ROMANIAN MATHEMATICAL COMPETITIONS 2016 · PDF fileROMANIAN MATHEMATICAL COMPETITIONS 2016 Edited by MARIEAN ANDRONACHE, ... nian Olympiad and the partitipation of the Romanian Teams

ROMANIAN MATHEMATICAL COMPETITIONS

2016

Edited by MARIEAN ANDRONACHE, MIHAIL BALUNA, RADU GOLOGANANDREI ECKSTEIN, MARIUS PERIANU, CALIN POPESCU, DINU SERBANESCU

with the cooperation of Dan Schwarz

Technical Editor ALEXANDRU NEGRESCU

TABLE OF CONTENTS

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

1.1. 2016 Romanian Mathematical Olympiad – District Round . . . . . . . . . . . . . . . . . . . . . .1

1.2. 2016 Romanian Mathematical Olympiad – Final Round . . . . . . . . . . . . . . . . . . . . . . .13

1.3. Shortlisted problems for the 2016 Romanian NMO . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1.4. Selection tests for the 2016 BMO and IMO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

1.5. Selection tests for the 2016 JBMO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

1.6. 2013 – 2014 Local Mathematical Competitions

1.6.1. The 2013 DANUBE Mathematical Competition . . . . . . . . . . . . . . . . . . . . . . . .77

1.6.2. The 2013 Tenth IMAR1 Mathematical Competition . . . . . . . . . . . . . . . . . . . 81

1.6.3. The 2013 Sixth STARS OF MATHEMATICS Competition . . . . . . . . . . . . . . . . 87

1.6.4. The 2016 Sixth ROMANIAN MASTER OF MATHEMATICS . . . . . . . . . . . . . 103

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

iii

iv

FOREWORD

The 21th 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.

The most part of the problems are discussed in detail, and alternative solutions orgeneralizations 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 National Education for constant involvement in support-ing the Olympiads and the participation of our teams in international events.

Special thanks are due to companies who were involved in sponsoring the Roma-nian Olympiad and the partitipation of the Romanian Teams to International Compe-titions: Emag Foundation and Tuca Zbarcea and Associates.

Radu GologanPresident of the Romanian Mathematical Society

v

vi

THE 65th ROMANIAN MATHEMATICAL OLYMPIAD

DISTRICT ROUND

7th GRADE

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

8th GRADE

Problem 1.

Solution.

Problem 2.

1

2 2016 ROMANIAN MATHEMATICAL OLYMPIAD – DISTRICT ROUND

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

9th GRADE

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

2016 ROMANIAN MATHEMATICAL OLYMPIAD – DISTRICT ROUND 3

10th GRADE

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

11th GRADE

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

4 2016 ROMANIAN MATHEMATICAL OLYMPIAD – DISTRICT ROUND

12th GRADE

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

THE 65th ROMANIAN MATHEMATICAL OLYMPIAD

FINAL ROUND

7th GRADE

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

8th GRADE

Problem 1.

Solution.

5

6 2016 ROMANIAN MATHEMATICAL OLYMPIAD – FINAL ROUND

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

9th GRADE

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

2016 ROMANIAN MATHEMATICAL OLYMPIAD – FINAL ROUND 7

10th GRADE

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

11th GRADE

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

8 2016 ROMANIAN MATHEMATICAL OLYMPIAD – FINAL ROUND

12th GRADE

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

SHORTLISTED PROBLEMS FOR THE 65th NMO

JUNIORS

1.

SENIORS

1.

PUTNAM SENIORS

1.

CONTRIBUTORS

????????????????????????????????????? Aurel Barsan, Nicolae Bourbacut, PetruBraica, Gheorghe Bumbacea, Alexandru Buna-Marginean, Lucian Dragomir, Bog-dan Enescu, Marin Ionescu, Gheorghe Iurea, Cristinel Mortici, Nicolae Musuroia,Ion Nedelcu, Radu Pop, Vasile Pop, Claudiu Stefan Popa, Florin Stanescu, TraianTamaian.

9

10 SHORTLISTED PROBLEMS FOR THE 2016 ROMANIAN NMO

THE 65th NMO SELECTION TESTS FOR THE BALKAN

AND INTERNATIONAL MATHEMATICAL OLYMPIADS

FIRST SELECTION TEST

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

SECOND SELECTION TEST

Problem 5.

