imo book 2012

2012

South Africa

and the 53rd

International

Mathematical

Olympiad

South Africa

and the 53rd

International

Mathematical

Olympiad

South Africa and the 53rdInternational Mathematical Olympiad

Published by:South African Mathematics Foundation

Private Bag X173Pretoria

0001

Tel: +27 (0) 12 392 9372

Fax: +27 (0) 12 392 9312

Website: www.samf.ac.za

SAMF, Pretoria 2012

ISBN: 1-920080-26-0

Editor: Dr JV van Zyl

SOUTH AFRICA ANDTHE 53rd INTERNATIONAL

MATHEMATICAL OLYMPIAD

Contents

1 Preface 1

2 Contributors 4

3 The Talent Search 2011 5

4 University of Stellenbosch Senior Camp: December 2011 7

5 The Monthly Problem Sets 12

6 Rhodes University Camp: April 2012 16

7 The IMO in Mar Del Plata, Argentina: July 2012 21

8 Solutions to the problems 24

8.1 Solutions to the Talent Search . . . . . . . . . . . . . . . . . . 24

8.2 Solutions to the University of Stellenbosch Senior Tests . . . . 31

8.3 Solutions to the Monthly Problems . . . . . . . . . . . . . . . 43

8.4 Solutions to the Rhodes University Tests . . . . . . . . . . . . 61

8.5 Solutions to the Pure Joy Tests . . . . . . . . . . . . . . . . . 76

8.6 Solutions to the IMO . . . . . . . . . . . . . . . . . . . . . . . 89

1 Preface

The program to identify, select and train South African teams for the In-ternational Mathematics Olympiad (IMO) began in 1991 when South Africawas invited to send an Observer to the IMO in Sweden. In July 2012 SouthAfrica participated in its 53rd IMO in Mar Del Plata, Argentina.

This book provides a comprehensive account of the 18-month Talent Search,camps and training program leading up to the 2012 International Mathemat-ical Olympiad. Due to political instability in North Africa, the Pan AfricanMathematical Olympiad did not take place in 2012.

Talent Search: January 2011 – November 2011

Young mathematicians all over the country were invited to take part in theTalent Search. A pamphlet containing five elementary problems was circu-lated to all schools with the entry form of the South African MathematicsOlympiad. Students participated by mailing their solutions to these problemsto the Talent Search office and were enrolled into a correspondence program.The program works with students receiving problem sets to solve at theirown pace. Their solutions are graded and they receive a copy of the modelanswers with the next set of problems.

No student is politely turned away: the only requirement for staying in theprogram is enthusiasm. IMO and PAMO participation is the climax of theprogram and the general motivating factor.

Stellenbosch Camp: December 2011

Around forty students nationwide had emerged as front-runners based ontheir performance in the South African Mathematics Olympiad and TalentSearch program and were invited to participate in a five-day training campat the University of Stellenbosch. The program for each day consisted of amorning test, followed by lectures on selected topics, problem-solving sessionsand friendly team competitions in the evenings. IMO and PAMO veterans,able to relate to the young people, acted as team leaders and coaches.

1

Monthly Problem Sets: January 2012 – June 2012

The Stellenbosch campers continued their training in the same format asthe talent search but with more advanced monthly problems sets. Studentswho performed well in these extended problem sets were invited to anothertraining camp in April.

Rhodes Camp: April 2012

The top sixteen performing students were invited to a final selection andtraining camp at Rhodes University. The camp format was similar to that ofthe Stellenbosch Camp but with more advanced lecturers and harder prob-lems. The morning tests were set up in a format similar to the one in theIMO/PAMO, with 4 questions in four and a half hours to get the studentsused to writing these types of papers and concentrating for long periods oftime. As a solution writing exercise the students were also asked to performdemos. They are given problems with solutions to present in pairs to theremaining students. The IMO team of six and PAMO team of four werechosen based on these camp results.

Pure Joy Camp: July 2012

Directly before the Olympiad, the IMO squad met for a final preparationcamp at the Pure Joy Resort outside Pretoria. The camp consisted of morn-ing tests, lectures and demo sessions, but focusing more on IMO-standardproblem solving to prepare the students for the Olympiad.

The Book

The aim of this book is to provide a source of problems to challenge andtrain contenders for future IMO and PAMO teams. The best way to developthe skills of problem solving is to solve problems. Many of the problemsin this book need perseverance and perseverance takes time. The readershould take heart in the fact that his time is well spent. Only after a seriousattempt has been made on a problem should the solution be consulted andthen only briefly to get a hint at the method, before going back and solvingthe problem independently. The solutions presented in this book are often

2

the most direct and elegant solutions, so a lot can be learned from them.Bythe nature of training program the problems are also well graded in order ofdifficulty. This provides a good basis for developing one’s training.

Thanks

Harmony

South Africa’s participation in the IMO, as well as the training camps, wasmade possible by the sponsorship of the Harmony Gold Mining Company.

SAMF

The South African Mathematics Foundation is the organising body for SouthAfrica’s participation in the IMO and PAMO, the South African MathsOlympiad, Talent Search and the training camps.

Camp Hosts

The University of Stellenbosch and Rhodes University provided lecture rooms,computer and photocopying services, as well as accommodation.

The IMO community

Each year at the IMO leaders and deputies exchange their national compe-tition papers and other materials. Many problems from those sources havefound their way into these pages.

Contributors

South Africa’s IMO program is now run almost entirely by IMO veterans.They have provided valuable resources in setting papers, marking solutions,organising and running camps.

JV van Zyl, June 2013

3

2 Contributors

Problems and SolutionsMr Dirk Basson, University of StellenboschMr Phil Labuschagne, University of Cape TownMr Maciej Stankiewicz, University of Cape TownDr JV van Zyl, Rhodes UniversityProf. Stephan Wagner, University of Stellenbosch

TypesettingMr Dirk Basson, University of StellenboschMr Phil Labuschagne, University of Cape TownMr Maciej Stankiewicz, University of Cape TownDr JV van Zyl, Rhodes University

ProofreadingMr Dirk Basson, University of StellenboschMr Phil Labuschagne, University of Cape TownMr Maciej Stankiewicz, University of Cape TownDr JV van Zyl, Rhodes University

This book was typeset using the LATEX 2ε mathematical typesetting pack-age, with amsmath, amsfonts and graphicx extension packages. The dia-grams were created with Dirk Laurie’s NAPOLEON package, which producesMETAPOST figures.

4

3 The Talent Search 2011

Talent Search 2011: Senior round 1

1.1. Divide a regular hexagon into exactly 8 identical shapes.

1.2. In a convex quadrilateral ABCD the sides AB, BC, CD are equal andAC = BD = AD. Find the angles of ABCD.

1.3. Solve the system of equations:

xy = x+ y; x2 + y2 = 1

1.4. Prove that the equation x5 + y3 = z2 has infinitely many solutions inthe natural numbers.

1.5. Prove that(x3 + x2 + 3)2 > 4x3(x− 1)2

for all real x.

1.6. There are 10 numbers written around the circumference of a circle.Some of them are positive and others are negative. During one movewe are allowed to change the sign of the numbers in three successivepositions, each of the three numbers changes sign. Prove that we canmake all the numbers positive after a finite number of moves.

5


2.1. Can you enumerate the edges of a tetrahedron with numbers from 1 to6 so that all four triangular faces have the same sum, where the sumof a face is calculated by adding the numbers on its edges?

2.2. Are there infinitely many integers whose square ends in three 4s, i.e.. . . 444?

2.3. There are 10 fencers taking part in a tournament. Each fencer duelswith each other exactly once. Is it possible that such a moment existswhen no two fencers have participated in the same number of duels?

2.4. If x2 + y2 = 1 and x, y > 0, prove that x3 + y3 >√

2xy.

2.5. B1 and C1 are marked on the bisector of angle A in triangle ABC sothat BB1 ⊥ AB, CC1 ⊥ AC. Let M be the midpoint of B1C1. Provethat MB = MC.

2.6. B1 and C1 are marked on the bisector of angle A in triangle ABC sothat BB1 ⊥ AB, CC1 ⊥ AC. Let M be the midpoint of B1C1. Provethat MB = MC.

6

4 University of Stellenbosch Senior Camp: De-

cember 2011

University of Stellenbosch Senior Camp December 2011: Test 1Time: 4 hours

1. An equilateral triangle and a regular hexagon have equal perimeters.What is the ratio of the area of the triangle to the area of the hexagon?

2. Solve k(k + 5)2 + 12 = n3 for positive integers k and n.

3. Prove that for all positive real numbers x and y:

x4 + y3 + x2 + y + 1 > 9

2xy.

When does equality hold?

4. The cells of a 10 × 10 square are susceptible to infection. In a unit oftime, the cells with 2 or more infected neighbours (having a commonside) become infected. Is it possible to start the infection in

(a) nine cells,

(b) ten cells,

in such a way that the infection spreads to the whole square?

5. Given a scalene triangle ABC with |AB| + |CA| = 2|BC|, show thatthe line joining the incenter and the centroid of the triangle is parallelto BC.

7


1. If x1, x2 and x3 are the roots of x3 − ax2 + bx− c = 0, prove that

(x1 − x2)2 + (x2 − x3)

2 + (x3 − x1)2 = 2a2 − 6b.

2. A little boy Mak has drawn a rectangle of size 20× 15, and divided itby straight lines into unit squares. How many squares (of any size) arethere in Mak’s picture?

3. For each positive integer a we consider the sequence (an) with a0 = aand an = an−1+40n! for n > 0. Prove that every such sequence containsinfinitely many numbers that are divisible by 2009.

4. Let a be a positive integer and a > 1. Prove that, for every positiveinteger n, the number

n(2n+ 1)(3n+ 1) . . . (an+ 1)

is divisible by all prime numbers smaller than a.

5. In triangle XY Z, L is a variable point on a fixed line passing throughX. LZ meets XY at P and LY meets XZ at Q. Show that PQ passesthrough a fixed point on Y Z.

8


1. A sequence (an) is defined by

a1 = 1, an = 3an−1 + 2n−1, for n > 2.

Find a formula for the general term an in terms of n.

2. Find all pairs (m,n) of positive integers for which 6m + 2n + 2 is aperfect square.

3. Let CH be the altitude of an acute angled triangle ABC, and let Obe the centre of its circumcircle. If T is the foot of the perpendiculardrawn from vertex C to the line AO, prove that the line TH bisectsthe side BC.

4. Represent the number 989 · 1001 · 1007 + 320 as the product of threeprimes.

5. On a 5 × 5 board, two players alternately mark numbers on emptycells. The first player always marks 1’s, the second 0’s. One number ismarked per turn, until the board is filled. For each of the nine 3 × 3squares the sum of the nine numbers on its cells is computed. Denoteby A the maximum of these sums. How large can the first player makeA, regardless of the responses of the second player?

9


1. Let a1, a2, ..., an, b1, b2, ..., bn be the numbers 1, 2, ..., 2n in someorder. Suppose that a1 < a2 < · · · < an and that b1 > b2 > · · · > bn.Prove that

n∑i=1

|ai − bi| = n2.

2. Let a1, a2, ..., am and b1, b2, ..., bn be two sets of reals with the samesum. Prove that you can find an m×n array so that the row sums area1, a2, ..., am and the column sums are b1, b2, ..., bn.

3. Given a regular hexagon, a point inside is chosen and lines to eachvertex of the hexagon are drawn, dividing it into six triangles. If thetriangles are alternately shaded grey and white, show that the area ofthe grey triangles is the same as the area of the white triangles.

4. Find all triangular numbers which are also perfect squares.

5. Let ω be a circle, and let A and B be points on a line not intersect-ing ω. Given a point X0 on ω, define a sequences X0, X1, X2, . . . andY0, Y1, Y2, . . . as follows: Yn is the second intersection of the line AXn

with ω and Xn+1 is the second intersection of the line BYn with ω.Prove that if for some integer k, it happens that Xk = X0 for somechoice of X0 then Xk = X0 for any choice of X0.

10


1. Find all functions f : N → N satisfying f(n + 1)f(n + 2) = f(n)2 forall n ∈ N.

2. Consider all the subsets of 1, 2, ..., N that do not contain two consec-utive numbers. For each subset, calculate the product of the membersof the set. What is the sum of the squares of these products?

Note: the product of the elements of an empty set is one.

3. From a point P , draw tangents PB and PT to a circle. Let A be thepoint on the circle such that AB is a diameter, and let H be the footof the perpendicular from T onto AB. Prove that AP bisects TH.

4. Let a, b, c, d > 0. Find all possible values of the sum

S =a

d+ a+ b+

b

a+ b+ c+

c

b+ c+ d+

d

c+ d+ a.

5. A triangular grid is obtained by tiling an equilateral triangle of sidelength n into n2 equilateral triangles of side length 1. Determine thenumber of parallelograms bounded by the line segments of the grid.

6. If n ∈ N and 3n+ 1 and 4n+ 1 are perfect squares, show that 56|n.

11

5 The Monthly Problem Sets

1. Two evenly matched teams are engaged in a series of games, whichends as soon as a team wins four games. Is it more likely for the seriesto end in exactly six games or in exactly seven games?

2. Let p be an odd prime number and let (a1, a2, . . . , ap) and (b1, b2, . . . , bp)be arbitrary arrangements of the p-tuple (0, 1, . . . , p − 1). For eachi, let ci be the non-negative remainder when the product aibi is di-vided by p. Show that (c1, c2, . . . , cp) cannot be a rearrangement of(0, 1, . . . , p− 1).

3. Let ABCD be a square with circumcircle Γ. Let M be on minor arcCD of Γ. Let BD and AM intersect at P , CD and AM intersect atR, BM and AC intersect at Q and BM and DC intersect at S. Showthat PS ⊥ QR.

4. Dylan has a list of all 25-digit numbers consisting of the digits 1, 2, 3and 4 such that there are an equal number of 1s and 2s. Robert hasa list of all 50-digit numbers consisting of 25 digits 1 and 25 digits 2.Show that the number of numbers on each list is the same.

5. Let PQ be the diameter of semicircle H. Circle ω is internally tangentto H and tangent to PQ at C. Let A be a point on H and B a pointon PQ such that AB is perpendicular to PQ and tangent to ω. Provethat AC bisects ∠PAB.

6. Let the sequence (un) be defined recursively by

u0 = 0, u1 = 1, un = 2011un−1 − un−2 for n > 2.

Find all the values of n for which un is prime.

7. Prove that the positive integer n is the product of exactly two primesthat differ by 2 if and only if

φ(n)σ(n) = (n− 3)(n+ 1).

(Recall that φ(n) equals the number of positive integers less than orequal to n, relatively prime to n, and σ(n) equals the sum of the positivedivisors of n, including 1 and n.)

12

8. An altitude from A to the triangle ABC intersects the side BC in D.A circle touches BC in D, intersects AB in M and N and intersectsAC in P and Q. Prove that

AM + AN

AC=AP + AQ

AB.

9. Let f : Q+ → R+ be a function that satisfies

f(x+ y)− f(x− y) = 4√f(x)f(y)

for all 0 < y < x.

(a) Prove that f(2x) = 4f(x) for all x ∈ Q+.

(b) Find all such functions.

10. The real-valued function f satisfies

f(tan 2x) = tan4 x+ cot4 x

for all real x. Prove that, for all real x,

f(sinx) + f(cosx) > 196.

11. Each positive integer is coloured red or blue. A function f from the setof positive integers to itself has the following two properties:

(a) if x 6 y, then f(x) 6 f(y); and

(b) if x, y and z are (not necessarily distinct) positive integers of thesame colour and x+ y = z, then f(x) + f(y) = f(z).

Prove that there exists a positive number a such that f(x) 6 ax for allpositive integers x.

12. Show that for every natural number n, there exists a natural numberk such that the number kn starts with the digits 2011.

13. Let P be a point inside triangle ABC. Construct the points P1, P2, P3

such that PP1 ⊥ BC, PP2 ⊥ CA, PP3 ⊥ AB and BP3 = BP1,CP2 = CP1. Prove that AP3 = AP2.

13

14. Let A, B, C be three distinct points in the plane for which AB = AC.Describe the locus of the points P for which ∠APB = ∠APC.

15. Prove that the following equation holds for all non-negative integers n:

b√n+√n+ 1 +

√n+ 2c = b

√9n+ 8c.

16. The acute-angled triangle ABC has circumcentre O and orthocentre H.The perpendicular bisector of BH meets AB at Q, the perpendicularbisector of CH meets AC at P and these two bisectors meet at R.

(a) Prove that the circumcircles of AOP , POR, ROQ and QOA haveequal radii. When are they equal in radius to the circumcircle oftriangle ABC?

(b) Prove that if PQ passes through O, then it also passes throughH. What can be said about ∠BAC in this case?

17. The diameter of a subset of the plane is the maximum distance betweenany two of its points. For example, the diameter of the unit square isequal to

√2.

