rmc brosura 2011 final

ROMANIAN MATHEMATICAL COMPETITIONS2011

Editorial Board of the RMC series

Marian AndronacheNational College “Sf. Sava” Bucharest

Mihai BalunaNational College “Mihai Viteazul” Bucharest

Mircea BecheanuUniversity of Bucharest

Bogdan EnescuNational College “ B.P. Hasdeu” Buzau

Radu Gologan – Coordinator of the seriesInstitute of Mathematics andUniversity “Politehnica” Bucharest

Calin PopescuInstitute of Mathematics

Dan SchwarzIntuitext Company

Dinu SerbanescuNational College “Sf. Sava” Bucharest

Copyright c© 2011 Societatea de Stiinte Matematice din Romania

All rights reserved. No part of this publication may be reproduced or distributed in anyform or by any means, or stored in a database or retrieval system, without the priorwritten permission of the publisher.

Societatea de Stiinte Matematice din RomaniaStr. Academiei 14, 010014 Bucuresti, Romania

http://www.rms.unibuc.rotel/fax: +40213124072

ISSN 2066–6446

ROMANIAN MATHEMATICAL COMPETITIONS

2011

Edited by MIHAIL BALUNA and RADU GOLOGAN

Contributors MARIAN ANDRONACHE, MIHAIL BALUNA, ANDREI ECKSTEIN

RADU GOLOGAN, CEZAR LUPU, CALIN POPESCU, DINU SERBANESCU

TABLE OF CONTENTS

Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

1.1. 2011 Romanian Mathematical Olympiad – District Round . . . . . . . . . . . . . . . . . . . . . .1

1.2. 2011 Romanian Mathematical Olympiad – Final Round . . . . . . . . . . . . . . . . . . . . . . .13

1.3. Shortlisted problems for the 2011 Romanian NMO . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.4. Selection tests for the 2011 BMO and IMO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

1.5. Selection tests for the 2011 JBMO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

1.6. 2011 Balkan Mathematical Olympiad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

1.7. 2010 – 2011 Local Mathematical Competitions

1.7.1. The 2010 DANUBE Mathematical Competition . . . . . . . . . . . . . . . . . . . . . . . .63

1.7.2. The 2010 Eighth IMAR1 Mathematical Competition . . . . . . . . . . . . . . . . . . 67

1.7.3. The 2010 Fourth STARS OF MATHEMATICS Competition . . . . . . . . . . . . . . .73

1.7.4. The 2011 Fourth ROMANIAN MASTER OF MATHEMATICS . . . . . . . . . . . . . 81

1.7.5. The 2011 CLOCK-TOWER SCHOOL Seniors Competition . . . . . . . . . . . . . . 93

1.6.6. The 2011 CLOCK-TOWER SCHOOL Juniors Competition . . . . . . . . . . . . . . 97

1INSTITUTE OF MATHEMATICS OF THE ROMANIAN ACADEMY. One-day IMO-type contest.

iii

FOREWORD

The 18th volume of the ”Romanian Mathematical Contests” series contains morethan 200 problems, submitted at different stages of the Romanian Mathematical Olym-piad, other Romanian Contests, and some international ones. Most of them are origi-nal, but some problems from other sources were used as well during competition.

A significant part of the problems are discussed in detail, and alternative solutionsor generalizations are given. Some of the solutions belong to students and were givenwhile they sat the contest; we thank them all.

We thank the Ministry of Education, Research, Youth and Sports for constant in-volvement in supporting the Olympiads and the participation of our teams in interna-tional events.

Special thanks are due to the ”Dinu Patriciu” Foundation, to WBS, and to theCouncil of the District of Dambovita – all sponsors of the Romanian IMO team.

The Editors

v

THE 62nd ROMANIAN MATHEMATICAL OLYMPIAD

DISTRICT ROUND

7th GRADE

Problem 1. In a square of side length 60, 121 distinct points are given. Show thatamong them there exists three points which are vertices of a triangle with an area notexceeding 30.

Solution. Divide the square into 60 rectangles 5×12. By the pigeonhole principle,there are three points among the given ones inside one of the rectangles. The area ofthis triangle does not exceed half of the area of the rectangle, that is 5 · 12/2 = 30, asneeded.

Problem 2. Consider an isosceles trapezoid ABCD with perpendicular diago-nals. The parallel from the intersection point of the diagonals meets the non-parallelsides [BC] and [AD] at points P and R respectively. Point Q is the mirror image ofP across the midpoint of [BC]. Show that

a) QR = AD; b) QR ⊥ AD.

Claudiu Popa

Solution. Let the diagonals AC and BD meet at point O and let M be the mid-point of the line segment [BC].a) Since OM joins the midpoints of two sides ofthe triangle PQR, MO ‖ RQ and OM = RQ

2 .On the other hand, [OM ] is a median of the right-angled triangle BOC, hence OM = 1

2BC =12AD. Consequently, RQ = AD.

b) Let the lines MO and AD meet at T . Then ∠MBO = ∠MOB = ∠DOT .Since ∠OCB = ∠TDO, it follows that ∠OTD = ∠BOC = 90, so MT ⊥ AD.

1

2 2011 ROMANIAN MATHEMATICAL OLYMPIAD – DISTRICT ROUND

Problem 3. A positive integer N has the digits 1, 2, 3, 4, 5, 6 and 7, so that eachdigit i, i ∈ 1, 2, 3, 4, 5, 6, 7 occurs 4i times in the decimal representation of N .Prove that N is not a perfect square.

Solution. N has 1 ·4 = 4 digits equal to 1, 2 ·4 = 8 digits equal to 2, . . ., 7 ·4 = 28

digits equal to 7, so the sum of its digits equals S = 4(12 + 22 + · · · + 72) = 560.Since 560 = 3 · 186 + 2, the number N is not a square, and the remainder left by aperfect square upon division by 3 cannot be equal to 2.

Problem 4. Find the sum of the elements of the set

M =n

2+

m

5| m,n = 0, 1, 2, . . . , 100

.

Gazeta Matematica

Solution. Consider the set A = 2a + 5b | a, b = 1, 2, . . . , 100 and notice that1 /∈ A and 3 /∈ A.

The largest even number from A is equal to 700 and it is obtained for a = b = 100.The number 698 is obtained for a = 99, b = 100. The largest odd number from A

is equal to 695 and it is obtained for b = 99, a = 100, implying that 697 /∈ A and699 /∈ A.

We claim that all the integers between 4 and 695 belong to A.Let y ≤ 500 and let r be the remainder left by y upon division by 5. Write

y = 5c + r with 0 ≤ c ≤ 100 and 0 ≤ r ≤ 4. If r is even, then y = 5c + 2k, wherer = 2k. If r is odd, then y ≥ 5, so c ≥ 1 and y = 5(c − 1) + 2(k + 3), wherer = 2k + 1.

For y > 500, write y = 500+ z with z ≥ 200. If z is even, then y = 5 · 100+ 2k,where z = 2k. If z is odd, then y ≤ 695, so z ≤ 195 and y = 5 ·99+2(k+3), wherez = 2k + 1. A quick inspection of all the above cases shows that the claim holds.

The sum of all the elements of M is equal to

S =1

10((1 + 2 + · · ·+ 700)− (1 + 3 + 697 + 699) = 350 · 697) = 35 · 697.

8th GRADE

Problem 1. Find the real numbers x and y such that

(x2 − x+ 1)(3y2 − 2y + 3)− 2 = 0.Petre Batranetu

2011 ROMANIAN MATHEMATICAL OLYMPIAD – DISTRICT ROUND 3

Solution. From x2−x+1 = (x− 12 )

2+ 34 ≥ 3

4 , 3y2−2y+3 = 3(y− 13 )

2+ 83 ≥ 8

3

we derive that (x2 − x+ 1)(3y2 − 2y + 3) ≥ 34 · 8

3 = 2, for any real numbers x andy. Equality holds for (3y − 1)2 = 0 and (x− 1

2 )2 = 0, that is x = 1

2 and y = 13 .

Problem 2. a) Show that m2−m+1 is an element of the set n2+n+1 | n ∈ N,for any positive integer m.

b) Let p be a perfect square, p > 1. Prove that there exists positive integers r andq such that p2 + p+ 1 = (r2 + r + 1)(q2 + q + 1).

Mircea Fianu

Solution. a) Since m2 − m + 1 = (m − 1)2 + (m − 1) + 1 and m − 1 ≥ 0, itfollows that m− 1 ∈ N and consequently m2 −m+ 1 ∈ n2 + n+ 1 | n ∈ N.

b) Write p = k2, where k is an integer. Since p > 1, we have k ≥ 2. Nowp2 + p+ 1 = k4 + k2 + 1 = (k2 + 1)2 − k2 = (k2 − k + 1)(k2 + k + 1). Numbersr = k and q = k − 1, both positive integer, satisfy the claim.

Problem 3. Consider a regular prism ABCA′B′C ′. A plane α containing pointA meet the rays (BB′ and (CC ′ at points E and F such that

area [ABE] + area [ACF ] = area [AEF ].

Find the angle determined by the planes (AEF ) and (BCC ′).

Gazeta Matematica

Solution. Let M be the midpoint of BC and let u be the angle determined by theplanes (AEF ) and (BCC ′). Since triangle MEF is the projection of the triangleAEF onto (BCC ′), we have

cosu =[MEF ]

[AEF ]=

[BCFE]

2[AEF ]=

[BCFE]

2([ABE] + [ACF ])=

=[BCFE]

4([MBE] + [MCF ])=

[BCFE]

2[BCFE]=

1

2,

hence u = 60.

Problem 4. Find all positive integers m such that

√m =

√m+ 2011.

Alexandru Blaga


Solution. Rewrite the equation√m − [

√m] =

√m+ 2011 − [

√m+ 2011] to

get√m+ 2011 −

√m = [

√m+ 2011] − [

√m] = p ∈ N. Squaring

√m+ 2011 =

p +√m yields 2011 = p2 + 2p

√m ∈ N, hence m = k2, k ∈ N∗. The relation

2011 = p(p + 2k) gives p = 1 and p + 2k = 2011, since 2011 is prime. Hencek = 1005 and m = 10052.

9th GRADE

Problem 1. Points M , N , P , Q are given on the sides AB, BC, CD, DA of a

parallelogram ABCD such that−−→MN +

−−→QP =

−→AC. Prove that

−−→PN +

−−→QM =

−−→DB.

Solution. Denote m = AMAB , n = BN

BC , p = DPDC , q = AQ

AD , to get−−→MN =

−−→AN −

−−→AM =

−−→AB+

−−→BN −

−−→AM = (1−m)

−−→AB+n

−−→AD,

−−→QP = p

−−→AB+ (1− q)

−−→AD.

The given condition becomes (1−m+p)−−→AB+(1− q+n)

−−→AD =

−→AC =

−−→AB+

−−→AD.

Notice that vectors−−→AB and

−−→AD have distinct directions to deduce that m = p and

n = q. On the other hand,−−→PN = (1−p)

−−→AB−(1−n)

−−→AD and

−−→QM = m

−−→AB−q

−−→AD,

implying−−→PN +

−−→QM = (1− p+m)

−−→AB − (1− n+ q)

−−→AD =

−−→AB −

−−→AD =

−−→DB, as

needed.

Problem 2. For each positive integer n consider the set An of all the numbersobtained by choosing signs in ±1 ± 2 ± · · · ± n; for instance, A2 = −3,−1, 1, 3and A3 = −6,−4,−2, 0, 2, 4, 6. Find the cardinal of the set An.

Solution. The largest element of the set is 1 + 2 + 3 · · ·+ n = n(n+1)2 , while the

smallest is −1− 2− · · · − n = −n(n+1)2 .

The difference of any two elements of the set is even, implying that all elementshave the same parity.

We claim that all numbers between −n(n+1)2 and n(n+1)

2 , sharing the same paritywith them belong to the set An – and only them. Indeed, let x ∈ An, x < n(n+1)

2 . If(one of) its representation begins with −1, changing into +1 gives x+ 2 ∈ An. Sup-pose that all its representations start with +1. There exists a first negative – otherwisex = 1 + 2 + · · ·+ n = n(n+1)

2 ; denote it by −k:

x = +1 + 2 + · · ·+ (k − 1)− k ± · · · ± n.

Switching the signs of the summands k − 1 and k we derive that x+ 2 ∈ An.Consequently, the set An has n(n+1)

2 + 1 elements.


Problem 3. Let f : R → R be a function such that its 2-fold composition is equalto the floor function, i.e. f(f(x)) = x, for any real number x. Prove that there existdistinct real numbers a and b such that |f(a)− f(b)| ≥ |a− b|.

Gazeta Matematica

Solution. We claim that f(n) ∈ Z, for any integer n. Indeed, write f(f(f(x))) =f([x]) = [f(x)] to derive that f(n) = [f(n)] for any integer n, implying f(n) ∈ Z.

Suppose that for all a, b ∈ Z we have |f(a) − f(b)| < |a − b|. Then |f(n +

1) − f(n)| < 1, for any integer n. Since both f(n) and f(n + 1) are integers wederive that f(n) = f(n + 1), hence f(n) = f(0), for any integer n. It follows thatn = f(f(n)) = f(f(0)) = 0, a contradiction.

Second solution. Notice that f(0) = f(1), otherwise 0 = f(f(0)) = f(f(1)) =

1, false. If |f(0) − f(1)| ≥ |0 − 1|, we are done; if else, from |0 − 1| = |f(f(0)) −f(f(1))| we obtain |f(f(0))− f(f(1))| > |f(0)− f(1)|. The distinct numbers f(0)and f(1) fulfill the claim.

Problem 4. Let a be a real number with a+ 1a = 1. Prove that

an+

1

an

= 1,

for every positive integer n, where x denotes the fractional part of a real number x.Dorel Mihet

Solution. Since a + 1a = a +

⌊1a

⌋+ 1, it follows that a + 1

a ∈ Z. For eachpositive integer n let sn = an + 1

an . We have s1sn = sn+1 + sn−1 for any integern ≥ 2. Notice that s2 = a2 + 1

a2 = (a+ 1a )

2 − 2 ∈ Z, hence, by induction, sn ∈ Z,for all positive integers n.

This implies an+ 1an = an + 1

an − an −⌊

1an

⌋= sn − an −

⌊1an

⌋∈ Z,

hence an+

1an

∈ 0, 1, for x ∈ [0, 1).

If an+

1an

= 0, then both numbers an and 1

an are integers, implying an = 1

and furthermore a = ±1. Then a +

1a

= 0 = 1, a contradiction. Consequently

an+

1an

= 1, for any positive integer n.

Remark. A number under the given condition is a solution to a + 1a = m ∈ Z,

implying a = m±√m2−42 , with |m| > 2. An alternative solution can be obtained

noticing that the Newton sums sn = an1+an2 of the roots of the equation a2−ma+1 =

0 satisfy the relation sn+2 −msn+1 + sn = 0, which, in fact, is an equivalent formof s1sn = sn+1 + sn−1.


10th GRADE

Problem 1. Prove that, if a, b, c are positive real numbers, then the equationax + bx = cx has at most a real solution.

Gazeta Matematica

Solution. Rewrite the equation as f(x) : =(ac

)x+(bc

)x= 1.

If (ac − 1)( bc − 1) ≤ 0, the equation has neither positive, nor negative solutions.If (ac −1)( bc−1) > 0, the function f is monotone, therefore the equation f(x) = 1

has at most one solution.

Problem 2. a) Let z1, z2, z3, z4 be distinct complex numbers of zero sum, havingequal absolute values. Prove that the points of affixes z1, z2, z3, z4 are the vertices ofa rectangle.

b) Let x, y, z, t be real numbers such that sinx + sin y + sin z + sin t = 0 andcosx+ cos y + cos z + cos t = 0. Prove that, for every integer n,

sin(2n+ 1)x+ sin(2n+ 1)y + sin(2n+ 1)z + sin(2n+ 1)t = 0.

Aurel Barsan

Solution. a) The equality z1 + z2 + z3 + z4 = 0 implies z1 + z2 + z3 + z4 = 0

and furthermore 1z1

+ 1z2

+ 1z3

+ 1z4

= 0 (1), for |z1| = |z2| = |z3| = |z4| = 0.Suppose z1 + z2 = −z3 − z4 = 0. The relation (1) gives z1z2 = z3z4, so

z1, z2 = −z3,−z4. On the other hand, if z1 + z2 = 0, then z3 + z4 = 0. In bothcases the numbers z1, z2, z3, z4 form two pair of equal sum, hence the conclusion.

b) Let z1 = cosx + i sinx, z2 = cos y + i sin y, z3 = cos z + i sin z and z4 =

cos t+ i sin t to get z1 + z2 + z3 + z4 = 0 and |z1| = |z2| = |z3| = |z4| = 1.As before, the numbers z1, z2, z3, z4 form two pairs of opposite numbers, so the

same goes for numbers z2n+11 , z2n+1

2 , z2n+13 , z2n+1

4 . Therefore z2n+11 + z2n+1

2 +

z2n+13 + z2n+1

4 = 0, implying the claim.

Problem 3. Let a and b be two complex numbers. Prove that the following state-ments are equivalent:

1) The absolute values of the roots of the equation x2−ax+b = 0 are respectivelyequal to the absolute values of the roots of the equation x2 − bx+ a = 0.

2) a3 = b3 or b = a.

Mihail Baluna


Solution. Let |x1| = |x3|, |x2| = |x4| (1) and notice that |a| = |x3x4| = |x1x2| =|b| to derive that |x1 + x2| = |x3 + x4| (2). The relations (1) and (2) show that thereexists a number k ∈ C such that x2 = kx1, x4 = kx3 or x2 = kx1, x4 = kx3.

In the first case we have a = kx23 = (1 + k)x1 and b = kx2

1 = (1 + k)x3, soa3 = k(1 + k)2x2

1x23 = b3.

In the latter case we have a = kx23 = (1 + k)x1 and b = kx2

1 = (1 + k)x3. Itfollows that x2

1x1 = x3x23, so x1 = x3 or a = b = 0, and furthermore x2 = x4, hence

a = b.Conversely, if b = a, then x1+x2 = x3+x4, x1x2 = x3x4, implying x1, x2 =

x3, x4. If a3 = b3, then a = εb, ε3 = 1. The roots satisfy the relations x1 + x2 =

ε(x3 + x4), x1x2 = ε2x3x4. Both cases lead to |x1|, |x2| = |x3|, |x4|, asneeded.

Problem 4. a) Show that if a, b > 1 are distinct real numbers then

loga(loga b) > logb(loga b).

b) Let a1 > a2 > . . . > an > 1 be real numbers, n ≥ 2. Prove that

loga1(loga1

a2)+loga2(loga2

a3)+. . .+logan−1(logan−1

an)+logan(logan

a1) > 0.

Solution. a) For a < b, loga(loga b) = (loga b)(logb(loga b)) > logb(loga b)

because logb(loga b) > 0 and loga b > 1.For a > b, the claim is reached from loga b < 1 and logb(loga b) < 0.b) Induct on n. For n = 2,

loga1(loga1

a2)+loga2(loga2

a1) > loga2(loga1

a2)+loga2(loga2

a1) = loga21 = 0.

Assume now that the claim holds for some n and consider the numbers a1 > a2 >

. . . > an+1 > 1. Then

loga1(loga1

a2) + . . .+ logan(logan

an+1) + logan+1(logan+1

a1) =

= loga1(loga1

a2) + . . .+ logan−1(logan−1

an) + logan(logan

a1)+

+ logan(logan


a1)− logan(logan

a1) >

> logan+1(logan


a1)− logan(logan

a1) =

= logan+1(logan

a1)− logan(logan

a1) > 0,

because logana1 > 1 and an+1 < an.


11th GRADE

Problem 1. Let α be an irrational number. For any n ∈ N∗ let an = nα anddefine the sequence (xn)n≥1 by xn = (a2 − a1)(a3 − a2) · · · (an+1 − an). Show thatthe sequence is convergent and find its limit.

Marian Cucoanes

Solution. We claim that x + y − y is equal to x or x − 1, for any realnumbers x and y. Write x + y = [y] + [x] + x + y to get x + y − y =

x + y − y. Hence if x + y < 1 then x + y − y = x, while if2 > x+ y ≥ 1 then x+ y − y = x − 1, as claimed.

Apply the above result to infer that either |an+1 − an| = α or |an+1 − an| =1− α. Set b = maxα, 1− α and notice that 0 < b < 1 (for α is irrational)to derive that |xn| ≤ bn, implying lim

n→∞|xn| = 0 and furthermore lim

n→∞xn = 0.

Problem 2. Consider the matrices A ∈ Mm,n(C), B ∈ Mn,m(C) with n ≤ m.Given that rank (AB) = n and (AB)2 = AB, find BA.

Marian Andronache

Solution. Left–multiply with B and right–multiply by A the equality (AB)2 =

AB to obtain (BA)3 = (BA)2. Recall that the rank of a matrix product does notexceed the rank of its factors to derive from ABAB = AB that rankBA ≥ n, hencerankBA = n.

The square matrix BA of order n is thus invertible, hence (BA)3 = (BA)2, whichimplies BA = In.

Problem 3. Let A,B ∈ M2(C) be two non-zero matrices with AB +BA = O2

and det(A+B) = 0. Prove that tr(A) = tr(B) = 0.

Gazeta Matematica

Solution. From AB+BA = O2 we obtain (A+B)2 = A2+B2 and (A−B)2 =

A2 + B2. Consequently, det(A − B) = 0. The characteristic equations for A + B

and A−B yield

A2 +B2 − tr(A+B)(A+B) = O2,

A2 +B2 − tr(A−B)(A−B) = O2,


so, subtracting we get tr(A)B = tr(B)A.

If tr(A) = 0, then tr(B) = 0 for else A = O2, a contradiction.

Hence A = λB, so AB+BA = O2 leads to λB2 = O2. Therefore λ = 0, whichyields tr(A) = tr(B) = 0.

Problem 4. Find all functions f : [0, 1] → R satisfying for all x, y ∈ [0, 1] theinequality |x− y|2 ≤ |f(x)− f(y)| ≤ |x− y|.

Nicoale Bourbacut

Solution. The condition |f(x)− f(y)| ≤ |x− y| is sufficient for the continuity off . From |x − y|2 ≤ |f(x) − f(y)| we derive that f is one-to-one, implying that f isa strictly monotonic function. We may assume that f is strictly increasing, because f

can be replaced by −f .

Set x = 0 and y = 1 to derive that 0 ≤ f(1)− f(0) ≤ 1, hence f(1) = f(0) + 1.

For x ≥ y we get f(x) − f(y) ≤ x − y or y − f(y) ≤ x − f(x), implying thatthe function g(x) = x− f(x) + f(0) is increasing.

Since g(0) = g(1) = 0, the function g is the identically null. Consequently, thefunctions are f±

a given by f±a (x) = ±x+ a, with a ∈ R.

Alternative solution. For x = 0 and y = 1 we get |f(1)−f(0)| = 1. For x ∈ [0, 1]

we obtain 1 = |f(1)− f(0)| ≤ |f(1)− f(x)|+ |f(x)− f(0)| ≤ |1− x|+ |1− x| =1− x+ x− 0 = 1.

Thus f(x) lies between f(0) and f(1); moreover, we have |f(1)− f(x)| = 1− x

and |f(x) − f(0)| = x. It follows that points (0, f(0)), (x, f(x)) and (1, f(1)) arecollinear, hence f(x) = f(0)± x.

12th GRADE

Problem 1. Show that1

π

∫ cos π13

sin π13

√1− x2 dx is a rational number.

Gazeta Matematica

Solution. Consider the function F : [0, π/2] → R, F (t) =

∫ cos t

sin t

√1− x2 dx.

The function is differentiable and the derivative is equal to

F ′(t) = (− sin t)√1− (cos t)2−(cos t)

√1− (sin t)2 = −(sin t)2−(cos t)2 = −1.


Thus F (t) = −t+k, k ∈ C. Since F (π/4) = 0, it follows that k = π/4, implyingF (t) = π/4 − t. Consequently, F (π/13)/π = (π/4 − π/13)/π = 9/52, which is arational number.

Problem 2. Let G be the set of matrices(a b

0 1

), a, b ∈ Z7, a = 0.

a) Show that G is a group with respect to the matrix multiplication.

b) Prove that there exists no proper homomorphism from G to Z7.

Solution. a) Consider A =

(a b

0 1

)and B =

(x y

0 1

)two elements of G. Then

AB =

(ax ay + b

0 1

)∈ G, for ax = 0.