11

12 SELECTION TESTS FOR THE 2016 BMO AND IMO

Solution.

Problem 6.

Solution.

Problem 7.

Solution.

Problem 8.

Solution.

THIRD SELECTION TEST

Problem 9.

Solution.

Problem 10.

Solution.

Problem 11.

Solution.

Problem 12.

Solution.

SELECTION TESTS FOR THE 2016 BMO AND IMO 13

FOURTH SELECTION TEST

Problem 13.

Solution.

Problem 14.

Solution.

Problem 15.

Solution.

Problem 16.

Solution.

FIFTH SELECTION TEST

Problem 17.

Solution.

Problem 18.

Solution.

Problem 19.

Solution.

Problem 20.

Solution.

14 SELECTION TESTS FOR THE 2016 BMO AND IMO

THE 65th NMO SELECTION TESTS FOR THE JUNIOR

BALKAN MATHEMATICAL OLYMPIAD

FIRST SELECTION TEST

Problem 1. Let ABC be a non-equilateral triangle such that m(∠A) = 60◦. LetD and E be the intersection points of the Euler line of triangle ABC and the sides ofthe angle ∠BAC. Prove that the triangle ADE is equilateral.

∗ ∗ ∗

Solution. et H and O be the orthocenter and the circumcenter of ABC, R theradius of the circumcenter; then OH meets the line AB at D and the line AC at E.

Let B′ be the foot of the altitude from B and C ′ be the foot of the altitude from C.Since BCC ′B′ is a cyclic quadrilateral we have ∠AB′C ′ ≡ ∠ABC. Hence∆AB′C ′ ∼ ∆ABC, having the similarity ratio AC′

AC = cos(BAC) = 12 . It follows

that the similarity ratio is the same with the ratio of the diameters of the circumcirclesof triangles AB′C ′ and ABC, so AH

2R = 12 , which leads to AH = R = AO. (1)

It is known that the rays (AH and (AO are isogonal, so ∠BAO ≡ ∠CAH . (2)From (1) it results that ∠AOH ≡ ∠AHO, so ∠AOD ≡ ∠AHE. Using (1) and(2), it follows that triangles AOD and AHE are congruent, so AD = AE and theconclusion follows.

15

16 SELECTION TESTS FOR THE 2016 JBMO

Problem 2. Let m,n be positive integers and x, y, z ∈ [0, 1] be real numbers.Prove that

0 ≤ xm+n + ym+n + zm+n − xmyn − ymzn − zmxn ≤ 1

and find when equality holds.

adapted from Lituanian Olympiad

Solution. (Dan Schwarz)Without loss of generality, we my assume that x is the largest of x, y, z; then we have:xm+n + ym+n + zm+n − xmyn − ymzn − zmxn = (xm − zm)(xn − yn) + (ym −zm)(yn − zn) ≥ 0.

The minimum is 0 and is obtained if x = max{y, z} and y = z, so if x = y = z.

The maximum is achieved for x = 1, and its value is1 +ym+n + zm+n−yn−ymzn− zm = 1−yn(1−ym)− zm(1− zn)−ymzn ≤ 1,

with equality only if one of y or z is 0 and the other one is 0 or 1.

Therefore, there are 6 equality cases, namely(x, y, z) ∈ {(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1)}.

Problem 3. Let M be the set of the natural numbers k for which it exists n ∈ Nsuch that the remainder of 3n when divided by n is k. Show that M is infinite.

Ioan-Laurentiu Ploscaru

SELECTION TESTS FOR THE 2016 JBMO 17

Solution. Let j be a fixed positive integer and p > 2 a prime so that 2jp > 32j

.Then 32

j

(32j(p−1) − 1) ≡ 0 (mod 2jp), because 2ϕ(2jp) = 2j(p − 1). Hence

32jp ≡ 32

j

(mod 2jp), so, for n = 2jp, one has rn = 32j

. It follows that 32j ∈M ,

for all j ∈ N∗, so M is an infinite set.

Alternative solution. Choosing n = 2 · 3j , the remainder rn of 3n when dividedby n verifies rn 6= 0, rn ≡ 0 (mod 3j), so rn ≥ 3j (in fact rn = 3j). Therefore, Mcontains arbitrarily large numbers, so it is infinite.