(a) Prove that the unit square can be covered by three sets of diameter

not exceeding√

658

.

(b) Prove that the unit square can not be covered by three sets of

diameter less than√

658

.

18. Given a non-isosceles triangle ABC, let D, E and F denote the mid-points of the sides BC, CA and AB respectively. The circumcircle oftriangle BCF and the line BE meet again at P , and the circumcircleof ABE and the line AD meet again in Q. Finally, the lines DP andFQ meet at R. Prove that the centroid G of the triangle ABC lies onthe circle PQR.

19. Find all triples (p, q, r) of prime numbers which satisfy

(p+ 1)(q + 2)(r + 3) = 4pqr.

20. A sequence (an)∞n=1 of natural numbers has the property that for anyn > 1, an+1 = an + bn, where bn is the number having the same digitsas an, but in the reverse order (unlike an, the number bn may start withone or more zeroes in the decimal notation). For instance, if a1 = 170we have a2 = 170 + 71 = 241, a3 = 241 + 142 = 383, and so on.

14

(a) Can the number a6 be prime?

(b) Can the number a7 be prime?

21. Show that there are infinitely many positive integers n such that n2 +1has two positive divisors whose difference is n.

22. Given a finite number of boys and girls, a sociable set of boys is a set ofboys such that every girl knows at least one boy in that set. A sociableset of girls is a set of girls such that every boy knows at least one girlin that set. Prove that the number of sociable sets of boys and thenumber of sociable sets of girls have the same parity. (Acquaintance isassumed to be mutual.)

23. The set T consists of 66 points in the plane, and set P consist of 16lines in the plane. We say that a point A ∈ T and a line ` ∈ P form anincident pair if A ∈ `. Show that the number of incident pairs cannotexceed 159, and that there is such a configuration with exactly 159incident pairs.

15

6 Rhodes University Camp: April 2012

Rhodes University Camp April 2012: Test 1Time: 41

2hours

1. Let a and b be two integers with a > b. If ab−1 and a+b are relativelyprime, and ab+ 1 and a− b are relatively prime, prove that

(ab+ 1)2 + (a− b)2

is not a perfect square.

2. A school has n students and there are some extra classes provided forthem so that each student can participate in any number of them. Weknow that there are at least two participants in any class. We alsoknow that if two different classes have two common students, then thenumbers of their participants are different. Prove that the total numberof classes is not greater than (n− 1)2.

3. Prove that for all positive numbers x, y and z,

x+ 1

y + 1+y + 1

z + 1+z + 1

x+ 16 x

y+y

z+z

x.

4. In triangle ABC, ∠A = 60. Let E and F be points on the extensionsof AB and AC such that BE = CF = BC. The circumcircle of ACEintersects EF in K (different from E). Prove that K lies on the bisectorof ∠BAC.

16


2hours

1. Let Γ1, with diameter PQ, and Γ2, with diameter QR, be two circlesexternally tangent at Q. Two points A and B, distinct from P andQ, are chosen on Γ1. Let AQ and BQ intersect Γ2 again at C and D,respectively. Let PA and RD intersect in U and PB and RC intersectin V . Prove that there is a point T in the plane such that the line UVpasses through T for all possible choices of A and B.

2. Show that for every natural number n the product(4− 2

1

)(4− 2

2

)(4− 2

3

). . .

(4− 2

n

)is an integer.

3. Let f : R→ R be such that for all x, y ∈ R,

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

Prove that f(x+ y) = f(x) + f(y) for all x, y ∈ R.

4. Suppose that n points are given on a circle and nk + 1 of the chordsbetween these points are drawn, where 2k + 1 < n. Prove that it ispossible to select k+1 of the chords such that no two of them intersect.

17


2hours

1. Does there exist a natural number N which is a power of 2 such thatthe digits of N can be permuted to form a power of 2 different fromN?

2. Let n be a positive integer. A train stops at 2n stations, including thefirst and last ones, numbered in order from the first to the 2nth. Itis known that on a certain car, for each pair of integers i, j such that1 6 i < j 6 2n, exactly one seat has been reserved for the trip fromthe ith station to the jth. Find the minimum number of seats that mustbe available in that car.

3. Find all functions f : R+ → R+ such that for all x, y > 0,

f(yf(x))(x+ y) = x2(f(x) + f(y)).

4. Let O be the intersection point of the diagonals of the convex quadri-lateral ABCD, with AO = OC. Points P and Q are marked on thesegments AO and CO, respectively, such that PO = OQ. Let N bethe intersection of AB and DP , and K be the intersection of CD andBQ.

Prove that the points N , O and K are collinear.

18


2hours

1. Prove that if x 6 y 6 z are real numbers satisfying xy + yz + zx = 1,then xz < 1

2. Is it possible to replace 1

2with a smaller number?

2. Let ABC be a triangle with orthocentre H and let P be a point on itscircumcircle. The line through A parallel to BP meets CH at Q andthe line through A parallel to CP meets BH at R. Prove that QR isparallel to AP .

3. In the plane there are 2012 congruent triangles of area 1 which all havethe same orientation. Each of these triangles contains the centroids ofall the other triangles.

(a) Show that the intersection of the 2012 triangles is a triangle similarto them with area at least 1

9.

(b) Show that the area of the union of these triangles is less than 229

.

4. Prove that for every natural number n, there exists a polynomial p(x)with integer coefficients such that p(1), p(2), . . . , p(n) are all distinctpowers of 2.

19


2hours

1. Let a1, a2, . . . , an be real numbers such that

am + am+1 + . . . an > m+ (m+ 1) + · · ·+ n

for every m = 1, 2, . . . , n. Prove that

a21 + a2

2 + · · ·+ a2n >

n(n+ 1)(2n+ 1)

6.

2. Let circles Γ1 and Γ2 intersect at D and P . The common tangent ofthe two circles closest to the point D touches Γ1 at A and Γ2 at B. Theline AD intersects Γ2 for the second time in C. Let M be the middleof the line segment BC. Prove that ∠DPM = ∠BDC.

3. A poker set contains three chips of each of 2n different colours for acomplete set of 6n chips. Let pn be the number of ways in which thechips can be partitioned into two piles of 3n chips each in such a waythat no pile contains three chips of the same colour. Prove that pn isodd if and only if n is a power of 2.

4. Find all primes p and q such that

2qp − pq = 7.

20

inputpta/master.tex

7 The IMO in Mar Del Plata, Argentina: July 2012

IMO in Mar Del Plata, Argentina 2012: Day 1Time: 4.5 hours

1. Given triangle ABC the point J is the centre of the excircle oppositethe vertex A. This excircle is tangent to the side BC at M , and tothe lines AB and AC at K and L, respectively. The lines LM andBJ meet at F , and the lines KM and CJ meet at G. Let S be thepoint of intersection of the lines AF and BC, and let T be the pointof intersection of the lines AG and BC.

Prove that M is the midpoint of ST .

(The excircle of ABC opposite the vertex A is the circle that is tangentto the line segment BC, to the ray AB beyond B, and to the ray ACbeyond C.)

2. Let n > 3 be an integer, and let a2, a3, . . . , an be positive real numberssuch that a2a3 . . . an = 1. Prove that

(1 + a2)2(1 + a3)

3 . . . (1 + an)n > nn.

3. The liars guessing game is a game played between two players A andB. The rules of the game depend on two positive integers k and nwhich are known to both players.

At the start of the game A chooses integers x and N with 1 6 x 6 N .Player A keeps x secret, and truthfully tells N to player B. Player Bnow tries to obtain information about x by asking player A questionsas follows: each question consists of B specifying an arbitrary set Sof positive integers (possibly one specified in some previous question),and asking A whether x belongs to S. Player B may ask as many suchquestions as he wishes. After each question, player A must immediatelyanswer it with yes or no, but is allowed to lie as many times as shewants; the only restriction is that, among any k+1 consecutive answers,at least one answer must be truthful.

21

After B has asked as many questions as he wants, he must specify aset X of at most n positive integers. If x belongs to X, then B wins;otherwise, he loses. Prove that:

(a) If n > 2k, then B can guarantee a win.

(b) For all sufficiently large k, there exists an integer n > 1.99k suchthat B cannot guarantee a win.

22

IMO in Mar Del Plata, Argentina 2012: Day 2Time: 4.5 hours

4. Find all functions f : Z → Z such that, for all integers a, b and csatisfying a+ b+ c = 0, the following equality holds:

f(a)2 + f(b)2 + f(c)2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).

5. Let ABC be a triangle with ∠BCA = 90, and let D be the foot ofthe altitude from C. Let X be a point in the interior of the segmentCD. Let K be the point on the segment AX such that BK = BC.Similarly, let L be the point on the segment BX such that AL = AC.Let M be the point of intersection of AL and BK.

Show that MK = ML.

6. Find all positive integers n for which there exist non-negative integersa1, a2, . . . , an such that

1

2a1+

1

2a2+ . . . 1

2an=

1

3a1+

2

3a2+ · · ·+ n

2an= 1.

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8 Solutions to the problems

8.1 Solutions to the Talent Search


1.1. The following image is a possible dissection of a hexagon into 8 identi-cal shapes:

1.2. Notice that 4ABC ≡ 4BCD (s, s, s) and 4ABD ≡ 4DCA (s, s, s).Let α = ∠CBD = ∠BCA = ∠CAB = ∠CDBand β = ∠ACD = ∠ADC = ∠BAD = ∠ABD.

B C

A D

It can then be seen that 3α + β = 180 from triangle ABC and4β + 2α = 360 from quadrilateral ABCD. Solving these two simulta-neous equations gives α = 36 and β = 72. Hence ∠A = 72,∠B =108,∠C = 108,∠D = 72.

1.3. We are trying to solve the equations

xy = x+ y (1)

1 = x2 + y2. (2)

24

Squaring both sides of (1) and using (2) to simplify gives

x2y2 = x2 + 2xy + y2 = 1 + 2xy ⇔ (xy)2 − 2(xy)− 1 = 0.

This gives xy = 1±√

2 or y = 1±√

2x

.

Notice that from equation (1), we can write y = xx−1

.Combining this with the previous result gives

x2 − x(1±√

2) + 1±√

2 = 0.

If x2 − x(1 +√

2) + 1 +√

2 = 0, then

x =(1 +

√2)±

√(1 +

√2)2 − 4(1 +

√2)

2

=(1 +

√2)±

√−1− 2

√2

2.

which is non-real.

If x2 − x(1−√

2) + 1−√

2 = 0, then

x =(1−

√2)±

√(1−

√2)2 − 4(1−

√2)

2

=(1−

√2)±

√2√

2− 1

2.

This presents a viable solution. Notice that if (x, y) is a solution, then(y, x) is also a solution. The two roots of the above equation are thedesired values (since they satisfy equation (1) by Vieta’s formulas) andwe can have (x, y) = (y, x) = (r1, r2) where

r1 =1−√

2 +√

2√

2− 1

2and r2 =

1−√

2−√

2√

2− 1

2.

1.4. Put x = k6, y = 2k10 and z = 3k15 for some natural number k. Then

x5 + y3 = (k6)5 + (2k10)3 = k30 + 8k30 = 9k30 = (3k15)2 = z2

as required. Since there are infinitely many possible values for k, thereare infinitely many triples (x, y, z).

25

1.5. First observe that if x is negative the inequality is trivial since the lefthand side is a square and hence larger than or equal to zero. The righthand side would be negative if x is negative. Now assume x ∈ R+.Apply the AM-GM inequality in x3 and (x− 1)2:

√x3(x− 1)2 6 x3 + (x− 1)2

2⇐⇒ 4x3(x− 1)2 6 (x3 + (x− 1)2)2.

Now, notice that 0 < x3 + (x − 1)2 = x3 + x2 − 2x + 1 < x3 + x2 + 3for positive x. This means

4x3(x− 1)2 6 (x3 + (x− 1)2)2 < (x3 + x2 + 3)2.

1.6. Label the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 in a clockwise directionaround the circle. Notice that if we switch a certain value twice, itremains in its original state. If we switch and element an odd num-ber of times we change it sign. To solve this problem it will sufficeto show that we can switch one number’s sign and leave all the re-maining numbers as they were after a finite number of moves. With-out loss of generality, we shall show that the number labelled 0 canbe switched and the remaining numbers left unchanged. If we switch(9, 0, 1), (0, 1, 2), (8, 9, 0), (2, 3, 4), (8, 7, 6), (7, 6, 5), (5, 4, 3) notice that allpositions are switched twice except 0 which is switched 3 times andhence has it’s sign changed.

26


2.1. Label the vertices as shown.

A

B

C

D

Let f(XY ) denote the number on edge XY . Notice that

f(BA) + f(AD) + f(DB) = f(AD) + f(DC) + f(AC)

= f(BC) + f(CD) + f(BD)

= f(AB) + f(BC) + f(AC).

according to the condition. This means that f(BC)+f(CD) = f(BA)+f(AD) and f(AD) + f(DC) = f(AB) + f(BC). Subtrating these twoequalities from one another we find that f(BC)− f(AD) = f(AD)−f(BC). Hence f(AD) = f(BC). This means that the numbers cannotbe distinct and so it cannot be done.

Alternative Solution

Notice that each edge is counted twice in the sum of the four faces.Therefore we have the sum of the four faces equals 2(1+2+3+4+5+6) =42, which is not divisible by 4, contradiction.

2.2. Notice that 382 = 1444 which ends in . . . 444.Moreover, (1000k + 38)2 = 1000000k2 + 76000k + 1444 which ends in. . . 444 for any natural number k. Thus there are infinitely many suchnumbers.

2.3. Imagine the fencers as nodes in a graph. There is an edge betweentwo nodes if those fencers have duelled. If everyone is to have duelleda different number of opponents, then since there are only 10 fencers,they must have each duelled distinct number from 0, 1, 2, . . . , 8, 9. In

27

any graph, the sum of all the degrees of each node is even since eachedge is counted twice. However, the sum of all the degrees for the 10fencers would have to be 1 + 2 + . . . + 8 + 9 = 9×10

2= 45 which is

odd. Hence such a state cannot be achieved. Furthermore note that aneasier solution is to just note that you can’t simultaneously use 0 and9.

2.4. First notice that by the AM-GM inequality we have

x5y + x3y3

2> x4y2 and

y5x+ y3x3

2> y4x2.

Also note that

x6 + y6 > x5y + xy5

⇐⇒ x6 − x5y + y6 − xy5 > 0

⇐⇒ x5(x− y) + y5(y − x) > 0

⇐⇒ (x− y)(x5 − y5) > 0,

which is true when x > y and when y > x and is therefore true. Thus

x6 + x3y3 + x3y3 + y6 > (x5y + x3y3) + (y5x+ y3x3) > 2x4y2 + 2x2y4

⇐⇒ (x3 + y3)2 > 2x2y2(x2 + y2) = 2x2y2

Since x, y > 0 we can conclude that x3 + y3 >√

2xy as required.

2.5. Drop a perpendicular from C1 onto AB meeting AB at C2. Simi-larly define B2 such that B1B2 ⊥ AC. Let Mc be on AC such thatMMc ⊥ AC and Mb on AB such that MMb ⊥ AB. Without loss ofgenerality, assume that the points on the angle bisector of ∠A are inthe order as shown:

Now 4AB1B2 ∼ 4AMMc ∼ 4AC1C

28

and 4AC1C2 ∼ 4AMMb ∼ 4AB1B.

A CMcB2

B

Mb

C2

M

C1

B1

So by the Midpoint Theorem, B2Mc = McC and C2Mb = MbB.This means that B2M = CM and C2M = MB.Notice that ∠MB1B2 = ∠BB1M and by symmetry, BB1 = B1B2.This means that 4MB1B2 ≡ 4MB1B. Hence BM = MB2.Thus BM = MB2 = CM .

2.6. This problem can be solved by using the well know Graeco-Latin square.This square is an n× n grid to be filled with two sets of n symbols insuch a way that in each row and each column, every element from eachof the sets of symbols occurs exactly once. There are two symbolsin each square, one from each set. In addition to this, every pair ofsymbols in the squares occurs only once in the entire grid.

Here is a possible solution to the 4×4 grid with symbols in A,B,C,Dand α, β, γ, δ:

Dδ

Cγ

Bβ

Aα

Cβ

Dα

Aδ

Bγ

Bα

Aβ

Dγ

Cδ

Aγ

Bδ

Cα

Dβ

29

In order the solve the problem, we can set

A = 1, B = 2, C = 3, D = 5, α = 7, β = 11, γ = 13, δ = 17.

Instead of the pairs, we multiply the two numbers together in thesquares. We choose these to be all primes (except for 1), this meansthat since each pair occurs exactly once in each square, the numbersgenerated will all be distinct. Moreover, each row and column will havea product equal to ABCDαβγδ and also the maximum product in asquare is 5× 17 = 85 < 100.