To end the proof, recall that matrix multiplication is an associative law, the group

unit is I2 ∈ G and, finally, the inverse matrix of A is

(a−1 −a−1b

0 1

)∈ G.

b) Let A =

(a b

0 1

)be an element of G with a = 1. Since

Ak =

(ak b(ak−1 + ak−2 + · · ·+ 1)

0 1

), it follows that A6 = I2.

Let f be a group homomorphism mapping G to Z7. Notice that 0 = f(I2) =

f(A6) = 6f(A) to derive that f(A) = 0. Since ker(f) = X ∈ G | f(X) = 0is a subgroup of G with at least 36 elements, and G has 42 elements, we obtain thatker(f) = G. Therefore f(X) = 0, for any X ∈ G.

Problem 3. Consider an increasing continuous function f : [0, 1] → R and definethe sequence (an)n≥1 by an = 1

2n

∑2n

k=1 f(

k2n

), for all integers n ≥ 1.

a) Prove that the sequence (an)n≥1 is increasing.

b) Given that there exists p ∈ N∗ such that ap =

∫ 1

0

f(x)dx, prove that f is a

constant function.

Nicoale Bourbacut


Solution. a) Notice that

an+1 =1

2n+1

2n+1∑k=1

f

(k

2n+1

)=

1

2n+1

(2n∑k=1

f

(k

2n

)+

2n∑k=1

f

(2k − 1

2n+1

))

and f(2k−12n+1

)≤ f

(k2n

)to derive that

an+1 ≤ 1

2n

2n∑k=1

f

(k

2n

)= an.

b) Consider the numbers k ∈ 1, 2, . . . , 2p, c ∈(k − 1

2p,k

2p

), and a division of

the closed interval [0, 1]

∆ =

(0,

1

2p, · · · , k − 1

2p, c,

k

2p, · · · , 1

).

Denote by S the upper Darboux sum with relative to ∆ to get

ap − S =1

2pf

(k

2p

)− f(c)

(c− k − 1

2p

)− f

(k

2p

)(k

2p− c

)

=

(f

(k

2p

)− f(c)

)(c− k − 1

2p

)≥ 0.

Since∫ 1

0

f(x)dx ≤ S ≤ ap =

∫ 1

0

f(x)dx, it follows that ap = S and, further-

more, f(c) = f

(k

2p

). Therefore, f is a constant function on each of the intervals

(k − 1

2p,k

2p

], k = 1, 2, . . . , 2p, whose union is (0, 1]. Hence f is a constant function

on [0, 1], because the function f is continuous.

Problem 4. Let A be a ring and let a be an element of A. Prove thata) If A is commutative and a is nilpotent, then a+x is invertible for any invertible

element of x ∈ A.b) If A is finite and a+ x is invertible for any invertible element x ∈ A, then a is

nilpotent.(An element a of a ring is called nilpotent if there exists some positive integer n

such that an = 0.)

Marian Andronache


Solution. a) Let x be an invertible element and let n be a positive integer suchthat an = 0. Since a + x = x(x−1a + 1), it is enough to show that x−1a + 1 isinvertible. Set b = x−1a. Then bn = x−nan = 0, (because A is commutative), henceb2n+1 = 0. Consequently,

1 = b2n+1 + 1 = (b+ 1)(b2n − b2n−1 + · · · − b+ 1),

i. e., b+ 1 is invertible.b) Induct on n, n ≥ 1, to prove that an − 1 is invertible. For x = −1 we get

that a − 1 is invertible. Assume that b = an − 1 is invertible. From the hypothesis itfollows that a− b−1 is also invertible, and so is ab− 1 = (a− b−1)b. Moreover,

an+1 − 1 = a+ (a(an − 1)− 1) = a+ (ab− 1)

is invertible. Since A is finite, there exist two integers q > p ≥ 1 such that ap = aq ,i.e. ap(aq−p − 1) = 0. Since aq−p − 1 is invertible, it follows that ap = 0.

THE 62nd ROMANIAN MATHEMATICAL OLYMPIAD

FINAL ROUND

7th GRADE

Problem 1. Find all positive integers r with the property that there exists positiveprime numbers p and q so that p2 + pq + q2 = r2.

Aurel Barsan

Solution. The given relation is equivalent to (p+ q)2= r2 + pq, which can be

written (p+ q + r) (p+ q − r) = pq .The divisors of pq are 1, p, q and pq. Since p + q > max p, q, it follows that

p+ q − r = 1 and p+ q + r = pq.Adding the last two equalities yields 2p+2q = pq+1 , that is (p− 2) (q − 2) = 3.This leads to (p, q) ∈ (3, 5), (5, 3); in both cases r = 7.

Problem 2. The numbers x, y, z, t, a and b are positive integers, so that xt−yz =

1 and xy > a

b > zt . Prove that ab ≥ (x+ z) (y + t).

Dan Nedeianu

Solution. From xy > a

b follows that xb > ya, hence xb−ya ≥ 1. In the same way,at− bz ≥ 1. Multiplying the first inequality by t, the second one by y and adding thetwo relations yields bxt− byz ≥ t+ y, that is b ≥ t+ y.

In the same way, a ≥ x+ z. These two inequalities lead to the conclusion.Second solution. If d = g.c.d.(a, b), then a = da1 and b = db1, with a1, b1 ∈ N∗,

(a1, b1) = 1 and xy > a1

b1. It follows that b1x− a1y = u ≥ 1 and a1t− b1z = v ≥ 1,

with u, v ∈ N. The first relation gives b1xz − a1yz = uz, and the second givesa1xt− b1zx = vx. Adding these relations yields a1 = uz+ vx ≥ z+ x. In the sameway, b1 = ut+ vy ≥ t+ y. Since ab ≥ a1b1, the conclusion is proven.

13

14 2011 ROMANIAN MATHEMATICAL OLYMPIAD – FINAL ROUND

Problem 3. The convex quadrilateral ABCD has ∠BCD = ∠ADC ≥ 90. Thebisectors of the angles ∠BAD and ∠ABC meet at a point M , placed on the line CD.

Prove that M is the midpoint of the segment [CD].Maria Mihet

Solution. Case I: AD and BC have a common point E. Then M is the incenterof the triangle ABE, hence (EM is the bisector of the angle ∠AEB.

Since ∠ECD = ∠EDC, the triangle EDC is isosceles with base [DC]. There-fore [EM ] is a median in triangle EDC, so M is the midpoint of the segment CD.

Case II: AD ‖ BC. Then ∠CAB = ∠ABD = 90, hence ∠MAB+∠MBA =

90, that is triangle MAB has a right angle in M .Denote N the midpoint of the segment [AB]. Then triangle NAM is isosceles

with base [AM ], so ∠NMA = ∠NAM = ∠MAD, hence MN ‖ AD. It followsthat MN is the central median of the trapezoid ABCD, whence the conclusion.

Problem 4. Triangle ABC has ∠ABC = 60. Points M and D are placed onthe sides (AC) and (AB) respectively, so that ∠BCA = 2∠MBC and BD = MC.

Find the measure of the angle ∠DMB.Gheorghe Bumbacea

Solution. Take G so that CG = CM , C ∈ (BG) and denote x = ∠MBC. Since∠MCB is exterior to the isosceles triangle MCG, ∠MGC = x. This means thattriangle MBG is isosceles and similar to triangle MCG. Thus, BG

MG = MBMC = MG

BD .Denote H the common point of the line AB and

the perpendicular bisector of the segment [BG]. Thentriangle HBG is isosceles and has an angle of 60, soit is equilateral. This yields

BG

MG=

GH

MB=

MG

BD.

2011 ROMANIAN MATHEMATICAL OLYMPIAD – FINAL ROUND 15

Since ∠MGH = ∠MBD = 60 − x, triangles MGH and DBM are similar,therefore we get the answer ∠DMB = ∠MHG = 30.

8th GRADE

Problem 1. Find all real numbers x, y, z, t ∈ [0,∞) so that

x+ y + z ≤ t, x2 + y2 + z2 ≥ t and x3 + y3 + z3 ≤ t.

Cristian Lazar

Solution. Adding x+ y+ z ≤ t, −2x2− 2y2− 2z2 ≤ −2t and x3+ y3+ z3 ≤ t,one gets x (1− x)

2+ y (1− y)

2+ z (1− z)

2 ≤ 0.

Since x, y, z ∈ [0,∞), it follows x, y, z ∈ 0, 1.If x = y = z = 0, then t = 0.If exactly two of the numbers x, y, z are nil, then 1 ≤ t and 1 ≥ t, therefore t = 1,

(x, y, z) ∈ (1, 0, 0) , (0, 1, 0) , (0, 0, 1).If exactly one of the numbers x, y, z is nil, then 2 ≤ t and 2 ≥ t, so t = 2,

(x, y, z) ∈ (1, 1, 0) , (1, 0, 1) , (0, 1, 1).If x = y = z = 1, then t = 3.

Problem 2. Let a, b, c be distinct positive integers.a) Prove that a2b2 + a2c2 + b2c2 ≥ 9.b) If, moreover, ab+ ac+ bc+ 3 = abc > 0, show that

(a− 1)(b− 1) + (a− 1)(c− 1) + (b− 1)(c− 1) ≥ 6.

Gabriel Popa

Solution. a) At least one of the numbers a, b, c has modulus at least 2, whencea2b2 + a2c2 + b2c2 ≥ 1 · 1 + 1 · 4 + 1 · 4 = 9.

b) The required inequality can be successively written

ab+ ac+ bc− 2(a+ b+ c) + 3 ≥ 6

ab+ ac+ bc− 3 ≥ 2(a+ b+ c)

(ab+ ac+ bc− 3)(ab+ ac+ bc+ 3) ≥ 2abc(a+ b+ c)

(ab+ ac+ bc)2 − 9 ≥ 2abc(a+ b+ c)

a2b2 + a2c2 + b2c2 ≥ 9,

that is exactly a).


Problem 3. Let V ABC be a regular pyramid with the base ABC having centerO. Denote I and H the incenter and the orthocenter of the triangle V BC and supposethat AH = 3OI . Find the measure of the angle made by a lateral edge of the pyramidwith the plane of the base.

Mircea Fianu

Solution. Denote M the midpoint of the edge [BC]. Since triangle V BC is isosce-les with V B = V C, the points V,H, I and M are collinear.

From AC⊥OB and AC⊥OV follows AC⊥(BOV ), hence V B⊥AC, which, to-gether with V B⊥CH leads to V B⊥(ACH), whence AH⊥V B. In the same wayAH⊥V C, therefore AH⊥(V BC), so AH⊥VM .

Denote I ′ the projection of O onto theplane (V BC); then I ′ belongs to VM andOI ′ ‖ AH . This leads to

OI ′

AH=

MO

MA=

1

3,

hence AH = 3OI ′. This shows that OI

equals the distance from O to (V BC), therefore I = I ′.From OI⊥(V BC) and I = incenter of the triangle V BC follows that O has equal

distances to the lines V B and BC.Denote J the projection of O onto the line V B. Then OJ = OM , hence the right

triangles OJB and OMB are congruent.This leads to m(∠ (V B, (ABC))) = m(∠V BO) = m(∠MBO) = 30.

Problem 4. A positive integer will be called typical if the sum of its decimal digitsis a multiple of 2011.

a) Show that there are infinitely many typical numbers, each having at least 2011multiples which are also typical numbers.

b) Does there exist a positive integer such that each of its multiples is typical?

***

Solution. a) All the numbers made up by 2011n digits 1 (where n is a positiveinteger) are typical. Each such number, with 2011k digits, has among its multiples thenumbers made up by 2011kn digits, which are typical.

b) We will show that the answer is ’NO’.


We notice that each positive integer A has a multiple of the form 99 . . . 900 . . . 0.Indeed, if the decimal expansion of 1/A has a period of p digits and the initial non-periodical part has q digits, then 1

A = BC , where C = 99 . . . 900 . . . 0 has p digits 9

and q digits 0.Suppose now that n is a positive integer. Take a multiple of n of the form m =

99 . . . 900 . . . 0, with p nines and q zeros and consider the multiple of n

M = (10p+1 − 1)m = 10p+1m−m = 99 . . . 98900 . . . 0100 . . . 0,

with p − 1 nines at the beginning. Then the sums of the digits of these numbers ares(M) = s(m) + 9, so they cannot be both multiples of 2011.

9th GRADE

Problem 1. Let n be a positive integer and a1, a2, . . . , an be real numbers suchthat am + am+1 + · · ·+ an ≥ m+ (m+ 1) + · · ·+ n, for every m = 1, 2, . . . , n.

Prove that a21 + · · ·+ a2n ≥ n(n+ 1)(2n+ 1)

6.

Romeo Raicu

Solution. Denote bk = ak−k, for k = 1, 2, . . . , n. Then bm+bm+1+· · ·+bn ≥ 0

for m = 1, 2, . . . , n. It followsn∑

i=1

a2i =

n∑i=1

b2i +

n∑i=1

2ibi +

n∑i=1

i2 ≥n∑

i=1

i2 =n(n+ 1)(2n+ 1)

6,

becausen∑

i=1

ibi = (b1 + · · ·+ bn) + (b2 + · · ·+ bn) + · · ·+ bn ≥ 0.

Problem 2. Let n be a positive integer. Prove that every integer between 1 and n!

can be written as the sum of at most n distinct positive divisors of n!.

* * *

Solution. We use induction on n; the base case n = 1 is obvious.Suppose now that the conclusion holds for some positive integer n and consider

m ≤ (n+ 1)!. Then m = (n+ 1)q + r, q, r ∈ N, 0 ≤ r ≤ n. It is easily noticed thatq ≤ n!, therefore q can be written as a sum of (at most) n divisors of n! – denote themd1, . . . , dk, k ≤ n, whence

m = (n+ 1)d1 + (n+ 1)d2 + · · ·+ (n+ 1)dk + r


and (n+ 1)di are divisors of (n+ 1)!.If r = 0, the induction step is finished. Otherwise r > 0, r divides (n + 1)! and

r < n+1 ≤ (n+1)di, therefore m can be written as the sum of k+1 distinct divisorsof (n+ 1)!, with k + 1 ≤ n+ 1 and we are done.

Problem 3. Given an integer n > 0, prove that at least one of the numbers [2n√2],

[2n+1√2], . . . , [22n

√2] is even ([a] denotes the integral part of the real number a).

Radu Gologan

Solution. Suppose, by contradiction, that all the numbers are odd. Then thereexists a positive integer a such that

2a− 1 < 2n√2 < 2a. (1)

It follows that 4a− 2 < 2n+1√2 < 4a and, since [2n+1

√2] is odd,

4a− 1 < 2n+1√2 < 4a.

Continuing in the same way we arrive at 2n+1a − 1 < 22n√2 < 2n+1a, whence

2n+1a− 22n√2 < 1, which leads to

2n+1(a2 − 22n−1) < a+ 2n−1√2. (2)

On the other hand, 2a > 2n√2 gives a2 > 22n−1, implying a2 − 22n−1 ≥ 1.

Thus (2) leads to

a > 2n+1 − 2n−1√2 = 2n−1(4−

√2) > 2n−1(

√2 + 1) > 2n−1

√2 + 0.5,

which contradicts (1).

Problem 4. Let ABC be a triangle and Ia be the excenter corresponding to A.Let P and Q be the tangency points of the excircle of center Ia with the lines AB andAC. Line PQ meets lines IaB and IaC at the points D and E respectively. DenoteA1 the common point of the lines DC and BE; define in the same way points B1 andC1.

Prove that the lines AA1, BB1, CC1 are concurrent.

Severius Moldoveanu


Solution. We firstly prove that A1 is the orthocenter of the triangle IaBC.In order to do this, we prove that the qua-

drilateral BEIaP is cyclic. Indeed,

∠PEIa= 180 − ∠PQA− ∠ECQ

= 180−(90− ∠A

2

)−(90− ∠C

2

)

=∠A2

+∠C2

= 90 − 1

2∠B

= ∠PBIa.

So the quadrilateral BEIaP is cyclic and, since ∠BPIa = 90, BE⊥CIa.This leads to BA1 ‖ CI , where I is the incenter of triangle ABC. In the same

way CA1 ‖ BI , therefore BICA1 is a parallelogram.Analogously AIBC1 is a parallelogram, hence AC1A1C is also a parallelogram,

that is the segments [AA1] and [CC1] have the same midpoint. In the same way,segments [BB1] and [AA1] have the same midpoint, whence the conclusion.

10th GRADE

Problem 1. Let f : R → R be a function such that, for every x, y ∈ R:

|f(x+ y) + sinx+ sin y| ≤ 2.

a) Prove that |f(x)| ≤ 1 + cosx, for every x ∈ R.b) Give an example of such a function, which vanishes nowhere in the interval

(−π, π).***

Solution. a) x = t− π2 , y = π

2 yields f(t)− cos t+ 1 ≤ 2, for every t ∈ R.Also, x = t+ π

2 , y = −π2 leads to f(t) + cos t− 1 ≥ −2, for every t ∈ R.

The conclusion is obtained putting toghether the above.b) An example is f(x) = 2 − 2| sin x

2 |. A heuristic approach: the hypothesis isequivalent to

2 ≥ maxu∈R

(f(t) + 2 sin

t

2cos

u

2

)= f(t) + 2| sin t

2|

−2 ≤ minu∈R

(f(t) + 2 sin

t

2cos

u

2

)= f(t)− 2| sin t

2|,


that is |f(x)| ≤ 2− 2| sin x2 |, ∀x ∈ R.

Problem 2. Find all positive integers n for which there exists three complex rootsof order n of the unity, not necessarily different, adding up to 1.

Mihai Piticari

Solution. If n is odd, then −1, 1, 1 are three complex roots of order n of the unity,adding up to 1.

On the other hand, if x, y, z ∈ C, xn = yn = zn = 1 and x + y + z = 1,then |x| = |y| = |z|, hence x + y + z = 1/x + 1/y + 1/z = 1, which leads toxy + xz + yz = xyz.

Replacing z = 1 − x − y gives (x + y)(1 − x)(1 − y) = 0, whence one of thenumbers x, y, z is 1 and the other two are opposite.

In the case n = odd, there are no opposite roots, so the final answer is: n = even.

Problem 3. Let a, b, c be three positive real numbers. Prove that the function

f : R → R, f(x) =ax

bx + cx+

bx

ax + cx+

cx

ax + bxis increasing on [0,∞) and

decreasing on (−∞, 0].

Solution. We use straightforward computation: if x ≤ y are real numbers, then

f(y)− f(x) =∑cyc

ay(bx + cx)− ax(by + cy)

(bx + cx)(by + cy)

=∑cyc

(aybx − axby)

(1

(bx + cx)(by + cy)− 1

(ax + cx)(ay + cy)

)

=∑cyc

axbx(ay−x − by−x

)·

· ((ax+y − bx+y) + cx(ay − by) + cy(ax − bx))

(bx + cx)(by + cy)(ax + cx)(ay + cy),

so f(y) − f(x) can be written as a sum of products of the form (ap − bp)(aq − bq)

multiplied with positive coefficients, where p = y − x ≥ 0 and all the q-s have thesame sign as x, y.

We finish now by noticing that

(ap − bp)(aq − bq) = bp+q((a

b

)p

− 1)((a

b

)q

− 1)≥ 0 if q ≥ 0

≤ 0 if q ≤ 0

hence f(y)− f(x) ≥ 0 if y ≥ x ≥ 0 and f(y)− f(x) ≤ 0 if 0 ≥ y ≥ x.


Problem 4. a) Show that, for every positive integer n, there exists uniquely deter-mined positive integers xn, yn such that (1 +

√33)n = xn + yn

√33.

b) Prove that if xn, yn are defined as above and p is a positive prime, then at leastone of the numbers yp−1, yp, yp+1 is divisible by p.

Dorel Mihet

Solution. a) The equality is obtained for xn =(n0

)+ 33

(n2

)+ 332

(n4

)+ . . . and

yn =(n1

)+ 33

(n3

)+ 332

(n5

)+ . . ..

Also, if a + b√33 = c + d

√33, a, b, c, d ∈ N and b = d, then

√33 = a−c

d−b isrational – false – hence b = d and a = c, which proves that the above representationis unique.

b) If p = 2, 3 or 11, then yp is divisible by p.In the other cases, we start noticing that xn+1 = xn + 33yn, yn+1 = xn + yn.Also, xp ≡ 1 (mod p) and yp ≡ 33

p−12 (mod p), henceforth y2p ≡ 1 (mod p).

This gives p | x2p − y2p = (xp − yp)(xp + yp) = 32yp−1yp+1 and, since p is prime

and p = 2, p | yp−1 or p | yp+1.

11th GRADE

Problem 1. Call a row of a matrix in Mn(C) permutable if, for any permutationof its entries, the value of the determinant does not change. Prove that any matrix thathas two permutable rows is singular.

Marian Andronache

Solution. Consider A ∈ Mn(C) such that row l is permutable. Denote by Γli

the algebraic complement of ali, i = 1, ..., n. Suppose i, j, k, p ∈ 1, 2, ..., n aresuch that ali = alj and Γlk = Γlp. Consider matrices B and C obtained from A bypermuting elements in row l, such that B contains ali in place (l, k) and element aljin place (l, p) and, matrix C contains alj in place (l, k) and ali in place (l, p); the twomatrices having in the other positions the initial elements.

Developing determinants along row l, we get det(B)−det(C) = (ali−alj)(Γlk−Γlp) = 0, in contradiction with the hypothesis.

We deduce that row l of A is permutable if and only if all its elements are equalor, if all algebraic complements of elements in line l are equal.

Consider now a matrix A that admits two permutable rows. Then:


• if both rows are constant det(A) = 0;• if both rows have constant algebraic complements then A∗ has two constant

rows; that is det(A∗) = 0, which implies det(A) = 0;• if one row has all elements equal to a, and another row has all algebraic com-

plements equal to b, then from AA∗ = det(A)In we deduce ab = 0, so a = 0 orb = 0.

Thus, in every case, det(A) = 0.

Problem 2. Let u : [a, b] → R be a continuous function which has at each pointx ∈ (a, b] a finite left derivative, denoted by u′

s(x). Prove that the function u isincreasing if and only if u′

s(x) ≥ 0, for all x ∈ (a, b].

Mihail Baluna

Solution. We shall use the following results which can be viewed as generaliza-tions of the theorems of Rolle and Lagrange.

Lemma 1. If a function u : [α, β] → R has left and right finite derivatives at anypoint in [α, β] andi u(α) = u(β), then there is c ∈ [α, β] such that u′

s(c) ≤ 0.Proof. It is obvious that u is continous on [α, β], so it has a global minimum which

belongs to the interior of the interval or coincides with on of the extremities; this pointcan be chosen as c.

Lemma 2. If a function u : [α, β] → R has right and left finite derivatives ateach point in the interval [α, β], then there is c ∈ [α, β] such that u(β) − u(α) ≥(β − α)u′

s(c).Proof. Define v : [α, β] → R, v(x) = u(x)− u(β)−u(α)

β−α x. The function v satisfiesthe conditions in Lemma 1, so there is a point c ∈ [α, β] such that v′s(c) ≤ 0.

Returning to the solution of the problem, let a ≤ x < y ≤ b. By Lemma 1,there exists c ∈ [x, y] such that u(y) − u(x) ≥ (y − x)u′

s(c), which implies that u isincreasing.

The reverse condition is easily implied by the definition of the left derivative.

Problem 3. Let g : R → R be a continuous, decreasing function, such thatg(R) = (−∞, 0). Prove that there are no continuous functions f : R → R such thatthe equality f f ... f︸︷︷︸

k times

= g is true for some integer k ≥ 2.

Dorel Mihet


Solution. Suppose that such a function f exists. Injectivity of g implies the in-jectivity of the continuous function f , which in turn is strictly monotone. As g isincreasing we conclude that f is increasing and k is an odd number. Moreover, f isnot surjective.

Denote by f [k] = f f ... f (k times f ). Because f(R) is an interval with(−∞, 0) = g(R) = f(fk−1(R)) ⊂ f(R), we deduce f is bounded from above.

Let m ∈ R be such that f(x) < m, ∀ x ∈ R. Then f [k−1](x) < m, for allx ∈ R, so g(x) = f(f [k−1](x)) > f(m), for all x ∈ R, in contradiction withg(R) = (−∞, 0).