Problem 4. Let ABC be an acute triangle with AB 6= AC. The incircle ω ofthe triangle touches the sides BC, CA and AB in D, E and F , respectively. Theperpendicular line erected from C to BC intersects EF at M , and, similarly, theperpendicular line erected at B to BC intersects EF at N . The line DM meets ω

again in P , and the line DN meets ω again in Q. Prove that DP = DQ.

Ruben Dario, Peru and Leonard Giugiuc, Romania

Solution. Let {T} = EF ∩ BC. Applying Menelaus’ theorem to the triangleABC and the transversal line E − F − T we obtain: TB

TC ·ECEA ·

FAFB = 1, i.e. TB

TC ·s−cs−a ·

s−as−b = 1, or TB

TC = s−bs−c , where the notations are the usual ones (1).

This means that triangles TBN and TCM are similar, therefore TBTC = BN

CM .From (1) it follows that BN

CM = s−bs−c ,

BDCD = s−b

s−c , and ∠DBN = ∠DCM = 90◦,which means that triangles BDN and CDM are similar, hence angles BDN andCDM are equal. This leads to the arcs DQ and DP being equal, and finally toDP = DQ.

Alternative solution. Let S be the intersection point of the altitude from A withthe line EF . Lines BN, AS, CM are parallel, therefore triangles BNF and ASF

are similar, as are triangles ASE and CME. We obtain BNAS = BF

FA and ASCM = AE

EC .

Multiplying these two together, we obtain BNCM = BF

FA ·AEEC = BF

EC = BDDC . (We have

used that AE = AF , BF = BD and CE = CD.)It follows that the right triangles BDN and CDM are similar (SAS), which leads tothe same ending as in the first proof.


Problem 5. The unit squares of a n× n board, n ≥ 2, are colored either black orwhite so that any black square has at least three white neighbors (a neighbor is a unitsquare with a common side). What is the maximum number of black unit squares?

Andrei Eckstein

Solution. The answer is n2−12 if n is odd and n2−4

2 if n is even.

Notice that:- the unit squares from the corners are white (because they only have two neighbors,not three);- the other squares from the border of the board have only three neighbors, so therecan’t be two consecutive black squares on the border;- any 2 × 2 square contains at most two black squares (if not, a black square wouldalready have two black neighbors, so it would have at most two white neighbors).

If n is odd, color the board like a chessboard with white corners. This coloring sat-isfies the requirements of the problem and it has n2−1

2 black squares, so the maximumnumber of black squares is at least n2−1

2 .On the other hand, split the board in 4 parts: one upper-left (n − 1) × (n − 1)

square, a unit square in the right bottom corner, a (n − 1) × 1 rectangle on the rightside and a 1×(n−1) rectangle on the bottom side. Tiling the (n−1)×(n−1) squarewith 2× 2 squares and the two rectangles with 2-square dominoes, according with theremarks made at the beginning, one can have at most (n−1)2

2 + n−12 + n−1

2 +0 = n2−12


black squares.It follows that if n is odd, the maximum number of black squares is n2−1

2 .

If n is even, again consider a chessboard-like coloring. This makes two whitecorners and two black corners; recolor the black corners as white. The configurationwe have now satisfies the requirements of the problem and it has n2−4

2 black squares,so the maximum number of black squares is at least n2−4

2 .

Split the board in 9 parts: one (n − 2) × (n − 2) square in the middle, fourunit squares in the corners and four (n − 2) × 1 or 1 × (n − 2) rectangles on theborders. Tiling the (n − 2) × (n − 2) square with 2 × 2 squares and the rectangleswith 2-square dominoes, the remarks made at the beginning lead to a maximum of(n−2)2

2 + 4 · n−22 + 4 · 0 = n2−42 black squares. Hence, if n is even, the maximum

number of black squares is n2−42 .

SECOND SELECTION TEST

Problem 6. Let O be the circumcircle of a triangle ABC. A circle k is tangentto the lines BC, CA, AB at points D, E, F , respectively, such that A is on the otherside of the line BC with respect to the circle. Suppose the circle k is equal to thecircumcircle of the triangle. Prove that lines OD and EF are perpendicular.

Solution. Let Ia be the center of the circle k and let T be the midpoint of the arcBC - not containing A - of the circumcircle ABC. Notice that OT is the perpen-dicular bisector of the line segment BC to deduce that OT ⊥ BC. As IaD ⊥ BC

and OT = IaD = R, the quadrangle ODIaT is a parallelogram. ConsequentlyOD ‖ TIa, or, equivalently, OD ‖ AIa.On the other hand, since AE = AF and (AIa bisects angle ∠FAE, the lines AIa

and EF are perpendicular, and so are lines OD and EF .