30

8.2 Solutions to the University of Stellenbosch SeniorTests

University of Stellenbosch Senior Camp December 2011: Test 1Solutions

1. Let the perimeter be 6k. The equilateral triangle will have side length2k, and will have an area four times the area of a triangle with sidelength k.

The hexagon will have side length k. By joining the vertices of thehexagon to the centre, it can be divided exactly into six equilateraltriangles of side length k.

The ratio of the area of the triangle to the area of the hexagon is then4 : 6 = 2 : 3.

2. Since n and k are integers, their difference will also be an integer. Sincek(k + 5)2 = k3 + 10k2 + 25k,

(k + 2)3 = k3 + 6k2 + 12k + 8 < k(k + 5)2 + 12

(k + 4)3 = k3 + 12k2 + 48k + 64 > k(k + 5)2 + 12.

The only way to get equality is if n = k + 3. In that case:

n3 = k3 + 9k2 + 27k + 27 = k3 + 10k2 + 25k + 12,

or 0 = k2 − 2k − 15 = (k − 5)(k + 3). As k is a positive integer, k = 5and n = 8 is the only solution.

3. By using the AM-GM, x4 + 1 > 2x2, and y3 + y > 2y2. Then

x4 + y3 + x2 + y + 1 > 3x2 + 2y2 > 2√

6xy.

As 4√

6 =√

96 > 9, this expression is greater than 92xy, and the

inequality is sharp (equality never occurs).

4. It is easy to observe that the infection can never spread outside of arectangle that bounds it. A less obvious but more useful observation isthat the total perimeter of the infected area will never increase.

Let infected squares be black and non-infected squares be white. It isclear that the spread can be considered on a square-by-square basis,as additional infected squares will not hamper the infection of others.

31

We need then to consider the cases in which a square may get infected.There are four ways in which it can have two or more black neighbours:

If it has exactly two infected neighbours, the black perimeter does notchange, if it has three infected neighbours, the black perimeter willdecrease by 2, and if all the neighbours are infected, the perimeter willdecrease by 4.

To solve the problem, nine cells have a maximal perimeter of 36, whereasthe total 10 × 10 square has a perimeter of 40, so it is not possible toinfect the entire square. If there are ten cells, which all lie along themain diagonal of the square, the infection will spread everyhere.

5. Adopting the usual notation, 2a = b+ c, hence s = a+ b+ c = 32a. The

area of the triangle is K = sr = 32ar = 1

2aha, where ha is the altitude

from A onto BC. Therefore ha = 3r, and so the incentre is a third ofthe way to the vertex. The centroid is also a third of the way to thevertex; therefore the line joining the incenter and centroid is parallelto BC.

32


1. Using Vieta’s formula, the sum of the roots is x1 + x2 + x3 = a, andthe sum of the pairwise products is x1x2 + x2x3 + x3x1 = b. Startingwith the right hand side:

2a2 − 6b = 2(x1 + x2 + x3)2 − 6(x1x2 + x2x3 + x3x1)

= 2(x21 + x2

2 + x23 + 2(x1x2 + x2x3 + x3x1))− 6(x1x2 + x2x3 + x3x1)

= 2(x21 + x2

2 + x23)− 2(x1x2 + x2x3 + x3x1)

= (x1 − x2)2 + (x2 − x3)

2 + (x3 − x1)2.

2. Consider a square of side length d, and the position of its bottom-rightunit square. It has d− 1 squares above, and d− 1 squares to the left ofit. Hence there are (21 − d)(16 − d) possibilities for the bottom-rightcorner, resulting in (21− d)(16− d) squares of size d appearing in thegrid. The total number of squares is

S =15∑d=1

(21− d)(16− d)

=15∑d=1

(336− 37d+ d2)

= 15 · 336− 3715∑d=1

d+15∑d=1

d2

= 15 · 336− 3715 · 16

2+

15 · 16 · 31

6= 1840.

3. Since gcd(40, 2009) = 1, we have 40k·ϕ(2009) ≡ 1 (mod 2009) for allnatural numbers k. For n > ϕ(2009) the exponent n! is certainlya multiple of ϕ(2009), and therefore an+1 ≡ an + 1 (mod 2009). Thismeans that all values modulo 2009 are taken cyclically and periodically,and all values are obtained infinitely often. Since this also for the value0, the proof is complete.

33

4. Let p < a be a prime. If p|n, the claim holds.

Suppose then that n is not divisible by p. Then the numbers 2n + 1,3n + 1, ..., (p + 1)n + 1 give different remainders when divided by p,which are denoted, respectively, by r1, r2, ..., rp. Indeed, if ri = rj,then

p|((i+ 1)n+ 1)− ((j + 1)n+ 1) = (i− j)n,

which implies that i = j. So each of the p possible remainders appearsexactly once. In particular, one of these remainders must be zero,meaning that the one of the factors in the product is divisible by p,which finishes the problem.

5. Let R be the intersection of Y Z and PQ. We wish to prove that R isconstant. Let R′ be the intersection of LQ and XR. Then it is knownthat L, Y,R′, Q form a harmonic range, i.e. (L, Y ;R′, Q) = −1. Thecross ratio depends on the angles ∠LXY,∠Y XR′,∠R′XZ. Since XL,XY and XZ are fixed, this implies that XR′ is fixed. Hence R is fixedsince it is the intersection of XR′ and Y Z.

L

P

Q

R

X Z

Y

R′

34


1. Evaluating the first few terms, one finds a1 = 1, a2 = 5, a3 = 19,a4 = 65, a5 = 211 and a6 = 665. The values always increase by a factorof 3, plus a little bit. A guess is that the terms are similar to 3n, andcomputing the difference one fits it to be 2n. We’ll show by inductionthat an = 3n − 2n. The base case (n = 1) works. Then, for all n ∈ N,

an+1 = 3an + 2n = 3(3n − 2n) + 2n = 3n+1 − 3 · 2n + 2n = 3n+1 − 2n+1,

as required, so the induction holds and an = 3n − 2n for all n ∈ N.

2. If both m and n are greater than one, then 6m+2n+2 ≡4 0+0+2 ≡4 2,which is not a perfect square. So at least one of m and n has to beexactly 1.

• If m = 1, then we want 2n + 8 to be a square. This has a solutiononly for n = 3. If n > 4 the expression will be divisible by 8 andnot 16, so it cannot be a square.

• If n = 1, then we want 6m + 4 to be a square. Considering thisexpression mod 7, we have (−1)m + 4, which will be 3 or 5 (mod7). However, perfect squares are only 0, 1, 2 or 4 (mod 7), so thereare no solutions.

So the unique solution to the problem is (m,n) = (1, 3).

3. Let TH cut BC at P . We want to show that P bisects BC. Since∠AHC = ∠ATC = 90, the points A, H, T and C lie on the samecircle.

A BH

P

T

C

O

Now ∠PHC = ∠TAC = 12(180 − ∠AOC) = 90 − ∠ABC = 90 −

∠HBC = ∠BCH = ∠PCH, so triangle PCH is isosceles. In addition,∠PBH = 90 − ∠BCH = 90 − ∠PHC = ∠PHB, so the trianglePBH is isosceles as well. Therefore PC = PH = PB and the claim isproved.

35

4. The three numbers have average 999. Let k = 999, and write:

989 · 1001 · 1007 + 320 = (k − 10)(k + 2)(k + 8) + 320

= k3 − 84k − 160 + 320

= k3 − 84k + 160

= (k + 10)(k − 2)(k − 8)

= 1009 · 997 · 991.

5. First, notice that player two can always ensure that A 6 6. The squaresof the grid can be partially tiled with 2 × 1 tiles, as shown below.Whenever a 1 is placed inside a tile, player two can ensure that it alsocontains a 0. As every 3 × 3 square contains three complete tiles, itwill have at least three zeros in it so, A 6 6.

1 2 3 2 12 4 6 4 23 6 9 6 32 4 6 4 21 2 3 2 1

Label each individual cell with the number of 3×3 squares that containit. By going for the highest-placed numbers, player 1 can get a totalof 45, which spread across 9 squares gives exactly 5. Let player 1 startby placing a 1 in the middle, and wlog, player two plays in the leftside of the board. By placing a 1 directly right of the middle (value 6),player two is forced to take the right-most cell in the middle row (elseplayer 1 will, and there will exist a 3 × 3 square with three 1s and 6empty blocks). By taking the block directly below the middle (value6), player one leaves the other stranded: if he does not place a 0 in thebottom-rightmost 3× 3 square, player 1 will get six 1s in it. If he does,player 1 will take the last square of value 6, and will finish with celltotals greater than 45, which will give A > 6. Hence A = 6.

36


1. Note that for a given i, ai and bi are never both in one of 1, 2, ..., n orn + 1, ..., 2n. Suppose, wlog, that ai, bi 6 n. Then a1, a2, ..., ai 6 n,and bi, bi+1, ..., bn 6 n, giving n + 1 elements less than or equal to n,which is impossible. Then every |ai − bi| is a difference between anelement in n+ 1, ..., 2n and an element in 1, 2, ..., n. Therefore:

∑i=1n

∣∣ai − bi∣∣ =n∑i=1

(n+ i)−n∑i=1

i = n2.

2. Let S =∑ai =

∑bi, and let ω = an + bn − S. Create the table

0 0 · · · 0 a1

0 0 · · · 0 a2...

.... . .

......

b1 b2 · · · bn−1 ω

The sum of the first m− 1 rows and n− 1 columns clearly satisfies theconditions, and the last row sum is b1+b2+·+bn−1+ω =

∑bi+an−S =

an as required. Likewise for the last column sum.

3.

Expand the hexagon to make a big equilateral triangle, and drop per-pendiculars from the point inside to the sides of the triangle. Thearea of the unshaded region of the hexagon, which is three triangles ofequal bases, is equal to 1

2side length multiplied by the the sum of the

heights. But in an equilateral triangle, no matter where the interiorpoint is chosen, the sum of these dropped perpendiculars is constant

37

(equal to the height of the triangle). So the unshaded and the shadedregions have equal area.

4. Suppose that n2 = Tm = m(m+1)2

. Multiplying by 8,

8n2 = 4m2 + 4m = (2m+ 1)2 − 1 ⇔ (2m+ 1)2 − 8n2 = 1.

Letting A = 2m+1 and B = 2n, this gives Pell’s equation, A2−2B2 =1. The first solution is A = 3, B = 2, and the general solution is givenby ak +

√2bk = (3 + 2

√2)k. The solution is

ak =(3 + 2

√2)k + (3− 2

√2)k

2and bk =

(3 + 2√

2)k − (3− 2√

2)k

2√

2.

The squares themselves are x2 = 132

((3 + 2√

2)k − (3 − 2√

2)k)2 fork ∈ N.

5. Consider circles ω1 and ω2 centred at A and B respectively such thatboth ω1 and ω2 are orthogonal to ω. Let I be one of the intersectionpoints of ω1 and ω2. Note that I must exist because AB is outside ω.Then invert the diagram through a circle centred at I with any radius.ω′1 and ω′2 (the images of ω1 and ω2) are straight lines passing throughthe centre of ω′ (since they are orthogonal and angles between circles arepreserved by inversion). ω′1 passes through the centre of circle AX ′iQ

′i

and ω′2 passes through the centre of circle BX ′iQ′i for the same reason.

Hence Q′i is X ′i reflected in line ω′1 and X ′i+1 is the reflection of Q′i inthe line ω′2. Thus ∠X ′iOX ′i+1 = 2θ where θ is the angle between thelines ω′1 and ω′2. Thus X ′0 = X ′k is equivalent to kθ = 2nπ, for someinteger n. This means that consecutive terms in the sequence of pointsX ′i are determined by a rotation through a constant angle. Hence thisresult will be true irrespective of the choice of X ′0. There is a one-to-one correspondence between Xi and X ′i and hence the same is true ifX0 = Xk for some positive integer k.

38

A B

I

X0

Y0

X1

ω

ω1ω2

A′

B′

O

X ′0

Y ′0

X ′1

I

ω′

ω′1

ω′2

39


1. Suppose that the function is non-constant. Then it will have a leastelement f(k). Then f(k+1) 6= f(k), or the function would be constant,

and by definition f(k+1) > f(k). But f(k+2) = f(k)2

f(k+1)= f(k) f(k)

f(k+1)<

f(k), which is a contradiction. So f(n) constant is the only allowedfunction.

2. Let P (N) be the sum of squares of the product of these subsets. ForN = 0 there is only the empty subset, so P (0) = 1. For N = 1, we addthe subset 1, so P (N) = 2. For N = 2, we add an extra set 2, soP (2) = 6. Evaluating P (3) = 24, one can guess that P (N) = (N + 1)!.We show this by induction:

To find P (N + 1), we have all the sets without element N + 1, whichhas sum P (N), and those with N + 1, which cannot contain N . Theproduct of each of those sets increases by N + 1, so the recurrence isP (N+1) = P (N)+N2P (N−1). The claim holds for small N . Supposeit works for some integer n. Then

P (n+ 1) = P (n) + (n+ 1)2P (n− 1) = (n+ 1)! + (n+ 1)2n! = (n+ 2)!,

so the induction holds, and P (N) = (N + 1)! for all integers N .

3. Construct P ′ on AT such that OP ′ ⊥ AB. Then OP ′ ‖ HT . SinceBT is perpendicular to both OP and AT , we have OP ‖ AT , and bythe midpoint theorem OP ′ bisects AP .

B

P

TA

H

O

P ′

Therefore OAP ′P is a parallelogram. Hence OP ′ is bisected by AP ,and since triangles HAT and OAP ′ are similar, AP bisects HT asrequired.

40

4. Observe that

S >a

a+ b+ c+ d+

b

a+ b+ c+ d+

c

a+ b+ c+ d+

d

a+ b+ c+ d= 1,

S <a

a+ b+

b

a+ b+

c

c+ d+

d

c+ d= 2.

The function changes smoothly as we vary a, b, c and d. We will provethat it comes arbitrarily close to 1 and 2, and therefore it assumes everyvalue in the interval (1,2). Taking a = b and letting c = d gives

S1(a, c) =2a

2a+ c+

2c

a+ 2c, ⇒ lim

c→0S1(a, c) = 1.

Similarly, taking a = c and b = d gives

S2(a, b) =2a

a+ 2b+

2b

2a+ b⇒ lim

b→0S2(a, b) = 2.

So the expression takes on all values between 1 and 2.

5. The parallelograms can be naturally partitioned into three sets, basedon the orientation of their sides in the grid. By symmetry there willbe the same number of parallelograms of each orientation, so we canrestrict ourselves to the ‘diamond’ case (as in the diagram).

a b c d

Extend the triangular grid by one. The sides of the parallelogram canthen be extended until they cut the new bottom row of the grid. Itis easy to see that every parallelogram will correspond to a differentset of four intersection points (marked a, b, c and d for the case of theshaded parallelogram), and that every set of four points will correspondto a parallelogram. This bijection makes it easy to count – there aren + 2 points on the bottom row, so there are

(n+2

4

)ways of making a

parallelogram. Taking into account the orientations, the final answeris 3(n+2

4

).

41

6. Let 3n + 1 = x2 and 4n + 1 = y2. Then y2 − x2 = n. As y isodd, y2 ≡8 1, so n has to be even, therefore x is odd, and thereforey2 − x2 ≡8 1− 1 = 0, hence 8|n.

Now notice that 4x2 − 3y2 = 1, so setting w = 2x we obtain Pell’sequation w2 − 3y2 = 1. This has solutions (2 +

√3)m = wm +

√3ym,

but only the odd m solutions correspond to an even wm. Starting withthe base solution 2 +

√3, and multiplying through by (2 +

√3)2 we get

the recursions

wm+1 = 2xm+1 = 14xm + 12ym, and ym+1 = 8xm + 7ym.

Then xm+1 = 7xm+6ym ≡7 −ym, and ym+1 = 8xm+7ym ≡7 xm, hencey2m+1 − x2

m+1 ≡7 x2m − y2

m, and by induction, x2m − y2

m ≡7 0. Hence 7divides n = y2

m − x2m.

42

8.3 Solutions to the Monthly Problems

1. Let the two teams be A and B, and represent each possible series as abit string, where a 1 represents a win by team A and a 0 a win by teamB. Let S be the set of all possible bit strings. We need to determinewhether there are more bit strings of length 6 or length 7 in S. Foreach bit string s ∈ S, the bit string obtained from s be interchanging 0and 1 is also in S, so we may restrict our attention to bit strings endingin a 1.

The number of bit strings of length 6 is equal to(53

)(exactly 3 of

the first 5 bits must be a 1 for A to win the tournament in the sixthgame), while the number of bit strings of length 7 is equal to

(63

). Since(

63

)>(53

), we conclude that the tournament is more likely to end in

exactly 7 games.

2. If ai = bj = 0, then ci = cj = 0. If (c1, c2, . . . , cp) is a rearrangementof (0, 1, . . . , p− 1), then no entry can appear more than once. Hencei = j.