Problem 4. Let A,B ∈ M2(C) such that A2 + B2 = 2AB. Prove AB = BA

and tr A = tr B.

Nicolae Bourbacut

Solution. a) We firstly prove that (AB−BA)2 = 0. Define the quadratic functiong by f(x) = det(A2 +B2 + x(AB −BA)) = det(A2 +B2) +mx+ x2 det(AB −BA). As f(−i) = det((A + iB)(A − iB)), f(i) = det((A − iB)(A + iB)), sof(−i) = f(i), we get m = 0. Moreover, from f(0) = det(A2 + B2) = det(2AB)

and f(−2) = det(2BA) = det(2AB) = f(0), we deduce that f is constant anddet(AB − BA) = 0. The characteristic equation for AB − BA (tr(AB) = tr(BA))gives then immediately (AB−BA)2 = 0. As AB−BA = (A−B)2 by hypothesis,we obtain 0 = (AB − BA)2 = (A− B)4. As the matrices are of order two we musthave (A−B)2 = O2, so AB = BA.

b) From (A − B)2 = O2 and 2 det(A − B) = det(A − B), the characteristicequation for A−B gives (tr(A)− tr(B))(A−B) = O2, which finishes the solution.

12th GRADE

Problem 1. Let m be a positive integer and p a prime. Suppose A is a unitary ringhaving exactly m invertible elements and such that 1 + 1 + · · ·+ 1︸︷︷︸

p times

= 0.

Prove that A has non-zero nilpotent elements if and only if p divides m.

Mihai Piticari, Sorin Radulescu

Solution. Let a ∈ A, a = 0, be nilpotent element and k ∈ N∗, such that apk

= 0.From (a + 1)p

k

= apk

+ 1 = 1, we have that a + 1 is invertible and the order of


a + 1 in the group of units U(A) is a power of p. Because the order of a + 1 dividesthe order m of U(A), we deduce that p divides m. Conversely, by Cauchy’s theorem,there is an element a of order p in U(A). Due to (a − 1)p = ap − 1 = 0 and a = 1,we conclude that a a− 1 is a non-zero nilpotent.

Problem 2. Consider n ∈ N, n ≥ 2, and A a unitary ring with n elements , suchthat the equation xn+1 + x = 0 has in A \ 0 the unique solution x = 1. Prove thatA is a field.

Ioan Baetu

Solution. Because 1 satisfies xn+1 + x = 0, we have 1 + 1 = 0, so all non-zeroelements of the additive group (A,+) are of order 2. By Cauchy’s theorem n = 2m,m ∈ N∗.

Let a ∈ A, a = 0. The ring A being finite, there exists p < q, such that ap = aq .By succesive multiplications with aq−p, aq = a(k+1)q−kp, k ≥ 1. Take k ∈ N∗, suchthat r = (k + 1)q − kp > 2q. Multiplication by ar−2q gives ar−q = a2(r−q). Letb = ar−q . As b2 = b, we get bn+1 = b, so bn+1 + b = 0, that is b ∈ 0, 1.

If b = ar−q = 0, take s ∈ N, s ≥ 2, such that as−1 = 0 si as = 0. Considerc = as−1. Then c2 = 0 and, as a consequence (c + 1)n+1 = (c + 1)2

m

(c + 1) =

(c2m

+ 1)(c+ 1) = c+ 1. By hypothesis c+ 1 ∈ 0, 1, which is a contradiction. Inconclusion ar−q = 1, i. e. a is invertible.

Problem 3. Let f : [0, 1] → (0,∞) be a continuous function. For n ∈ N, n ≥ 2,consider 0 = t0 < t1 < · · · < tn = 1, such that

∫ t1

t0

f(t) dt =∫ t2

t1

f(t) dt = · · · =∫ tn

tn−1

f(t) dt.

Compute limn→∞

n1

f(t1)+

1

f(t2)+ · · ·+ 1

f(tn)

.

Viorel Vajaitu

Solution. Define F : [0, 1] → [0, I], by F (x) =∫ x

0f(t) dt, where I =

∫ 1

0f(t) dt.

As f is pozitive, F is strictly increasing, so one-to-one. Because F is continuous,F (0) = 0 and F (1) = I , F is surjective also and F−1(kI/n) = tk, k = 0, 1, · · · , n,n ≥ 2.


Denote by (xn)n≥2 the given sequence. We have

1

xn=

1

n

n∑k=1

1

f(tk)=

1

n

n∑k=1

1

f(F−1(kI/n))=

1

n

n∑k=1

1

F ′(F−1(kI/n))

=1

n

n∑k=1

(F−1)′(kI/n).

As (F−1)′ is continous

limn→∞

1

xn=

∫ 1

0

(F−1)′(Ix) dx =1

IF−1(Ix)

∣∣∣1

0=

1

I.

Problem 4. Let f : R → R be a non-decreasing function and F : R → Ra function having right and left finite derivatives at any point in R and F (0) = 0.Suppose that lim

x↑x0

f(x) ≤ F ′s(x0) and lim

x↓x0

f(x) ≥ F ′d(x0), for any x0 ∈ R. Prove

that F (x) =∫ x

0f(t) dt, x ∈ R.

Mihail Baluna

Solution. Consider the function G : R → R, given by G(x) = F (x)−∫ x

0

f(t) dt.

Let x0 ∈ R, x = x0. Then

F (x) =F (x)− F (x0)

x− x0· (x− x0) + F (x0).

As F has right and left finite derivatives in x0, the function x → F (x)−F (x0)x−x0

, x = x0,

is bounded around x0, so limx→x0

F (x) = F (x0). So G is continuous.

Let us show that G has finite right and left derivatives in x0 ∈ R, G′s(x0) ≥ 0 and

G′d(x0) ≤ 0. Let x > x0. Because limx↓x0 f(x) ≤ f(t) ≤ f(x), x0 < t ≤ x, we get

limx↓x0

f(x) ≤ 1

x− x0

∫ x

x0

f(t) dt ≤ f(x),

solimx↓x0

1

x− x0

∫ x

x0

f(t) dt = limx↓x0

f(x).

Consequently

limx↓x0

G(x)−G(x0)

x− x0= lim

x↓x0

(F (x)− F (x0)

x− x0− 1

x− x0

∫ x

x0

f(t) dt)

= F ′d(x0)− lim

x↓x0

f(x) ≤ 0.


Analogously, limx↑x0

G(x)−G(x0)

x− x0= F ′

s(x0)− limx↑x0

f(x) ≥ 0.

In conclusion, 0 ≤ G′s(x) < ∞ and −∞ < G′

d(x) ≤ 0, for any real x, therefore

limε↓0

G(x− ε)−G(x− ε/2)

ε= −1

2G′

s(x) ≤ 0, x ∈ R,

andlimε↓0

G(x+ ε)−G(x+ ε/2)

ε=

1

2G′

d(x) ≤ 0, x ∈ R.

Let us show that G is constant. Consider a < b, such that G(a) = G(b).If G(a) < G(b), consider λ > (G(a)−G(b))/(b− a). The function H : [a, b] →

R, H(x) = G(x) + λx, is continuous, H(a) < H(b) and

limε↓0

H(x+ ε)−H(x+ ε/2)

ε=

1

2G′

d(x) +λ

2≤ λ

2< 0, x ∈ R.

So, for ay real x, there is δ(x) > 0, such that H(x+ ε) < H(x+ ε/2), 0 < ε < δ(x).Let c ∈ [a, b], such that H(c) = min H(x) : a ≤ x ≤ b. Because H(a) < H(b),we have a ≤ c < b. Consider a number ε < min(b− c, δ(c)). Then

H(c) ≤ H(c+ ε) < H(c+ ε/2) < · · · < H(c+ ε/2n) < · · · ≤ H(c),

(the last inequality is a consequence of the continuity of H), which is a contradiction.If G(a) > G(b), proceed analogously: consider λ < (G(a) − G(b))/(b − a).

In this case the corresponding continuous function H(x) = G(x) + λx, satisfiesH(a) > H(b) and

limε↓0

H(x− ε)−H(x− ε/2)

ε= −1

2G′

s(x)−λ

2≤ −λ

2< 0, x ∈ R.

That is, for any real x, there exists δ(x) > 0, such that H(x− ε) < H(x− ε/2),0 < ε < δ(x). Let c ∈ [a, b], such that H(c) = min H(x) : a ≤ x ≤ b. BecauseH(a) > H(b), we get a < c ≤ b. Fix a real positive number ε < min(c − a, δ(c)).Then

H(c) ≤ H(c− ε) < H(c− ε/2) < · · · < H(c− ε/2n) < · · · ≤ H(c),

which is a contradiction.We thus proved that G is constant. Taking into account G(0) = 0, we get G(x) =

0, for any realx .

SHORTLISTED PROBLEMS FOR THE 62nd NMO

JUNIORS

1. Prove that if n and p are integers, 1 < p < n, then the number

(2p+ 1)n3 + 6n(12 + 22 + 32 + ...+ p2)

can be written as the sum of 2p+ 1 different perfect cubes.

2. Given an equilateral triangle ABC and the points M ∈ [BC], N ∈ [AC],P ∈ [AB] such that BM = MC, 3AN = NC and 2BP = AP , find the measure of∠NMP .

3. In a triangle ABC it is known that ∠A = 120, ∠C = 20, (AD is thebisector of ∠BAC (where D ∈ (BC)) and the point E is taken so that E ∈ (AC)

and [BD] ≡ [CE]. Find the measure of ∠EBC.

4. The convex quadrilateral ABCD has [AB] ≡ [BC], m(∠DCB) < 90 andm(∠DCB) + m(∠DAB) = 180. The point E is taken on the segment (DC) suchthat DA+DC = 2DE. Prove that BE ⊥ DC.

5. The convex pentagon ABCDE has m(∠A) = m(∠C) = 90, AB = 3,BC = 5, CD = 10, DE = 8 and the reflection of C about BD is on the line AE.The line MA is perpendicular on the plane of the pentagon and MA = 6. Find themeasure of the angle of the lines ME and AB.

6. The parallelogram ABCD has center O, AD = DB and BA > BC. Thepoint V is outside the plane (ABC) such that V D ⊥ (ABC). Point T is the footof the perpendicular from D onto AB and E,F,G are the feet of the bisectors of theangles ∠V DA, ∠V DB, ∠V DC respectively (where E ∈ V A, F ∈ V B,G ∈ CV ).Prove that:

27

28 SHORTLISTED PROBLEMS FOR THE 2011 ROMANIAN NMO

i) EF ‖ (ABC);ii) m(∠((V AB), (ABC))) = 45 if and only if V D = DT ;iii) if V D = DT , then tan(∠((GEF ), (ABC))) = 1

2 if and only if BABC = 6

5 .

7. Find all the perfect squares whose product of their decimal digits is a prime.

8. Given a prime number p, p ≥ 3 and d a square free number (that is d’s primedecomposition contains no repeated factors), find the number of the elements of theset

Ap =

x =

n√d+

n

p

− n

√d | n ∈ N

,

where a denotes the fractional part of the real number a.

9. Prove that, if x, y > 0, then 0 <x

x+ y4+

y

y + x4− 1

x2y2 + 1< 1.

10. Let x, y be real numbers such that x+y, x3+y3, x5+y6 and y5 are rational.Prove that x and y are rational.

SENIORS

1. An infinite set A of real numbers contains at least an irrational number. Provethat for each positive integer n it is possible to find n elements of A whose sum isirrational.

2. Given integers n ≥ 3 and k ≥ 1, find all positive integers a1 < a2 < . . . < an

such that1

ak1+

1

ak2+ . . .+

1

akn= 1.

3. Let p be a positive integer and Ap = x ∈ R | pxx = (p+ 1)[x]. Find thecardinals of the sets Ap and A1 ∪A2 ∪ ... ∪Ap.

4. Solve the equation (ax + bx)2011 =(a2011 + b2011

)x, where a, b are positive

reals.

5. Find all the functions f : Z → N fulfilling the conditions:i) f(m+ n) = f(m) + f(n) + 2mn, for every integers m,n;ii) f(f(x))− f(1) is a perfect square for each integer x.

6. Find all positive integers n for which there exists a set S ⊂ C with n, suchthat i) each element of S has modulus 1, ii)

∑z∈S

z = 0 and iii) z + w = 0, for each

z, w ∈ S.

SHORTLISTED PROBLEMS FOR THE 2011 ROMANIAN NMO 29

7. Given in the plane equilateral triangles ABC and BDE, with C ∈ (BD)

and A, E are on different sides of the line BD. Denote M,N ,P the midpoints ofthe segments (AB), (CD), (BE) respectively. Find the measures of the angles of thetriangle MNP .

PUTNAM SENIORS

1. Find all polynomials P,Q with real coefficients, such that, for infinitely manypositive integers n, P (1)P (2) . . . P (n) = Q(n!).

2. The sequence (an)n≥1 of real numbers is such that the sequence (xn)n≥1 de-fined by xn = maxan, an+1, an+2 is convergent and the sequence (yn)n≥1 definedby yn = an+1 − an has limit 0. Prove that the sequence (an)n is convergent.

3. Let A be a n× n matrix with complex entries.a) Prove that there exists a n×n matrix with complex entries B such that AB = 0n

and rank A+ rank B = n.b) If 1 < rank A < n, prove that there exists a n× n matrix with complex entries

C such that AC = 0n, CA = 0n and rank A+ rank C = n.

4. Let f : R → R be a continuous function such that, on each non degeneratedinterval I , the function reaches its maximum or its minimum in an interior point of I .Prove that f is a constant.

5. Let f : R → R be a function with the property: |f(x)−f(y)| ≤ | sinx−sin y|,for each x, y ∈ R.

a) Prove that there exists an unique c ∈ R such that f(c) = c.b) Consider the sequence (xn)n∈N with x0 = 0 and xn+1 = f(xn) for every

n ∈ N. Prove that limn→∞

xn = c.

6. Given a continuous function f : R → R, denote, for each interval [a, b],mab = minx∈[a,b] f(x) and Mab = maxx∈[a,b] f(x). Find all the continuous func-

tions f : R → R such that, for every a < b, f(a+ b

2

)=

mab +Mab

2.

7. The matrix A ∈ Mk(R) (k ≥ 2) has the property: for every positive integer n,det(n2Ik −A2) ≥ nk (det(nIk −A) + det(nIk +A)− 1) . Prove that tr(A) = 0.

30 SHORTLISTED PROBLEMS FOR THE 2011 ROMANIAN NMO

8. The function f : [0, 1] → R is differentiable and

∫ 1

0

f(x) dx =

∫ 1

0

xf(x) dx = 0.

Prove that there exists c ∈ (0, 1) such that f ′(c) = 0.

9. Let f : [0, 1] → R be a function which is continuous on [0, 1] and differentiablein 0. Consider the function s : [0, 1] → R,

s(x) = supc ∈ [0;x] |∫ x

0

f(t) dt = xf(c).

a) Prove that the function f s is differentiable in 0 and (f s)′(0) = 12f

′(0).b) If f ′(0) = 0, prove that s is differentiable in 0 and s′(0) = 1

2 .

10. Find all the functions f :[−π

2 ;π2

]→ R such that:

a) f is differentiable on[−π

2 ,π2

];

b) there exists a antiderivative F of f such that F (x) + f ′(x) ≤ 0, ∀x ∈[−π

2 ,π2

]and F

(−π

2

)= F

(π2

)= 0.

11. Let (K,+, ·) be a finite field. Prove that:a) if K has 4k+1 elements, then the polynomial f = X4+4 has four roots in K;b) the polynomial g = X8 − 16 has at least a root in K.

12. Prove that ∫ 1

0

eex

dx ≥ e(e2 − 3)

2.

Contributors

Gheorghe Molea, Dan Nedeianu, Petre Batranetu, Cosmin Manea & Dragos Petri-ca, Dan Nedeianu, Alexandru Blaga, Aurel Barsan, Cecilia Deaconescu & DumitruDobre, Dan Nedeianu, Catalin Cristea, Dinu Teodorescu, Dorel Mihet, Cecilia Dea-conescu & Dumitru Dobre, Aurel Barsan, Marian Ionescu, Nicolae Bourbacut, DanNedeianu, Ovidiu Furdui, Cristinel Mortici, Vasile Pop, Radu Gologan, Nicolae Bour-bacut, ***, Nelu Chichirim, Cezar Lupu, Dan Marinescu, Dorel Mihet, Dorel Mihet,Dan Nedeianu.

THE 62nd NMO SELECTION TESTS FOR THE BALKAN

AND INTERNATIONAL MATHEMATICAL OLYMPIADS

FIRST SELECTION TEST

Problem 1. Determine all real-valued functions f on the set of real numbers sat-isfying the condition

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

for all real numbers x and all non-negative real numbers y.

SEEMOUS 2011 Short List

Solution. Clearly, any constant function is a solution to the problem. Conversely,assume without loss of generality that f(0) = 0. Fix a positive real number y andlet x = ny, where n is a non-negative integer number, to obtain 2f(ny) = f((n +

1)y) + f((n + 2)y) and thereby get f(ny) = f(y)(1 − (−2)n)/3. Consequently,f(y) = −f(2y) = f(4y) = −5f(y), which shows that f vanishes identically on thenon-negative real ray.

Finally, if x is any real number, write 2f(x) = f(x + |x|) + f(x + 2|x|) = 0 toconclude that f vanishes identically on the set of real numbers.

Problem 2. Prove that the set S = nπ : n = 0, 1, 2, 3, · · · contains arithmeticprogressions of any finite length, but no infinite arithmetic progressions.

Vasile Pop, SEEMOUS 2011 Short List

Solution. If x is a real number, let x = x − x denote the fractional part ofx. Given an integer number m ≥ 3, there exists a positive integer number n such

31

32 SELECTION TESTS FOR THE 2011 BMO AND IMO

that nπ < 1/m, for the set kπ : k = 0, 1, 2, 3, · · · is dense in the closed unitinterval [0, 1]. Consequently,

knπ = knπ+ knπ = knπ+ knπ = knπ, k = 1, 2, · · · ,m,

so nπ, 2nπ, · · · , mnπ are m numbers in S in arithmetic progression with rationπ.

Suppose, if possible, that S contains an infinite arithmetic progression nkπ,k = 0, 1, 2, 3, · · · , with (integral) ratio r, where the nk form a strictly increasingsequence of positive integer numbers. Write nkπ = n0π + kr + nkπ to deducethat r is positive and

nk+1 − nk =r + nk+1π − nkπ

π∈(r − 1

π,r + 1

π

).

The length of this interval is less than 1, so nk+1 − nk = n for some positive integern and all indices k. Hence, nk = n0 + kn, so nk/k

k→∞−→ n. On the other hand,nk/k

k→∞−→ r/π, so π = r/n wich contradicts irrationality of π.

Problem 3. Let ABC be a triangle such that AB < AC. The perpendicularbisector of the side BC meets the side AC at the point D, and the (interior) bisectrix ofthe angle ADB meets the circumcircle ABC at the point E. Prove that the (interior)bisectrix of the angle AEB and the line through the incentres of the triangles ADE

and BDE are perpendicular.

Solution. The lines BC and DE are parallel, so the angles BED and DAE areequal. Then so are the angles AED and DBE. Let I and J be the incentres of thetriangles ADE and BDE, respectively. It follows that the triangles DIE and DJB

are similar, so DI/DE = DJ/DB. Since the angles IDJ and EDB are equal, thetriangles DIJ and DEB are similar, so the angles DIJ and DEB are equal. Let theline BE meet the line IJ at the point F . Notice that the quadrangle DIEF is cyclicto deduce that the angles EFI and EDI are both equal to one half of the angle ACB.Consequently, the line IJ is parallel to the exterior bisectrix of the angle AEB. Theconclusion follows.

Problem 4. Given an integer number n ≥ 2, evaluate the sum∑σ∈Sn

(sgnσ)n(σ),

SELECTION TESTS FOR THE 2011 BMO AND IMO 33

where Sn is the set of all n-element permutations, and (σ) is the number of disjointcycles in the standard decomposition of σ.

SEEMOUS 2011 Short List

Solution. The sum in question is fn(n) = n!, where

fn(x) =∑σ∈Sn

(sgnσ)x(σ) = x(x− 1) · · · (x− n+ 1).

The latter is a straightforward consequence of the following recurrence formula:

fn(x) = xfn−1(x)− (n− 1)fn−1(x) = (x− n+ 1)fn−1(x), n ≥ 3,

where f2(x) = x(x− 1). To establish the recurrence formula, consider the decompo-sition of a permutation σ in Sn into disjoint cycles,

σ = (· · · ) · · · (· · · )(n) or σ = (· · · ) · · · (· · · n · · · ) · · · (· · · ),

along with the permutation σ′ in Sn−1, obtained from σ by deleting n :

σ′ = (· · · ) · · · (· · · )(n) or σ = (· · · ) · · · (· · · n · · · ) · · · (· · · ).

Clearly, sgnσ′ = sgnσ and (σ′) = (σ)−1 in the former case, and sgnσ′ = −sgnσand (σ′) = (σ) in the latter.

Conversely, any given permutation σ′ in Sn−1 extends to a unique permutation σ

in Sn which fixes n (the first case above), and to exactly n − 1 permutations σ in Sn

which do not fix n — for the latter can be inserted in precisely n− 1 ways into σ′ (thesecond case above). The conclusion follows.

SECOND SELECTION TEST

Problem 5. A square with side length is contained in a unit square whose centreis not interior to the former. Show that ≤ 1/2.

Marius Cavachi

Solution. Proof from The Book. Notice that there is a line through the centre ofthe unit square separating the square with side length and a standard quarter of theunit square (i. e., a square with side length 1/2 and one vertex at the centre of the unit


square). To conclude, apply the celebrated theorem of Erdos stating that the sum ofthe side lengths of two squares packed into a unit square does not exceed 1.

Solution 1. The square with side length is the intersection of two strips, S andS′, of breadth . Since the centre O of the unit square is not interior to the squarewith side length , it is not interior to at least one of the two strips, say S. Clearly, thedistance from O to the farthest component ′ of the boundary of S is at least .

Suppose, if possible, that > 1/2. Then the line ′ meets the boundary of theunit square at precisely two points, X and Y , situated on consecutive sides. Let Adenote the vertex of the unit square shared by those sides, and let M and N denotetheir midpoints. Without loss of generality, we may (and will) assume that the cir-cular labelling around the boundary of the unit square is M , X , A, Y , N . Sincedist (O, ′) ≥ > 1/2, the line t, parallel to ′ and tangent to the quarter MN of theincircle of the unit square, meets the segments MX and NY : the former at X ′, andthe latter at Y ′. Clearly, X ′Y ′ > XY ≥ . Finally, to reach a contradiction, let Tdenote the point of contact of t and the quarter MN of the incircle of the unit square,and write successively:

1 = 1/2 + 1/2 = MA+AN = (MX ′ +X ′A) + (AY ′ + Y ′N)

= MX ′ + (X ′A+AY ′) + Y ′N = X ′Y ′ + (X ′A+AY ′)

> 2 ·X ′Y ′ > 2 ·XY ≥ 2.

Solution 2. Begin by noticing that if the vertices M , N , P , Q of a rectangle lieon the boundary of a triangle ABC ( M ∈ AB, N ∈ AC and P,Q ∈ BC), thenMN/BC +MQ/AA′ = AM/AB + BM/AB = 1, where A′ is the perpendicularfoot dropped from the vertex A onto BC.

Hence if a triangle ABC, whose internal angles at B and C are not obtuse, con-tains a square with side length with a pair of opposite sides parallel to BC, then1/BC + 1/AA′ ≤ 1/, where A′ is the perpendicular foot dropped from the vertexA onto BC : extend the farthest side of the square, which is parallel to BC, to meetthe sides AB and AC at M and N , respectively, drop the perpendicular feet M ′ andN ′ from M and N , respectively, onto the side BC, notice that ≤ min (MN,MM ′)

and apply the above result.Back to the problem, let O be the centre of the unit square and let A, B, C, D

be a circular labelling of its vertices around the boundary. Draw a line d through O,parallel to a pair of opposite sides of the square with side length . The latter lies in


one of the closed half-planes determined by d. Clearly, ≤ 1/2 if d is parallel to apair of opposite sides of the unit square.