Problem 7. Let a, b, c be positive numbers with abc ≥ 1. Prove that 1a3+2b3+6 +

1b3+2c3+6 + 1

c3+2a3+6 ≤13 .

Ionut Grigore

Solution. Notice that a3 + b3 + 1 ≥ 3ab and b3 + 1 + 1 ≥ 3b to obtainthat 1

a3+2b3+6 ≤1

3ab+3b+3 , hence it is sufficient to show that 1ab+b+1 + 1

bc+c+1 +1

ca+a+1 ≤ 1.

To this end, observe that 1ab+b+1 + 1

bc+c+1 + 1ca+a+1 = 1

ab+b+1 + abab2c+abc+ab +

babc+ab+b

abc≥1≤ 1

ab+b+1 + abb+1+ab + b

1+ab+b = 1+ab+bab+b+1 = 1.

Solution. Subtract 1/6 from each of the left hand-side summands and write suc-cessively

∑cyc

(1

a3+2b3+6 −16

)≤ 1

3 −12 , then

∑cyc

−a3−2b36(a3+2b3+6) ≤ −

16 or further∑

cyc

a3+2b3

a3+2b3+6 ≥ 1.

It is sufficient to prove that∑cyc

a3

a3+2b3+6 ≥13 and

∑cyc

b3

a3+2b3+6 ≥13 .

To this end, notice that∑cyc

a3

a3+2b3+6 =∑cyc

a4

a4+2ab3+6a

CBS≥

(a2+b2+c2)2

a4+b4+c4+2(ab3+bc3+ca3)+6(a+b+c)

(1)

≥ 13 .

The inequality (1) rewrites 3(a4+b4+c4)+6(a2b2+b2c2+c2a2) ≥ a4+b4+c4+


2(ab3+bc3+ca3)+6(a+b+c), and follows from 2(a4+b4+c4) ≥ 2(ab3+bc3+ca3)

and 6(a2b2+b2c2+c2a2) ≥ 6(ab·bc+bc·ca+ca·ab) = 6abc(a+b+c) ≥ 6(a+b+c).

On the other hand,∑cyc

b3

a3+2b3+6 =∑cyc

b4

ba3+2b4+6b

CBS≥

(a2+b2+c2)2

2(a4+b4+c4)+(ba3+cb3+ac3)+6(a+b+c)

(2)

≥ 13 . The inequality (2) rewrites 3(a4 +

b4 +c4)+6(a2b2 +b2c2 +c2a2) ≥ 2(a4 +b4 +c4)+(ba3 +cb3 +ac3)+6(a+b+c),and follows from a4 + b4 + c4 ≥ ba3 + cb3 + ac3 and 6(a2b2 + b2c2 + c2a2) ≥6(ab · bc + bc · ca + ca · ab) = 6abc(a + b + c) ≥ 6(a + b + c).

Solution. As a3 + 2b3 = a3 + b3 + b3 ≥ 3ab2, it suffices to prove that 1ab2+2 +

1bc2+2+ 1

ca2+2 ≤ 1. Rewrite the inequality as− ab2

2(ab2+2)−bc2

2(bc2+2)−ca2

2(ca2+2) ≤ 1− 32 ,

or, equivalently, ab2

ab2+2 + bc2

bc2+2 + ca2

ca2+2 ≥ 1.

Recall that abc ≥ 1, and write ab2

ab2+2abc + bc2

bc2+2abc + ca2

ca2+2abc ≥ 1 to observe thatit is enough to show that b

b+2c + cc+2a + a

a+2b ≥ 1. Indeed, bb+2c + c

c+2a + aa+2b =

b2

b2+2bc + c2

c2+2ca + a2

a2+2ab

CBS≥ (a+b+c)2

b2+2bc+c2+2ca+a2+2ab = 1, which concludes theproof.

Problem 8. Let n be an integer greater than 2 and consider the set A = {2n −1, 3n − 1, . . . , (n − 1)n − 1}. Given that n does not divide any element of A, provethat n is a square-free number. Does it necessarily follow that n is a prime number?