Without loss of generality we may suppose that ap = bp = 0 so thatcp = 0. Then we have

(p− 1)! ≡p a1a2 . . . ap−1 ≡p b1b2 . . . bp−1 ≡p c1c2 . . . cp−1

and so

(p− 1)! ≡p c1c2 . . . cp−1

≡p (a1a2 . . . ap−1)(b1b2 . . . bp−1)

≡p ((p− 1)!)2,

which implies that (p− 1)! ≡p 1. However, since p is odd, this contra-dicts Wilson’s Theorem, and we conclude that (c1, c2, . . . , cp) is not arearrangement of (0, 1, . . . , p− 1).

3. Place the square in the coordinate plane withA(−1, 1), B(1, 1), C(1,−1)and D(−1,−1). The equation of Γ is x2 + y2 = 2, the equation of theline AC is y = −x and the equation of the line BC is y = x. Let Mhave coordinates (m,n), where m2 + n2 = 2. The lines AM and BMhave the following equations:

AM :y =n− 1

m+ 1(x+ 1) + 1

43

BM :y =n− 1

m− 1(x− 1) + 1.

The coordinates of the four given intersection points are then

P

(m+ n

2 +m− n,

m+ n

2 +m− n

), Q

(n−m

m+ n− 2,

m− nm+ n− 2

),

R

(1 + 2m+ n

1− n,−1

), S

(2m− n− 1

1− n,−1

).

We now calculate the product of the gradients of the two lines PS andQR:

mPSmQR

=2(m+ 1)(1− n)

(m+ n)(1− n)− (2m− n− 1)(2 +m− n)

× 2(1− n)(m− 1)

(1− n)(n−m)− (1 + 2m+ n)(m+ n− 2)

=4(m+ 1)(m− 1)(n− 1)2

4(−n2 −m2 +mn−m+ n+ 1)(−n2 −m2 −mn+m+ n+ 1)

=(m+ 1)(m− 1)(n− 1)2

(m+ 1)(n− 1)(m− 1)(1− n)(since m2 + n2 = 2)

= −1,

which shows that the two lines are perpendicular.

4. Let D be the set of 25-digit numbers with the same number of 1s and2s, and let R be the set of 50-digit numbers with 25 digits 1 and 25digits 2. We define a bijection between the two sets.

Let d = d1d2 . . . d25 ∈ D. We define a function on the digits of d asfollows:

f(di) =

11 if di = 122 if di = 212 if di = 321 if di = 4

By replacing each digit di in d with f(di) we obtain a number in R;indeed, if di equals 3 or 4, then f(di) contains the same number of 1sand 2s, and since there are equal numbers of 1s and 2s in d, there willbe equal numbers of the digit pairs 11 and 22. Hence each number in d

44

corresponds to a number in R - clearly two different numbers in d willcorrespond to two different numbers in R.

Conversely, using the inverse of f , we can map each number in R toa number in D. A similar argument shows that this mapping is well-defined and injective as well, showing that the two sets have an equalnumber of elements.

5. In version centred at C with any radius gives 4CAP ∼ 4CP ′A′ and4CAB ∼ 4CB′A′. Clearly the line PQ inverts to itself. ω invertsto ω′, a line. H inverts to semicircle H ′ tangent to ω′ with diameterP ′Q′. AB inverts to arc A′B′ of a circle with diameter CB′ tangentto ω′. Now observe that arc A′Q′ and arc CA′ are symmetrical w.r.tthe perpendicular bisector of CQ′. Which implies ∠CP ′A′ = ∠CB′A′.Notice that AC bisects ∠PAB iff ∠CAP = ∠CAB iff ∠CP ′A′ =∠CB′A′ and so we are done.

6. For generality, let x = 2011. Expanding the first few terms of thesequence with the recurrence un = xun−1 − un−2 we find

u0 = 0, u1 = 1, u2 = x

u3 = x2 − 1 = u22 − u2

1

u4 = x3 − 2x = 2u3u2 − xu22

u5 = x4 − 3x2 + 1 = u23 − u2

2

u6 = x5 − 4x3 + 2x = 2u4u3 − xu23.

From this we can conjecture that u2n = 2unun+1 − xu2n and u2n+1 =

u2n+1−u2

n for n > 0. We now prove this by induction. Firstly note thatit holds for n = 1. Now assume it holds up to n, and we will show thatit holds for n+ 1.

u2n+2 = xu2n+1 − u2n

= x(u2n+1 − u2

n)− (2unun+1 − xu2n)

= xu2n+1 − 2unun+1

= 2xu2n+1 − 2unun+1 − xu2

n+1

= 2un+1(xun+1 − un)− xu2n+1

= 2un+1un+2 − xu2n+1

and

u2n+3 = xu2n+2 − u2n+1

45

= x(2un+1un+2 − xu2n+1)− (u2

n+1 − u2n)

= −x2u2n+1 + x(2un+1(xun+1 − un) + u2

n − u2n+1

= x2u2n+1 − 2xun+1un + u2

n − u2n+1

= (xun+1 − un)2 − u2n+1

= u2n+2 − u2

n+1.

The result thus follows by induction.

The above identities suggest that we may be able to find explicit fac-torisations, but for a factorisation to prove that a number is not primeit is necessary for the factors to be greater than 1. So let us show thatui − ui−1 > 2 for all i > 2. For i = 2 the result holds by inspection.Suppose it holds for some i. Then ui+1 − ui = 2011ui − ui−1 − ui =2010ui−ui−1 > ui−ui− 1 > 2. The result again follows by induction.

Inspecting the first few terms, we see that u2 = 2011 is prime. Letus show that there are no other primes. u0 and u1 are non-prime byinspection. For n > 1, u2n+1 = u2

n+1 − u2n = (un+1 − un)(un+1 + un).

By the inequality above, both factors are at least 2 and so u2n+1 is notprime. For n > 1, u2n = un(2un+1 − 2011un). The first term is clearlygreater than 1, and the second equals un+1 − un−1 > un+1 − un > 2.

Thus, u2 = 2011 is the only prime in the sequence.

7. Let φ(n)σ(n) = (n − 3)(n + 1), and let prime p divide n such thatn = pam for some m not divisible by p. Then we have

φ(n) = φ(pa)φ(m) = pa−1(p− 1)φ(m)

σ(n) = σ(pa)σ(m) =

(pa+1 − 1

p− 1

)σ(m),

soφ(n)σ(n) = pa−1(pa+1 − 1)φ(m)σ(m).

If a > 2, it follows that, modulo pa−1,

0 ≡ φ(n)σ(n) = (n− 3)(n+ 1) ≡ −3.

Hence a = 2 and p = 3. Thus either a = 1 or a = 2 and p = 3.

However, if n = 9 does not satisfy the given equality, and if n =9p1p2 . . . pk, then

φ(n)σ(n) = 78(p21 − 1)(p2

2 − 1) . . . (p2k − 1) ≡9 0,

46

since p21 ≡3 1 for every prime pi 6= 3, but

(n− 3)(n+ 1) ≡9 −3.

Hence n = p1p2 . . . pk for some distinct primes pi.

We now prove the following: if p is a prime not dividing m and

φ(m)σ(m) 6 (m− 3)(m+ 1),

thenφ(pm)σ(pm) < (pm− 3)(pm+ 1).

Indeed,

(pm− 3)(pm+ 1)− φ(pm)σ(pm)

8. Let P (X) denote the power of point X with respect to circle M,N,D.Then

AB(AM + AN) = AB · AM + AB · AN= (AN +NB)AM + AB(AB −BN)

= AN · AM +BN · AM + AB2 − AB ·BN= P (A) + AD2 +BD2 −BN(AB − AM)

= P (A) + AD2 + P (B)−BN ·BM= P (A) + AD2 + P (B)− P (B)

= P (A) + AD2.

A similar argument shows that AC(AP + AQ) = P (A) + AD2, whichconcludes the proof.

9. Choose any positive rational p, q and let x = p+q2

and y = p−q2

. Then

f(p)− f(q) = 4√f(x)f(y) > 0,

so f(p) > f(q) and hence f is strictly increasing.

Now we show that f has values arbitrarily close to 0 i.e., for any ε ∈ Q+,we can find x such that f(x) < ε. Substituting x = 2y gives

f(3y)− f(y) = 4√f(2y)f(y) > 4

√f(y)f(y) = 4f(y),

47

and hence f(3y) > 5f(y) or f(y) < 3y5

. Substituting this into itself

gives f(y) < f(32y)52 , and repeating gives f(y) < f(3ky)

5k . Now let y = 13k ,

giving f( 13k ) < f(1)

5k . Since f(1) does not depend on k, we can choose

k sufficiently large so that f(1)5k is less than ε and hence y satisfies the

condition.

Consider the original equation again, and consider what happens asf(y) tends to zero. f(x) stays fixed so 4

√f(x)f(y) can be made arbi-

trarily small. Also, f(x) lies between f(x− y) and f(x+ y), so for anyδ we can choose a y such that f(x) − δ < f(x − y) < f(x) < f(y) <f(x) + δ.

Now let x = z + ε and x = z − ε. The original equation becomes

f(2z) = 4√f(z + ε)f(z − ε) + f(2ε).

By choosing ε sufficiently small, we can ensure that f(ε) is arbitrarilyclose to zero and f(z + ε) and f(z − ε) are arbitrarily close to f(z).The right hand side can thus be made arbitrarily close to 4f(z), and itfollows that the left hand side must be 4f(z).

We can extend f(2z) = 4f(z) to f(nz) = n2f(z) for all positive integersn. For n = 1 and n = 2 we already have it. Suppose it holds up tosome n. We now show that it also holds for n + 1. Let x = nx, y = x,giving

f((n+ 1)x) = 4√f(nx)f(x) + f((n− 1)x)

= 4√n2f(x)f(x) + (n− 1)2f(x)

= 4nf(x) + (n− 1)2f(x)

= (n+ 1)2f(x).

The result follows by induction.

We can further extend this to positive rational coefficients. Let n,mbe positive integers. n2f(x) = f(m · n

m· x) = m2f( n

m· x), from which

it follows that f( nm· x) = (m

n)2.

Now let f(1) = a. By the above, f(x) = ax2 for all positive rationalsx. Conversely, it is trivial to substitute this into the original equationto confirm that it is a valid solution, for any a ∈ Q+.

10. Note that since tan 2θ attans all real numbers in the interval(−π

4, π

4

),

we may restrict our attention to this domain. Also,

f(− tan 2θ) = f(tan 2(−θ))

48

= tan4(−θ) + cot4(−θ)= tan4 θ + cot4 θ

= f(tan 2θ),

so f is an even function, and we may restrict θ to the interval (0, π4).

Next, let 0 < x < y, and let x = tan 2α, y = tan 2β, where 0 < α <β < π

4. Then

f(x)− f(y) = f(tan 2α)− f(tan 2β)

= tan4 α + cot4 α− tan4 β − cot4 β

= (tan4 α− tan4 β)−(

1

tan4 β− 1

tan4 α

)= (tan4 α− tan4 β)

(1− 1

tan4 α tan4 β

).

Now, tan θ is increasing on the interval (0, π4), and for θ ∈ (0, π

4), 0 <

tan θ < 1, hence tan4 α − tan4 β < 0 and tan4 α tan4 β < 1 =⇒1− 1

tan4 α tan4 β< 0, hence f(x)−f(y) > 0, showing that f is decreasing.

Next, note that both cot θ and tan θ are concave up on the interval(0, π

4), so we have that tan 2α+tan 2β

2> tan(α+β), by Jensen’s Inequality.

Hence, setting x = tan 2α and y = tan 2β, and recalling that f isdecreasing, we have

f

(x+ y

2

)= f

(tan 2α + tan 2β

2

)6 f

(tan 2

(α + β

2

))= tan4

(α + β

2

)+ cot4

(α + β

2

)6 tan4 α + tan4 β

2+

cot4 α + cot4 β

2(Jensen’s Inequality)

=f(tan 2α) + f(tan 2β)

2

=f(x) + f(y)

2.

This shows that f is concave up.

Now let x = sinα and y = cosα. Since f is even, we may assume that

both x and y are positive, and hence x+y26√

x2+y2

2. Using Jensen’s

49

Inequality and recalling that f is decreasing, we have that

f(x) + f(y) > 2f

(x+ y

2

)> 2f

(√x2 + y2

2

)= 2f

(1√2

).

Now, let tan 2θ = 1√2. Then tan θ =

√3 −√

2 and cot θ =√

3 +√

2,and so

f

(1√2

)= f(tan 2θ)

= tan4 θ + cot4 θ

= (√

3−√

2)4 + (√

3 +√

2)4

= 2(√

34

+ 6(√

2√

3)2 +√

24)

= 98.

Hence f(cos θ) + f(sin θ) > 2(98) = 196.

11. For integers x and y, by a segment [x, y] we mean the set of all integerst such that x 6 t 6 y; the length of this segment is y − x.

If, for every two positive integers x and y of the same colour we havef(x)x

= f(y)y

, then one can choose a = maxf(r)r, f(b)

b, where r and b are

arbitrary red and blue numbers, respectively. So we can assume thatthere are two red numbers x and y such that f(x)

x6= f(y)

y.

Set m = xy. Then each segment of length m contains a blue number.Indeed, assume that all the numbers on the segment [k, k+m] are red.Then

f(k +m) = f(k + xy) = f(k + x(y − 1)) + f(x) = · · · = f(k) + yf(x),

f(k +m) = f(k + xy) = f(k + (x− 1)y) + f(y) = · · · = f(k) + xf(y),

so yf(x) = xf(y), a contradiction. Now we consider two cases.

Case 1. Assume that there exists a segment [k, k + m] of length mconsisting of blue numbers. Define D = maxf(k), . . . , f(k+m). Weclaim that f(z) − f(z − 1) 6 D, whenever z > k, and the conclusionfollows. Consider the largest blue number b1 not exceeding z, so z−b1 6m, and some blue number b2 in the segment [b1 + k, b1 + k + m], sob2 > z. Write f(b2) = f(b1) + f(b2 − b1) 6 f(b1) + D to deduce thatf(z + 1)− f(z) 6 f(b2)− f(b1) 6 D, as claimed.

Case 2. Each segment of length m contains numbers of both colours.Fix any red number R > 2m such that R + 1 is blue and set D =

50

maxf(R), f(R+1). Now we claim that f(z+1)−f(z) 6 D, wheneverz > 2m. Consider the largest red number r not exceeding z and thelargest blue number b smaller than r; then 0 < z−b = (z−r)+(r−b) 62m, and b + 1 is red. Let t = b + R + 1; then t > z. If t is blue, thenf(t) = f(b)+f(R+1) 6 f(b)+D, and f(z+1)−f(z) 6 f(t)−f(b) 6 D.Otherwise, f(t) = f(b+1)+f(R) 6 f(b+1)+D, hence f(z+1)−f(z) 6f(t)− f(b+ 1) 6 D, as claimed.

Romanian Masters 2012

12. Suppose we wish to find a solution where kn has d+ 4 digits. Then wemust satisfy 2011 · 10d 6 kn < 2012 · 10d, or equivalently

k ∈[

2011

n· 10d,

2012

n· 10d

).

The size of this interval is 10d

n. Clearly, we can choose d sufficiently

large such that the interval has size greater than 1, in which case itmust contain an integer, and this integer will be a solution.

13. Place the points A, B and C in the complex plane such that P is atthe origin. Then we have the following equations:

PP1 ⊥ BC =⇒ B − CP1

=C − BP1

(3)

PP2 ⊥ AC =⇒ A− CP2

=C − AP2

(4)

PP3 ⊥ AB =⇒ B − AP3

=A− BP3

(5)

BP1 = BP3 =⇒ (B − P1)(B − P1) = (B − P3)(B − P3)

=⇒ BP3 + BP3 − P3P3 = BP1 + BP1 − P1P1 (6)

CP1 = CP2 =⇒ (C − P1)(C − P1) = (C − P2)(C − P2)

=⇒ CP2 + CP2 − P2P2 = CP1 + CP1 − P1P1 (7)

Then,

AP 22 − AP 2

3

= (A− P2)(A− P2)− (A− P3)(A− P3)

= AP3 + AP3 − AP2 − AP2 + P2P2 − P3P3

= BP3 + BP3 − CP2 − CP2 + P2¯P − 2− P3P3 (using (4) and (5))

= (BP3 + BP3 − P3P3)− (CP2 + CP2 − P2P2

51

= (BP1 + BP1 − P1P1)− (CP1 + CP1 − P2P1 (using (6) and (7))

= (B − C)P1 − (C − B)P1

= 0, (using (3))

as required.

Xu Vol 1 p62

14. Clearly every point on the perpendicular bisector of BC satisfies therequirements of the problem. We now show that no other point P does.