Otherwise, we may assume that d meets the side AB at E and the extension of theside AD beyond D at F , and the triangle AEF contains the square with side length. We prove that < 1/2. To this end, drop the perpendicular foot A′ from A onto thesegment EF . By the preceding, 1/AA′+1/EF ≤ 1/, so it is sufficient to show that

1/AA′ + 1/EF > 2. (∗)

Let O′ be the perpendicular foot dropped from O onto the side AB, and let EO′ =

x/2 and ∠AEF = θ. Clearly, sin θ = 1/√1 + x2 and cos θ = x/

√1 + x2, so

EF =AE

cos θ=

(1 + x)√1 + x2

2xand AA′ = AE sin θ =

1 + x

2√1 + x2

,

and (∗) is equivalent to x/(1 + x)√1 + x2 +

√1 + x2/(1 + x) > 1, x > 0, which is

easily established.

Problem 6. Let A0A1A2 be a non-equilateral triangle. The incircle of the triangleA0A1A2 touches the side AiAi+1 at the point Ti+2 (indices are reduced modulo 3).Let Xi be the perpendicular foot dropped from the point Ti onto the line Ti+1Ti+2.Show that the lines AiXi are concurrent at a point situated on the Euler line of thetriangle T0T1T2.

Gazeta Matematica

Solution. The lines AiAi+1 and XiXi+1 are parallel, for they are both antiparal-lel to the line TiTi+1. Hence the triangles A0A1A2 and X0X1X2 are homologus: thethree lines AiXi are concurrent at the homology centre which lies on the homologyline. The latter passes through the incentres of the two triangles: one is the circum-centre of the triangle T0T1T2 and the other the orthocentre. The conclusion follows.

Problem 7. Given a positive integer number n, determine the maximum numberof edges a simple graph on n vertices may have in order that it contain no cycles ofeven length.

American Mathematical Monthly

Solution. The required maximum is 3(n− 1)/2. It is achieved, for instance, byan (n− 1)/2 arm wind-mill if n is odd, and an (n− 2)/2 arm wind-mill with an extraedge joined at the hub if n is even.


To show that a simple graph on n vertices with no cycles of even length has at most3(n− 1)/2 edges, let G be one such with a maximal edge-set E. By maximality, G isconnected. Let T be a spanning tree (that is, a maximal connected acyclic subgraph)of G, and let E′ denote the edge-set of T ; it is well-known that |E′| = n − 1. Theend-points of any edge e in E \ E′ are joined by a unique simple path α in T . SinceT is acyclic, if e and e′ are distinct edges in E \E′, then the edges the correspondingpaths in T , α and α′, may share form a path; and since G has no cycles of even length,the paths α and α′ are actually edge-disjoint — otherwise, at least one of the cyclesα + e, α′ + e′ and α + e + α′ + e′ (mod 2) would have an even length. It followsthat the total length of the cycles α + e, e ∈ E \ E′, does not exceed |E|. Thereare |E \ E′| = |E| − |E′| = |E| − n + 1 such cycles, each of length at least 3, so3(|E| − n+ 1) ≤ |E|; that is, |E| ≤ 3(n− 1)/2.

Problem 8. Show that:a) There are infinitely many positive integer numbers n such that there exists a

square equal to the sum of the squares of n consecutive positive integer numbers. (Forinstance, 2 and 11 are such: 52 = 32 + 42 and 772 = 182 + 192 + · · ·+ 282.)

b) If n is a positive integer number which is not a perfect square and if x0 is aninteger number such that x2

0 + (x0 + 1)2 + · · · + (x0 + n − 1)2 is a perfect square,then there are infinitely many positive integer numbers x such that x2 + (x + 1)2 +

· · ·+ (x+ n− 1)2 is a perfect square.American Mathematical Monthly

Solution. a) We must show that there are infinitely many positive integer numbersn such that the equation y2 = x2+(x+1)2+ · · ·+(x+n− 1)2 has positive integralsolutions. To this end, rewrite the equation in the form

y2 = n((x+ (n− 1)/2)2 + (n2 − 1)/12). (1)

We show that if n is a square greater than 25 not divisible by 2 or 3, then (1) has apositive but finite number of solutions in positive integers x and y. Since n is a perfectsquare, the expression in the outmost parantheses in (1) must be a perfect square z2,i. e., there is a positive integer z such that

z2 − (x+ (n− 1)/2)2 = (n2 − 1)/12. (2)

Thus, z ± (x + (n − 1)/2) must be complementary even divisors of (n2 − 1)/12

differing by more than n − 1; hence there are only finitely many solutions to (2) in


positive integers. One such is obtained by taking

z − (x+ (n− 1)/2) = 2 and z + (x+ (n− 1)/2) = (n2 − 1)/24,

i. e., by letting x and z have the positive integer values

x = (n− 25)(n+ 1)/48 and z = 1 + (n2 − 1)/48.

b) Rewrite the equation y2 = x2 + (x+ 1)2 + · · ·+ (x+ n− 1)2 in the form

(2y)2 − n(2x+ n− 1)2 = (n− 1)n(n+ 1)/3 (3)

and let Tn(x) = x2 + (x+ 1)2 + · · ·+ (x+ n− 1)2. We may assume n > 1. SinceTn(x) = Tn(−x − n + 1), we may further assume that x0 ≥ −(n − 1)/2. SupposeTn(x0) = y20 , i. e., (2y0)2 − n(2x0 + n− 1)2 = (n− 1)n(n+ 1)/3, where we mayassume y0 > 0 (and hence y0 > (n− 1)/2). By the theory of the Pell equation, thereare infinitely many pairs of positive integers u, v such that

u2 − nv2 = 1. (4)

Clearly, u is odd if n is even, so that (n− 1)(u− 1) is even. We now use the identity

(2y0 + (2x0 + n− 1)√n)(u+ v

√n) = 2y + (2x+ n− 1)

√n, (5)

where

x = x0u+ y0v + (n− 1)(u− 1)/2 and y = y0u+ x0nv + (n− 1)nv/2. (6)

Multiplying (5) by the identity obtained from (5) by replacing√n by −

√n, we find

that if x and y are given by (6) and u and v satisfy (4), then

(2y)2 − n(2x+ n− 1)2 = ((2y0)2 − n(2x0 + n− 1)2)(u2 − nv2)

= (2y0)2 − n(2x0 + n− 1)2 = (n− 1)n(n+ 1)/3.

Thus, if x, y are given by (6), they satisfy (3), so that Tn(x) is a perfect square.Further, since x0 ≥ −(n− 1)/2 and y0 > (n− 1)/2, it follows that y ≥ y0u > 0 and

x ≥ y0v−u(n−1)/2+(n−1)(u−1)/2 = y0v− (n−1)/2 ≥ y0− (n−1)/2 > 0.

This ends the proof.

Remark. Write Tn(x) = n(x+ (n− 1)/2)2 + (n− 1)n(n+ 1)/12 to derive a)from b). To this end, notice that if (n+ 1)/12 is a perfect square, say n = 12m2 − 1,then (1) has the obvious solution x = −6m2 +m + 1 and y = m(12m2 − 1), so b)applies to show that (1) has infinitely many solutions in positive integers.


FOURTH – ALL GEOMETRY – SELECTION TEST2

Problem 9. Let ABCD be a cyclic quadilateral so that BC and AD meet at apoint P . Consider a point Q, different from P , on the line BP so that PQ = BP andconstruct the parallelograms CAQR and DBCS. Prove that the points C, Q, R, Sare concyclic.

BMO Shortlist 2011, United Kingdom

Solution. It is enough to prove that ∠RQC = ∠RSC. (†)

Now ∠RQC = ∠ACQ = ∠ACB = ∠ADB. (1)

Take T such that QABT is a parallelogram. Then−→BT =

−→AQ =

−→CR and

−−→BD =

−→CS imply ∆BTD ≡ ∆CRS, so ∠RSC = ∠TDB. (2)

On the other hand, since P is the midpoint of the segment BQ and ABTQ is aparallelogram, P is also the midpoint of the segment AT , so ∠TDB = ∠ADB.

Combining this with (1) and (2), we get (†).

Problem 10. Let ABCD be a convex quadrangle such that AB = AC = BD

(vertices are labelled in circular order). The lines AC and BD meet at point O, thecircles ABC and ADO meet again at point P , and the lines AP and BC meet at pointQ. Show that the angles COQ and DOQ are equal.

BMO Shortlist 2011, Bulgaria

Solution. We shall prove that the circles ADO and BCO meet again at the incen-tre I of the triangle ABO, so the line IO is the radical line of the circles ADO andBCO. Noticing further that the lines AP and BC are the radical lines of the pairs of

2The third test was the Balkan Mathematical Olympiad


circles (ABC,ADO) and (ABC,BCO), respectively, it follows that the lines AP ,BC and IO are concurrent (at point Q), whence the conclusion.

To show that the point I lies on the circle ADO, notice that

∠AIO = 90 +1

2∠ABO = 90 +

1

2∠ABD = 90 +

1

2(180 − 2∠ADB)

= 180 − ∠ADB = 180 − ∠ADO.

Similarly, the point I lies on the circle BCO, for

∠BIO = 90 +1

2∠BAO = 90 +

1

2∠BAC = 90 +

1

2(180 − 2∠ACB)

= 180 − ∠ACB = 180 − ∠BCO.

Remark. We may consider the corresponding configuration derived from fourgeneric points in the plane, A, B, C, D, subject only to AB = AC = BD. Theargument applies mutatis mutandis to show that the point Q always lies on one of thetwo bisectrices of the angle COD.

Problem 11. Given a triangle ABC, let D be the midpoint of the side AC and letM be the point that divides the segment BD in the ratio 1/2; that is, MB/MD = 1/2.The rays AM and CM meet the sides BC and AB at points E and F , respectively.Assume the two rays perpendicular: AM ⊥ CM . Show that the quadrangle AFED

is cyclic if and only if the line of support of the median from A in triangle ABC meetsthe line EF at a point situated on the circle ABC.

BMO Shortlist 2011, Saudi Arabia

Solution. Denote by a, b, c the sidelengths, and by ma, mb, mc the lengths ofthe medians of the triangle ABC. Since MD is median in the right-angled triangleAMC, it follows that 2mb/3 = MD = AD = CD = b/2, so mb = 3b/4, whence(3b/4)2 = m2

b = (a2 + c2)/2− b2/4; that is, 13b2 = 8(a2 + c2).


Next, apply the Menelaus theorem to get EC/EB = 4 = FA/FB and deducethereby that the lines AC and EF are parallel. The quadrangle AFED is therefore atrapezium; it is cyclic if and only if AF = DE.

Express the two in terms of a, b and c. Recall that FA/FB = 4 to obtain AF =

4c/5. Next, apply Stewart’s theorem in triangle BCD to get DE2 = b2/2− 4a2/25.By the preceding, the quadrangle AFED is cyclic if and only if 25b2 − 8a2 = 32c2.Recall that 13b2 = 8(a2 + c2) to express b and c in terms of a: b = 2a

√2/3 and

c = 2a/3.

Finally, let N be the midpoint of the side BC and let the lines AN and EF meetat P . Notice that EN = a/2− a/5 = 3a/10, and the triangles ANC and PNE aresimilar, to obtain NP = 3ma/5, so

NA ·NP = 3m2a/5 = 3(2(b2 + c2)− a2)/20 = a2/4 = NB ·NC.

The conclusion follows.

FIFTH SELECTION TEST

Problem 12. Show that there are infinitely many positive integer numbers n suchthat n2 + 1 has two positive divisors whose difference is n.

Solution. Define the sequence (ak)k≥0 by a0 = 1, a1 = 2 and ak+2ak = a2k+1 +

1, k = 0, 1, 2, · · · , and check inductively that the ak are all positive integer numbers,the nk = ak+1 − ak form a strictly increasing sequence of positive integer numbers,and ak and ak+1 both divide n2

k + 1.


Problem 13. Let n be an integer number greater than 2, let x1, x2, · · · , xn be n

positive real numbers such that

n∑i=1

1

xi + 1= 1,

and let α be a real number greater than 1. Show that

n∑i=1

1

xαi + 1

≥ n

(n− 1)α + 1

and determine the cases of equality.

BMO Shortlist 2011, Serbia

Solution. Let yi = 1/(xi+1), i = 1, 2, · · · , n, so the yi are positive real numbersthat add up to 1. Upon substitution, the left-hand member of the required inequalitybecomes

n∑i=1

yαi(1− yi)α + yαi

=

n∑i=1

yαi(∑j =i yj

)α

+ yαi

,

the latter on account of y1 + y2 + · · · + yn = 1. Apply Jensen’s inequality to theconvex function t → tα, t > 0, to get

(∑j =i

yj

)α

≤ (n− 1)α−1∑j =i

yαj , i = 1, 2, · · · , n,

son∑

i=1

yαi(∑j =i yj

)α

+ yαi

≥n∑

i=1

yαi(n− 1)α−1

∑j =i y

αj + yαi

.

Now write zi = yαi , i = 1, 2, · · · , n, z = z1 + z2 + · · ·+ zn and a = (n− 1)α−1 totransform the right-hand member of the above inequality to

n∑i=1

zi(1− a)zi + az

.

Finally, notice that the function t → t

(1− a)t+ az, t < az/(a − 1), is convex, to

conclude by Jensen’s inequalty:

n∑i=1

zi(1− a)zi + az

≥ n ·1n

∑ni=1 zi

(1− a) 1n∑n

i=1 zi + az=

n

(n− 1)a+ 1.


Clearly, equality holds if and only if the zi are all equal; tracing back, this is thecase if and only if the xi are all equal to n− 1.

Remark. Since the inequality in the statement can be rewritten as

n∑i=1

1

xi + 1· xi + 1

xαi + 1

≥

n∑i=1

xi

xi + 1+ 1

(n∑

i=1

xi

xi + 1

)α

+ 1

,

it is tempting to consider the real-valued function f(x) = (x + 1)/(xα + 1), x ≥ 0,and try to apply Jensen’s inequality. Unfortunately, f is concave on the closed unitinterval [0, 1] and convex on the ray x ≥ 1, as shown by the second derivative:

f ′′(x) =αxα−2

(xα + 1)3

((α− 1)xα+1 + (α+ 1)xα − (α+ 1)x− (α− 1)

).

The sign of f ′′ is given by

g(x) = (α− 1)xα+1 + (α+ 1)xα − (α+ 1)x− (α− 1), x ≥ 0,

whose first two derivatives are

g′(x) = (α+ 1)((α− 1)xα + αxα−1 − 1

)

and

g′′(x) = α(α2 − 1)xα−2(x+ 1) > 0, x > 0.

Hence g′ is strictly increasing. Since g′(0) = −(α + 1) < 0, and g′(1) = 2(α2 −1) > 0, it follows that g′ has a unique zero x0 ∈ (0, 1). Consequently, g is strictlydecreasing on the closed interval [0, x0] and strictly increasing on the ray x ≥ x0.Since g(0) = 1− α < 0, and g(1) = 0, the conclusion follows.

Problem 14. Given a set L of lines in general position in the plane (no two linesin L are parallel and no three lines are concurrent) and another line , show that thetotal number of edges of all faces in the corresponding arrangement interesected by

is at most 6|L|.

Chazelle et al., Edelsbrunner et al.


Solution. Assume without loss of generality that is horizontal and does not passthrough any vertex of the arrangement of lines in L.

First, we shall bound the total number of edges of the upper parts of all facesintersected by , that is, those parts that lie above . The boundary of the upper partof such a face K consits of two convex chains of edges, the left and right chain, and aportion of . If the upper part of K is bounded, then the left and right chains meet atthe topmost vertex of K. Otherwise, the last (topmost) edges of these chains are half-lines. The edges belonging to the left (respectively, right) chain, with the exception ofthe topmost edge, are called the left (respectively, right) edges of K.

We claim that every line in L contains at most one left edge. Suppose, if possible,that some line in L has two portions e and e′ that are left edges of K and K ′, respec-tively, where e′ is above e. Then the line supporting the topmost edge of the left chainof K would cross K ′, contradicting the fact that K ′ is a face of the arrangement oflines in L. Hence, the total number of left (respectively, right) edges of the upper partsof the faces intersected by is at most |L|. Taking into account the topmost edgesof the chains, the total number of edges of the upper parts of the faces intersected by is at most 4|L|. The same argument applies verbatim for the lower parts of thesefaces. Consequently, the total number of edges of the faces intersected by is at most8|L| − 2|L| = 6|L|; the second term is due to the fact that the edges crossed by arecounted twice: once in the upper parts and once in the lower parts.

SIXTH SELECTION TEST

Problem 15. Given a positive integer number k, define the function f on the setof all positive integer numbers to itself by

f(n) =

1, if n ≤ k + 1,

f(f(n− 1)) + f(n− f(n− 1)), if n > k + 1.

Show that preimage of every positive integer number under f is a finite non-empty setof consecutive positive integer numbers.


Solution. It is sufficient to show that the difference ∆(n) = f(n)− f(n− 1) is 0or 1 for all integer numbers n ≥ 2, and f is unbounded.

Clearly, ∆(n) = 0, 2 ≤ n ≤ k + 1, provides the basis for an inductive proof. Ifn > k + 1, apply the recurrence. By the induction hypothesis, ∆(n− 1) is 0 or 1 and


1 ≤ f(n − 1) ≤ n − 1. Notice that ∆(n) = ∆(n − f(n − 1)) if ∆(n − 1) = 0 and∆(n) = ∆(f(n− 1)) if ∆(n− 1) = 1 to conclude that ∆(n) is indeed 0 or 1.

To show that f is unbounded, suppose, if possible, that f achieves its maximumvalue N for the first time at m to deduce recursively that N = f(m+ n) = f(N) +

f(m − N + n) for all positive integers n; that is, f(m − N + n) = N − f(N), forall positive integers n. Setting n ≥ N yields a contradiction: N = N − f(N). Theconclusion follows.

Problem 16. Given a prime number p congruent to 1 modulo 5 such that 2p + 1

is also prime, show that there exists a matrix of zeros and ones containing exactly 4p

(respectively, 4p + 2) ones no submatrix of which contains exactly 2p (respectively,2p+ 1) ones.


Solution. Let p = 5q + 1, q ≥ 2, and write 4p = 5(4q + 1) − 1. Form a 5-by-(4q + 1) matrix consisting of ones only except for a single entry. Such a matrixhas exactly 4p ones. A submatrix comprising r rows, 1 ≤ r ≤ 5, and s columns,1 ≤ s ≤ 4q + 1, contains rs or rs − 1 ones. In the former case, rs = 2p impliesr ≥ p or s ≥ p, because p is prime; in the latter, rs− 1 = 2p implies rs = 2p+ 1, sor = 2p+ 1 or s = 2p+ 1, because 2p+ 1 is prime. Both cases contradict the size ofthe matrix, so no submatrix contains exactly 2p ones.

Next, write 4p+ 2 = 5(4q+ 1) + 1. Now form a 5-by-(4q+ 1) matrix consistingof ones only except for one column that contains only a single one. Such a matrixhas exactly 4p + 2 ones. A submatrix comprising r rows, 1 ≤ r ≤ 5, and s all-onecolumns, 0 ≤ s ≤ 4q+1, contains rs or rs+1 ones. In the former case, rs = 2p+1

implies r = 2p+1 or s = 2p+1 because 2p+1 is prime; in the latter, rs+1 = 2p+1,so either r ≥ p or s ≥ p since p is prime. Again, this contradicts the size of the matrixand the conclusion follws.

Problem 17. The incircle of a triangle ABC touches the sides BC, CA, AB atpoints D, E, F , respectively. Let X be a point on the incircle, different from the pointsD, E, F . The lines XD and EF , XE and FD, XF and DE meet at points J , K, L,respectively. Let further M , N , P be points on the sides BC, CA, AB, respectively,such that the lines AM , BN , CP be concurrent. Prove that the lines JM , KN andLP are concurrent.

Dinu Serbanescu


Solution. Let the lines KL and NP , LJ and PM , JK and MN meet at pointsQ, R, S, respectively. By Desargues’ theorem on perspective triangles, the lines JM ,KN and LP are concurrent if and only if the points Q, R and S are collinear.

Without loss of generality, we may assume that X lies on the arc FD of the incirclethat does not contain E. Consequently, K lies on the side FD, while L and J lie onthe respective extensions of the sides DE and EF .

Consider the cyclic quadrangle DEFX : The diagonals meet at K, and the exten-sions of the opposite sides meet at L and J , respectively, so the line LJ is the polarof K with respect to the incircle – in what follows, all polar lines are considered withrespect to the incircle. Since the line FD is the polar of B, and K lies on the line FD,it follows that B lies on the line LJ . Similarly, C lies on the line JK, and A lies onthe line KL. Consequently, the lines AQ and CS meet at K.

Projectively, the lines FD and LJ meet at some point T . Notice that T lies on thepolar lines of B and K to deduce that the line BK is the polar of T , so the cross-ratio(TFKD) is harmonic. Let further the lines BK and PM meet at U . Read fromB, the cross-ratio (RPUM) equals (TFKD), so it is harmonic; that is, U is theharmonic conjugate of R relative to M and N . Consequently, the points Q, R and S


are collinear if and only if the lines MQ, NU and PS are concurrent.To prove the lines MQ, NU and PS concurrent, simply check that

QN

QP· UP

UM· SMSN

= 1.

To this end, write

QN = NA · sin(∠NAQ), QP = PA · sin(∠PAQ),

UP = PB · sin(∠PBU), UM = MB · sin(∠MBU),

SM = MC · sin(∠MCS), SN = NC · sin(∠NCS)

to get upon rearrangement of factors

QN

QP· UP

UM· SMSN

=

(NA

NC· PB

PA· MC

MB

)

(sin(∠NAQ)

sin(∠PAQ)· sin(∠PBU)

sin(∠MBU)· sin(∠MCS)

sin(∠NCS)

).

Finally, notice that the lines in both triples

(AM,BN,CP ) and (AQ,BU,CS)

are concurrent (the former by hypothesis, and the latter concur at K) to infer that theproducts in the parantheses above both equal 1 and thereby conclude the proof.

Remark. Clearly, the problem is of projective character: all we need is a tritangentconic and two pairs of suitably perspective triangles:

Let ABC be a triangle and let γ be a conic tangent at points D, E, F to the linesBC, CA, AB, respectively. Let further J , K, L be points on the lines EF , FD, DE,respectively, and let M , N , P be points on the lines BC, CA, AB, respectively. Iftriangles JKL and DEF are perspective from some point on γ, and triangles MNP

and ABC are perspective, then triangles JKL and MNP are perspective.The proof goes along the same lines.

SELECTION TESTS FOR THE 2011 JBMO SELECTION TESTS FOR THE 2011 JBMO

THE 62nd NMO SELECTION TESTS FOR THE JUNIOR

BALKAN MATHEMATICAL OLYMPIAD

FIRST SELECTION TEST

Problem 1. Call a positive integer balanced if the number of its distinct primefactors is equal to the number of its digits in the decimal representation; for example,the number 385 = 5 · 7 · 11 is balanced, while 275 = 52 · 11 is not. Prove that thereexist only a finite number of balanced numbers.

Solution. Let p1 = 2, p2 = 3, p3 = 5, . . . be the sequence of primes. Anybalanced number a with n digits satisfies a ≥ p1p2 · · · pn. Since p1p2 · · · p11 =

2 · 3 · 5 · · · 29 · 31 => 1011 and pk > 10, for any k > 11, it follows that there are nobalanced numbers having more than 10 digits.

Problem 2. Consider a convex pentagon A0A1A2A3A4 so that the rays (AiAi+1

and (Ai+3Ai+2 meet at Bi+4, for each i = 0, 1, 2, 3, 4 – indices are consideredmodulo 5. Show that

4∏i=0

AiBi+3 =

4∏i=0

AiBi+2.

Solution. Let Ci be the projection of Bi on the line Ai+2Ai+3, i = 0, 1, 2,

3, 4. Notice that the right–angled triangles (possibly degenerate) AiBi+3Ci+3 and

AiBi+2Ci+2 are similar, for angles Ai are vertical. HenceAiBi+3

AiBi+2=

Bi+3Ci+3

Bi+2Ci+2,

implying4∏

i=0

AiBi+3

AiBi+2=

4∏i=0

Bi+3Ci+3

Bi+2Ci+2= 1,

as claimed.

47

48 SELECTION TESTS FOR THE 2011 JBMO

Remark. The claim holds for any n-gon with n ≥ 5 sides, within the given condi-tions.

Problem 3. Let n be a positive integer and let x1, x2, . . . , xn and y1, y2, . . . , yn

be real numbers. Prove that there exists a number i, i = 1, 2, . . . , n, such thatn∑

j=1

|xi − xj | ≤n∑

j=1

|xi − yj |.