Marius Bocanu

Solution. Suppose not and write n = pa for a prime p and a number a > 1 withp | a. Notice that (a+1)n−1 = a((a+ 1)n−1 +(a+ 1)n−2 + . . .+ 1) and a+ 1 ≡ 1

(mod p) to infer that n divides (a + 1)n − 1, a contradiction.Further, n needs not be a prime number; take for example n = 15 = 3 · 5.

Remark.All primes n meet the requirements of the problem, and the value n = 15 shows thatnot only the primes satisfy the conditions. Furthermore, not all square-free numbershave the property from the problem, as n = 6 = 2 · 3 shows.

Problem 9. All the 16 squares of a 4 × 4 array are white. Define a move byselecting a rectangle 1 × 3 or 3 × 1 and switching the colors of each of its squaresfrom white to black or from black to white. Is it possible that all squares turn blackafter a sequence of moves?


Solution. Fill the array with numbers as follows:

1 2 3 1

2 3 1 2

3 1 2 3

1 2 3 1

to observe that a move will change colors in one square of each number. Asinitially there are six squares labeled 1, an even number of moves is required to turnblack all these squares. On the other hand, five squares where labeled with 2 at thestart, requiring an odd number of moves to turn all black, hence the answer is negative.

THIRD SELECTION TEST

Problem 10. For n ∈ N, consider the system

(Sn) :

{x2 + ny2 = z2

nx2 + y2 = t2, x, y, z, t ∈ N.

If M1 = {n ∈ N | system (Sn) has infinitely many solutions}, andM2 = {n ∈ N | system (Sn) has no solutions}, prove that:a) 7 ∈M1, 10 ∈M2,b) sets M1 and M2 are infinite.

Solution. a) Notice that x = 1, y = 3, z = 8, t = 4 is a solution to the system(S7), and so is (k, 3k, 8k, 4k), for any k ∈ N, hence 7 ∈M1.If (x, y, z, t) is a solution to the system (S10), it would follow that 11(x2 + y2) =

z2 + t2. From 11 | z2 + t2 we get 11 | z and 11 | t. Then 11 | x2 + y2, hence 11 | xand 11 | y. Therefore, there exist x1, y1, z1, t1 ∈ N such that x = 11x1, y = 11y1,z = 11z1, t = 11t1. This leads to 11(x2

1 + y21) = z21 + t21. By continuing thisprocedure (,,infinite descent”), we get that x is a multiple of any power of 11, and, asx 6= 0, we arrive to a contradiction. In conclusion, the system (S10) has no solutions,i.e. 10 ∈M2.b) It is easy to see that any number n of the form m2 − 1, m ∈ N, belongs to M1: we


can choose x = y and notice that (k, k,mk,mk), k ∈ N, are solutions to the system(Sm2−1), hence m2 − 1 ∈M1, ∀m ∈ N.Applying the same steps as we did above for n = 10, it is easy to prove that if p is aprime of the form 4m + 3, and n = p− 1, then n ∈M2. As there are infinitely manysuch primes, the conclusion follows readily.One can prove that actually any number congruent to 2 (mod 4) belongs to M2.

Problem 11. Let x and y be be real nonzero numbers, such that x3+y3+3x2y2 =

x3y3. Determine the set of the possible values of E = 1x + 1

y .

Solution. Rewrite the given condition successively (x + y)3 − 3xy(x + y) =

x3y3 − 3x2y2, i.e. (x + y)3 − (xy)3 = 3xy(x + y)− 3x2y2, or (x + y − xy)(x2 +

2xy+y2 +x2y+xy2 +x2y2) = 3xy(x+y−xy). We either have x+y = xy, whichleads to E = 1 (obtained for x = y = 2), or x2+2xy+y2+x2y+xy2+x2y2 = 3xy.The last equality, multiplied by 2, can be written equivalently x2(y + 1)2 + y2(x +

1)2 + (x − y)2 = 0, which is possible if and only if x = y = −1 (x = y = 0 isforbidden). In this case, E = −2. In conclusion, the set of the possible values of E is{−2, 1}.

Problem 12. Let ABCD be a cyclic quadrilateral whose diagonals are not per-pendicular and intersect at X . Let A′, C ′ be the projections of A and C onto the lineBD and let B′, D′ be the projections of B and D onto AC. Prove that:a) the perpendicular lines drawn from the midpoints of the sides onto the oppositesides are concurrent at a point called Mathot’s point;b) points A′, B′, C ′, D′ are cocyclic;c) if O′ is the circumcenter of A′B′C ′, then O′ is the midpoint of the line segmentdetermined by the orthocenters of triangles XAB and XCD;d) O′ is the Mathot point of the quadrilateral ABCD.