Let D be the midpoint of BC, and suppose that there exists a point Pnot on the line AD such that ∠APB = ∠APC. Let Q be the reflectionof P about the line AD; then Q also satisfies ∠AQB = ∠AQC. HenceA, P , Q and C are concyclic, and so are A, P , Q and B. However,this means that both B and C lie on the circumcircle of triangle APQ,which is clearly impossible.

15. For each non-negative integer n,

2√n+ 1 >

√n+√n+ 2.

Indeed,

√n+ 1−

√n =

1√n+ 1 +

√n

>1√

n+ 2 +√n+ 1

=√n+ 2−

√n+ 1,

from which the result follows.

We now prove the required equality. It is true for n = 0 and n = 1, byinspection. Let n > 2. By the above result,

√n+√n+ 1 +

√n+ 2 < 3

√n+ 1 =

√9n+ 9.

We now show that√n+√n+ 1 +

√n+ 2 >

√9n+ 8. First note that,

for n > 2,

n(n+ 2)−(n+

7

9

)2

=4

9n− 49

81> 8

9− 49

81> 0

52

which implies that √n(n+ 2) > n+

7

9.

Hence,

[√n+√n+ 1 +

√n+ 2]2 >

[3

2(√n+√n+ 2

]2

=9

4

[n+ (n+ 2) + 2

√n(n+ 2)

]>

9

4

[2n+ 2 + 2

(n+

7

9

)]=

9

4

[4n+

32

9

]= 9n+ 8.

Hence we have

√9n+ 8 <

√n+√n+ 1 +

√n+ 2 <

√9n+ 9.

Since 9n+ 8 and 9n+ 9 are consecutive, no perfect square lies betweenthem, and so no integer lies strictly between

√9n+ 8 and

√9n+ 9.

This is enough to conclude the required equality.

16. (a) Note that R is the circumcentre of triangle BHC, hence OR bi-sects BC. Also, ∠BRC = 2∠(180 − ∠BHC) = 2∠QRP =2∠A = ∠BOC, so it follows that R is the reflection of O acrossthe line BC.

Next, since QR is perpendicular to BH, which is perpendicular toAC, we have that AC ‖ QR, and hence ∠BQR = ∠A = ∠BOR,which shows that BQOR is a cyclic quadrilateral. Using the sinerule in triangle BQR we see that the circumradius of triangleBQR is equal to BR

2 sinBQR= R

2 sinA, where R is the circumradius

of triangle ABC. Similarly, CPOR is a cyclic quadrilateral withcircumradius R

2 sinA.

Let the line through O perpendicular to BH meet BH in K andAB in L, let the foot of the altitude from B be E and let themidpoint of BH be M . It is well known that AH = OR, andhence, since AH ‖ OR and QR ‖ OK ‖ AC, it follows thatHE = MK =⇒ KE = MH = BM =⇒ AL = BQ.Hence L and Q are symmetric about the midpoint of AB, andso OQ = OL. Hence, since OL ‖ AC, ∠OQA = ∠OLQ = ∠A.

53

Using the sine rule in triangle OAQ shows that it has circumradiusAO

2 sin ∠AQO = R2 sinA

. Similarly, ∠OPA = ∠A and triangle OAP has

circumradius R2 sinA

.

These radii are equal to R if and only if sinA = 12, i.e. if and only

if ∠A = 30.

48 New Problems in Euclidean Geometry

Problem H5

The acute-angled triangle ABC has circumcentre O and orthocentre H. Theperpendicular bisector of BH meets AB at Q, the perpendicular bisector ofCH meets AC at P and these two bisectors meet at R.

(a) Prove that the circumcircles AOP,POR,ROQ,QOA have equal radii.When are they equal in radius to the circumcircle of ABC?

(b) Prove that if PQ passes through O then it also passes through H. Whatcan be said about ∠BAC in this case ?

(c) Suppose circles AOP and ROQ meet again at S. What can be saidabout circle PRS?

A

B C

H

Q

P

R

O

S

(b) Since QR is the perpendicular bisector of BH, we have that∠HQR = ∠BQR = ∠A. Considering the opposite angles ofthese two angles and recalling that ∠AQO = ∠A, we have that∠HQO + 180 = 3∠A =⇒ ∠HQO = 3∠A − 180. Next, since∠OQA = ∠A = ∠OPQ, it follows that ∠POQ = 360 − 3∠A.Hence ∠POQ+HQO = 360− 3∠A+ 3∠A− 180 = 180, whichshows that OP ‖ QH. Hence, if PQ passes through O, it alsopasses through H. In this case, ∠HQO = 0 = 3∠A − 180,implying that ∠A = 60.

New problems in Euclidean geometry H5

17. (a) Set the square on the coordinate plane with vertices (±12,±1

2).

Then the three circles with diameter√

658

and centres (0, 716

), (14,− 1

16)

and (−14,− 1

16) covers the unit square.

54

(b) Label the square ABCD, and suppose that three sets S1, S2, S3 of

diameter less than√

658

covers it. Then two of the vertices, withoutloss of generality A and B, belong to the same set, say S1. Thenclearly C and D do not belong to S1.

Suppose that C and D are both elements of S2. Consider edgeAD. Since a + 1

82 = 6564

, S1 can only contain points on AD lessthan distance 1

8from A, and similarly S2 can only contain points

on AD less than distance 18

from D, and so neither the midpointsof AD nor BC belongs to S1 ∪ S2. Hence both midpoints mustbelong to the set S3, but a similar argument as above shows thatS3 can cover at most 2 × 1

8= 1

4units of segment AD. Since

2× 18

+ 14

= 12< 1, this configuration cannot cover ABCD. Hence

we may suppose that C ∈ S2 and D ∈ S3.

As before, S1 can only contain points at a distance less than 18

from A, S3 must contain the rest of segment AD. In particular, ifX is the point on AD such that AX = 1

8, then X ∈ S1. Let E be

the midpoint of CD. By Pythagoras, the distance EX = 658

, andhence E cannot belong to S3. A similar argument shows that Ecannot belong to S2 either, and clearly E does not belong to S1.Hence it is impossible to cover all of ABCD with sets of diameterless than

√658

.

18. We will use the following lemma:

Lemma 1. Let AD be a median in triangle ABC. Then cot∠BAD =2 cotA+ cotB and cot∠ADC = 1

2(cotB − cotC).

Proof. Let CC1 and DD1 be the perpendiculars from C and D to AB.Using signed lengths, we write

cotBAD =AD1

DD1

=12(AC1 + AB)

12CC1

=CC1 cotA+ CC1(cotA+ cotB)

CC1= 2 cotA+ cotB.

Similarly, denoting by A1 the projection of A onto BC, we get

cotADC =DA1

AA1

55

=12BC − A1C

AA1

=12(AA1 cotB + AA1 cotC)− AA1 cotC

AA1

=1

2(cotB − cotC).

Turning to the given problem, by the lemma we get

cotBPD = 2 cotBPC + cotPBC

= 2 cotBFC + cotPBC (from circle BFPC)

= 2 · 1

2(cotA− cotB) + 2 cotB + cotC

= cotA+ cotB + cotC.

Similarly, cotGQF = cotA + cotB + cotC, so ∠GPR = ∠GQF andGPRQ is cyclic.

Remark: The angle ∠GPR = ∠GQF is the Brocard angle.


19. Dividing both sides of the equation by pqr, we obtain(1 +

1

p

)(1 +

2

q

)(1 +

3

r

)= 4.

If p, q, r > 5, then(1 +

1

p

)(1 +

2

q

)(1 +

3

r

)6 6

5· 7

5· 8

5< 4,

hence at least one of p, q or r is less than 5, and since they’re all prime,we have the following cases:

Case 1: p = 2. Then 3(q+ 2)(r+ 3) = 8qr =⇒ (5q− 6)(5r− 9) = 144which has the solution (q, r) = (3, 5).

Case 2: p = 3. Then 4(q + 2)(r + 3) = 12qr =⇒ (q − 1)(2r − 3) = 9,which has no solutions.

Case 3: q = 2. Then 4(p + 1)(r + 3) = 8pr =⇒ (p − 1)(r − 3) = 6,which has no solutions.

Case 4: q = 3. Then 5(p+1)(r+3) = 12pr =⇒ (7p−5)(7r−15) = 180which has the two solutions (p, r) = (5, 3), (2, 5).

56

Case 5: r = 2. Then 5(p+ 1)(q+ 2) = 8pq =⇒ (3p− 5)(3q− 10) = 80which has the solution (p, q) = (7, 5).

Case 6: r = 3. Then 6(p + 1)(q + 2) = 12pq =⇒ (p − 1)(q − 2) = 4which has the solution (p, q) = 5, 3).

Hence the equation has the three solutions (p, q, r) = (2, 3, 5), (5, 3, 3)and (7, 5, 2).

Czech and Slovak MO 2011

20. Let an = an,1an,2 . . . an,kn be the decimal representation of an. Notethat if kn is even, then an ≡11 an,1 − an,2 + · · · − an,kn and bn ≡11

an,k − an,k−1 + · · · − an,1 ≡ −an. Hence an+1 = an + bn ≡11 0, so an+1

is divisible by 11. Also note that if an is divisible by 11, then bn is alsodivisible by 11, and hence an+1 is divisible by 11. In order to provethat a7 cannot be prime it suffices to prove that one of a1, a2, . . . , a6

has an even number of digits.

For a contradiction, suppose that all six numbers have an odd numberof digits. Note that an+1 can have at most one more digit than an, soin fact a1, . . . , a6 all have the same number of digits.

Let a1,1 = a and a1,k1 = d. Then the first digit of a2 is either a + d ora+ d+ 1, and since a2 has the same number of digits as a1, a+ d < 10.Hence the units digit of a2 equals a+ d.

Thus the first digit of a3 is at least 2(a + d) < 10, and the final digitof a3 is at least 2(a + d). Continuing in this way, we see that the firstdigit of a4 is at least 4(a+d), the first digit of a5 is at least 8(a+d) andthe first digit of a6 is at least 16(a+ b). However, 16(a+ d) is certainlynot less than 10, which implies that a6 must have more digits than a1,a contradiction. This shows that a7 must be divisible by 11, and hencecannot be prime.

Starting with a1 = 10, 220 yields a6 = 185, 767, which is prime.

Czech and Slovak MO 2006

21. We wish to show that there are infinitely many positive integers n suchthat the equation n2 + 1 = p(p + n) has a solution for p a positiveinteger. Rewrite this as p2 + pn− n2 − 1 = 0, or

p =−n+

√n2 + 4n2 + 4

2=

√5n2 + 4− n

2.

57

Note that if 5n2 +4 is a perfect square, then p will be a positive integer,so it suffices to show that there are infinitely many n such that 5n2 + 4is a perfect square.

Note that 5(1)2 + 4 = 32. We now construct an infinite sequence ofdifferent solutions.

Suppose that m and n are positive integers such that 5n2 + 4 = m2.Then m > n and m and n have the same parity. Hence m+3n

2is an

integer greater than n. I claim that N = m+3n2

is also a number suchthat 5N2 + 4 is a perfect square. Indeed,

5

(m+ 3n

2

)2

+ 4 =5(m2 + 6mn+ 9n2) + 16

4

=5m2 + 30mn+ 45n2 + 16

4

=5(5n2 + 4) + 30mn+ 9(5n2 + 4)− 20

4

=25n2 + 30mn+ 9m2

4

=

(5n+ 3m

2

)2

.

Sincem and n have the same parity, 5n+3m2

is an integer. Hence, startingwith n = 1, we can construct an increasing sequence of positive integersn such that 5n2 + 4 is a perfect square, and for these n, n2 + 1 can bewritten as the product of two numbers with difference n.

22. A set X of boys is separated from a set Y of girls if no boy in X isan acquaintance of a girl in Y . Similarly, a set Y of girls is separatedfrom a set X of boys if no girl in Y is an acquaintance of a boy in X.Since acquaintance is assumed mutual, separation is symmetric: X isseparated from Y if and only if Y is separated from X.

This enables counting the number n of ordered pairs (X, Y ) of separatedsets X, of boys, and Y , of girls, in two ways and thereby showing thatit is congruent modulo 2 to both numbers in question.

Given a set X of boys, let YX be the largest set of girls separatedfrom X, to deduce that X is separated from exactly 2|YX | sets of girls.Consequently, n =

∑X 2|YX | which is clearly congruent modulo 2 to

the number of sociable sets of boys.

The same argument applies to show that n is congruent modulo 2 tothe number of sociable sets of girls.

58


23. Denote by A1, A2, . . . , A66 the points from T and by ai the numberof lines from P containing Ai. Then the number of pairs of lines in-tersecting at Ai equals

(ai

2

), and the number of incident pairs I =∑66

i=1 ai. Since any two lines meet in at most one point, we have∑66i=1

(ai

2

)6(162

)= 120. Let bk be the number of points from T

which lie on exactly k lines from P . Then∑bk = 66,

∑(k2

)bk 6 120

and I =∑kbk 6

∑12

(3 +

(k2

))= 1

2(3 · 66 + 120) = 159, because

3 +(k2

)> 2k. Equality is attained when bk = 0 for k /∈ 2, 3, b2 = 39

and b3 = 27 - i.e. when the lines from P determine exactly 39 doubleand 27 triple intersection points.

, g

A1 A2 A3

B1B2B3

K

An example of a configuration with 159 incident pairs can be con-structed using Pappus’ Theorem. Take points A1, A2, A3 on a line aand B1, B2, B3 on line b ‖ a, then draw 9 lines AjBj, i, j ∈ 1, 2, 3.For instance, in the diagram we set A1A2 : A2A3 : B1B2 : B2B3 =2 : 2 : 3 : 6, so among these lines no two are parallel and no threeconcurrent. By Pappus’ Theorem, the 98 lines determine 18 intersec-tion points which are collinear in triples - so these determine another 6lines. Together with these 6 lines, we have 15 lines and 24 triple inter-sections. Moreover, three lines obtained by Pappus’ Theorem meet ina point (denoted by K), which gives us 25 triple intersections. Drawone more line through two double intersection points only. The set Pof the 16 drawn lines and set T consisting of the 27 triple intersectionpoints and the 39 remaining double intersection points determine 159incident pairs.

59

Serbian MO 2011

60

8.4 Solutions to the Rhodes University Tests

Rhodes University Camp April 2012: Test 1Solutions

1. We know that

(a− b)2 + (ab+ 1)2 = a2b2 + a2 + b2 + 1 = (a2 + 1)(b2 + 1),

so it will be enough to show that a2 + 1 and b2 + 1 are relatively prime.Suppose that there is a prime number p that divides both a2 + 1 andb2 + 1. Then p divides a2− b2, and hence p divides one of a+ b or a− b.If p divides a− b, then p divides ab− b2, and since p divides b2 + 1, itdivides ab−b2 +b2 +1 = ab+1, which is impossible, since it is assumedthat a− b and ab+ 1 are relatively prime. If p divides a+ b, a similarcontradiction is derived.

Iranian Mathematical Olympiad 2011

2. Let Ai, 2 6 i 6 n, be the set of all classes with i participants. We willshow that |Ai| 6 n(n−1)

i(i−1). Since every pair of students is together in at

most one class of each Ai, we have that(i2

)|Ai| 6

(n2

), giving

|Ai| 6(n2

)(i2

) =n(n− 1)

i(i− 1).

Now, the total number of classes equals

|A2|+ |A3|+ · · ·+ |An|

6 n(n− 1)

(1

2(2− 1)+

1

3(3− 1)+ · · ·+ 1

n(n− 1)

)= n(n− 1)

(1− 1

n

)= (n− 1)2,

as required.


3. Consider the quantity xy− x+1

y+1= x−y

y(y+1). We need to show that

x− yy(y + 1)

+y − zz(z + 1)

+z − x

x(x+ 1)> 0,

61

or, equivalently, that

x

y(y + 1)+

y

z(z + 1)+

z

x(x+ 1)> y

y(y + 1)+

z

z(z + 1)+

x

x(x+ 1).

However, for any ordering of x, y and z, the numbers 1x(x+1)

, 1y(y+1)

and1

z(z+1)are reversely ordered (since x, y and z are positive), so the last

inequality above follows from the rearrangement inequality.

Russian Mathematical Olympiad 2011

4.

F

A

C

B

K

T

E

Let T be the intersection of BF and CE. We have that ∠BCE +∠BEC = ∠ABC, and since BC = BE, it follows that ∠BCE =12∠ABC. Similarly, ∠CBF = 1

2∠ACB. Hence

∠CTF = ∠BCE + ∠CBF =1

2(∠ABC + ∠ACB) = 60,

and so ABTC is a cyclic quadrilateral. Hence ∠EBF = ∠ACE =∠AKE and ∠ABF = 180−∠EBF = 180−∠AKE = ∠AKF , whichimplies that ABKF is cyclic. Hence we have that ∠EBK = ∠CFK,and since ACKE is cyclic, we have ∠BEK = ∠FCK. Together withBE = FC it follows that triangles KBE and KFC are congruent.HenceKC = KE, which means that ∠CAK and ∠BAK are subtendedby equal chords in the circumcircle of ACE, and hence are equal, asrequired.

