G be the points of intersection of Γ and Ω, such that
A, F , B, C and G lie on Ω in this order. Let K be the second point of
intersection of the circumcircle of triangle BDF and the segment AB.
Let L be the second point of intersection of the circumcircle of triangle
CGE and the segment CA.
Suppose that the lines F K and GL are different and intersect at the
point X. Prove that X lies on the line AO.

Problem 5. Let R denote the set of real numbers. Determine all

functions f : R → R satisfying the equation
f (x + f (x + y)) + f (xy) = x + f (x + y) + yf (x)
for all real numbers x and y.

Problem 6. The sequence a1 , a2 , . . . of integers satisfies the following

conditions:
(i) 1 ≤ aj ≤ 2015 for all j ≥ 1;
(ii) k + ak =  + a for all 1 ≤ k < .
Prove that there exist two positive integers b and N such that
 
 n 
  
 (aj − b) ≤ 10072

j=m+1 
for all integers m and n satisfying n > m ≥ N .

Language: English Time: 4 hours and 30 minutes

Each problem is worth 7 points

53
 Mathematics Competitions Vol 28 No 2 2015

2 Results

Mark Distribution by Question

Mark Q1 Q2 Q3 Q4 Q5 Q6
0 93 256 408 91 153 521
1 89 151 122 36 255 11
2 5 77 12 61 34 15
3 21 27 1 18 90 6
4 72 8 3 11 8 3
5 12 13 0 1 4 3
6 20 14 1 8 3 7
7 265 31 30 351 30 11
Total 577 577 577 577 577 577
Mean 4.3 1.4 0.7 4.8 1.5 0.4

Some Country Totals Some Country Totals

Rank Country Total Rank Country Total
1 USA 185 17 Poland 117
2 China 181 18 Taiwan 115
3 South Korea 161 19 Mexico 114
4 North Korea 156 20 Hungary 113
5 Vietnam 151 20 Turkey 113
6 Australia 148 22 Brazil 109
7 Iran 145 22 Japan 109
8 Russia 141 22 United Kingdom 109
9 Canada 140 25 Kazakhstan 105
10 Singapore 139 26 Armenia 104
11 Ukraine 135 27 Germany 102
12 Thailand 134 28 Hong Kong 101
13 Romania 132 29 Bulgaria 100
14 France 120 29 Indonesia 100
15 Croatia 119 29 Italy 100
16 Peru 118 29 Serbia 100

54
 Mathematics Competitions Vol 28 No 2 2015

Distribution of Awards at the 2015 IMO

Country Total Gold Silver Bronze H.M.
Albania 37 0 0 0 3
Algeria 60 0 1 1 2
Argentina 70 0 0 1 4
Armenia 104 0 1 5 0
Australia 148 2 4 0 0
Austria 63 0 0 3 1
Azerbaijan 73 0 0 2 4
Bangladesh 97 0 1 4 1
Belarus 84 0 0 3 3
Belgium 67 0 1 0 3
Bolivia 5 0 0 0 0
Bosnia and Herzegovina 76 0 0 2 4
Botswana 1 0 0 0 0
Brazil 109 0 3 3 0
Bulgaria 100 0 2 1 2
Cambodia 24 0 0 0 2
Canada 140 2 0 4 0
Chile 12 0 0 0 1
China 181 4 2 0 0
Colombia 72 0 0 4 0
Costa Rica 53 0 0 2 2
Croatia 119 1 3 1 0
Cuba 15 0 0 1 0
Cyprus 58 0 1 0 2
Czech Republic 74 0 0 3 3
Denmark 52 0 0 2 1
Ecuador 27 0 0 0 2
El Salvador 14 0 0 0 0
Estonia 51 0 0 1 3
Finland 26 0 0 0 1
France 120 0 3 3 0
Georgia 80 0 1 3 1
Germany 102 0 2 3 0
Ghana 5 0 0 0 0
Greece 71 0 1 2 2
Hong Kong 101 0 2 3 1

55
 Mathematics Competitions Vol 28 No 2 2015

Distribution of Awards at the 2015 IMO

Country Total Gold Silver Bronze H.M.
Hungary 113 0 3 3 0
Iceland 41 0 0 0 3
India 86 0 1 2 3
Indonesia 100 0 2 4 0
Iran 145 3 2 1 0
Ireland 37 0 0 0 3
Israel 83 1 0 2 2
Italy 100 1 2 0 0
Japan 109 0 3 3 0
Kazakhstan 105 1 1 2 2
Kosovo 24 0 0 0 1
Kyrgyzstan 17 0 0 0 0
Latvia 36 0 0 0 3
Liechtenstein 18 0 0 1 0
Lithuania 54 0 0 1 1
Luxembourg 12 0 0 0 1
Macau 88 0 1 2 3
Macedonia (FYR) 45 0 0 1 1
Malaysia 66 0 0 3 1
Mexico 114 1 2 3 0
Moldova 85 0 1 2 3
Mongolia 74 0 0 2 4
Montenegro 19 0 0 1 0
Morocco 27 0 0 0 1
Netherlands 76 0 0 3 1
New Zealand 72 0 0 2 4
Nicaragua 26 0 0 0 3
Nigeria 22 0 0 0 2
North Korea 156 3 3 0 0
Norway 54 0 1 0 2
Pakistan 25 0 0 1 0
Panama 9 0 0 0 0
Paraguay 53 0 0 3 0
Peru 118 2 2 1 0
Philippines 87 0 2 2 1
Poland 117 1 1 4 0

56
 Mathematics Competitions Vol 28 No 2 2015

Distribution of Awards at the 2015 IMO

Country Total Gold Silver Bronze H.M.
Portugal 70 0 0 3 1
Puerto Rico 18 0 0 1 0
Romania 132 1 4 1 0
Russia 141 0 6 0 0
Saudi Arabia 81 0 1 3 2
Serbia 100 1 1 2 2
Singapore 139 1 4 1 0
Slovakia 97 0 2 3 0
Slovenia 46 0 0 1 1
South Africa 68 0 0 1 2
South Korea 161 3 1 2 0
Spain 47 0 0 1 2
Sri Lanka 51 0 0 0 4
Sweden 63 0 0 2 2
Switzerland 74 0 0 3 2
Syria 69 0 1 1 3
Taiwan 115 0 4 1 1
Tajikistan 57 0 1 1 2
Tanzania 0 0 0 0 0
Thailand 134 2 3 1 0
Trinidad and Tobago 26 0 1 0 0
Tunisia 41 0 0 1 2
Turkey 113 0 5 0 0
Turkmenistan 64 0 0 2 2
Uganda 6 0 0 0 0
Ukraine 135 2 3 1 0
United Kingdom 109 0 4 1 1
United States of America 185 5 1 0 0
Uruguay 16 0 0 0 1
Uzbekistan 64 0 0 3 2
Venezuela 13 0 0 0 1
Vietnam 151 2 3 1 0
Total (104 teams, 577 contestants) 39 100 143 126

N.B. Not all countries sent a full team of six students.

57
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