Solution. a) Let O be the circumcenter of ABCD. It is well known that the mid-points of the sides of a quadrilateral ABCD are the vertices of a parallelogram, hencethe line segments joining the midpoints of two opposite sides have the same midpoint,G. The perpendicular lines from O to AB and CD pass through the midpoints ofthese sides, therefore the perpendiculars dropped from O and from the midpoints of


two opposite sides onto the their opposite side form a parallelogram whose center isG. It follows that the two perpendicular lines dropped from the midpoints of two op-posite sides onto their opposite side intersect at the reflection of O in G. The othertwo perpendiculars intersect at the same point.b) We assume the angle AXB to be acute, the other case being similar. The quadri-laterals ABA′B′, CDC ′D′ and ABCD being cyclic, we have∠XD′C ′ ≡ ∠XDC ≡ ∠XAB ≡ ∠XA′B′, hence A′B′C ′D′ is cyclic.c) Let H1 and H2 be the orthocenters of triangles XAB, and XCD, respectively.If O′′ is the midpoint of [H1H2], as O′′ belongs to the midsegment of the trapezoidH1B

′H2D′, O′′ belongs to the perpendicular bisector of the line segment [B′D′].

Similarly, O′′ belongs to the midsegment of the trapezoid A′H1C′H2, hence to the

perpendicular bisector of [A′C ′]. As A′C ′ and B′D′ are not parallel, it follows thatO′′ is precisely the circumcenter of A′B′C ′D′, i.e. O′′ coincides with O′.d) We have ∠A′B′X ≡ ∠ABX ≡ ∠DCX , hence A′B′ ‖ CD. If N is the midpointof [AB], then NA′ = NB′, hence N belongs to the perpendicular bisector of [A′B′].But so does O′, therefore it follows that NO′ ⊥ CD. Similarly, O′ belongs to theperpendicular dropped from the midpoint of [CD] on AB, hence O′ is the Mathotpoint of the quadrilateral.

Problem 13. In every unit square of an n × n board a positive integer is written.A move consists of choosing a 2 × 2 square and adding 1 to exactly three of the four


number written in the chosen square. We call a positive integer n good if starting fromany initial numbers, there is a sequence of moves that makes all the numbers from theboard equal.a) Prove that n = 6 is not good.b) Prove that n = 4, and n = 1024 are good.

Solution. a) Let us notice that the sum of the 36 numbers written on the boardis invariant modulo 3. When all the numbers on the board are equal, their sum is amultiple of 36, hence a multiple of 3. But this requires that the initial sum is also amultiple of 3. In conclusion, if the sum of the initial numbers is not a multiple of 3,there is no succession of moves that makes the numbers equal. In conclusion, 6 is notgood.b) For n=4, we chose an arbitrary unit square and prove that we can increase by 1 allthe other 15 numbers from the board. For example, if the chosen square lies in theupper left 2× 2 corner, we successively increment the numbers in the squares markedwith B, then those marked by C, D, and E. Finally, we increment 3 of the four numbersin the squares marked with A (with the exception of the chosen one). Thus, combin-ing five moves, we have obtained a move that is equivalent to decreasing the chosennumber by 1.

A A B B

A A C B

D C C E

D D E E

Repeating this new move and choosing each time the largest number (or one of thoseif there are several largest ones), we can make the numbers equal.For n a power of 2, in particular for n = 210 = 1024, we prove by induction that any2k × 2k board (k ≥ 1) from which an arbitrary unit square has been removed, canbe tiled with non-overlapping L-shaped trominos (figures obtained by removing a unitsquare from a 2× 2 square).The possibility of tiling such ,,deficient squares” with L-shaped trominos is a classicalproblem due to Golomb. We reproduce the proof fromhttp://www.cut-the-knot.org/Curriculum/Geometry/Tromino.shtmlFor k = 1 the statement is obvious.