Solution. Without the loss of generality, suppose x1 ≤ x2 ≤ . . . ≤ xn. For eachk = 1, 2, . . . , n we have |x1 − xk|+ |xn − xk| = |x1 − xn| ≤ |x1 − yk|+ |xn − yk|,hence

n∑k=1

|x1 − xk|+n∑

k=1

|xn − xk| ≤n∑

k=1

|x1 − yk|+n∑

k=1

|xn − yk|.

The claim holds for i = 1 or i = n.

Problem 4. Let k and n be integer numbers with 2 ≤ k ≤ n − 1. Consider a setA of n real numbers such that the sum of any k distinct elements of A is a rationalnumber. Prove that all elements of the set A are rational numbers.

Solution. The difference of any two elements from A is a rational number. Toshow this, let x = y ∈ A and choose other k − 1 elements of A – the choice canbe made, for k − 1 ≤ n − 2. Denote s the sum of the k − 1 elements and apply thehypothesis to infer that x+ s and y + s are both rational numbers. Subtracting w getx− y ∈ Q, as claimed.

Now let α ∈ A. Rewrite the elements of A as α + qi, qi ∈ Q. The sum of kdistinct elements is equal to kα+

∑qi ∈ Q, hence α ∈ Q, as needed.

Problem 5. Consider n persons, each of them speaking at most 3 languages. Fromany 3 persons there are at least two which speak a common language.

i) For n ≤ 8, exhibit an example in which no language is spoken by more thantwo persons.

ii) For n ≥ 9, prove that there exists a language which is spoken by at least threepersons.

Solution. i) Split the 8 persons in two groups of 4. Set any pair of persons in eachgroup to speak a different language for a total of 6 + 6 = 12 languages, each spokenby 2 persons, each person speaking 3 languages.

SELECTION TESTS FOR THE 2011 JBMO 49

For n ≤ 7, just remove 8− n persons.ii) Assume by contrary that each language is spoken by at most two persons. Then

each person A can speak with at most three others, for otherwise, by pigeon–hole prin-ciple, there exists a language spoken by other two persons besides A, a contradiction.Let B, C, D the persons which whom A can speak. Likewise, E can speak with (atmost) three others, namely F, G, H . There is left at least another person, say Z, andin the group A, E, Z no language is spoken in common, a contradiction.

SECOND SELECTION TEST

Problem 6. Determinea) the smallest numberb) the biggest number

n ≥ 3 of non-negative integers x1, x2, ... , xn, having the sum 2011 and satisfying:x1 ≤ |x2 − x3 | , x2 ≤ |x3 − x4 | , ... , xn−2 ≤ |xn−1 − xn | , xn−1 ≤ |xn − x1 |and xn ≤ |x1 − x2 | .

Solution. Let x1, x2, ... , xn non-negative integers satisfying the conditions fromthe statement. Let us imagine them as being arranged on a circle, in this order. Wedenote with M the biggest of the numbers x1, ... , xn. Without loss of generality, wemay suppose that x1 = M . Obviously M = 0. From x1 ≤ |x2−x3 | and the choiceof x1 it follows that x2, x3 = M, 0. Therefore any M is followed by an M anda 0 on the circle, either in the succession M −M − 0 or the succession M − 0−M .By induction after the last M known on the circle, we find that on the circle there areonly the numbers 0 and M . Moreover

• there are no the neighboring zeros (or else the number preceding them wouldalso be a 0 and, inductively, all the numbers on the circle would be zeros, whichcontradicts the fact that their sum is 2011)

• there are no three consecutive M -s. Since 2011 is prime and x1+x2+...+xn =

kM , where k is the number of the M -s, we find that M = 1 and k = 2011.a) The smallest value of n is obtained when the number of zeros is the smallest

possible. Out of three consecutive numbers on the circle, at least one is 0, therefore wehave at least 1006 zeros, which means at least 3017 numbers. A possible configurationfulfilling the conditions of the statement is x3k = 0, k = 1, 2, ..., 1005, x3017 = 0,and the rest of the xj = 1.


b) The biggest n is obtained when the number of zeros is the largest possible.Since there can be no neighboring zeros, we have at most 2011 zeros. This happens ifwe place alternatively 0 and 1 on the circle, configuration that satisfies the conditionsfrom the statement. Thus, the biggest value of n is 4022.

Problem 7. We consider an n × n (n ∈ N, n ≥ 2) square divided into n2 unitsquares. Determine all the values of k ∈ N for which we can write a real number ineach of the unit squares such that the sum of the n2 numbers is a positive number,while the sum of the numbers from the unit squares of any k × k square is a negativenumber.

Solution. We will prove that the desired numbers k are those that are not factorsof n.

If k | n that we can tile the n × n square with k × k squares and the total sumshould simultaneously be positive and negative.

If k | n then n = kq + r, where 0 < r < k. We fill the unit squares with a

(to be chosen conveniently later on) in the positions (ik, jk) with i, j = 1, ..., q andwith 1 in the other positions. Every k × k square contains exactly one unit squareof the form (ik, jk), therefore the sum in every k × k square is a + k2 − 1. Thetotal sum is q2a+ n2 − q2. We will choose a arbitrarily from the non-empty interval(1− n2

q2 , 1− k2).

Remark: For the case k | n there are many other ways of choosing the numbersfrom the unit squares. Another choice is to fill all the unit squares of the columns jk,j = 1, ..., q, with a convenient a and the other ones with 1.

Problem 8. a) Prove that if the sum of the non-zero digits a1, a2, ... , an is amultiple of 27, then it is possible to permute these digits in order to obtain an n-digitnumber that is a multiple of 27.

b) Prove that if the non-zero digits a1, a2, ... , an have the property that every n-digit number obtained by permuting these digits is a multiple of 27, then the sum ofthese digits is a multiple of 27.

Andrei Eckstein

Solution. a) Obviously n ≥ 3. If n = 3 then a1 = a2 = a3 = 9 and 999... 27.

Suppose n ≥ 4. Having the sum of its digits a multiple of 9, each of the numbersformed with the digits a1, a2, ... , an is a multiple of 9, therefore division with 27will yield one of the remainders 0, 9 or 18.


• If all the digits give the same remainder r when divided by 3:

− if r = 0 then we choose N = a1a2...an. We have thatN

3=

a13

a23

...an3

has

the sum of its digits a multiple of 9, therefore the numberN

3is a multiple of 9, hence

N is a multiple of 27.

− if r ∈ 1, 2, then from a1 + a2 + ... + an... 27 and ai ≡ 0 (mod 3), for all i

it follows that nr ≡ 0 (mod 3), hence n is a multiple of 3. Then: a1a2...an − (a1 +

a2 + ...+ an) = 9(a1a1...a1︸︷︷︸n−1 cifre

+ a2a2...a2︸︷︷︸n−2 cifre

+ ...+ an−1

︸︷︷︸M

).

Then M ≡ (n−1)a1+(n−2)a2+...+an−1 ≡ r((n−1)+(n−2)+...+2+1

)≡

r · n(n− 1)

2≡ 0 (mod 3). Therefore, in this case, a1a2...an and a1 + a2 + ... + an

give the same remainder at the division by 27.• If not all of the digits a1, a2, ... , an have the same reminder when divided by

3, then we choose three of them such that their sum is not a multiple of 3. Let a, b, cbe these three digits. Then the numbers x = ...abc, y = ...bca and z = ...cab givedifferent reminders at the division by 27. Indeed, x− y = abc− bca = 108a− 81b−9(a+b+c) = M27−9(a+b+c) = M27. Since the possible reminders at the divisionby 27 are only 0, 9 and 18, it follows that (exactly) one of the numbers x, y, z is amultiple of 27.

b) The differece between two numbers obtained one from the other by swappingtwo neighboring digits a and b is 9(a− b) · 10k and needs to be a multiple of 27. Weobtain that each of the digits a1, ..., an has to give the same remainder when dividedby 3. From here one continues as by a) in order to show that, in this case, the numbersa1a2...an and a1 + a2 + ...+ an give the same remainder when divided by 27.

Problem 9. The measure of the angle A of the acute triangle ABC is 60, andHI = HB, where I and H are the incenter and the orthocenter of the triangle ABC.Find the measure of the angle B.

Solution. We have m(∠BIC) ≡ m(∠BHC) = 120, hence B, H, I, C aresituated on a circle.

If m(∠B) > 60 then m(∠HBI) = m(∠ABI) − m(∠ABH) = 12m(∠B) −

30.

But m(∠HBI) = m(∠HIB) = m(∠HCB) = 90−m(∠B), hence m(∠B) =


80. It is easy to see that the case m(∠B) ≤ 60 is not possible.

THIRD SELECTION TEST

Problem 10. It is said that a positive integer n > 1 has the property (p) if in itsprime factorization

n = pα11 · ... · pαj

j

at least one of the prime factors p1, ... , pj has the exponent equal to 2.

a) Find the largest number k for which there exist k consecutive positive integersthat do not have the property (p).

b) Prove that there is an infinite number of positive integers n such that n, n + 1

and n+ 2 have the property (p).

Dorel Mihet

Solution. a) Among any 8 consecutive integers there exists one of the form 8j+4.This number has the property (p) because the factor 2 from its prime factorization hasthe exponent 2. Therefore there can be at most 7 consecutive positive integers that donot have the property (p). Since none of the numbers 29, 30, 31, 32, 33, 34, 35 hasthe property (p), the largest k is k = 7.

b) Since the numbers 98 = 2 ·72, 99 = 32 ·11, and 100 = 22 ·52 have the property(p), so will the numbers 98+(7 ·3 ·2)3 ·k, 99+(7 ·3 ·2)3 ·k, and 100+(7 ·3 ·2)3 ·k.Alternative solution. By the Chinese remainder theorem, there exist an infinite num-ber of solutions for the system of simultaneous congruences: n ≡ 4 (mod 8), n ≡8 (mod 27), n ≡ 23 (mod 125). Then n, n + 1, and n + 2 all have the property(p) because the factor 2, 3, and 5, respectively, has the exponent 2 in their primefactorizations.

Problem 11. Find all the finite sets A of real positive numbers having at least twoelements, with the property that a2 + b2 ∈ A for every a, b ∈ A with a = b.

Solution. If a1 < a2 < ... < an are the elements of A, then a21 + a22, a21 + a23,

a21+a2n, a22+a2n,..., a2n−1+a2n belong to A, and a21+a22 < a21+a23 < ... < a21+a2n <

a22 + a2n < ... < a2n−1 + a2n, which implies that 2n − 3 ≤ n; hence A has at most 3elements.


If n = 3 and A = a, b, c with a < b < c, then a2 + b2 = a, b2 + c2 =

c, a2+ c2 = b, hence a2− c2 = a− c, and a+ c = 1. Substituting in a2+ c2 = b, we

obtain b = 2a2 − 2a+ 1 = 2(−b2) + 1, and therefore b =1

2. Then a2 − a+

1

4= 0

leads to a =1

2, which contradicts a < b.

Hence n = 2, and A = a, b with a2 + b2 = a, i.e., for a ∈ (0, 1), we getb =

√a− a2. In conclusion, the desired sets are a,

√a− a2, with a ∈ (0, 1),

a = 1

2.

Problem 12. Let ABC be a triangle, Ia the center of the excircle at side BC, andM its reflection across BC. Prove that AM is parallel to the Euler line of the triangleBCIa.

Solution. Let I be the incenter of ABC, Ha the orthocenter of IaBC, and Oa themidpoint of the segment [IIa]. Then BOa = IOa = IaOa = COa, therefore Oa isthe circumcenter of triangle IaBC. Moreover, IB ‖ CHa (both are perpendicular onBIa) and, similarly, IC ‖ BHa, hence BICHa is a parallelogram.Let P denote the projection of Ia onto BC and T be the midpoint of [AIa]. If r, raare the inradius and the radius of the excircle at side BC, respectively, then we have:IA

AIa=

r

ra=

HaP

IaP, hence

HaIaIaP

=IIaIaA

=2IaOa

2IaT=

IaOa

IaT. This proves that

OaHa is parallel to TP . But TP ‖ AM , therefore AM is parallel to OaHa, i.e. it isparallel to the Euler line of the triangle IaBC.

Problem 13. Let m be a positive integer. Determine the smallest positive integern for which there exist real numbers x1, x2, . . . , xn ∈ (−1, 1) such that |x1|+ |x2|+· · ·+ |xn| = m+ |x1 + x2 + · · ·+ xn|.

Solution. Let us consider x1, x2, ... , xn a solution of the equation above. Wemay suppose, without loss of generality, that x1 + x2 + ... + xn ≥ 0 (otherwise wechange the signs of all the numbers) and that x1 ≤ ... ≤ xp ≤ 0 < xp+1 ≤ ... ≤ xn.Then the equation becomes −2(x1+x2+...+xp) = m. From xi ∈ (−1, 1) we obtain

that m < 2p, i.e. p >m

2. Moreover, n−p > xp+1+ ...+xn ≥ −x1− ...−xp =

m

2.

If m is even then p ≥ m

2+1, n−p ≥ m

2+1, and therefore n ≥ m+2. A convenient

choice for n = m+ 2 is x1 = ... = xm2 +1 = − m

m+ 2, xm

2 +2 = ... = xn =m

m+ 2.

If m is odd, from p ≥ m+ 1

2, n− p ≥ m+ 1

2, we obtain n ≥ m+ 1. A convenient


choice for n = m+1 is x1 = ... = xm+12

= − m

m+ 1, xm+1

2 +1 = ... = xn =m

m+ 1.

In conclusion, the smallest value of n for which the equation has solutions isn = m+ 1 if m is odd and n = m+ 2 if m is even.

FOURTH SELECTION TEST

Problem 14. For every positive integer n let τ(n) denote the number of its positivefactors. Determine all n ∈ N that satisfy the equality τ(n) =

n

3.

Solution. If n ∈ N satisfies the condition τ(n) =n

3, then 3 | n. Put n = 3k,

k ∈ N. If k is even thenk

2=

n

6is a factor of n. Even if all the positive numbers

smaller thann

6are factors of n and the numbers

n

5,n

4, ...,

n

1are also factors of n, we

haven

3= τ(n) ≤ n

6+ 5. Hence n ≤ 30. Checking the numbers 6, 12, 18, 24 and 30

we find that 18 and 24 satisfy the desired condition.In the case when k is odd we can proceed similarly, obtaining n = 9, or, alterna-

tively, we can use the fact that a number having an odd number of positive factors is a

perfect square. If n = m2, then n has at most m+1 factors, hence m+1 ≥ m2

3. We

obtain that m ≤ 3 and the conclusion.In conclusion, the problem admits three solutions: 9, 18 and 24.Alternative solution: Since 3 | n, it follows that n has a prime factorization of

the form n = 3a · pα11 · ... · pαj

j and the number of his positive factors is τ(n) =

(a+ 1)(α1 + 1)...(αj + 1). The condition τ(n) =n

3leads to 3a−1 · pα1

1 · ... · pαj

j =

(a + 1)(α1 + 1)...(αj + 1). Since pαii ≥ 2αi ≥ αi + 1, in order for n to satisfy the

equation from the statement, it is necessary that a+ 1 ≥ 3a−1, hence a = 1 or a = 2.If in the prime factorization of n there is a prime pi > 3, then pαi

i > 4αi ≥ 2αi+2

and the equality τ(n) =n

3can not take place. If a = 1 then n = 3 · 2m and the

equality τ(n) =n

3reduces to 2(m + 1) = 2m. We find m = 3 (for m ≥ 4 we have

2m > 2m+ 2), hence n = 24.Similarly, if a = 2 then n = 32 · 2m and from 3(m + 1) = 3 · 2m we obtain

m ∈ 0, 1, i.e. n ∈ 9, 18.

Problem 15. Let ABC be a triangle with circumcenter O. The points P and Q


are interior points of the sides CA and AB respectively. Let K, L and M be themidpoints of the segments BP , CQ and PQ, respectively, and let Γ be the circlepassing through K, L and M . Suppose that the line PQ is tangent to the circle Γ.Prove that OP = OQ.

Sergei Berlov, Russia, IMO 2009

Solution. From AB ‖ KM and AC ‖ LM we obtain that ∠KMQ ≡ ∠AQP

and ∠LMP ≡ ∠APQ. If PQ is tangent to the circumcircle of the triangle KLM

then ∠KLM ≡ ∠KMQ, hence ∠KLM ≡ ∠AQP . Then ∆APQ ∼ ∆MKL,

henceAQ

ML=

AP

MK. It follows that

AQ

PC=

AP

BQ, i.e. AQ · BQ = AP · PC. This

means that P and Q have equal powers with respect to the circumcircle of the triangleABC. As they are both situated inside the circle, they are at equal distance from thecenter of the circle, O.

Problem 16. a) Find the largest possible value of the number

x1x2 + x2x3 + ...+ xn−1xn,

if x1, x2, ... , xn (n ≥ 2) are non-negative integers and their sum is 2011.b) Find the numbers x1, x2, ... , xn for which the maximum value determined at

a) is obtained.

Dorel Mihet

Solution. a) Let x1, x2, ... , xn be non-negative integers satisfying the conditionsfrom the statement. We put M = max

1≤i≤nxi. If xj = M then x1x2 + x2x3 + ... +

xn−1xn ≤ x1xj+x2xj+ ...+xj−1xj+xjxj+1+xjxj+2+ ...+xjxn = xj(2011−xj) = M(2011 − M) ≤ 1005 · 1006. Indeed, the last inequality comes to (M −1005)(M − 1006) ≥ 0 which is true for any integer M . The largest possible value is1005 · 1006 because this value can be obtained by choosing, for example, x1 = 1005,x2 = 1006 and xk = 0 for k ≥ 3.

b) For n = 2 we have x1x2 = 1005 · 1006 ⇔ x1(2011 − x1) = 1005 · 1006 ⇔(x1 − 1005)(x1 − 1006) = 0 ⇔ (x1, x2) ∈ (1005, 1006), (1006, 1005).For n = 3, x1x2 + x2x3 = 1005 · 1006 ⇔ x2(x1 + x3) = 1005 · 1006 comes, asabove, to x2 = 1005, x1 + x3 = 1006 or x2 = 1006, x1 + x3 = 1005. We obtain(x1, x2, x3) ∈ (k, 1005, 1006 − k) | k = 0, 1, ..., 1006 ∪ (k, 1006, 1005 − k) |k = 0, 1, ..., 1005.


For n ≥ 4, we denote by j the smallest index for which xj > 0. Then, repla-cing xj by 0 and xj+2 by xj+2 + xj , increases the value of the sum by xjxj+3.Using this remark it is easy to see that if x1x2 + x2x3 + ... + xn−1xn = 1005 ·1006, then at most three of the terms can be non-zero. We obtain (x1, ..., xn) ∈(0, ..., 0, k, 1005, 1006−k, 0, ..., 0) | k = 0, 1, ..., 1006∪(0, ..., 0, k, 1006, 1005−k, 0, ..., 0) | k = 0, 1, ..., 1005, where the group of the three non-zero componentscan be located anywhere.

Problem 17. Show that there is an infinite number of positive integers t such thatnone of the equations x2 + y6 = t, x2 + y6 = t+ 1, x2 − y6 = t, x2 − y6 = t+ 1

has solutions (x, y) ∈ Z× Z.

Dorel Mihet

Solution. If x is a positive integer, then either x12 ≡ 0 (mod 13) or x12 ≡ 1 (mod13), hence x6 is congruent with −1, 0 or 1 modulo 13. Therefore, if t is congruentwith 6 (mod 13), then ±x6+ t is congruent with 5, 6 or 7 (mod 13), while ±x6+ t+1

is congruent with 6, 7 or 8 (mod 13).On the other hand, perfect squares are congruent with 0, 1, 4, 9, 3, 12 or 10 (mod

13). Therefore a perfect square can not be equal to a number of the form ±x6 + t or±x6 + t+1 if t ≡ 6 (mod 13). In conclusion, all the numbers that are congruent with6 modulo 13 have the required property.

THE 28th BALKAN MATHEMATICAL OLYMPIAD

Iasi, Romania, May 4th – May 8th, 2011

Problem 1. Let ABCD be a cyclic quadrilateral which is not a trapezoid andwhose diagonals meet at E. The midpoints of AB and CD are F and G respectivelyand is the line through G parallel to AB. The feet of the perpendiculars from E

onto and CD are H and K, respectively. Prove that the lines EF and HK areperpendicular.

United Kingdom

Solution. The points E, K, H , G are on the circle of diameter GE, so the anglesEHK and EGK are equal.

57

58 2011 BALKAN MATHEMATICAL OLYMPIAD

Also, from ∠DCA = ∠DBA and CE/CD = BE/BA follows

CE

CG=

2CE

CD=

2BE

BA=

BE

BF,

so the triangles CGE and BFE are similar. In particular, the angles EGC and BFE

are equal, and therefore so are the angles EHK and BFE.But the lines EH and BF are perpendicular and so, since EF and HK are ob-

tained by rotations of these lines by the same (directed) angle, the lines EF and HK

also perpendicular.

Problem 2. Given three real numbers x, y, z such that x+ y + z = 0, show that

x(x+ 2)

2x2 + 1+

y(y + 2)

2y2 + 1+

z(z + 2)

2z2 + 1≥ 0.

When does equality hold?

Greece

Solution. The inequality is clear if xyz = 0, in which case equality holds if andonly if x = y = z = 0.

Henceforth assume xyz = 0 and rewrite the inequality as

(2x+ 1)2

2x2 + 1+

(2y + 1)2

2y2 + 1+

(2z + 1)2

2z2 + 1≥ 3.

Notice that (exactly) one of the products xy, yz, zx is positive, say yz > 0, to get

(2y + 1)2

2y2 + 1+

(2z + 1)2

2z2 + 1≥ 2(y + z + 1)2

y2 + z2 + 1(by Jensen)

=2(x− 1)2

x2 − 2yz + 1(for x+ y + z = 0)

≥ 2(x− 1)2

x2 + 1. (for yz > 0)

Here equality holds if and only if x = 1 and y = z = −1/2. Finally, since

(2x+ 1)2

2x2 + 1+

2(x− 1)2

x2 + 1− 3 =

2x2(x− 1)2

(2x2 + 1)(x2 + 1)≥ 0, x ∈ R,

the conclusion follows. Clearly, equality holds if and only if x = 1, so y = z = −1/2.Therefore, if xyz = 0, equality holds if and only if one of the numbers is 1, and theother two are −1/2.

2011 BALKAN MATHEMATICAL OLYMPIAD 59

Solution 2. Rewrite the inequality as in Solution 1,

(2x+ 1)2

2x2 + 1+

(2y + 1)2

2y2 + 1+

(2z + 1)2

2z2 + 1≥ 3,

and use the condition x+ y + z = 0 to replace 3 in the right-hand member by

(2x+ 1)2

43 (x

2 + y2 + z2) + 1+

(2y + 1)2

43 (x

2 + y2 + z2) + 1+

(2z + 1)2

43 (x

2 + y2 + z2) + 1,

and thereby deduce that it is sufficient to show that

(2x+ 1)2

2x2 + 1≥ (2x+ 1)2

43 (x

2 + y2 + z2) + 1(∗)

and the like. Unless x = −1/2, in which case equality holds trivially, use again thecondition x + y + z = 0 to transform (∗) into the equivalent and obvious inequality(y−z)2 ≥ 0, which yields the equality case y = z. This proves the required inequalityand shows that equality holds if and only if the variables all vanish simultaneously ortwo equal −1/2 and the third equals 1.

Problem 3. Let S be a finite set of positive integer numbers which has the fol-lowing property: if x is a member of S, then so are all positive divisors of x. Call asubset T of S good (respectively, bad ) if it is non-empty and, whenever x and y aremembers of T , and x < y, the ratio y/x is (respectively, is not) a power of a primenumber; agree that a singleton (one-element) subset is both good and bad. Show thata maximal good subset of S has as many elements as a minimal partition of S into badsubsets.

Bulgaria

Solution. Notice first that a bad subset of S contains at most one element from agood one, to deduce that a partition of S into bad subsets has at least as many membersas a maximal good subset.

Notice further that the elements of a good subset of S must be among the termsa geometric sequence whose ratio is a prime: if x < y < z are elements of a goodsubset of S, then y = xpα and z = yqβ = xpαqβ for some primes p and q and somepositive integers α and β, so p = q for z/x to be a power of a prime.