Alternative solution by Juhno Son

Let L be the intersection of EF and the angle bisector of ∠BAC. LetBC = a, AC = b and AB = c. Then EB = FC = a.

Using the cosine rule in triangle ABC, we obtain a2 = b2 + c2− bc, andusing the cosine rule in triangle AEF , we obtain

EF 2 = (a+ c)2 + (a+ b)2− (a+ c)(a+ b) = a2 + b2 + c2 + ab+ ac− bc.

62

Since AL bisects angle AEF , it follows that ELFL

= AEFA

= a+ca+b

, which

yields FL = a+b2a+b+c

FE. Now,

FL · FE − FC · FA =a+ b

2a+ b+ cFE · FE − a(a+ b)

=a+ b

2a+ b+ c(a2 + b2 + c2 + ab+ ac− bc)− a(a+ b)

=a+ b

2a+ b+ c(b2 + c2 − bc− a2)

= 0 (cosine rule in triangle ABC).

Hence FL ·FE = FC ·FA, which shows that ACLE is a cyclic quadri-lateral, i.e. L = K.

F

A

C

B

LE

c

a

a

a

b


63


1.

P R

A

U

V

CB

D

Since PQ and QR are diameters of Γ1 and Γ2, we have ∠PAQ =∠QCR = 90, so PA ‖ RC, from which it follows that PV ‖ RU .Similarly, PU ‖ RV , and so PURV is a parallellogram. Since thediagonals of a parallellogram bisect each other, UV passes through themidpoint of PR.

Austrian Mathematical Olympiad 2011

2. We have(4− 2

1

)(4− 2

2

). . .

(4− 2

n

)=

n∏k=1

(4k − 2

k

)=

n∏k=1

2

(2k − 1

k

)=

n∏k=1

(2k − 1

k· 2k

k

)=

(2n)!

(n!)2=

(2n

n

),

which is an integer for each n.

Czech and Slovak Mathematical Olympiad 1999

3. Suppose that f(a + b) 6= f(a) + f(b) for some real numbers a and b.From the given equation it then follows that f(a+ b) = −f(a)− f(b).

If f(a + b) = 0, then f(a) + f(b) = −f(a + b) = 0 = f(a + b),contradicting our choice of a and b, so we may assume that f(a+b) 6= 0.Then |f(2a+ 2b)| = 2|f(a+ b)| 6= 0 and

|f(2b+ 2b)| = |f(a) + f(a+ 2b)|

64

= |f(a) + f(a+ b) + f(b)| or |f(a)− f(a+ b)− f(b)|= 0 or 2|f(a)|= 2|f(a)|,

since |f(2a+ 2b)| 6= 0.

Similarly, |f(2a + 2b)| = 2|f(b)|, from which it follows that |f(a)| =|f(b)|.If f(a) = −f(b), then f(a+ b) = −f(a)− f(b) = 0, a contradiction. Iff(a) = f(b), then |f(a)| = |f(a + b)| = |f(a) + f(b)| = 2|f(a)|, whichmeans that |f(a+ b)| = |f(a)| = 0, a contradiction. Hence we concludethat no such numbers a and b exist, and f(x+ y) = f(x) + f(y) for allreal x and y.

Thai Mathematical Olympiad 2011

4. Without loss of generality, we may assume that the n points form aregular n-gon. We will partition all the possible chords of the n-goninto n different sets such that all the chords in each set are parallel.We distinguish two cases:

If n is odd, then each chord is parallel to exactly one of the sides of then-gon, so define Ai to be the set of chords parallel to side i.

If n = 2k is even, let the n-gon be P1P2 . . . P2k. Then each chord isparallel to a pair of sides, or parallel to a chord of the form PiPi+2 forsome i. For 1 6 i 6 k, let Ai be the set of chords parallel to side PiPi+1,and for k < i 6 2k, let Ai be the set of chords parallel to Pi−2Pi.

Since we have nk + 1 chords, at least k + 1 of these chords must lie inone of the n sets Ai. Since the chords in each set Ai are parallel, notwo of them intersect.


65


1. Suppose that the digits of 2a can be rearranged to form 2b, with a > b.Then, since the two numbers have the same digit set, it follows thatthey’re congruent modulo 9. Hence 9 | 2a − 2b = 2b(2a−b − 1) and so2a−b ≡9 1. However, the smallest positive power of 2 with this propertyis 26, so a − b > 6. But in that case, 2a > 2b+6 = 64 · 2b > 10 · 2b,which means that 2a and 2b can’t have the same number of digits, acontradiction. So no such pair exists.


2. When the train leaves station k, the passengers on board the train arethose that are travelling from station i to j, with i 6 k < j. Hence thereare k(2n−k) people on board the train at that point, with a maximumof n2 between stations n and n + 1. Moreover, at station k there arek−1 passengers disembarking, and 2n−k passengers boarding, so thereneeds to be

(2n− k)− (k − 1) = 2(n− k) + 1

seats free in the car for 1 6 k 6 n. After station n, 2(k−n)−1 seats areleft free. So in order for everyone to have a seat up to station n, thereneeds to be at least

∑nk=1[2(n− k) + 1] = n2 seats on the train. Since

the number of passengers never exceed n2, this number will suffice.

Italian Mathematical Olympiad 2011

3. Setting y = x, we obtain 2xf(xf(x)) = 2x2f(x), and so f(xf(x)) =xf(x) since x > 0.

Now suppose f(x) = f(y). Then

x2(f(x) + f(y)) = f(yf(x))(x+ y) = f(yf(y))(x+ y) = yf(y)(x+ y)

and so

2x2f(x) = (xy+y2)f(y) =⇒ 2x2−xy−y2 = (2x+y)(x−y) = 0 =⇒ x = y,

since all the quantities in the equations are positive and so y 6= −2x. Sof is injective. Morever, setting x = y = 1, we obtain f(f(1)) = f(1),and so injectivity implies that f(1) = 1.

Now set x = 1 in the given equation:

f(y)(y + 1) = 1 + f(y) =⇒ f(y) =1

y,

66

and it is a simple matter to check that the solution f(x) = 1x

satisfiesthe given equation.

Swiss Mathematical Olympiad 2005

4.B

D

A C

N

KL

M

Q

P

Draw NM ‖ KL ‖ AC as in the figure. From the similarity of thetriangles BOQ and BLK it follows that

LK

OQ=BL

BO= 1 +

LO

BO,

From the similarity of the triangles DOC and DLK it folllows that

LK

OC=DL

DO= 1− LO

DO.

Therefore

LK

(1

OQ− 1

OC

)= LO

(1

BO+

1

DO

).

Hence

LK

LO=DO +BO

DO ·BO· OC ·OQOC −OQ

=BD

DO ·BO· OC ·OQ

QC.

Similarly we obtain

NM

OM=

BD

DO ·BO· OC ·OQ

QC.

Therefore, since ∠NMO = ∠KLO and LKLO

= NMOM

, triangles ONM andOKL are similar. Hence ∠NOM = ∠KOL. Since MOL is a straightline, it follows that so is NOK.

Belarusian Mathematical Olympiad 2007

67


1. If x and z have opposite signs, or if one of them is 0, then the inequalityis obviously true, so x, y and z are either all positive or all negative.If they are all negative, we may replace x, y and z with −x, −y and−z without changing the conditions of the problem, so we can assumethat x, y and z are all positive.

In that case, 0 < xy 6 xz 6 yz, and so 1 = xy+ yz+xz > xy+ 2xz >2xz which yields xz < 1

2.

Setting x = y = 1n

for some natural number n > 2, the equation

xy + yz + xz = 1 implies that z = n2−12n

> 1n

if n > 2. Hence xz =n2−12n2 = 1

2− 1

2n2 , which can be made arbitrarily close to 12. This shows

that 12

cannot be replaced by any smaller number.

Mediterranean Mathematical Competition 2007

2.A

B

Q

C

P

F

R

H

T

Let T be the intersection of BH and PC and F be the intersection ofAB and CH. Then

∠HQA = 90 − ∠BAQ= 90 − ∠ABP (since PB ‖ AQ)

= 90 − ∠ACP= ∠RTC= ∠ART (since PC ‖ AR)

68

which shows that AHQR is a cyclic quadrilateral. Hence

∠ARQ = ∠AHF = ∠ABC = ∠APC.

Now, since AR ‖ PC, it follows that AP ‖ QR.

New Problems in Euclidian Geometry

3. (a) Let AsBsCs be the given triangles, with s ∈ 1, 2, . . . , 2012 andlet XY Z stand for the half-plane with the boundary line XYand interior point Z. Each of the triangles is the intersection of thethree half-planes AsBsCs, BsCsAs and CsAsBs, hence theintersection of all the triangles is the intersection of all such half-planes. Since the half-planes AsBsCs, 1 6 s 6 2012 differ onlybe translation, their intersection is the half-plane AiBiCi for acertain i. Similarly, the intersection of the half-planes BsCsAsis the half-plane BjCjAj for some j and the intersection ofthe half-planes CsAsBs is the half-plane CkAkBk for some k.Hence the intersecion of all the half-planes is the intersection ofthe three half-planes AiBiCi, BjCjAj and CkAkBk, whichis a triangle ABC which is similar to the original triangles, sinceits sides are parallel to the sides of the original triangles. Let thecoefficient of similarity be λ. Clearly 0 < λ 6 1; we wish to showthat λ > 1

3.

Let v be the altitude from the vertex Ci to the side AiBi in tri-angle AiBiCi. Since the line AiBi coincides with the line AB,the distance of the centroid Gi of triangle AiBiCi from the lineAB equals 1

3v. Since every triangle contains the centroid of all

the triangles, triangle ABC contains all of them as well, and inparticular, ABC contains the centroid Gi, which means that thedistance between C and the line AB is at least 1

3v. Comparing

the altitudes from the vertices Ci and C in the similar trianglesAiBiCi and ABC, we thus obtain immediately that λ > 1

3, and

so the area of ABC is at least 19.

(b) Let AsBsCs be any one of the given triangles. The triangle ABC iscontained in it, and so the vertex As lies in the half-plane BCAat a distance at most v from the line BC, where v is the length ofthe altitude from the vertex As in triangle AsBsCs. On the otherhand, the distance of the side BsCs from A is also at most v. Sincethe triangle ABC contains centroids of all the given triangles, thedistance between the side BsCS and BC cannot exceed 1

3v. The

distance of the vertex A from the side BC equals λv, so altogether

69

the distance between the parallel lines BC and BsCs is at mostmin1

3, 1−λ·v. All the given triangles thus lie in the belt bounded

by the two parallel lines a0 and a1 parallel to BC whose distancefrom BC is v and min1

3, 1−λ·v, respectively. A similar assertion

clearly holds for the other two sides AC and AB, so the union ofthe 2012 triangles lie in the intersection of the three belts.

We now distinguish two cases, depending on the value of min13, 1−

λ.If 1

36 λ < 2

3, then the intersection of the belts is a hexagon

which is obtained from the triangle T determined by the tripleof lines (a1, b1, c1) upon removing the three small triangles Ta, Tband Tc determined by the triples of lines (a0, b1, c1), (a1, b0, c1) and(a1, b1, c0) respectively. From the way the lines a0, b0, c0, a1, b1, c1are defined, it follows that the T is similar to the given trian-gles with coefficient of similarity equal to 1 + λ and the trianglesTa, Tb, Tc are also similar to the given triangles with coefficient ofsimilarity λ− 1

3.

The area of the hexagon is therefore given by (recalling that λ < 23)

S = (1 + λ)2 − 3

(λ− 1

3

)2

=8

3− 2(λ− 1)2 <

8

3− 2

9=

22

9.

If 236 λ 6 1, then the intersection of the three belts is again a

hexagon, and the corresponding triangle T is similar to the given

70

triangles with coefficient of similarity 3 − 2λ, and the trianglesTa, Tb, Tc have coefficients of similarity 1 − λ. The area of thehexagon in this case is given by

(3− 2λ)2 − 3(1− λ)2 = (λ− 3)2 − 3 6 49

9− 3 =

22

9,

with equality if λ = 23.

However, since we only start with a finite number of triangles,the side a0 of the hexagon can only contain a finite number ofvertices of the given triangles (and cannot coincide with one ofthese triangles’ sides), and so the entire area of the hexagon cannotbe covered with only finitely many of these triangles, hence theinequality is strict.


4. We prove the assertion by induction on n. Put p1(x) = x + 1 andp2(x) = 6− 2x, which satisfies the conditions of the problem for n = 1and n = 2.

Suppose that pn(x) is polynomial such that pn(1), pn(2), . . . , pn(n) aredistinct powers of 2. We claim that gcd(pn(n + 1), n!) is a power of 2.Indeed, if there is a prime number 2 < q 6 n that divides pn(n + 1),then q divides pn(n+ 1− q) (a power of 2), which is a contradiction.

Let 2m be the greatest power of 2 that divides n!. So if a = pn(n+ 1),then gcd(a, n!

2m ) = 1. By Euler’s theorem, there exists a natural numberk such that ak − 1 is divisible by n!

2m , so ak − 1 = t·n!2m for some integer

t. Now define

pn+1(x) = 2mpn(x)k − t(x− 1)(x− 2) . . . (x− n).

For 1 6 i 6 n, we have pn+1(i) = 2mpn(i)k, which is a power of 2, and

pn+1(n+ 1) = 2mak − t · n! = 2m(ak − t · n!

2m

)= 2m.

71

Since the pn(i), 1 6 i 6 n are distinct powers of 2 and k is positive, itfollows that the pn+1(i), 1 6 i 6 n+ 1 are all distinct powers of 2.


72


1. Make the substitution ak = bk + k for k = 1, 2, . . . , n. The conditionsof the problem then becomes bm + bm+1 + · · · + bn > 0 for every m =1, 2, . . . , n, while the required inequlity becomes

n(n+ 1)(2n+ 1)

66 (b1 + 1)2 + (b22 + 2)2 + · · ·+ (bn + n)2

= (b21 + b22 + · · ·+ b2n) + (12 + 22 + · · ·+ n2) + 2(b1 + 2b2 + · · ·+ nbn).

But b21 + . . . b2n > 0, 12 + 22 + · · ·+ n2 = n(n+1)(2n+1)6

while

b1+2b2+· · ·+nbn = bn+(bn−1+bn)+(bn−2+bn−1+bn)+· · ·+(b1+b2+· · ·+bn) > 0

by the (restated) given condition.

Romanian Mathematical Olympiad 2011

2.

P

AB

S

C

MD

Let S be the intersection of PD and AB. Then S lies on the radicalaxis of the two circles, so AS = SB, that is, PS is a median in trianglePAB.

Now, by the tan-chord theorem,

∠BAP = 180 − ∠ADP = ∠CDP = ∠CBP.

Also, ∠ABP = ∠BCP , which implies that 4PAB ||| 4PBC. SincePM is a median in triangle BPC, it follows that ∠DPB = ∠MPC.Hence

∠DPM = ∠DPB+∠BPM = ∠MPC+∠BPM = ∠BPC = ∠BDC.

73

Dutch IMO Team Selection Test 2010

3. No three chips of the same colour should be in the same pile, so each pilemust contain a chip of each colour. So essentially we need to partition2n chips into two piles of n each. This can be done in pn = 1

2

(2nn

)ways.

Now,

pn =1

2

(2n

n

)=

(2n)!

2(n!)2

= 1 · 3 · · · · · (2n− 1)2 · 4 · · · · · 2n

2(n!)2

= 1 · 3 · · · · · (2n− 1)2nn!

2(n!)2

= 1 · 3 · · · · · (2n− 1)2n−1

n!.

Now, let m be a positive integer such that 2m 6 n < 2m+1. Then thegreatest integer a such that 2a divides n! is

a = bn2c+ b n

22c+ b n

23c+ · · ·+ b n

2mc

6 n

2+n

22+n

23+ · · ·+ n

2m

= n− n

2m6 n− 1.

Now, pn is odd exactly when a = n−1, which happens if only if n = 2m.


4. It is easy to see that p is odd and p 6= q, so p > 3 and (p, q) = 1.

If q = 2, then 2p+1 = 7+p2. The only solution is p = 3, as 2n+1 > 7+n2

for n > 4. For q > 3, by Fermat’s Little Theorem we get that qp ≡p q,so p | 2qp − 7 ≡ 2q − 7 and similarly, q | p+ 7. Let p+ 7 = kq for somepositive integer k.

If 2q − 7 6 0, we have q = 3 and p | −1, which is false.

If 2q − 7 > 0, then 2q − 7 > p, so 2q > p + 7 > kq, therefore k = 1 ork = 2.