Assume the statement true for n = 2k−1 and let us prove it for n = 2k. Draw thecenter lines of the 2k × 2k board. The cut-off square lies in exactly one of the fourthus obtained 2k−1×2k−1 boards. We may remove one square from each of the otherthree boards by placing a tromino at the center of the 2k × 2k board. The result is atromino and four 2k−1× 2k−1 boards, each with one square removed - just a situationto apply the inductive hypothesis.This tiling allows us to construct a sequence of moves that increases by 1 the numberin every square with the exception of one square that can be arbitrarily chosen. Byrepeatedly using such sequences we can make all the numbers of the board equal.

FOURTH SELECTION TEST

Problem 14. The altitudes AA1, BB1, CC1 of the acute triangle ABC intersect atH . Let A2 be the reflection of point A in the line B1C1 and let O be the circumcenterof triangle ABC.a) Prove that the points O,A2, B1, C are cocyclic.b) Prove that the points O,H,A1, A2 are cocyclic.

∗ ∗ ∗

Solution. a) The angles ∠ABC and ∠AB1C1 are equal, therefore so are theircomplementary angles, ∠BAA1 and ∠A2AC. It follows that the rays (AH and (AA2

are isogonal, hence A2 ∈ (AO. As AO = CO, we have ∠ACO ≡ ∠OAC ≡∠AA2B1, therefore points O,A2, B1, C are cocyclic. (The arguments above hold inthe case A2 ∈ (AO) as well as in the case O ∈ (AA2).)b) From the power of the point A with respect to the circles through O,A2, B1, C andH,A1, C,B1 respectively, it follows that AB1 · AC = AO · AA2 and AB1 · AC =

AH ·AA1. Since AO ·AA2 = AH ·AA1, from the converse of the power of the pointtheorem, it follows that the points O,A2, H,A1 are cocyclic.


Alternative solution: for b)Consider {A3} = (AA2 ∩B1C1 and let O1 be the midpoint of [AH]. Then O1 is thecircumcenter of triangle AB1C1. As [O1A3] is a midsegment in triangle AHA2, wehave ∠HA2A ≡ ∠O1A3A. But the similarity of triangles ABC and AB1C1 leadsto the equality of the corresponding angles ∠AA1O and ∠AA3O1, which means that∠AA2H ≡ ∠AA1O and the conclusion.Remark: The result remains valid even when the triangle ABC is not acute.

Problem 15. We are given an m×n grid and three colors. We wish to color eachsegment of the grid with one of the three colors so that each unit square has two sidesof one color and two sides of a second color. How many such colorings are possible?

∗ ∗ ∗

Solution. We label the lines from top to bottom and the columns from left to right.The leftmost side of the unit square in the upper-left corner can be colored in 3 ways.Subsequently, there are 3 ways of choosing the side of this unit square that is to re-ceive the same color as the first side. The remaining two sides of the square have to becolored with the same color. This color can be chosen in 2 ways. In conclusion, theunit square in the upper-left corner can be colored in 3 · 3 · 2 = 18 ways.Next, we successively color the unit squares in the first line, from left to right. Everytime, the leftmost side of the square has already been colored. Thus, there are 6 ways


of coloring each of these squares. We proceed similarly on the first column. We colorthe unit squares from top to buttom and for each of these squares there are 6 ways ofcoloring.Now we move on to color the squares situated on rows 2, 3, . . . ,m and columns2, 3, . . . , n. We color them from top to bottom, from left to right. Thus, when inturn to be colored, each unit square has already two sides colored: the upper side andthe leftmost side. If these sides have received different colors, then the colors of thissquare have already been decided; there are 2 ways of coloring the remaining sideswith these two colors. If the two sides that are already colored, have the same color,then the remaining sides will receive the same color. This color can be chosen in 2ways. Therefore, for each of these squares, there are 2 ways of coloring.In conclusion, there are 18·6m−1 ·6n−1 ·2(m−1)(n−1) = 3m+n ·2mn ways of coloring.

Problem 16. Real numbers a, b and c are such that a, b ≥ 1 ≥ c ≥ 0 anda + b + c = 3.a) Show that 2 ≤ ab + bc + ca ≤ 3.

b) Prove the inequality 24a3+b3+c3 + 25

ab+bc+ca ≥ 14 and determine the equality cases.