Next, let P = 2, 3, 5, 7, 11, · · · denote the set of all primes, let

m = max expp x : x ∈ S and p ∈ P,


where expp x is the exponent of the prime p in the canonical decomposition of x, andnotice that a maximal good subset of S must be of the form a, ap, · · · , apm forsome prime p and some positive integer a which is not divisible by p. Consequently,a maximal good subset of S has m + 1 elements, so a partition of S into bad subsetshas at least m+ 1 members.

Finally, notice by maximality of m that the sets

Sk = x : x ∈ S and∑p∈P

expp x ≡ k (mod m+ 1), k = 0, 1, · · · ,m,

form a partition of S into m+ 1 bad subsets. The conclusion follows.

Problem 4. The opposite sides of a convex hexagon of unit area are pairwiseparallel. The lines of support of three alternate sides meet in pairs to determine thevertices of a triangle. Similarly, the lines of support of the other three alternate sidesmeet in pairs to determine the vertices of another triangle. Show that the area of atleast one of these two triangles is greater than or equal to 3/2.

Bulgaria

Solution. Unless otherwise stated, throughout the proof indices take on valuesfrom 0 to 5 and are reduced modulo 6. Label the vertices of the hexagon in circularorder, A0, A1, · · · , A5, and let the lines of support of the alternate sides AiAi+1 andAi+2Ai+3 meet at Bi. To show that the area of at least one of the triangles B0B2B4,B1B3B5 is greater than or equal to 3/2, it is sufficient to prove that the total area ofthe six triangles Ai+1BiAi+2 is at least 1:

5∑i=0

areaAi+1BiAi+2 ≥ 1.

To begin with, reflect each Bi through the midpoint of the segment Ai+1Ai+2

to get the points B′i. We shall prove that the six triangles Ai+1B

′iAi+2 cover the

hexagon. To this end, reflect A2i+1 through the midpoint of the segment A2iA2i+2

to get the points A′2i+1, i = 0, 1, 2. The hexagon splits into three parallelograms,

A2iA2i+1A2i+2A′2i+1, i = 0, 1, 2, and a (possibly degenerate) triangle, A′

1A′3A

′5.

Notice first that each parallelogram A2iA2i+1A2i+2A′2i+1 is covered by the pair of

triangles (A2iB′2i+5A2i+1, A2i+1B

′2iA2i+2), i = 0, 1, 2. The proof is completed

by showing that at least one of these pairs contains a triangle that covers the trian-gle A′

1A′3A

′5. To this end, it is sufficient to prove that A2iB

′2i+5 ≥ A2iA

′2i+5 and

2011 BALKAN MATHEMATICAL OLYMPIAD 61

A2j+2B′2j ≥ A2j+2A

′2j+3 for some indices i, j ∈ 0, 1, 2. To establish the first

inequality, notice that

A2iB′2i+5 = A2i+1B2i+5, A2iA

′2i+5 = A2i+4A2i+5, i = 0, 1, 2,

A1B5

A4A5=

A0B5

A5B3and

A3B1

A0A1=

A2A3

A0B5,

to get2∏

i=0

A2iB′2i+5

A2iA′2i+5

= 1.

Similarly,2∏

j=0

A2j+2B′2j

A2j+2A′2j+3

= 1,

whence the conclusion.

Solution 2. With reference to Solution 1, suppose, if possible, that the six tri-angles AiAi+1B

′i+5 fail to cover the hexagon A0A1 · · ·A5. Then some point X of

the hexagon falls outside each of those triangles. Up to an affine transformation, wemay (and will) assume that the triangles B0B2B4 and B1B3B5 are both equilateral,so the hexagon is equiangular. We may further assume that the angle XA0A1 doesnot exceed 60. Since X is not covered by the triangle A0A1B

′5, the angle XA1A0

exceeds 60, so XA0 > XA1 and the angle XA1A2 is less than 60. Next, the tri-angle A1A2B

′0 does not cover X , so the angle XA2A1 exceeds 60, XA1 > XA2

and the angle XA2A3 is less than 60. Continuing, we obtain XAi > XAi+1, i =0, 1, · · · , 5, and thereby reach the contradiction XA0 > XA1 > · · · > XA5 > XA0.

Solution 3. In the setting of Solution 2 – the triangles B0B2B4 and B1B3B5 areboth equilateral etc. – let ai denote the length of the segment AiAi+1, let 0 and 1


respectively denote the side lengths of the equilateral triangles B0B2B4 and B1B3B5

and notice that ai−1 + ai + ai+1 = i mod 2 to get a0 + a2 + a4 = 21 − 0 anda1 + a3 + a5 = 20 − 1.

Incidentally, these relations show that the given configuration is geometrically pos-sible and non-degenerate if and only if 1/2 < 0/1 < 2.

Now let A, A0, A1 respectively denote the areas of the hexagon A0A1 · · ·A5 andthe triangles B0B2B4 and B1B3B5. Then

A0 −A =∑

i=0,1,2

areaA2i+1B2iA2i+2 =

√3

4(a21 + a23 + a25)

≥√3

12(a1 + a3 + a5)

2 =

√3

12(20 − 1)

2.

Similarly, A1 −A ≥ (21 − 0)2√3/12, so

A0 +A1 − 2A ≥√3

12

((20 − 1)

2 + (21 − 0)2)

=

√3

12

(20 + 21 + 4(0 − 1)

2)≥

√3

12

(20 + 21

)

=1

3(A0 +A1) .

Consequently, A0 +A1 ≥ 3A and the conclusion follows.

Solution 4. With reference again to the notation and conventions in Solution 1, letr = B2iB2i+2/B2i+3B2i+5 denote the ratio of similarity of the triangles B0B2B4

and B3B5B1, let ri = Ai+1Ai+2/Bi−2Bi+2 denote the ratio of similarity of thetriangles Ai+1BiAi+2 and Bi−2BiBi+2, and notice that rr2i = 1 − r2i−1 − r2i+1

and r2i+1 = r(1−r2i−r2i+2) to obtain r0+r2+r4 = 2−1/r and r1+r3+r5 = 2−r.

As already noticed in Solution 3, these relations show that the given configurationis geometrically possible and non-degenerate if and only if 1/2 < r < 2.

Now let A0 and A1 respectively denote the areas of the triangles B0B2B4 andB1B3B5 and recall that the hexagon A0A1 · · ·A5 has unit area, to write

1

A0= 1− (r20 + r22 + r24) ≤ 1− 1

3(r0 + r2 + r4)

2 = 1− 1

3

(2− 1

r

)2

,

1

A1= 1− (r21 + r23 + r25) ≤ 1− 1

3(r1 + r3 + r5)

2 = 1− 1

3(2− r)

2,

and thereby conclude that A0 ≥ 3/2 if 1 ≤ r < 2, and A1 ≥ 3/2 if 1/2 < r ≤ 1.

THE DANUBE MATHEMATICAL COMPETITION

Calarasi, October 2010

Problem 1. Determine all integer numbers n ≥ 3 such that the regular n-gon canbe decomposed into isosceles triangles by noncrossing diagonals.

Solution. The required numbers are of the form n = 2r(2s+1), where r and s arenonnegative integer numbers which do not vanish simultaneously. Clearly, any suchn works.

To establish the converse, let K be a regular n-gon, n ≥ 4, which can be decom-posed into isosceles triangles by noncrossing diagonals. Begin by noticing that eachedge e of K must be an edge of a unique isosceles triangle Te in the decomposition.Two cases are possible: either e is opposite the apex of Te or e and one of the adjacentedges of K are the edges of Te issuing from the apex. (Since n ≥ 4, Te cannot beequilateral, so the apex is well defined.)

If n is even, no vertex of K lies on the perpendicular bisector of an edge of K, sothe first case is ruled out. Consequently, the decomposition must contain exactly oneof the two bracelets of n/2 isosceles triangles clipped off by short diagonals joiningconsecutive vertices of K of likewise parity. These short diagonals are the edges ofa regular n/2-gon which is also decomposed into isosceles triangles by noncrossingdiagonals and the conclusion follows by induction.

If n is odd, then K has a unique edge e opposite the apex of Te: Since n is oddand each short diagonal clips off two edges of K, at least one such e exists. The apexof Te lies on the perpendicular bisector of e, so it must be the vertex of K oppositee. Uniqueness of e should now be clear: were there another such e′, the interiors ofTe and Te′ would overlap. Consequently, e is unique and K splits into Te and two

63

64 2010 DANUBE MATHEMATICAL COMPETITION

polygons L and L′ which are reflections of one another in the perpendicular bisectorof e.

To complete the proof, it is sufficient to show that the number of vertices of Lis one plus a power of 2. Begin by noticing that L inherits by restriction a de-composition into isosceles triangles by noncrossing diagonals. Let x0, . . . , xm bea circular labelling of the vertices of L around the boundary, where x0 is the ver-tex of K opposite e and xm is a vertex of e. Since dist(xi, xj) < dist(x0, xm) ifi, j = 0,m, it follows that x0xm is an edge of an isosceles triangle with apexat some xk, 0 < k < m. Notice that dist(x0, xi) < dist(xi, xm) if 0 < i < m/2

and dist(x0, xi) > dist(xi, xm) if m/2 < i < m, to deduce that m must be even,k = m/2, and L splits into an isosceles triangle, x0xm/2xm, and two polygons,x0 · · ·xm/2 and xm/2 · · ·xm, which are reflections of one another in the perpendic-ular bisector of the segment x0xm. Now we are essentially back in the situation thatarose above. Repeat the same argument verbatim to infer that m/2 must be even andso on all the way down to conclude that m must be a power of 2.

Problem 2. Given a triangle ABC, let A′, B′ and C ′ be the perpendicular feetdropped from the centroid G of the triangle ABC on the sides BC, CA and AB,respectively. Reflect A′, B′ and C ′ through G to A′′, B′′ and C ′′, respectively. Provethat the lines AA′′, BB′′ and CC ′′ are concurrent.

Solution. Let A0 be the perpendicular foot dropped from A on the line BC, letA1 be the midpoint of the side BC, and let A2 be the point where the line AA′′ meetsthe line BC; define the points Bi and Ci, i = 0, 1, 2, similarly. We shall prove that(AA0, AA2), (BB0, BB2) and (CC0, CC2) are pairs of isotomic lines; that is, A0

and A2 are reflections of one another through A1 and the like. It will then follow thatthe lines AA′′ = AA2, BB′′ = BB2 and CC ′′ = CC2 are concurrent at the isotomicconjugate of the orthocentre of the triangle ABC.

Clearly, it is sufficient to show the lines AA0 and AA2 isotomic. To this end,notice first that A′A1/A0A1 = 1/3. On the other hand, Menelaus’ theorem applied totriangle GA′A1 and transversal AA′′A2 yields A1A2/A

′A2 = 3/4, so A′A1/A2A1 =

1/3. Consequently, A0 and A2 are indeed reflections of one another through A1.

Problem 3. All sides and diagonals of a convex n-gon, n ≥ 3, are coloured oneof two colours. Show that there exist (n + 1)/3 pairwise disjoint monochromatic

2010 DANUBE MATHEMATICAL COMPETITION 65

segments. (Two segments are disjoint if they do not share an endpoint or an interiorpoint.)

Solution. If all sides are monochromatic, then the assertion is clearly true. Oth-erwise, delete a vertex incident with two sides of different colours together with itsneighbours, delete all sides and diagonals incident with these three vertices and applyinduction.

Problem 4. Given a prime number p congruent to 3 modulo 4, show that w2p +

x2p + y2p = z2p for no integer numbers w, x, y, z whose product is not divisible byp.

Solution. Suppose there are four such numbers. Without loss of generality, wemay (and will) assume that they are jointly coprime: (w, x, y, z) = 1. Reduction mod-ulo 4 shows that z and exactly one of the numbers w, x, y, say y, must be odd. Writew2p + x2p = z2p − y2p =

(z2 − y2

) (∑p−1k=1 z

2(p−k−1)(z2k − y2k

)− pz2(p−1)

)to

deduce that the second factor above is congruent to 3 modulo 4 and infer thereby thatin its decomposition into prime factors some prime q ≡ 3 (mod 4) occurs with an oddexponent. Since −1 is a quadratic non-residue modulo q, it follows that w and x areboth divisible by q, and in the decomposition of w2p+x2p into prime factors, q occurswith an even exponent. Hence z2 − y2 is divisible by q, and therefore so is pz2(p−1).Notice that p = q (for q divides w, but p does not by assumption) to deduce that z isdivisible by q. Then so is y. Consequently, w, x, y, z all share the common factor q,in contradiction with their joint coprimality.

Problem 5. Given an integer number n ≥ 3, determine the real numbers x1, x2,· · · , xn minimising (n− 1)(x2

1 + x22 + . . .+ x2

n) + nx1x2 · · ·xn, subject to xk ≥ 0,k = 1, 2, . . . , n, and x1 + x2 + · · ·+ xn = n.

Solution. The minimum is n2 and is achieved when all xk = 1 or when exactlyone of the xk is 0 and other ones are all n/(n− 1). Notice that if some xk = 0, thenthe Cauchy-Schwarz inequality hints at an educated guess of the minimum; the guessis further supported by the fact that the same value is achieved when all xk = 1.

Let f(x1, x2, · · · , xn) = (n− 1)(x21 +x2

2 + . . .+x2n)+nx1x2 · · ·xn. Assuming

0 ≤ x1 ≤ x2 ≤ · · · ≤ xn, we first show that

f(x1, x2, · · · , xn) ≥ f(x1, (x2 + xn)/2, x3, · · · , xn−1, (x2 + xn)/2).

66 2010 DANUBE MATHEMATICAL COMPETITION

To begin with, notice that xn ≥ 1, to get

f(x1, x2, . . . , xn)− f(x1, (x2 + xn)/2, x3, . . . , xn−1, (x2 + xn)/2)

=1

4(x2 − xn)

2 (2(n− 1)− nx1x3 · · ·xn−1)

≥ 1

4(x2 − xn)

2 (2(n− 1)− nx1x3 · · ·xn−1xn) .

We now show that x1x3 · · ·xn−1xn ≤ 2(n − 1)/n, so the difference above isindeed non-negative. Notice first that

n = x1+x2+ · · ·+xn ≥ 2x1+x3+ · · ·+xn ≥ (n−1) (2x1x3 · · ·xn−1xn)1/(n−1)

,

so x1x3 · · ·xn−1xn ≤ nn−1/(2(n − 1)n−1). Since nn−1/(2(n − 1)n−1) ≤2(n − 1)/n, n ≥ 2, the conclusion follows. (The latter follows from the well-knownfact that the sequence un = (1 + 1/n)n+1, n ≥ 1, is decreasing, so un ≤ u1 = 4,n ≥ 1; that un decreases is easily seen by Bernoulli’s inequality: n2(n+1)/(n2 −1)n+1 = (1 + 1/(n2 − 1))n+1 ≥ 1 + (n+ 1)/(n2 − 1) = n/(n− 1), n ≥ 2.)

Consequently, the value of f does not increase when the largest and the secondsmallest of the xk is replaced by their arithmetic mean. It follows that f achieves itsminimum when n − 1 of the xk are all equal to some x, and the smallest of the xk isn − (n − 1)x. (That f actually achieves a minimum subject to the stated conditionsfollows by a standard compactness argument which may be assumed without proof.)Since the xk are all non-negative, and n − (n − 1)x is the smallest among them, itfollows that 1 ≤ x ≤ n/(n− 1).

So we have to minimise the polynomial function

g(x) = (n− 1)((n− (n− 1)x)2 + (n− 1)x2

)+ n(n− (n− 1)x)xn−1

on the closed interval 1 ≤ x ≤ n/(n − 1). To this end, recall the guess made in thebeginning, to evaluate

g(x)− n2 = n(n− (n− 1)x)(x− 1)2n−2∑k=1

(xk−1 + xk−2 + · · ·+ 1

)≥ 0,

1 ≤ x ≤ n/(n − 1). Equality holds if and only if x = 1 or x = n/(n − 1), whencethe conclusion.

THE Eighth IMAR MATHEMATICAL COMPETITION

Bucuresti, S. Haret National College, November 2010

Problem 1. Show that a sequence (εn)n∈N of plus and minus ones is periodicwith period a power of 2, if and only if εn = (−1)P (n), n ∈ N, where P is aninteger-valued polynomial with rational coefficients.

Solution. A polynomial P of degree at most k with complex coefficients is integer-valued if and only if

P =

k∑j=0

aj

(X

j

)=

k∑j=0

ajj!

X(X − 1) · · · (X − j + 1),

where the aj are all integer numbers, so its coefficients are rational.

We show that for such a P , the sequence ((−1)P (n))n∈N is periodic with period2r, where r = min s : 2s > k. To this end, it suffices to show that if j < 2s and m isinteger, then

(mj

)≡

(m+2s

j

)mod 2. These are the coefficients of Xj in the expansions

of (1 +X)m and (1 +X)m+2s , respectively. The congruence (1 +X)2s ≡ 1 +X2s

mod 2 follows easily by induction on s. Hence (1+X)m+2s = (1+X)m(1 +X2s

)mod 2. Since j is less than 2s, it is immediate that the coefficients of Xj in (1 +X)m

and (1 +X)m+2s have the same parity.

Conversely, let α0, · · · , α2r−1 be arbitrary integers, and let β0, · · · , β2r−1 be thesolution to the lower triangular system of linear equations

2r−1∑i=0

βi

(j

i

)= αj , j = 0, · · · , 2r − 1.

67

68 2010 IMAR MATHEMATICAL COMPETITION

Since the coefficient of βj in the j-th equation is 1, the βi are all integers. Thepolynomial

2r−1∑i=0

βi

(X

i

)=

2r−1∑i=0

βi

i!X(X − 1) · · · (X − i+ 1)

realizes the sequence (−1)αj , j = 0, · · · , 2r − 1, and its extension with period 2r.

Solution 2. Given an integer-valued polynomial P with rational coefficients, weshow that the sequence ((−1)P (n))n∈N is periodic with period a power of 2. Clearly,it is sufficient to show that, for some non-negative integer s, the integer numbers P (n)

and P (n + 2s) both have the same parity, whatever n ∈ N. To this end, consider apositive integer m such that Q = mP is a polynomial with integral coefficients (e.g.,let m be the least common multiple of the denominators of the coefficients of P whenwritten in lowest terms), let 2r be the highest power of 2 dividing m, and let s be aninteger greater than r. Write m = 2r(2m′ + 1) and fix a positive integer n. Since 2s

divides the difference

Q(n+ 2s)−Q(n) = m(P (n+ 2s)− P (n)) = 2r(2m′ + 1)(P (n+ 2s)− P (n)),

and s > r, the conclusion follows.Conversely, given a sequence (εn)n∈N of plus and minus ones which is periodic

with period 2r, let

ak =

0, if εk = 1,

1, if εk = −1,k = 0, · · · , 2r − 1,

and consider the polynomial (Lagrange)

P =

2r−1∑k=0

ak∏j =k

X − j

k − j

=

2r−1∑k=0

(−1)k+1ak

1

k!

∏0≤j<k

(X − j)

1

(2r − k − 1)!

∏k<j≤2r−1

(X − j)

=2r−1∑k=0

(−1)k+1ak

(X

k

)(X − k − 1

2r − k − 1

).

Clearly, P has rational coefficients, is integer-valued, and P (k) = ak, k =

0, . . . , 2r − 1, so εk = (−1)P (k), k = 0, · · · , 2r − 1. To prove that the latter ex-tends to all of N, it is sufficient to show that P (n) and P (n+ 2r) both have the same

2010 IMAR MATHEMATICAL COMPETITION 69

parity, whatever n ∈ N. This amounts to showing that if j,m ∈ N and j < 2r, then(mj

)and

(m+2r

j

)have the same parity. Recalling that

(2r

i

), i = 1, · · · , 2r − 1, are all

even, this follows for instance from the identity(m+ 2r

j

)=

j∑i=0

(m− i

j − i

)(2r

i

);

or, which is actually the same, from the argument in Solution 1.

Problem 2. Given a triangle ABC, let D be the point where the incircle of thetriangle ABC touches the side BC. A circle through the vertices B and C is tangentat point E to the incircle of the triangle ABC. Show that the line DE passes throughthe excentre of the triangle ABC corresponding to the vertex A.

Solution. Let I be the incentre of the triangle ABC, let IA be the excentre corre-sponding to the vertex A, and notice that the vertices B and C both lie on the circle ofdiametre IIA. The line DIA meets again the latter circle at point K, and the lines BC

and IK meet at point L (unless AB = AC in which case the conclusion is obvious).Notice that the line BC is the radical axis of the circles BEC and BIC to deduce thatLB ·LC = LI ·LK. On the other hand, LI ·LK = LD2, for K is the perpendicularfoot dropped from the right-angled vertex D of the triangle DIL. Consequently, thepoint L is the radical centre of the following three circles: the incircle of the triangleABC, the circle BEC, and the circle BIC. Since the common tangent at E of thefirst two circles is their radical axis, it must pass through L. It follows that E is thereflection of D across the line IL, so the lines DE and IL are perpendicular and weare done.

Problem 3. Given an integer number n ≥ 2, a positive real number A, and n +

1 distinct points in the plane, X0, X1, · · · , Xn, show that the number of trianglesX0XiXj of area A does not exceed 4n

√n.

70 2010 IMAR MATHEMATICAL COMPETITION

Solution. Suppose that for some integer n ≥ 2, there exist n + 1 distinct pointsin the plane, X0, X1, · · · , Xn, such that the number of triangles X0XiXj of area A

be greater than 4n√n. Choose the minimal such n, notice that n ≥ 4, and let G be

the graph whose vertices are X1, · · · , Xn and whose edges are the XiXj such thatareaX0XiXj = A. Then every vertex Xi of G is adjacent to at least 4

√n other

vertices, since otherwise, removing Xi would reduce the number of triangles by atmost 4

√n, and we would be left with a configuration of n distinct points such that the

number of triangles X0XjXk of area A is at least 4n√n− 4

√n > 4(n− 1)

√n− 1,

contradicting the minimal choice of n. Consequently, for each Xi, there are at least4√n points Xj such that the triangle X0XiXj has area A. These points lie on two

parallel lines to the line X0Xi. One of these linear sets of points, say Si, contains atleast 1

24√n points. Notice that at least n/2 of the Si are pairwise distinct. Without

loss of generality, we may (and will) assume that the first n/2 of the Si are amongthese. Finally, recall that n ≥ 4, so

√n ≤ n/2, and consider the points Xj on the first

√n lines Si, i = 1, · · · ,

√n, to get

n ≥∣∣S1 ∪ · · · ∪ S

√n

∣∣ ≥√n∑

i=1

|Si| −∑

1≤i<j≤√n

|Si ∩ Sj |

≥1

2√n 4

√n −

(√n2

),

which is false for n ≥ 4.

Problem 4. Let r be a positive integer and let Nr be the smallest positive integersuch that the numbers

Nr

n+ r

(2n

n

), n = 0, 1, 2, · · · ,

are all integer. Show that

Nr =r

2

(2r

r

).

Solution. We first show that

Nr ≤ r

2

(2r

r

)

by proving that

K(n, r) =r

2(n+ r)

(2n

n

)(2r

r

)

2010 IMAR MATHEMATICAL COMPETITION 71

is an integer for all n ≥ 0 and r ≥ 1. Notice that

K(0, r) =

(2r − 1

r

)and K(n, 1) =

1

n+ 1

(2n

n

),

the latter being a Catalan number, so K(0, r) and K(n, 1) are integers for all n and r.Next, the recurrence relation (which will be proved at the end of the solution)

K(n, r + 1)−K(n+ 1, r) = 2K(n, 1)K(0, r)

shows by induction on r that K(n, r) is indeed an integer for all n and r.Suppose now that, for some r, Nr < Mr = r

2

(2rr

). The integer Nr is the least

common multiple, over all n ≥ 0, of the denominators of the numbers 1n+r

(2nn

)when written in lowest terms. By the argument above, Mr is a multiple of all thesedenominators. Hence Nr divides Mr. By the definition of Nr, any prime p that dividesMr/Nr also divides K(n, r) for each n ≥ 0. Since K(n, s + 1) = K(n + 1, s) +

2K(n, 1)K(0, s), induction on m shows that p divides K(n,m) for all n ≥ 0 andm ≥ r.