For k = 1 we obtain p+ 7 = q, which is impossible for two odd primesp and q, so k = 2 and p + 7 = 2q. Suppose p > q. Since p, q > 3, weget qp > pq and then 7 = 2qp − pq > pq > 27, a contradiction. Thus

74

q > p and then p+ 7 = 2q > 2p which yields p = 3 or p = 5. For p = 3we have q = 5, while p = 5 gives q | 12, with no solution.

The two solutions are thus (p, q) = (3, 2) and (3, 5).

Romanian Mathematical Olympiad

75

8.5 Solutions to the Pure Joy Tests

Pure Joy Camp July 2012: Test 1Solutions

1. Since the equation is symmetric in m and n, we may assume thatm > n, and it suffices to show that m

n< 2.

Rewrite the equation as (x+m)(x+ n− 1) = n. Since x is an integer,this expresses n as a product of two integers, which means that bothfactors must lie in one of the intervals [−n,−1] or [1, n]. Either way,we have that (x+m)− (x+n− 1) 6 n− 1 which leads to m 6 2n− 2,from which it follows that m

n6 2− 2

n< 2.


2. The point D, being the foot of the perpendicular from L, is an interiorpoint of side AB, and similarly, E is an interior point of AC. The pointO is thus inside triangle ADE, hence ∠COB < ∠DOE.

α

A

B C

O

ED

L

O′

Let ∠CAB = α. Then ∠COB = 2∠CAB = 2α since O is the cir-cumcentre of triangle OBC. Similarly, since L is the circumcentre oftriangle DOE, ∠DOE = 180 − 1

2∠DLE. From the kite ADLE we

have ∠DLE = 180 − ∠DAE = 180 − α. Combining this, we obtain

2α = ∠COB < ∠DOE = 180 − 12(180 − α) = 90 +

α

2,

which implies α < 60.

Let O′ be the reflection of O in the line BC. In the quadrilateralABO′C we have

∠CO′B + ∠CAB = ∠COB + ∠CAB = 2α + α < 180,

76

so the point O′ is outside the circumcircle of ABC. Hence O and O′

are two points of ω such that one of them lies inside the circumcircle,while the other one lies outside. Therefore, the two circles intersect intwo points.

Proposed at IMO 2011

3. Note that the statement of the problem is invariant under translationsof x; hence without loss of generality we may suppose that the numbersd1, d2, . . . , d9 are positive.

We shall prove that N = d8 satisfies the desired property, where d =maxd1, d2, . . . , d9. Suppose that there is some integer x > N suchthat P (x) is composed of primes less than 20 only. Then for every indexi ∈ 1, 2, . . . , 9 the number x + di can be expressed as a product ofthe first 8 primes.

Since x + di > x > d8, there is some prime power fi > d that dividesx + di. Invoking the pigeonhole principle we see that there are twodistinct indices i and j such that fi and fj are powers of the sameprime; suppose that fi 6 fj. Now both the numbers x+ di and x+ djare divisible by fi and hence so is their difference di − dj. But as

0 < |di − dj| 6 maxdi, dj 6 d < fi,

this is impossible.


77


1. Substitute x = −t and y = 2t to obtain

g(f(t)) = f(−t). (8)

Next, let y = −x to obtain

g(f(0)) = f(x) + xg(−x)

=⇒ f(x) = g(f(0))− xg(−x)

=⇒ f(x) = c− xg(−x), (9)

where c = g(f(0)).

Substitute equation (8) with t = −x−y and (9) into the given equationto obtain

f(−x− y) = f(x) + (2x+ y)g(x)

=⇒ c+ (x+ y)g(x+ y) = c− xg(−x) + (2x+ y)g(y)

=⇒ (x+ y)g(x+ y) = (2x+ y)g(y)− xg(−x). (10)

Set y = 0 into this equation to obtain

xg(x) = 2xg(0)− xg(−x)

=⇒ g(−x) = 2g(0)− g(x) (11)

for x 6= 0, but x = 0 is readily seen to satisfy the equation as well.

Substitute (11) into (10) to obtain

(x+ y)g(x+ y) = (2x+ y)g(y)− x [2g(0)− g(x)] .

Exchanging x and y leaves the left-hand side of the equation unchangedand so we must have

(2x+ y)g(y)− x [2g(0)− g(x)] = (2y + x)g(x)− y [2g(0)− g(y)]

=⇒ 2xg(y)− 2xg(0) = 2yg(x)− 2yg(0)

=⇒ g(y)− g(0)

y=

g(x)− g(0)

x

for all x and y. Hence each side of the above equation must be constant,and so g(x) = ax+ b for some constants a and b.

78

Substitute this into (9) to obtain f(x) = ax2 − bx + c. Substitutingthese expressions for f and g into (8), we obtain

a(ax2−bx+c)+b = ax2+bx+c =⇒ a(a−1)x2−b(a+1)x+ca−c+b = 0

for all real numbers x. Hence a = 0 or a = 1. If a = 0, then b = c = 0,and f and g are identically 0, which satisfies the given equation.

If a = 1, then b = 0 and c ∈ R. The corresponding functions aref(x) = x2 + c and g(x) = x which satisfy the given equation.


2. Introduce a Cartesian coordinate system in the plane. Every circle hasan equation of the form p(x, y) = x2 + y2 + q(x, y) = 0, where q(x, y)is a polynomial of degree at most 1. For any point A = (xA, yA), wehave p(xA, yA) = d2 − r2, where d is the distance from A to the centreof the circle and r is the radius of the circle.

For each i ∈ 1, 2, 3, 4, let pi(x, y) = x2 + y2 + qi(x, y) = 0 be theequation of the circle with centre Oi and radius ri and let di be thedistance from Ai to Oi. Consider the equation

4∑i=1

pi(x, y)

d2i − r2

i

= 1. (12)

Since the coordinates of the points A1, A2, A3 and A4 satisfy (12) butthese four points do not lie on a circle or on a line, equation (12) definesneither a circle, nor a line. Hence, the equation is an identity and sothe coefficient of x2 + y2 has to be zero, i.e.

4∑i=1

1

d2i − r2

i

= 0.

Alternative solution

Let M be the intersection of the diagonals A1A3 and A2A4. On eachdiagonal, choose a direction and let x, y, z and w be the signed distancesfrom M to the points A1, A2, A3 and A4, respectively.

Let ω1 be the circumcircle of A2A3A4 and let B1 be the second intersec-tion point of ω1 and A1A3 (thus, B1 = A3 if and only if A1A3 is tangentto ω1). Since the expression O1A

21 − r2

1 is the power of the point A1

with respect to ω1, we get

O1A21 − r2

1 = A1B1 · A1A3.

79

On the other hand, from the equality MB1 ·MA3 = MA2 ·MA4 weobtain MB1 = yw

z. Hence we have

O1A21 − r2

1 =(ywz− x)

(z − x) =z − xz

(yw − xz).

Substituting the analogous expressions into the sought sum we get

4∑i=1

1

OiA2i − r2

i

=1

yw − xz

(z

z − x− w

w − y+

x

x− z− y

y − w

)= 0.

A2

A4

A1

A3

M

B1 x

y

z

w


3. A pair (a, n) satisfying the conditions of the problem will be called awinning pair. It is straightforward to check that the pairs (1, 1), (3, 1)and (5, 4) are winning pairs.

Now suppose that a is a positive integer not equal to 1, 3 or 5. Wewill show that there are no winning pairs (a, n) by distinguishing threecases.

Case 1: a is even. In this case we have a = 2αd for some positiveinteger α and some odd d. Since a > 2α, for each positive integer nthere exists an i ∈ 0, 1, . . . , a− 1 such that n + i = 2α−1e, where eis some odd integer. Then we have t(n+ i) = e and

t(n+ a+ i) = t(2αd+ 2α−1e) = 2d+ e ≡ e+ 2 mod 4.

So we get t(n+a+ i)− t(n+ i) ≡ 2 mod 4, and (a, n) is not a winningpair.

Case 2: a > 8 is odd. For each positive integer n, there exists ani ∈ 0, 1, . . . , a− 5 such that n+ i = 2d for some odd d. We get

t(n+ i) = d 6≡ d+ 2 = t(n+ i+ 4) mod 4,

80

and

t(n+ a+ i) = n+ a+ i ≡ n+ a+ i+ 4 = t(n+ a+ i+ 4) mod 4.

Therefore the integers t(n+a+i)−t(n+i) and t(n+a+i+4)−t(n+i+4)cannot both be divisible by 4, and therefore there are no winning pairs.

Case 3: a = 7. For each positive integer n, there exists an i ∈0, 1, . . . , 6 such that n + i is either of the form 8k + 3 or of theform 8k + 6, where k is a nonnegative integer. But we have

t(8k + 3) ≡ 3 6≡ 1 ≡ 4k + 5 = t(8k + 3 + 7) mod 4

andt(8k + 6) = 4k + 3 ≡ 3 6≡ 1 ≡ t(8k + 6 + 7) mod 4.

Hence, there are no winning pairs of the form (7, n).


81


1. The given relation implies that

f(f g(n)(n)

)< f(n+ 1) for all n, (13)

which will turn out to be sufficient to determine f .

Let k > 0 be an integer such that f(k) = minn>0f(n), and supposethat k 6= 1. Then, by (13), we have that

f(f g(k−1)(k − 1)

)< f(k),

contradicting the minimality of f(k). Hence f(1) is the minimum valueof f , and f(n) > f(1) for all n > 1.

Similarly, if k > 1 is an integer such that f(k) = minn>1f(n), thenk 6= 2 leads to f

(f g(k−1)(k − 1)

)< f(k), which implies that f g(k−1)(k−

1) = 1 6 f(1). This leads to k − 1 = 2, a contradiction. Hencef(1) < f(2) < f(n) for all n > 2. Inductively, one obtains that f isstrictly increasing, and in particular, f(n) > n.

Suppose that there is an integer m such that f(m) > m. Then f(n) > nfor all n > m, and hence

f g(m)+1(m) > f g(m)(m) > · · · > f(m).

However, by (13), we have f g(m)+1(m) < f(m + 1). Hence we havesqueezed a function value inbetween f(n) and f(n+1), a contradiction.Hence f(n) = n. Substituting this into the given equation, one obtains

n+ gn(n) = n+ 2− g(n+ 1) =⇒ g(n+ 1) = 2− gn(n) 6 1.

This implies that g(n) = 1 for all n. The pair (f, g) = (n, 1) satisfiesthe given equation.


2. If AB = AC, then the statement is trivial. So without loss of generalitywe may assume that AB < AC. Denote the tangents to Ω at points Aand X by a and x, respectively.

Let Ω1 be the circumcircle of triangle AB0C0. The circles Ω1 and Ωare homothetic with centre A, so they are tangent at A, and a is theirradical axis. Now, the lines a, x and B0C0 are the three radical axes of

82

the circles Ω, Ω1 and ω. Since a 6‖ B0C0, these three lines are concurrentat some point W .

B

X

C

T

W

O

A0

A

B0

G

C0

D

The points A and D are symmetric with respect to B0C0; hence WX =WA = WD. This means that W is the centre of the circumcircle γ oftriangle ADX. Moreover, we have ∠WAO = ∠WXO = 90, where Odenotes the centre of Ω. Hence ∠AWX + ∠AOX = 180.

Denote by T the second intersection point of Ω and line DX. Notethat O belongs to Ω1. Using the circles γ and Ω, we find that

∠DAT = ∠ADX − ∠ATD= 1

2(360 − ∠AWX)− 1

2∠AOX

= 180 − 12(∠AWX + AOX)

= 90.

So AD ⊥ AT , and hence AT ‖ BC. Thus, ATCB is an isoscelestrapezoid inscribed in Ω.

Denote by A0 the midpoint of BC, and consider the image of ATCBunder the homothety h with centre G and factor −1

2. We have h(A) =

A0, h(B) = B0 and h(C) = C0. From the symmetry about B0C0, wehave ∠TCB = ∠CBA = ∠B0C0A = ∠DC0B0. Using AT ‖ DA0, weconclude that h(T ) = D. Hence the points D, G and T are collinear,and X lies on the same line.


83

3. For m = 1, the answer is clearly equal to 12, so assume m > 1. In the

sequel, the word collision will be used to denote the meeting of exactlytwo ants, moving in opposite directions.

If at the beginning we place an ant on the southwest corner squarefacing east and an ant on the southeast corner square facing west, thenthey will meet in the middle of the bottom row at time m−1

2. After

the collision, the ant that moves to the north will stay on the boardfor another m − 1

2time units and thus the last ant falls off at time

m−12

+ m − 12

= 3m2− 1. We will now prove that this is the latest

possible moment.

Consider any collision of two ants a and a′. Let us change the rule forthis collision, and enforce these two ants to turn anticlockwise. Thenthe succeeding behaviour of all the ants does not change; the only differ-ence is that a and a′ swap positions. These arguments may be appliedto any collision separately, so we may assume that at any collision,either both ants rotate clockwise or both of them rotate anticlockwiseby our own choice.

In particular, we may assume that there are only two types of ants,depending on their initial direction: NE-ants, which move only north oreast, and SW-ants, moving only south and west. Then we immediatelyobtain that all ants will have fallen off the board after 2m − 1 timeunits. However, we can get a better bound by considering the lastmoment at which a given ant collides with another ant.

Choose a coordinate system such that the corners of the board are(0,0), (m, 0), (m,m) and (0,m). At time t, there will be no NE-antsin the region (x, y) |x + y < t + 1 and no SW-ants in the region(x, y) |x+ y > 2m− t− 1. So if two ants collide at (x, y) at time t,we have

t+ 1 6 x+ y 6 2m− t− 1. (14)

Analogously, we may change the rules so that each ant would movealternatingly north and west, or alternatingly south and east. By doingso, we find that apart from (14) we also have |y − x| 6 m − t − 1 foreach collision at point (x, y) at time t.

To visualise this, put

B(t) = (x, y) ∈ [0,m]2 | t+1 6 x+y 6 2m−t−1 and |x−y| 6 m−t−1..

The following picture displays B(t) for t = 12

and t = 72

in the casem = 6:

84

An ant can thus only collide with another ant at time t if it happensto be in the region B(t).

Now suppose that a NE-ant has its last collision at time t and thatit does so at point (x, y) (if the ant does not collide at all, it will falloff the board within m − 1

2< 3m

2− 1 times units, so this case can be

ignored). Then we have (x, y) ∈ B(t) and thus x + y > t + 1 andx− y > −(m− t− 1), so we get

x > (t+ 1)− (m− t− 1)

2= t+ 1− m

2.

By symmetry we also have y > t + 1 − m2

, and hence minx, y >t + 1 − m

2. After this collision, the ant will move directly to an edge,

which will take at most m−minx, y units of time. In sum, the totalamount of time the ant stays on the board is at most

t+ (m−minx, y) 6 t+m−(t+ 1− m

2

)=

3m

2− 1.

By symmetry, the same holds for SW-ants as well.


85


1. Since quadrilateral KBCM is cyclic, we have ∠CMB = ∠CKB andhence ∠AML = ∠AKL. However, since AMLK is cyclic, we have∠AML+∠AKL = 180, which leads to ∠CMB = ∠CKB = ∠AML =∠AKL = 90. Hence BM and CK are altitudes, L is the triangle’sorthocentre, and the triangle is necessarily acute. The two circumcir-cles of qudrilaterals KBCM and AKLM have the same radius, henceAL = BC.

Let the angles of triangle ABC be α, β and γ, denoted in the usualway. We have θ = ∠KCB = ∠KMB = ∠KML = ∠KAL, and so thetriangles KBC and KLA are congruent, which give CK = AK, and sotriangle AKC is right isosceles. This leads to α = 45 and γ = 45+θ =45+90−β = 135−β. Hence (α, β, γ) = (45, 45+θ, 90−θ), where0 < θ < 90.

Any triangle with angles (45, 45 + θ, 90 − θ), θ ∈ (0, 45) is seen tosatisfy the conditions of the problem.

A B

C

L

K

M


2. Let f be a function satisfying the given condition. For each integer m,the function g defined by g(x) = f(x) +m satisfies the same condition.Therefore we may assume that f(0) = 0 by subtracting f(0) from f(x)if necessary.

For any prime p, the condition on f with (x, y) = (p, 0) states that f(p)divides pn. Since the set of primes is infinite, there exist integers d andε with 0 6 d 6 n and ε ∈ −1, 1 such that for infinitely many primesp we have f(p) = εpd. Denote the set of these primes by P . Since afunction g satisfies the given condition if and only if −g satisfies thesame condition, we may suppose that ε = 1.

The case d = 0 can be ruled out, since 0 does not divide any nonzerointeger. Suppose d > 1 and write n = md + r, where m and r are

86

integers such that m > 1 and 0 6 r 6 d − 1. Let x be an arbitraryinteger. For each prime p ∈ P , the difference f(p)−f(x) divides pn−xn.Using the equality f(p) = pd, we get

pn − xn = pr(pd)m − xn ≡ prf(x)m − xn ≡ 0 mod pd − f(x).