Leonard Giugiuc

Solution. a) Denote ab+ bc+ ca = q and abc = p. As (1−a)(1− b)(1− c) ≥ 0,it follows that q− 2 ≥ p; but p ≥ 0, hence q ≥ 2. As 3(ab+ bc+ ca) ≤ (a+ b+ c)2,we obtain q ≤ 3.b) a3 + b3 + c3 = (a+ b+ c)3− 3(a+ b+ c)(ab+ bc+ ca) + 3abc = 3(9− 3q + p);from a), p ≤ q − 2⇒ 3(9− 3q + p) ≤ 3(7− 2q)⇒ 24

a3+b3+c3 ≥8

7−2q . It is enoughto show that 8

7−2q + 25q ≥ 14, which is equivalent to 7(2q− 5)2 ≥ 0, which is clearly

true.Equality holds if and only if p = q − 2 and q = 5

2 , p = 12 .

But p = q − 2 leads to (1 − a)(1 − b)(1 − c) = 0, i.e. at least one of the numbersa, b, c is equal to 1.Denote by x and y the two remaining numbers. Then x + y + xy = 5

2

xy = 12 .

Solving the system, we obtain a = 1 + 1√2

, b = 1 and c = 1− 1√2.


Problem 17. Let ABCD be a cyclic quadrilateral, and let E and F denote themidpoints of diagonals [AC] and [BD], respectively. If {G} = AB ∩ CD, {H} =

AD ∩BC, prove that:a) the intersection points of the angle bisectors of ∠AHB and ∠AGD with the sidesof the quadrilateral ABCD are the vertices of a rhombus;b) the center of this rhombus lies on the line EF .

adapted by Andrei Eckstein and Mircea Fianu

Solution. We are going to treat only the case when C ∈ (GD) and C ∈ (BH),all the other cases being similar.a) Let M and N be the intersection points of the angle bisector of angle ∠G with thesides BC and AD, respectively. Consider I , K the intersection points of the angle bi-sector of angle ∠H with sides CD and AB, respectively. Also, put {J} = MN∩IK.From triangle GBC we get ∠G = 180◦ − ∠GBC − ∠GCB = ∠B + ∠C − 180◦,hence ∠CMJ = 90◦ + 1

2 (∠B − ∠C). Similarly, ∠CIJ = 90◦ + 12 (∠D − ∠C).

From the quadrilateral CMJI , using the fact that ∠B + ∠D = 180◦, it follows that∠MJI = 90◦.In the triangles GIK and HMN , line segments [GJ ] and [HJ ] are angle bisectorsand altitudes, therefore they are also medians. Thus, the diagonals of the quadrilateralMKNI are perpendicular and bisect each other.b) The sides of the rhombus are parallel to AC and BD, respectively. Indeed, accord-ing to the angle bisector theorem, MB

MC = GBGC and BK

KA = HBHA . But from the power of

the point G with respect to the circle, GBGC = GD

GA . In order to prove that MK ‖ AC,i.e. BK

KA = BMMC , it is sufficient to prove that HA ·GD = HB ·GA. This follows from

[AHG] = HA·GD sinD2 = HB·GA sinB

2 and sinB = sinD. Thus, MK ‖ AC.Consider {L} = MK ∩ BF , {P} = AE ∩ KN , {O} = DF ∩ IN and {Q} =

MI ∩ CE. Line segments AE, BF , CE and DF are medians in triangles ABD,ABC, BCD, CDA, therefore LO and PQ meet in the center of the rhombus.Put {X} = LO ∩ EF and{Y } = QP ∩ EF . Then XE

XF = LBLF = BK

KA andY EY F = QE

QC = BMMC . As BK

KA = BMMC , it follows that points X and Y coincide, which

means that the lines EF , LO and PQ pass through the center of the rhombus.


THE DANUBE MATHEMATICAL COMPETITION

Calarasi, October 2013

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

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32 2011 DANUBE MATHEMATICAL COMPETITION

THE Tenth IMAR MATHEMATICAL COMPETITION

Bucuresti, S. Haret National College, November 2013

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

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34 2011 IMAR MATHEMATICAL COMPETITION

THE Sixth STARS OF MATHEMATICS COMPETITION

Bucuresti, ICHB High School, December 2013

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

JUNIORS

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Solution.

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36 2012 “STARS” MATHEMATICAL COMPETITION

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

THE Sixth ROMANIAN MASTER OF MATHEMATICS

Bucuresti, T. Vianu National College, March 2016

FIRST DAY

Problem 1.

Solution.

Problem 2.

Solution.

Problem 3.

Solution.

Problem 4.

Solution.

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