Now choose k such that pk ≥ r. Since p divides(pk

j

), j = 1, 2, · · · , pk − 1, the

identity(2nn

)=

∑nj=1

(nj

)2yields

(2pk

pk

)≡ 2 (mod p). Therefore p does not divide

12

(2pk

pk

)= K(0, pk). This contradicts the preceding paragraph, so Nr = r

2

(2rr

)for all

r.To prove the recurrence relation, notice that

(2m+ 2

m+ 1

)= 2

(2m+ 1

m

)= 2

2m+ 1

m+ 1

(2m

m

),

to get

K(n, r + 1)−K(n+ 1, r) =

r + 1

2(n+ r + 1)

(2n

n

)(2r + 2

r + 1

)− r

2(n+ r + 1)

(2n+ 2

n+ 1

)(2r

r

)=

1

n+ r + 1

(2n

n

)(2r

r

)(2r + 1− r

2n+ 1

n+ 1

)=

1

n+ 1

(2n

n

)(2r

r

)= 2K(n, 1)K(0, r).

THE Fourth STARS OF MATHEMATICS COMPETITION

Bucuresti, ICHB High School, December 2010

Problem 1. Let D := 1, 2, . . . , n × 1, 2, . . . , n. Prove there exists a set

S ⊂ D with |S| ≥⌊3

5n(n+ 1)

⌋, such that for any (x1, y1), (x2, y2) ∈ S we have

(x1 + x2, y1 + y2) ∈ S.

Solution. It is easy to find a weaker bound of |S| = n

⌈1

2n

⌉by taking S =

(x, y) ∈ D | x > n/2. To find the bound asked, we need look at the diagonals ofthe tableau!

n · · · u

u-1 u-2 · · · · · · · · · · · · · · · 3 2 1

1 2 3 · · · · · · · · · · · · · · · u-2 u-1 u · · · n

Diagram for the selection of Su (exact for n = 13).

73

74 2010 ”STARS” MATHEMATICAL COMPETITION

Define the diagonal ∆k = (x, y) ∈ N×N | x+ y = k, for all k ≥ 0. It is clearthat if (x1, y1) ∈ ∆k1 and (x2, y2) ∈ ∆k2 then (x1 + x2, y1 + y2) ∈ ∆k1+k2 .

Therefore Su :=⋃

u≤k≤2u−1

(∆k ∩ D), for u < n + 1 ≤ 2u − 1, has the defining

property. Since we have |∆k∩D| = k−1 for 2 ≤ k ≤ n+1, respectively |∆k∩D| =

2n − (k − 1) for n + 1 ≤ k ≤ 2n, it follows that |Su| =n∑

k=u

(k − 1) +

2u−1∑k=n+1

(2n −

(k − 1)) = −1

2(5u2 − (8n + 9)u + 2(n + 1)(n + 2)). Thus the maximal value for

|Su| is obtained at the nearest integer υ to8n+ 9

10, thus at υ =

⌊4n+ 7

5

⌋, for which

|Sυ| =⌊3

5n(n+ 1)

⌋.

REMARKS. The set D is a product poset of chains, also seen as being a gradedposet – with elements appearing in its Hasse diagram on levels by their rank; the rankbeing precisely the sum of the coordinates (as some justification).3

Problem 2. Let ABCD be a square, and the points M ∈ [BC], N ∈ [CD],P ∈ [DA], such that

∠(−−→AB,

−−→AM) = x, ∠(

−−→BC,

−−→MN) = 2x, ∠(

−−→CD,

−−→NP ) = 3x.

i) Show that, for any x ∈ [0, π/8], such a configuration uniquely exists, and P

ranges over the entire segment [DA];ii) Determine the number of angles x ∈ [0, π/8] for which ∠(

−−→DA,

−−→PB) = 4x.

Solution. i) Assume AB = 1, and denote t = tanx. Since

1 = tanπ

4=

2 tanπ

8

1− tan2π

8

it follows tanπ

8is the positive root of t2 + 2t− 1 = 0, and so tan

π

8=

√2− 1.

Now BM = t, hence M takes once and only once all values of interval [0,√2−1],

while CN = (1 − t) tan 2x =2t

1 + t, therefore N takes all values of the interval

3Actually it is a rank-unimodal and rank-symmetric poset; a seminal result is de Bruijn-Tengbergen-Kruyswijk theorem. In common algebraic-combinatorial jargon, the set S is sum-free, or (S+S)∩S = ∅(Minkowski sumset notation).

2010 ”STARS” MATHEMATICAL COMPETITION 75

[0, 2−√2]. Lastly, we also get DP =

(1− 2t

1 + t

)tan 3x =

1− t

1 + t· t(3− t2)

1− 3t2≥ 0,

whence PA = 1− 1− t

1 + t· t(3− t2)

1− 3t2=

(t2 + 2t− 1)(t2 + 1)

(t+ 1)(3t2 − 1)≥ 0, hence P takes all

values of the full interval [0, 1].Therefore such a configuration will uniquely exist for any x ∈ [0, π/8]. In fact the

points M , N and P , all three, vary monotonically.4

The position that must occur.

ii) Take now Q ∈ BC such that ∠(−−→DA,

−−→PQ) = 4x > 0. Then

BQ = (1− PA tan 4x)/ tan 4x

=

(1− (t2 + 2t− 1)(t2 + 1)

(t+ 1)(3t2 − 1)· 4t(1− t2)

1− 6t2 + t4

)1− 6t2 + t4

4t(1− t2)

=

(1− 4t(1− t)(t2 + 1)

(3t2 − 1)(t2 − 2t− 1)

)(t2 + 2t− 1)(t2 − 2t− 1)

4t(1− t2)

=t2 + 2t− 1

4t(1− t2)(3t2 − 1)(7t4 − 10t3 − 2t+ 1).

What we look for is Q ≡ B, i.e. BQ = 0 (as a matter of fact occurring forx = π/8, and seemingly no more realized).

But 7t4−10t3−2t+1 = (t2+2t−1)(7t2−24t+55)−136t+56 = (1−x)(2−7x3)− (3x3+1), thus decreasing on [0,

√2− 1], while 56 < 136(

√2− 1), hence the

polynomial considered above has just one root 0 < τ ≈ 0.3447 < 0.4142 ≈√2− 1,

corresponding to just one angle 0 < ξ ≈ 0.3319 < 0.3927 ≈ π/8.5

4An alternative proof uses the fact that such a configuration, with P ≡ A, can occur if (and only if)x = π/8 (hence P cannot bypass A for angles 0 ≤ x ≤ π/8). Limit cases are P ≡ D for x = 0 andP ≡ A for x = π/8, so by continuity P ranges over the entire segment [DA]. The monotony of thevariation of M , N and P must be argued by stronger methods than just continuity.

5For angle x ∈ (ξ, π/8) the line PQ does not meet anymore the segment (AB), going beyond B,since Q does not vary monotonically. This phenomenon, I humbly reckon, can only be fathomed throughanalytic methods.


Thus the two solutions x ∈ [0, π/8] are ξ and π/8.

Problem 3. Find the largest constant K ≥ 0 such that for any 0 ≤ k ≤ K, andfor any non-negative real numbers a, b, c, satisfying a2 + b2 + c2 + kabc = k + 3, tohave a+ b+ c ≤ 3.

Solution. Let us work from first principles. Whenever (at least) one variable iszero, say c, it follows a2 + b2 = k + 3, hence a + b + c ≤

√2(k + 3) ≤ 3 for

k ≤ 3/2, with equality holding for a = b = 3/2. We thus have a starting tentativelimiting bound of K = 3/2.

Let also notice that a = b = c = 1 checks for any value of k, while providing amaximal admissible value for a+ b+ c = 3.

Assume σ = a + b + c > 3. Notice that then (for some k > 0) k(1 − abc) =

a2 + b2 + c2 − 3 ≥ 1

3(a+ b+ c)

2 − 3 =1

3σ2 − 3 > 0, by Cauchy-Schwarz, and

so abc < 1. Now is the time for the key step. Since a + b + c > 3, assuming someordering of our variables, say 0 ≤ a ≤ b ≤ c, it will follow c > 1, and so we cancompute

k =a2 + b2 + c2 − 3

1− abc

=(a+ b)2 + c2 − 3− 2ab

1− abc

=(σ − c)2 + c2 − 3− 2ab

1− abc

=c(2c2 − 2σc+ σ2 − 3)− 2 + 2(1− abc)

c(1− abc)

=2

c+

2c3 − 2σc2 + (σ2 − 3)c− 2

c(1− abc).

Denote f(c) = 2c3 − 2σc2 + (σ2 − 3)c− 2. As (x− y)(f(x)− f(y))

= (x − y)2

(3

2

(x+ y − 2

3σ

)2

+1

2(x− y)2 +

1

3(σ2 − 9)

)≥ 0, f is increasing;

and as f(1) = (σ + 1)(σ − 3) > 0, it follows f(c) > 0. Therefore the minimal value

for k is reached when ab = 0, for a = 0, whence k ≥ b2 + c2 − 3 ≥ 1

2(b+ c)2 − 3 =

1

2σ2 − 3 >

3

2.

The issue of the extremal points, leading to equality when k = 3/2, will better beaddressed within the next alternative solution.


Alternative Solution. Let us apply the trusted method of Lagrange multipliers.Define

L(a, b, c) = a+ b+ c− λ(a2 + b2 + c2 + kabc− k − 3).

The analysis of the values on the border of the domain of L has been done above,leading to K ≤ 3/2 (we could now simplify our work to just k = 3/2, but it isenlightening to see it in full generality). The system of partial derivatives is

∂L

∂a= 1− λ(2a+ kbc)

∂L

∂b= 1− λ(2b+ kca)

∂L

∂c= 1− λ(2c+ kab)

Equaling the partial derivatives to zero forbids λ = 0. Then, from pairwise equal-ities, we get

λ(a− b)(2− kc) = λ(b− c)(2− ka) = λ(c− a)(2− kb) = 0.

One possibility is a = b = c = x, thus (from the constraint) 3x2+kx3 = k+3, or(x−1)(kx2+(k+3)x+(k+3)) = 0, with only non-negative real solution x = 1, sincethe coefficients of the quadratic factor are non-negative. Then a + b + c = 3x = 3,the admissible maximum.

The other possibility is for two variables to be equal, say a = b, but not equalto the third, hence needing a = b = 2/k, thus (from the constraint) 8/k2 + c2 +

4c/k = k + 3, or k2c2 + 4kc− (k3 + 3k2 − 8) = 0, with non-negative real solution

c =(k + 2)

√k − 1− 2

kfor k3+3k2−8 ≥ 0, i.e. k ≥ κ ≈ 1.3553. Then a+b+c =

(k + 2)√k − 1 + 2

k< 3 for k < 2, since it is equivalent to (k− 2)3 < 0. As k needs

be at most K ≤ 3/2, these points are critical, but not global maxima (in fact they turnto be global minima).6

6For the record, for the values k3 + 3k2 − 8 ≥ 0 we also have(k + 2)

√k − 1 + 2

k≤

√2(k + 3),

since it turns to be equivalent to

(2

k3 + 3k2 − 8

k√

2(k + 3) + 4

)2

≥ 0. Let κ ≈ 1.3553 be the unique real root

of k3 + 3k2 − 8 = 0, with κ ∈ (1, 3/2). Then for k ∈ (κ, 3/2) we have the points on the border beinglocal maxima, the other being global minima, with point (1, 1, 1) as unique global maximum.


Putting it all together, the largest admissible value for K turns to be 3/2, with themaximum value a+b+c = 3 being reached only at points (3/2, 3/2, 0), (3/2, 0, 3/2),(0, 3/2, 3/2) and (1, 1, 1). For 0 ≤ k < 3/2 the unique maximum is reached at(1, 1, 1). Notice the importance of examining the values on the border; without that,the other critical interior points found but (1, 1, 1) (which works for any k) achievea larger value than 3 for a + b + c only starting with k > 2, so would induce theerroneous bound K = 2.

Alternative Solution. The fact the value 3 for a+ b+ c is reached for k = 3/2

both at a = b = c = 1, and at a = b = 3/2 and c = 0 et al., suggests this is aSchur-type inequality. Indeed, assume 0 ≤ k ≤ 3/2 and

∑a > 3.

Then

>k + 3 =∑

a2 + kabc

≥∑

a2 +3kabc∑

a

=

(1− k

3

)∑a2 +

k

3

(∑a2 +

9abc∑a

)

≥(1− k

3

)∑a2 +

2k

3

∑ab

=

(1− 2k

3

)∑a2 +

k

3

(∑a2 + 2

∑ab)

=

(1− 2k

3

)∑a2 +

k

3(∑

a)2

≥(1

3

(1− 2k

3

)+

k

3

)(∑

a)2

=k + 3

9(∑

a)2

> k + 3

since inequality∑

a2 +9abc∑

a≥ 2

∑ab is in turn equivalent to

(∑a2)(∑

a) +

9abc ≥ 2 (∑

ab) (∑

a), then∑

a3 + 3abc ≥∑

a2b +∑

ab2, at last∑

a(a −

b)(a−c) ≥ 0, a basic form of Schur; while∑

a2 ≥ 1

3

(∑a)2

by Cauchy-Schwarz.

Problem 4. Let a, b, c be given positive integers. Prove there exists some positiveinteger N such that

a | Nbc+ b+ c

b | Nca+ c+ a


c | Nab+ a+ b

if and only if, denoting d = gcd(a, b, c) and a = dx, b = dy, c = dz, the positiveintegers x, y, z are pairwise co-prime, and also gcd(d, xyz) | x+ y + z.

Solution. The necessity of having x, y, z be pairwise co-prime is proved by say,assuming gcd(x, y) > 1.

Then a | Nbc+b+c becomes x | dNyz+y+z, and so we must have gcd(x, y) | z,absurd, since under this assumption it then follows gcd(x, y) | gcd(x, y, z) = 1.

On the other hand, if the co-primality condition holds, consider the integers

xyz −∑

x < 2xyz −∑

x < · · · <(∑

xy)xyz −

∑x.

These∑

xy integer numbers will yield different remainders modulo∑

xy, since ifixyz −

∑x ≡ jxyz −

∑x (mod

∑xy), then also

∑xy | |i − j|xyz, whence

i = j, since we have 0 ≤ |i − j| <∑

xy and gcd(xyz,∑

xy) = 1. Thereforethere will exist some (unique) 1 ≤ t ≤

∑xy such that

∑xy | txyz −

∑x, i.e.

txyz−∑

x = C∑

xy for some positive integer C, therefore txyz = C∑

xy+∑

x,so x | Cyz+y+ z et al. We found a suitable value C for the triplet x, y, z (for similarrelations with the ones sought for a, b, c). Then all the other suitable values must beof the form C ′ = C + Mxyz, since we need have xyz | (C ′ − C)

∑xy, while

gcd(xyz,∑

xy) = 1.

Now the time has come to analyze the last condition. In order to have a | Nbc +

b+c, and the similar others, equivalent to x | dNyz+y+z et al., we need have dN =

C + Mxyz for some non-negative integer M . Denote e = gcd(d, xyz); then e | d,so e | C +Mxyz. But then we also must have e | xyz | (C +Mxyz)

∑xy +

∑x,

hence e |∑

x.

Conversely, if e |∑

x, then e | xyz | (C +Mxyz)∑

xy +∑

x, so e | C∑

xy.

Since clearly gcd(e,∑

xy) = 1, this means e | C. Therefore we need haved

eN =

C

e+ M

xyz

e, and since clearly gcd

(d

e,xyz

e

)= 1, take M ≡ −C

e

(xyze

)−1

(mod d/e), wherefored

edivides

C

e+ M

xyz

e. Take now N =

C +Mxyz

d(of

course, N ′ = N +Mabc also works).


REMARKS. Notice there exist easy counterexamples for gcd(x, y, z) = 1, but x,y, z not pairwise co-prime; just take x = pq, y = qr, z = rp, with p, q, r prime.

Notice also there exist counterexamples to contradict the last condition, for exam-ple (a, b, c) = (6, 9, 15), when d = 3, e = 3,

∑x = 10, and so e

∑x.

For all such examples thus there exists no solution.

THE Fourth ROMANIAN MASTER OF MATHEMATICS

Bucuresti, T. Vianu National College, February 2011

FIRST DAY

Problem 1. Prove that there exist two functions f, g : R → R, such that f g isstrictly decreasing and g f is strictly increasing.

(Poland) Andrzej Komisarski & Marcin Kuczma

Solution. Let

• A =⋃k∈Z

([−22k+1,−22k

)⋃(22k, 22k+1

]);

• B =⋃k∈Z

([−22k,−22k−1

)⋃(22k−1, 22k

]).

Thus A = 2B, B = 2A, A = −A, B = −B, A∩B = ∅, and finally A∪B∪0 = R.Let us take

f(x) =

x for x ∈ A

−x for x ∈ B and g(x) = 2f(x).

0 for x = 0

Then f(g(x)) = f(2f(x)) = −2x and g(f(x)) = 2f(f(x)) = 2x.

Problem 2. Determine all positive integers n for which there exists a polynomialf(x) with real coefficients, with the following properties:

1) for each integer k, the number f(k) is an integer if and only if n does not divide k;2) the degree of f is less than n.

81

82 2011 ”MASTER” MATHEMATICAL COMPETITION

Solution. We will show that such polynomial exists if and only if n = 1 or n is apower of a prime.

We will use two known facts stated in LEMMATA 1 and 2.

LEMMA 1. If pa is a power of a prime and k is an integer, then

(k − 1)(k − 2) . . . (k − pa + 1)

(pa − 1)!

is divisible by p if and only if k is not divisible by pa.

Proof. First suppose that pa | k and consider

(k − 1)(k − 2) · · · (k − pa + 1)

(pa − 1)!=

k − 1

pa − 1· k − 2

pa − 2· · · k − pa + 1

1.

In every fraction on the right-hand side, p has the same maximal exponent in thenumerator as in the denominator. Therefore, the product (which is an integer) is notdivisible by p.

Now suppose that pa k. We have

(k − 1)(k − 2) · · · (k − pa + 1)

(pa − 1)!=

pa

k· k(k − 1) · · · (k − pa + 1)

(pa)!.

The last fraction is an integer. In fractionpa

k, denominator k is not divisible by pa.

LEMMA 2. If g(x) is a polynomial with degree less than n then

n∑=0

(−1)(n

)g(x+ n− ) = 0.

Proof. Apply induction on n. For n = 1 then g(x) is a constant and

(1

0

)g(x+ 1)−

(1

1

)g(x) = g(x+ 1)− g(x) = 0.

Now assume that n > 1 and the LEMMA holds for n− 1. Let h(x) = g(x+ 1)−g(x); the degree of h is less than the degree of g, so the induction hypothesis applies

2011 ”MASTER” MATHEMATICAL COMPETITION 83

for g and n− 1:

n−1∑=0

(−1)(n− 1

)h(x+ n− 1− ) = 0

n−1∑=0

(−1)(n− 1

)(g(x+ n− )− g(x+ n− 1− )

)= 0

(n− 1

0

)g(x+ n) +

n−1∑=1

(−1)((

n− 1

− 1

)+

(n− 1

))g(x+ n− )− (−1)n−1

(n− 1

n− 1

)g(x) = 0

n∑=0

(−1)(n

)g(x+ n− ) = 0.

LEMMA 3. If n has at least two distinct prime divisors then the greatest commondivisor of

(n1

),(n2

), . . . ,

(n

n−1

)is 1.

Proof. Suppose to the contrary that p is a common prime divisor of(n1

), . . . ,

(n

n−1

).

In particular, p |(n1

)= n. Let a be the exponent of p in the prime factorization of n.

Since n has at least two prime divisors, we have 1 < pa < n. Hence,(

npa−1

)and

(npa

)are listed among

(n1

), . . . ,

(n

n−1

)and thus p |

(npa

)and p |

(n

pa−1

). But then p divides(

npa

)−(

npa−1

)=

(n−1pa−1

), which contradicts LEMMA 1.

Next we construct the polynomial f(x) when n = 1 or n is a power of a prime.For n = 1, f(x) = 1

2 is such a polynomial.If n = pa where p is a prime and a is a positive integer then let

f(x) =1

p

(x− 1

pa − 1

)=

1

p· (x− 1)(x− 2) · · · (x− pa + 1)

(pa − 1)!.

The degree of this polynomial is pa − 1 = n− 1.The number (k−1)(k−2)···(k−pa+1)

(pa−1)! is an integer for any integer k, and, by LEMMA

1, it is divisible by p if and only if k is not divisible by pa = n.Finally we prove that if n has at least two prime divisors then no polynomial f(x)

satisfies (1, 2). Suppose that some polynomial f(x) satisfies (1,2), and apply LEMMA


2 for g = f and x = −k where 1 ≤ k ≤ n− 1. We get that(n

k

)f(0) =

∑0≤≤n, =k

(−1)k−

(n

)f(−k + ).

Since f(−k), . . . , f(−1) and f(1), . . . , f(n − k) are all integers, we conclude that(nk

)f(0) is an integer for every 1 ≤ k ≤ n− 1.By dint of LEMMA 3, the greatest common divisor of

(n1

),(n2

), . . . ,

(n

n−1

)is 1.

Hence, there will exist some integers u1, u2, . . . , un−1 for which u1

(n1

)+ · · · +

un−1

(n

n−1

)= 1. Then

f(0) =

(n−1∑k=1

uk

(n

k

))f(0) =

n−1∑k=1

uk

(n

k

)f(0)

is a sum of integers. This contradicts the fact that f(0) is not an integer. So suchpolynomial f(x) does not exist.

Alternative Solution. (I. Bogdanov) We claim the answer is n = pα for someprime p and nonnegative α.

LEMMA. For every integers a1, . . . , an there exists an integer-valued polyno-mial P (x) of degree < n such that P (k) = ak for all 1 ≤ k ≤ n.

Proof. Induction on n. For the base case n = 1 one may set P (x) = a1. For theinduction step, suppose that the polynomial P1(x) satisfies the desired property for all1 ≤ k ≤ n− 1. Then set P (x) = P1(x) + (an − P1(n))

(x−1n−1

); since

(k−1n−1

)= 0 for

1 ≤ k ≤ n− 1 and(n−1n−1

)= 1, the polynomial P (x) is a sought one.

Now, if for some n there exists some polynomial f(x) satisfying the problemconditions, one may choose some integer-valued polynomial P (x) (of degree < n −1) coinciding with f(x) at points 1, . . . , n − 1. The difference f1(x) = f(x) −P (x) also satisfies the problem conditions, therefore we may restrict ourselves to thepolynomials vanishing at points 1, . . . , n− 1 — that are, the polynomials of the formf(x) = c

∏n−1i=1 (x − i) for some (surely rational) constant c. Let c = p/q be its

irreducible form, and q =∏d

j=1 pαj

j be the prime decomposition of the denominator.

1. Assume that a desired polynomial f(x) exists. Since f(0) is not an integer, wehave q (−1)n−1(n− 1)! and hence p

αj

j (−1)n−1(n− 1)! for some j. Hence

n−1∏i=1

(pαj

j − i) ≡ (−1)n−1(n− 1)! ≡ 0 (mod pαj

j ),


therefore f(pαii ) is not integer, too. By the condition (i), this means that n | pαi

i , andhence n should be a power of a prime.

2. Now let us construct a desired polynomial f(x) for any power of a primen = pα. We claim that the polynomial

f(x) =1

p

(x− 1

n− 1

)=

n

px

(x

n

)

fits. Actually, consider some integer x. From the first representation, the denominatorof the irreducible form of f(x) may be 1 or p only. If pα x, then the prime decom-position of the fraction n/(px) contains p with a nonnegative exponent; hence f(x) isinteger. On the other hand, if n = pα | x, then the numbers x−1, x−2, . . . , x−(n−1)

contain the same exponents of primes as the numbers n−1, n−2, . . . , 1 respectively;hence the number (

x− 1

n− 1

)=

∏n−1i=1 (x− i)∏n−1i=1 (n− i)

is not divisible by p. Thus f(x) is not an integer.

Problem 3. A triangle ABC is inscribed in a circle ω. A variable line chosenparallel to BC meets segments AB, AC at points D, E respectively, and meets ω atpoints K, L (where D lies between K and E). Circle γ1 is tangent to the segmentsKD and BD and also tangent to ω, while circle γ2 is tangent to the segments LE andCE and also tangent to ω. Determine the locus, as varies, of the meeting point ofthe common inner tangents to γ1 and γ2.