Since we have r < d, for large enough primes p ∈ P we obtain

|prf(x)m − xn| < pd − f(x).

Hence prf(x)m − xn has to be zero. This implies r = 0 and xn =(xd)m = f(x)m. Since m is odd, we obtain f(x) = xd.

Hence, the only possible functions f are f(x) = εxd + c where ε ∈−1, 1, d is a positive divisor of n and c is an integer. These functionsall satisfy the condition of the problem.


3. There are various examples showing that k = 3 does have the desiredproperty. For example, one can take

A1 = 1, 2, 3 ∪ 3m |m > 4A2 = 4, 5, 6 ∪ 3m− 1 |m > 4A3 = 7, 8, 9 ∪ 3m− 2 |m > 4.

To check that this partition fits, we notice that the sums of two distinctelements of A1 represent all numbers n > 1 + 12 = 13, those of A2

represent all numbers n > 4 + 11 = 15 and those of A3 represent allnumbers n > 7 + 10 = 17. We are left to express 15 and 16 as sums oftwo elements of A3, and these are given by 15 = 7 + 8 and 16 = 7 + 9.

Let us now suppose that for some k > 4 there exist sets A1, A2, . . . , Aksatisfying the given property. Clearly, the sets A1, A2, A3, A4 ∪ . . . Akalso satisfy the same property, so we may assume that k = 4.

But Bi = Ai ∩ 1, 2, . . . , 23 for i = 1, 2, 3, 4. Now for any indexi each of the ten numbers 15, 16, . . . , 24 can be written as the sumof two distinct elements of Bi. Therefore, this set needs to contain atleast 5 elements. As we have |B1|+ |B2|+ |B3|+ |B4| = 24, there has tobe some index j for which |Bj| = 5. Let Bj = x1, x2, x3, x4, x5. Fi-nally, the sums of two distinct elements of Aj representing the numbers15, 16, . . . , 24 should be exactly all the pairwise sums of the elementsof Bj. Calculating the sum of these numbers in two different ways, wereach

4(x1 + x2 + x3 + x4 + x5) = 15 + 16 + · · ·+ 24 = 195,

87

which is impossible, since 195 is not divisible by 4. This contradictioncompletes our solution.


88

8.6 Solutions to the IMO

IMO in Mar Del Plata, Argentina 2012: Day 1Solutions

1. We have KM ⊥ BJ hence BM is parallel to the bisector of ∠ABC.Therefore ∠BMK = 1

2∠ABC and ∠FMB = 1

2∠ACB. Denote by X

the intersection of KM and FJ . From the triangle FXM we derive:

∠XFM = 90 − ∠FMB − ∠BMK =1

2∠BAC.

The points K and M are symmetric with respect to the line FJ hence∠KFJ = ∠JFM = 1

2∠BAC = ∠KAJ , therefore K, J , A, and F

belong to a circle which implies that ∠JFA = ∠JKA = 90 andKM‖AS. The quadrilateral SKMA is a trapezoid and from ∠SMK =∠AKM we obtain SM = AK. Similarly, we get AL = TM . SinceAK = AL (tangents from A to the excircle) we get SM = TM .

2. The inequality between arithmetic and geometric mean implies

(ak + 1)k =

(ak +

1

k − 1+

1

k − 1+ · · ·+ 1

k − 1

)k> kk · ak ·

1

(k − 1)k−1

=kk

(k − 1)k−1· ak

The inequality is strict unless ak = 1k−1

. Multiplying analogous in-equalities for k = 2, 3, · · · , n yields

(a2 + 1)2 · (a3 + 1)3 · · · (an + n)n >22

11· 33

22· 44

33· · · nn

(n− 1)n−1· a2 · a3 · · · an

= nn.

3. The game can be reformulated in an equivalent one: The player Achooses an element x from the set S (with |S| = N) and the player Basks the sequence of questions. The j-th question consists of B choosinga set Dj ⊆ S and player A selecting a set Pj ∈

Qj, Q

Cj

. The player

A has to make sure that for every j > 1 the following relation holds:

x ∈ Pj ∪ Pj+1 ∪ · · · ∪ Pj+k.

89

The player B wins if after a finite number of steps he can choose a setX with |X| 6 n such that x ∈ X.

(a) It suffices to prove that if N > 2k + 1 then the player B candetermine a set S ′ ⊆ S with |S ′| 6 N − 1 such that x ∈ S ′.Assume that N > 2n + 1. In the first move B selects any setD1 ⊆ S such that |D1| > 2k−1 and |DC

1 | > 2k−1. After receivingthe set P1 from A, B makes the second move. The player B selectsa set D2 ⊆ S such that |D2 ∩ PC

1 | > 2k−2 and |DC2 ∩ PC

1 | > 2k−2.The player B continues this way: in the move j he/she chooses aset Dj such that |Dj ∩ PC

j | > 2k−j and |DCj ∩ PC

j | > 2k−j.

In this way the player B has obtained the sets P1, P2, . . ., Pk suchthat (P1 ∪ · · · ∪ Pk)C > 1. Then B chooses the set Dk+1 to be asingleton containing any element outside of P1 ∪ · · · ∪ Pk. Thereare two cases now:

Case 1. The player A selects Pk+1 = DCk+1. Then B can take

S ′ = S \Dk+1 and the statement is proved.

Case 2. The player A selects Pk+1 = Dk+1. Now the playerB repeats the previous procedure on the set S1 = S \ Dk+1 toobtain the sequence of sets Pk+2, Pk+3, . . ., P2k+1. The followinginequality holds:

|S1 \ (Pk+2 · · ·P2k+1)| > 1,

since |S1| > 2k. However, now we have∣∣∣(Pk+1 ∪ Pk+2 ∪ · · · ∪ P2k+1)C∣∣∣ > 1,

and we may take S ′ = Pk+1 ∪ · · · ∪ P2k+1.

(b) Let p and q be two real numbers such that 1.99 < p < q < 2. Letus choose k0 such that(

p

q

)k06 2 ·

(1− q

2

)and pk − 1.99k > 1.

We will prove that for every k > k0 if |S| ∈(1.99k, pk

)then there

is a strategy for the player A to select sets P1, P2, . . . (based onsets D1, D2, . . . provided by B) such that for each j the followingrelation holds:

Pj ∪ Pj+1 ∪ · · · ∪ Pj+k = S.

90

Assuming that S = 1, 2, . . . , N, the player A will maintain thefollowing sequence of N -tuples: (x)∞j=0 =

(xj1, x

j2, . . . , x

jN

). Ini-

tially we set x01 = x0

2 = · · · = x0N = 1. After the set Pj is selected

then we define xj+1 based on xj as follows:

xj+1i =

1, if i ∈ Pj

q · xji , if i 6∈ Pj.

The player A can keep B from winning if xji 6 qk for each pair(i, j). For a sequence x, let us define T (x) =

∑Ni=1 xi. It suffices

for player A to make sure that T (xj) 6 qk for each j.

Notice that T (x0) = N 6 pk < qk.

We will now prove that given xj such that T (xj) 6 qk, and aset Dj+1 the player A can choose Pj+1 ∈

Dj+1, D

Cj+1

such that

T (xj+1) 6 qk. Let y be the sequence that would be obtained ifPj+1 = Dj+1, and let z be the sequence that would be obtained ifPj+1 = DC

j+1. Then we have

T (y) =∑

i∈DCj+1

qxji + |Dj+1|

T (z) =∑

i∈Dj+1

qxji +∣∣DC

j+1

∣∣ .Summing up the previous two equalities gives:

T (y) + T (z) = q · T(xj)

+N 6 qk+1 + pk, hence

min T (y) , T (z) 6 q

2· qk +

pk

26 qk,

because of our choice of k0.

91

IMO in Mar Del Plata, Argentina 2012: Day 2Solutions

4. Substituting a = b = c = 0 yields 3f(0)2 = 6f(0)2 which impliesf(0) = 0. Now we can place b = −a, c = 0 to obtain f(a)2 + f(−a)2 =2f(a)f(−a), or, equivalently (f(a)− f(−a))2 = 0 which implies f(a) =f(−a).

Assume now that f(a) = 0 for some a ∈ Z. Then for any b we havea+b+(−a−b) = 0 hence f(a)2+f(b)2+f(a+b)2 = 2f(b)f(a+b), whichis equivalent to (f(b)− f(a+ b))2 = 0, or f(a + b) = f(b). Thereforeif f(a) = 0 for some a 6= 0, then f is a periodic function with period a.

Placing b = a and c = 2a in the original equation yields f(2a) ·(f(2a)− 4f(a)) = 0. Choosing a = 1 we get f(2) = 0 or f(2) = 4f(1).

If f(2) = 0, then f is periodic with period 2 and we must have f(n) =f(1) for all odd n. It is easy to verify that for each c ∈ Z the function

f(x) =

0, 2 | n,c, 2 6| n

satisfies the conditions of the problem.

Assume now that f(2) = 4f(1) and that f(1) 6= 0. Assume thatf(i) = i2 · f(1) holds for all i ∈ 1, 2, . . . , n (it certainly does fori ∈ 0, 1, 2). We place a = 1, b = n, c = −n − 1 in the originalequation to obtain:

f(1)2 + n4f(1)2 + f(n+ 1)2 = 2n2f(1)2 + 2(n2 + 1)f(n+ 1)f(1)

⇔(f(n+ 1)− (n+ 1)2f(1)

)·(f(n+ 1)− (n− 1)2f(1)

)= 0.

If f(n+1) = (n−1)2f(1) then setting a = n+1, b = 1−n, and c = −2in the original equation yields

2(n− 1)4f(1)2 + 16f(1)2 = 2 · 4 · 2(n− 1)2f(1)2 + 2 · (n− 1)4f(1)

which implies (n−1)2 = 1 hence n = 2. Therefore f(3) = f(1). Placinga = 1, b = 3, and c = 4 into the original equation implies that f(4) = 0or f(4) = 4f(1) = f(2). If f(4) 6= 0 we get

f(2)2 + f(2)2 + f(4)2 = 2f(2)2 + 4f(2)f(4)

hence f(4) = 4f(2). We already have that f(4) = f(2) and this im-plies that f(2) = 0, which is impossible according to our assumption.

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Therefore f(4) = 0 and the function f has period 4. Then f(4k) = 0,f(4k+ 1) = f(4k+ 3) = c, and f(4k+ 2) = 4c. It is easy to verify thatthis function satisfies the requirements of the problem.

The remaining case is f(n) = n2f(1) for all n ∈ N, or f(n) = cn2 forsome c ∈ Z. This function satisfies the given condition.

Thus the solutions are: f(x) = cx2 for some c ∈ Z; f(x) =

0, 2 | n,c, 2 6| n

for some c ∈ Z; and f(x) =

0, 4 | n,c, 2 - n,

4c, n ≡ 2 (mod 4)for some c ∈ Z.

5. Since AL2 = AC2 = AD ·AB, the triangles ALD and ABL are similarand hence ∠ALD = ∠XBA. LetR be the point on the extension ofDCover point C such that DX ·DR = BD·AD. Since ∠BDX = ∠RDA =90 we conclude4RAD ∼ 4BXD, hence ∠XBD = ∠ARD, therefore∠ALD = ∠ARD and the points R, A, D, and L belong to a circle. Thisimplies that ∠RLA = 90 hence RL2 = AR2 − AL2 = AR2 − AC2.Analogously we prove that RK2 = BR2 − BC2 and ∠RKB = 90.Since RC ⊥ AB we have AR2 − AC2 = BR2 − BC2, therefore RL2 =RK2 hence RL = RK. Together with ∠RLM = ∠RKM = 90 weconclude 4RLM ∼= 4LKM hence MK = ML.

6. Let M = maxa1, . . . , an. Then we have 3M = 1 · 3M−a1 + 2 · 3M−a2 +

· · · + n · 3M−an ≡ 1 + 2 + · · · + n = n(n+1)2

(mod n). Therefore, the

number n(n+1)2

must be odd and hence n ≡ 1 (mod 4) or n ≡ 2 (mod4).

We will now prove that each n ∈ N of the form 4k + 1 or 4k + 2 (forsome k ∈ N) there exist integers a1, . . ., an with the described property.

For a sequence a = (a1, a2, . . . , an) let us introduce the following nota-tion:

L(a) =1

2a1+

1

2a2+· · ·+ 1

2anand R(a) =

1

3a1+

2

3a2+· · ·+ n

3an.

Assume that for n = 2m + 1 there exists a sequence a = (a1, . . . , an)of non-negative integers with L(a) = R(a) = 1. Consider the sequencea′ = (a′1, . . . , a

′n+1) defined in the following way:

a′j =

aj, if j 6∈ m+ 1, 2m+ 2

am+1 + 1, if j ∈ m+ 1, 2m+ 2.

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Then we have

L (a′) = L(a)− 1

2am+1+ 2 · 1

2am+1+1= 1

R (a′) = R(a)− m+ 1

3am+1+

m+ 1

3am+1+1+

2m+ 2

3am+1+1= 1.

This implies that if the statement holds for 2m + 1, then it holds for2m+ 2.

Assume now that the statement holds for n = 4m+ 2 for some m > 2,and assume that a = (a1, . . . , a4m+2) is the corresponding sequenceof n non-negative integers. We will construct a following sequencea′ =

(a′1, a

′2, . . . , a

′4m+13

)that satisfies L (a′) = R (a′) = 1 thus proving

that the statement holds for 4m+ 13. Define:

a′j =

am+2 + 2 if j = m+ 2aj + 1 if j ∈ 2m+ 2, 2m+ 3, 2m+ 4, 2m+ 5, 2m+ 6a j

2+ 1 if j ∈ 4m+ 4, 4m+ 6, 4m+ 8, 4m+ 10, 4m+ 12

am+2 + 3 if j ∈ 4m+ 3, 4m+ 5, 4m+ 7, 4m+ 9, 4m+ 11, 4m+ 13aj otherwise.

We now have

L (a′)

= L(a)− 1

2am+2−

6∑j=2

1

2a2m+j+

1

2am+2+2+

6∑j=2

1

2a2m+j+1+

6∑j=2

1

2a2m+j+1+ 6 · 1

2am+2+3

= 1.

It remains to verify that R (a′) = R(a) = 1. We write

R (a′)−R(a)

= R

(a′m+2, a′4m+3, a′4m+5, a′4m+7, a′4m+9, a′4m+11, a′4m+13

m+ 2, 4m+ 3, 4m+ 5, 4m+ 7, 4m+ 9, 4m+ 11, 4m+ 13

)−R

(am+2

m+ 2

)+

6∑j=2

(R

(a′2m+j, a′4m+2j

2m+ j, 4m+ j

)−R

(a2m+j

2m+ j

)),

where

R

(c1, . . . , ckd1, . . . , dk

)=

d1

3c1+ · · ·+ dk

3ck.

For each j ∈ 2, 3, 4, 5, 6 we have

R

(a′2m+j, a′4m+2j

2m+ j, 4m+ j

)−R

(a2m+j

2m+ j

)=

2m+ j

3a2m+j+1+

4m+ 2j

3a2m+j+1−2m+ j

3a2m+j= 0.

94

The first term in the expression for R (a′) − R(a) is also equal to 0because

R

(a′m+2, a′4m+3, a′4m+5, a′4m+7, a′4m+9, a′4m+11, a′4m+13

m+ 2, 4m+ 3, 4m+ 5, 4m+ 7, 4m+ 9, 4m+ 11, 4m+ 13

)−R

(am+2

m+ 2

)=

m+ 2

3am+2+2+

6∑j=1

4m+ 2j + 1

3am+2+3− m+ 2

3am+2

= 0.

Thus R (a′) = 0 and the statement holds for 4m + 13. It remains toverify that there are sequences of lengths 1, 5, 9, 13, and 17. One wayto choose these sequences is:

(1), (2, 1, 3, 4, 4), (2, 3, 3, 3, 3, 4, 4, 4, 4), (2, 3, 3, 4, 4, 4, 5, 4, 4, 5, 4, 5, 5),

(3, 2, 2, 4, 4, 5, 5, 6, 5, 6, 6, 6, 6, 6, 6, 6, 5).

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Results of the South African team

1 2 3 4 5 6 Total Position

Kira Dusterwald 7 0 0 7 2 0 16 253rd Bronze MedalAshraf Moolla 7 0 0 2 7 0 16 253rd Bronze medalDylan Nelson 7 0 0 2 2 0 11 339th Hon. mentionRobert Spencer 5 0 0 7 2 0 14 303rd Hon. mentionDalian Sunder 3 0 0 6 2 0 11 339th

Sean Wentzel 7 1 2 7 7 1 25 66th Silver medal

Total 36 1 2 31 22 1 93 41st

The medal cutoffs were 16 for bronze, 22 for silver and 28 for a gold.

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imo book 2012

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