(Russia) Vasily Mokin & Fedor Ivlev

Solution. Let P be the meeting point of the common inner tangents to γ1 and γ2.Also, let b be the angle bisector of ∠BAC. Since KL ‖ BC, b is also the anglebisector of ∠KAL.

Let H be the composition of the symmetry S with respect to b and the inversionI of centre A and ratio

√AK ·AL (it is readily seen that S and I commute, so since

S2 = I2 = id, then also H2 = id, the identical transformation). The elements of theconfiguration interchanged by H are summarized in Table I.

Let O1 and O2 be the centres of circles γ1 and γ2. Since the circles γ1 and γ2 aredetermined by their construction (in a unique way), they are interchanged by H, there-fore the rays AO1 and AO2 are symmetrical with respect to b. Denote by 1 and 2

the radii of γ1 and γ2. Since ∠O1AB = ∠O2AC, we have 1/2 = AO1/AO2. On


the other hand, from the definition of P we have O1P/O2P = 1/2 = AO1/AO2;this means that AP is the angle bisector of ∠O1AO2 and therefore of ∠BAC.

The limiting, degenerated, cases are when the parallel line passes through A –when P coincides with A; respectively when the parallel line is BC – when P coin-cides with the foot A′ ∈ BC of the angle bisector of ∠BAC (or any other point onBC). By continuity, any point P on the open segment AA′ is obtained for some posi-tion of the parallel, therefore the locus is the open segment AA′ of the angle bisectorb of ∠BAC.

point K ←→ point Lline KL ←→ circle ω

ray AB ←→ ray AC

point B ←→ point Epoint C ←→ point D

segment BD ←→ segment EC

arc BK ←→ segment EL

arc CL ←→ segment DK

Table 1: Elements interchanged by H.

SECOND DAY

Problem 4. Given a positive integer n =

s∏i=1

pαii , we write Ω(n) for the total

numbers∑

i=1

αi of prime factors of n, counted with multiplicity. Let λ(n) = (−1)Ω(n)

(so, for example, λ(12) = λ(22 · 31) = (−1)2+1 = −1).Prove the following two claims:i) There are infinitely many positive integers n such that λ(n) = λ(n+ 1) = +1;ii) There are infinitely many positive integers n such that λ(n) = λ(n+1) = −1.

(Romania) Dan Schwarz

Solution. Notice that we have Ω(mn) = Ω(m) + Ω(n) for all positive integersm,n (Ω is a completely additive arithmetic function), translating into λ(mn) = λ(m)·


λ(n) (so λ is a completely multiplicative arithmetic function), hence λ(p) = −1 forany prime p, and λ(k2) = λ(k)2 = +1 for all positive integers k.7

The start (first 100 terms) of the sequence S = (λ(n))n≥1 is

+1,−1,−1,+1,−1,+1,−1,−1,+1,+1,−1,−1,−1,+1,+1,+1,−1,−1,−1,−1,

+1,+1,−1,+1,+1,+1,−1,−1,−1,−1,−1,−1,+1,+1,+1,+1,−1,+1,+1,+1,

−1,−1,−1,−1,−1,+1,−1,−1,+1,−1,+1,−1,−1,+1,+1,+1,+1,+1,−1,+1,

−1,+1,−1,+1,+1,−1,−1,−1,+1,−1,−1,−1,−1,+1,−1,−1,+1,−1,−1,−1,

+1,+1,−1,+1,+1,+1,+1,+1,−1,+1,+1,−1,+1,+1,+1,+1,−1,−1,−1,+1.

i) The Pell equation x2−6y2 = 1 has infinitely many solutions in positive integers;all solutions are given by (xn, yn), where xn+yn

√6 = (5+2

√6)n. Since λ(6y2) = 1

and also λ(6y2 + 1) = λ(x2) = 1, the thesis is proven.

Alternative Solution. Take any existing pair with λ(n) = λ(n + 1) = 1. Thenλ((2n+1)2−1) = λ(4n2+4n) = λ(4)·λ(n)·λ(n+1) = 1, and also λ((2n+1)2) =

λ(2n+ 1)2 = 1, so we have built a larger (1, 1) pair.

ii) The equation 3x2 − 2y2 = 1 (again Pell theory) has also infinitely many solu-tions in positive integers, given by (xn, yn), where xn

√3+yn

√2 = (

√3+

√2)2n+1.

Since λ(2y2) = −1 and λ(2y2 + 1) = λ(3x2) = −1, the thesis is proven.

Alternative Solution. Assume (λ(n− 1), λ(n)) is the largest (−1,−1) pair, there-fore λ(n + 1) = 1 and λ(n2 + n) = λ(n) · λ(n + 1) = −1, therefore againλ(n2 + n + 1) = 1. But then λ(n3 − 1) = λ(n − 1) · λ(n2 + n + 1) = −1,and also λ(n3) = λ(n)3 = −1, so we found yet a larger such pair than the one westarted with, contradiction.

Alternative Solution. Assume the pairs of consecutive terms (−1,−1) in S arefinitely many.

7Also see Sloane’s Online Encyclopædia of Integer Sequences (OEIS), sequence A001222 for Ω andsequence A008836 for λ, which is called Liouville’s function. Its summatory function

∑d|n

λ(d) is equal to

1 for a perfect square n, and 0 otherwise. Polya conjectured that L(n) :=n∑

k=1

λ(k) ≤ 0 for all n, but this

has been proven false by Minoru Tanaka, who in 1980 computed that for n = 906, 151, 257 its value was

positive. Turan showed that if T (n) :=

n∑k=1

λ(k)

k≥ 0 for all large enough n, that will imply Riemann’s

Hypothesis; however, Haselgrove proved it is negative infinitely often.


Then from some rank on we only have subsequences (1,−1, 1, 1, . . . , 1,−1, 1).By ”doubling” such a subsequence (like at point ii)), we will produce

(−1, ?, 1, ?,−1, ?,−1, ?, . . . , ?,−1, ?, 1, ?,−1).

According with our assumption, all ?-terms ought to be 1, hence the produced subse-quence is

(−1, 1, 1, 1,−1, 1,−1, 1, . . . , 1,−1, 1, 1, 1,−1),

and so the ”separating packets” of 1’s contain either one or three terms. Now assumesome far enough (1, 1, 1, 1) or (−1, 1, 1,−1) subsequence of S were to exist. Since itlies within some ”doubled” subsequence, it contradicts the structure described above,which thus is the only prevalent from some rank on. But then all the positions of the(−1)-terms will have the same parity. However though, we have λ(p) = λ(2p2) =

−1 for all odd primes p, and these terms have different parity of their positions. Acontradiction has been reached.8

Alternative Solution. (I. Bogdanov) Take ε ∈ −1, 1. There obviously existinfinitely many n such that λ(2n + 1) = ε (just take 2n + 1 to be the product of anappropriate number of odd primes). Now, if either λ(2n) = ε or λ(2n + 2) = ε, weare done; otherwise λ(n) = −λ(2n) = −λ(2n+ 2) = λ(n+ 1) = ε. Therefore, forsuch an n, one of the three pairs (n, n+ 1), (2n, 2n+ 1) or (2n+ 1, 2n+ 2) fits thebill.

We have thus proved the existence in S of infinitely many occurrences of all pos-sible subsequences of length 1, viz. (+1) and (−1), and of length 2, viz. (+1,−1),(−1,+1), (+1,+1) and (−1,−1).9

Problem 5. For every n ≥ 3, determine all the configurations of n distinct pointsX1, X2, . . . , Xn in the plane, with the property that for any pair of distinct points Xi,Xj there exists a permutation σ of the integers 1, . . . , n, such that d(Xi, Xk) =

d(Xj , Xσ(k)) for all 1 ≤ k ≤ n. (We write d(X,Y ) to denote the distance betweenpoints X and Y .)

(United Kingdom) Luke Betts

8Using the same procedure for point i), we only need notice that λ((2k + 1)2) = λ((2k)2) = 1, andthese terms again are of different parity of their position.

9Is this true for subsequences of all lengths = 3, 4, etc.? If no, up to which length ≥ 2?


Solution. Let us first prove that the points must be concyclic. Assign to each pointXk the vector xk in a system of orthogonal coordinates whose origin is the point of

mass of the configuration, thus1

n

n∑k=1

xk = 0.

Then d2(Xi, Xk) = ||xi − xk||2 = 〈xi − xk, xi − xk〉 = ||xi||2 − 2 〈xi, xk〉 +

||xk||2, hencen∑

k=1

d2(Xi, Xk) = n||xi||2−2

⟨xi,

n∑k=1

xk

⟩+

n∑k=1

||xk||2 = n||xi||2+

n∑k=1

||xk||2 = n||xj ||2 +n∑

k=1

||xσ(k)||2 =

n∑k=1

d2(Xj , Xσ(k)), therefore ||xi|| = ||xj ||

for all pairs (i, j). The points are thus concyclic (lying on a circle centred at O(0, 0)).Let now m be the least angular distance between any two points. Two points

situated at angular distance m must be adjacent on the circle. Let us connect each pairof such two points with an edge. The graph G obtained must be regular, of degree

deg(G) = 1 or 2. If n is odd, sincen∑

k=1

deg(Xk) = n deg(G) = 2|E|, we must have

deg(G) = 2, hence the configuration is a regular n-gon.If n is even, we may have the configuration of a regular n-gon, but we also may

have deg(G) = 1. In that case, let M be the next least angular distance between anytwo points; such points must also be adjacent on the circle. Let us connect each pairof such two points with an edge, in order to get a graph G′. A similar reasoning yieldsdeg(G′) = 1, thus the configuration is that of an equiangular n-gon (with alternatingequal side-lengths).

Problem 6. The cells of a square 2011× 2011 array are labelled with the integers1, 2, . . . , 20112, in such a way that every label is used exactly once. We then identifythe left-hand and right-hand edges, and then the top and bottom, in the normal wayto form a torus (the surface of a doughnut). Determine the largest positive integer Msuch that, no matter which labelling we choose, there exist two neighbouring cellswith the difference of their labels at least M .10

(Romania) Dan Schwarz

Preamble. For a planar N × N array, it is folklore that this value is M = N ,with some easy models shown below. As such, the problem is mentioned in [BELA

BOLLOBAS - The Art of Mathematics], 21. Neighbours in a Matrix.10Cells with coordinates (x, y) and (x′, y′) are considered to be neighbours if x = x′ and y− y′ ≡ ±1

(mod 2011), or if y = y′ and x− x′ ≡ ±1 (mod 2011).


1 2 . . . NN+1 N+2 . . . 2N

......

. . ....

(N-1)N+1 (N-1)N+2 . . . N2

A planar parallel N ×N model array.

1 2 4 . . . N(N-1)/2 + 13 5 . . . N(N-1)/2 + 26 . . ....

......

. . ....

...N(N+1)/2 - 1 . . . N2 - 2

N(N+1)/2 . . . N2 - 1 N2

A planar diagonal N ×N model array.

Solution. For the toroidal case, it is clear the statement of the problem is referringto the cells of a ZN ×ZN lattice on the surface of the torus, labeled with the numbers1, 2, . . . , N2, where one has to determine the least possible maximal absolute valueM of the difference of labels assigned to orthogonally adjacent cells.

The toroidal N = 2 case is trivially seen to be M = 2 (thus coinciding with theplanar case).

1 23 4

The unique 2× 2 toroidal array.

For N ≥ 3 we will prove that value to be at least M ≥ 2N − 1. Consider such aconfiguration, and color all cells of the square in white. Go along the cells labeled 1,2, etc. coloring them in black, stopping just on the cell bearing the least label k which,after assigned and colored in black, makes that all lines of a same orientation (rows, orcolumns, or both) contain at least two black cells (that is, before coloring in black thecell labeled k, at least one row and at least one column contained at most one blackcell). Wlog assume this happens for rows. Then at most one row is all black, since


if two were then the stopping condition would have been fulfilled before cell labeledk (if the cell labeled k were to be on one of these rows, then all rows would havecontained at least two black cells before, while if not, then all columns would havecontained at least two black cells before).

Now color in red all those black cells adjacent to a white cell. Since each row,except the potential all black one, contained at least two black and one white cell, itwill now contain at least two red cells. For the potential all black row, any of theneighbouring rows contains at least one white cell, and so the cell adjacent to it hasbeen colored red. In total we have therefore colored red at least 2(N−1)+1 = 2N−1

cells.The least label of the red cells has therefore at most the value k + 1− (2N − 1).

When the white cell adjacent to it will eventually be labeled, its label will be at leastk + 1, therefore their difference is at least (k + 1)− (k + 1− (2N − 1)) = 2N − 1.

The models are kind of hard to find, due to the fact that the direct proof offers littleas to their structure (it is difficult to determine the equality case during the argumentinvolving the inequality with the bound, and then, even this is not sure to be prone tobeing prolonged to a full labeling of the array).

The weaker fact the value M is not larger than 2N is proved by the general modelexhibited below (presented so that partial credits may be awarded).

N+1 N+2 . . . 2N3N+1 3N+2 . . . 4N

......

. . ....

(2-1)N+1 (2-1)N+2 . . . 2N...

.... . .

...2kN+1 2kN+2 . . . (2k+1)N

......

. . ....

2N+1 2N+2 . . . 3N1 2 . . . N

A general model for M = 2N in a N ×N array.

By examining some small N > 2 cases, one comes up with the idea of spiralmodels for the true value M = 2N − 1. The models presented are for odd N (since2011 is odd); similar models exist for even N (but are less symmetric).


7 2 63 1 58 4 9

The spiral 3× 3 array.

23 16 7 15 2217 8 2 6 149 3 1 5 13

18 10 4 12 2124 19 11 20 25


47 40 29 16 28 39 4641 30 17 7 15 27 3831 18 8 2 6 14 2619 9 3 1 5 13 2532 20 10 4 12 24 3742 33 21 11 23 36 4548 43 34 22 35 44 49


(2n+1)2-2 (2n+1)2-9 . . . n(2n-1)+1 . . . (2n+1)2-10 (2n+1)2-3(2n+1)2-8 . . . n(2n-1)+2 n(2n-1) . . . (2n+1)2-11

......

. . ....

......

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

2n2 . . . 8 2 6 . . . 2n(n-1)+2 2n(n+1)+22n2 + 1 . . . 3 1 5 . . . 2n(n+1)+1

2n2+2 . . . 10 4 12 . . . 2n(n+1)...

.... . .

......

.... . .

......

(2n+1)2-7 . . . n(2n+1) n(2+1)+2 . . . (2n+1)2-4(2n+1)2-1 (2n+1)2-6 . . . n(2n+1)+1 . . . (2n+1)2-5 (2n+1)2

The general spiral N ×N array for N = 2n+ 1 ≥ 5.

THE CLOCK-TOWER SCHOOL SENIORS COMPETITION

Rm. Valcea, March 2011

Problem 1. Let n be a positive integer and a1 ≤ a2 ≤ · · · ≤ an be positive realnumbers. Show that(

n∑k=1

a2k

)(n∑

k=1

kak

)≤

(n∑

k=1

ak

)(n∑

k=1

ka2k

).

Solution. The hypothesis leads to (i−j)(ai−aj)aiaj ≥ 0 for every i and j, hence

0 ≤n∑

i,j=1

(i− j)(ai − aj)aiaj

=

n∑i,j=1

(ia2i aj − iaia

2j − ja2i aj + jaia

2j

)

=

(n∑

i=1

ia2i

)n∑

j=1

aj −

(n∑

i=1

iai

)n∑

j=1

a2j −

(n∑

i=1

a2i

)n∑

j=1

jaj +

(n∑

i=1

ai

)n∑

j=1

ja2j

=2

(n∑

k=1

ka2k

)n∑

k=1

ak − 2

(n∑

k=1

kak

)n∑

k=1

a2k,

which leads to the required relation.

Problem 2. Let f : R → R be a function such that:

f(xy + x+ y) + f(xy − x− y) = 2(f(x) + f(y)), ∀ x, y ∈ R. (1)

Prove that f fulfills the relation:

f(x+ y) = f(x) + f(y), ∀ x, y ∈ R. (2)

93

94 2011 ”CLOCK-TOWER” SENIOR MATHEMATICAL COMPETITION

Solution. x = y = 0 yields f(0) = 0 and y = 0 yields f(−x) = −f(x), x ∈ R.Then

y = 1 ⇒ f(2x+ 1) = 2f(x) + f(1) (3)

y = −1 ⇒ f(2x− 1) = 2f(x)− f(1) (4)

Denote x ∗ y = xy + x+ y and notice that (ASOC) : x ∗ (y ∗ z) = (x ∗ y) ∗ z.For x = 1, y ∗ 1 = 2y + 1, hence

f(y ∗ 1) = 2f(y) + f(1), ∀ y ∈ R (5)

therefore

f(x ∗ (y ∗ 1)) = f(x ∗ (2y + 1))

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

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

(3,4)= 2f(xy − y) + f(x) + 2f(y) + f(1) (6)

and

f((x ∗ y) ∗ 1) (5)= 2f(x ∗ y) + f(1)

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

= 2f(xy − x− y) + 2f(x) + 2f(y) + f(1). (7)

Relations (6), (7) and (ASOC) yield:

f(xy − x− y) + f(x) = f(xy − y), ∀ x, y ∈ R.

The last relation means that f(u + v) = f(u) + f(v) for all u, v ∈ R which canbe written in the form u = x, v = xy − x − y, that is there exists x, y ∈ R so thatx = u and y = v+u

u−1 ; this happens if u = 1 or u = 1 = −v.It remains to check that f(1 + v) = f(1) + f(v), ∀ v ∈ R. Indeed, (4) and

x → x+ 1 imply

f(2x+ 1) = 2f(x+ 1)− f(1)(3)= 2f(x) + f(1),

whence f(x+ 1) = f(x) + f(1), ∀ x ∈ R.

Problem 3. A certain language uses an alphabet containing three letters. Somesequences of two ore more letters are forbidden, and every two forbidden sequenceshave different lengths. Prove that there exists admissible words of every length.

2011 ”CLOCK-TOWER” SENIOR MATHEMATICAL COMPETITION 95

Solution. Let an be the number of admissible words with n letters; then a0 = 1

(the empty word), a1 = 3 and a2 = 8.Then, if we add a letter at the end of a correct word with n letters, we obtain either

a correct word with n + 1 letters, or a forbidden word of the form XY , with Y aforbidden sequence with k letters, 2 ≤ k ≤ n+1 and X a correct word with n−k+1

letters. So, the forbidden words with n+1 letters are at most a0+a1+a2+. . .+an−1,hence

an+1 ≥ 3an − (a0 + a1 + a2 + . . .+ an−1).

The above relation allows to prove inductively that an+1 > 2an (∗), for everyn ≥ 1. Indeed, the base case is obvious, and if (∗) is true for all the numbers from 0

to n− 1, n ≥ 2, then an ≥ 2kan−k, 0 ≤ k ≤ n− 1, whence

an+1 ≥ an

(3−

(1

2n+

1

2n−1+ . . .+

1

2

))> 2an.

This shows that there are at least 2n words of length n.

Problem 4. A point P is on the side AB of a convex quadrilateral ABCD. Let ωbe the incircle of the triangle CPD and I be its center. It is known that ω touches theincircles of the triangles APD and BPC at K, respectively L. Let E, respectively F bethe meetincg points of the lines AC and BD, respectively AK and BL. Prove that thepoints E, I and F are collinear.

96 2011 ”CLOCK-TOWER” SENIOR MATHEMATICAL COMPETITION

Solution. Denote J the center of the circle k placed in the half-plane (AB,C andtangent to AB, DA and BC. Denote a, respectively b the incircles of the triangles ADPand BCP. We will firstly prove that F ∈ IJ . The point A is the center of the homothetywhich transforms a into k; K is the center of the homothety with negative ratio whichtransforms a into ω. Denote F the center of the homothety which transforms ω into k.It is known that A, K and F are collinear. In the same way, F ∈ BL, hence F = F ,so F ∈ IJ .

We prove now that E ∈ IJ . Comparing the lengths of the tangents from A, P ,C, D to the circles k and a gives AP + DC = AD + PC. This shows that thequadrilateral APCD has an incircle d. Let X be the center of the homothety whichtransforms a into ω. The above reasoning, applied to the circles a, d and ω showsthat A,C and X are collinear. Consider now the circles a, ω and k. The point A is thecenter of the homothety which transforms a into k and X is the center of the homothetywhich transforms a into ω, hence XA contains the center E of the homothety whichtransforms ω into k, whence E ∈ AC. In the same way E ∈ BD, therefore E = E,so E ∈ IJ .

THE CLOCK-TOWER SCHOOL JUNIORS COMPETITION

Rm. Valcea, March 2011

Problem 1. Show that the number

aaa . . . a︸︷︷︸2011 times

0a

is not a perfect square for any non-nil digit a.

Solution. The last digit of a perfect square cannot be 2, 3, 7 or 8.If a = 5, the next to the last digit must be 2 – false.If a = 6, the number is of the form 4k + 2, k ∈ N, which cannot be a perfect

square.The cases a = 4 and a = 9 reduce to a number of the form 111 . . . 01 beeing a

pefect square, which is impossible, because such a number is of the form 8k + 5.

Problem 2. Find all positive integers a, b for which there exists sets A,B of po-sitive integers so that A∩B = ∅, A∪B = N∗ and aA = bB (if x is a number and M

is a set of numbers, xM = xm | m ∈ M).

Solution. We can assume that 1 ∈ A. Then a ∈ bB, hence there exists p ∈ B suchthat a = pb. Moreover, p ≥ 2, because 1 ∈ A.

Every pair (pb, b) , with b ∈ N∗, is a solution: we use the partition

A =p2nq | n ∈ N, q ∈ N∗, p q

B =p2n+1q | n ∈ N, q ∈ N∗, p q

Problem 3. A set D of n straight lines in a plane has the property that each lineof the set intersects exactly 2011 straight lines of D.

97

98 2011 ”CLOCK-TOWER” JUNIOR MATHEMATICAL COMPETITION

Find n.

Solution. Let d be a line from D and k be the number of the lines from D parallelto d. Then n = 2012 + k.

Since every line a from D different from d meets 2011 lines from D, a is parallelto n− 2012 = k lines from D.

This shows that each line from D is parallel to exactly k other lines, whence the nlines can be divided into groups of k + 1 mutually parallel lines.

So k + 1 | n, hence k + 1 | k + 2012, therefore k ∈ 0, 2010 , that is n can be2012 or 4022.

A possible configuration for n = 2012 is made up by the supports of the sides ofa regular 2012-gon.

A possible configuration for n = 4022 is made up by 2011 parallel lines, meetingother 2011 parallel lines.

Problem 4. The triangle ABC and the points M ∈ (BC), N ∈ (AC), P ∈ (AB)

fulfill the conditions ∠BMP ≡ ∠CNM ≡ ∠APN and BM = CN = AP .Prove that the triangle ABC is equilateral.

Solution. We start noticing that m(PMN) = 180 − BMP − NMC = 180 −CNM − NMC = NCM and, in the same way, MNP = NAP , so

∆ABC ∼ ∆NPM. (1)

Suppose now, without loss of generality, that C ≥ A ≥ B. (2)

Then AB ≥ BC ≥ AC, whence, from (1), NP ≥ PM ≥ MN. These, toghetherwith ∠BMP ≡ ∠CNM ≡ ∠APN and BM = CN = AP , lead to A ≥ B ≥C. (3)

Relations (2) and (3) show that A ≡ B ≡ C, whence the conclusion.

SPONSORS AND PARTNERS OF THE 2011 IMO ROMANIAN TEAM

Since 2009, the Dinu Patriciu Foundation

has been one of the main sponsors of theGazeta Matematică journal and of the Ro-manian National Mathematical Olympiad.

WE TAKE STOCK IN THE

FUTURE WBS, sole supplier of full brokerage serviceson international stock markets for Romanianclients, was established by a former member

of the Romanian Math Olympiad team. AtWBS we strongly believe that education isthe best investment and are therefore proudto support the top Math students of thecountry.

Since 2010, the Council of the Dâmboviţa

County has been providing full support fororganizing the qualification camps for theRomanian team prior the IMOs.

Paralela 45 Publishing House is one of the mostrepresentative Romanian suppliers of books, fromfiction to science.Starting with 2011, Paralela 45 is a partner of the

Romanian Mathematical Society in the graphicdesigning and printing the Gazeta Matematicajournal.A series of mathematical texts edited by the Soci-

ety will be also published under the partnership.

ISSN 2066-6446

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