MEMO 200717 EN with solutions 070914

What is the maximum possible number of points in a subset S ⊆ M which does not contain three distinct points being the vertices of a right triangle?... At a MEMO-like competition, there [r]

1 MEMO, Eisenstadt, Austria Individual competition, September 22,

2 A set of balls contains n balls which are labeled with numbers 1, 2, 3, , n Suppose

we are given k > 1 such sets We want to colour the balls with two colors, black andwhite, in such a way that

(a) the balls labeled with the same number are of the same colour,

(b) any subset of k + 1 balls with (not necessarily all different) labels a1, a2, ,

ak+1 satisfying the condition a1+ a2+ + ak = ak+1, contains at least one ball

of each colour

Find, depending on k, the greatest possible number n which admits such a colouring

3 Let k be a circle and k1, k2, k3 and k4 four smaller circles with their centres O1, O2,

O3 and O4 respectively on k For i = 1, 2, 3, 4 and k5 = k1 the circles ki and ki+1meet at Ai and Bi such that Ai lies on k The points O1, A1, O2, A2, O3, A3, O4, A4,lie in that order on k and are pairwise different Prove that B1B2B3B4 is a rectangle

4 Determine all pairs (x, y) of positive integers satisfying the equation

x! + y! = xy

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

Time: 5 hours

Time for questions: 45 min

1 MEMO, Eisenstadt, Austria Team competition, September 23, 2007

5 Let a, b, c, d be arbitrary real numbers from the closed interval [1

2,2] satisfyingabcd= 1 Find the maximal value of

µ

a+ 1b

¶ µ

b+ 1c

¶ µ

c+ 1d

¶ µ

d+ 1a

¶

6 For a set P of five points in the plane in general position, we denote the number

of acute-angled triangles with vertices in P by a(P ) (a set of points is said to be ingeneral position if no three points lie on a line) Determine the maximal value of

a(P ) over all possible sets P

7 Let s(T ) denote the sum of the lengths of the edges of a tetrahedron T We considertetrahedra with the property that the six lengths of their edges are pairwise differentpositive integers, where one of them is 2 and and another one of them is 3

1 Find all positive integers n for which there exists a tetrahedron T with s(T ) = n

2 How many such pairwise non congruent tetrahedra T with s(T ) = 2007 exist?Two tetrahedra are said to be non congruent, if one cannot be transformed by reflec-tions with respect to planes, translations and/or rotations into the other

(It is not necessary to prove that the tetrahedra are not degenerated i.e have positivevolume.)

8 Determine all positive integers k with the following property: there exists an integer

a such that (a + k)3

−a3

is a multiple of 2007

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

Time: 5 hours

Time for questions: 45 min

2nd MEMO, Olomouc,

Czech Republic Individual competition, September 6, 2008

I–3: Let ABC be an isosceles triangle with |AC| = |BC| Its incircle touches AB and BC

at D and E, respectively A line (different from AE) passes through A and intersectsthe incircle at F and G The lines EF and EG intersect the line AB at K and L,respectively Prove that |DK| = |DL|

I–4: Find all integers k such that for every integer n, the numbers 4n + 1 and kn + 1 arerelatively prime

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

Time: 5 hours

Time for questions: 45 min

2nd MEMO, Olomouc,

Czech Republic Team competition, September 7, 2008

T–1: Find all functions f : R → R (so, f is a function from the real numbers to the real

numbers) such that

xf(x + xy) = xf (x) + f (x2

)f (y)for all real numbers x and y

T–2: In a given n-tuple of positive integers with n ≥ 2, we choose in each step a pair of

numbers and replace each of them by their sum, i.e we make the transformation

( , a, , b, ) → ( , a + b, , a + b, )

Determine all values of n for which, after a finite number of steps, we can get ann-tuple of identical numbers from any initial n-tuple

T–3: Given an acute-angled triangle ABC, let E be a point situated on the different side of

the line AC than B, and let D be an interior point of the line segment AE Supposethat ∠ADB = ∠CDE, ∠BAD = ∠ECD and ∠ACB = ∠EBA Prove that B, Cand E are collinear

T–4: Let n be a positive integer Prove that if the sum of all positive divisors of n is a

perfect power of 2, then the number of these divisors is also a perfect power of 2

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

Time: 5 hours

Time for questions: 45 min

Find all functions f : R → R such that

f(xf (y)) + f (f (x) + f (y)) = yf (x) + f (x + f (y))for all x, y ∈ R, where R denotes the set of real numbers

ProblemI-2

Suppose that we have n > 3 distinct colours Let f(n) be the greatest integer with theproperty that every side and every diagonal of a convex polygon with f (n) vertices can becoloured with one of n colours in the following way:

• at least two distinct colours are used, and

• any three vertices of the polygon determine either three segments of the same colour or

of three different colours

ProblemI-4

Determine all integers k > 2 such that for all pairs (m, n) of different positive integers notgreater than k, the number nn−1−mm−1 is not divisible by k

Time: 5 hours

Time for questions: 30 min

Each problem is worth 8 points.

The order of the problems does not depend on their difficulty.

x4+ y4

+ z4+ 16(x2

+ y2+ z2) > 8(x3

+ y3+ z3) + 27and determine when equality holds

Problem T-2

Let a, b, c be real numbers such that for every two of the equations

x2+ ax + b = 0, x2+ bx + c = 0, x2+ cx + a = 0there is exactly one real number satisfying both of them

Determine all the possible values of a2

+ b2+ c2

Problem T-3

The numbers 0, 1, 2, , n (n > 2) are written on a blackboard In each step we erase aninteger which is the arithmetic mean of two different numbers which are still left on the black-board We make such steps until no further integer can be erased Let g(n) be the smallestpossible number of integers left on the blackboard at the end Find g(n) for every n

Problem T-4

We colour every square of the 2009 × 2009 board with one of n colours (we do not have

to use every colour) A colour is called connected if either there is only one square of that

colour or any two squares of the colour can be reached from one another by a sequence ofmoves of a chess queen without intermediate stops at squares having another colour (a chessqueen moves horizontally, vertically or diagonally) Find the maximum n, such that for everycolouring of the board at least one colour present at the board is connected

Problem T-6.

Suppose that ABCD is a cyclic quadrilateral and CD = DA Points E and F belong tothe segments AB and BC respectively, and ADC = 2EDF Segments DK and DM areheight and median of the triangle DEF , respectively L is the point symmetric to K withrespect to M Prove that the lines DM and BL are parallel

Problem T-8

Find all non-negative integer solutions of the equation

2x+ 2009 = 3y

5z

Time: 5 hours

Time for questions: 45 min

Each problem is worth 8 points.

The order of the problems does not depend on their difficulty.

OL YMPIAD

Find all functions f : R → R such that for all x, y ∈ R, we have

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

Problem I-2

All positive divisors of a positive integer N are written on a blackboard Two players A and

B play the following game taking alternate moves In the first move, the player A erases N Ifthe last erased number is d, then the next player erases either a divisor of d or a multiple of

d The player who cannot make a move loses Determine all numbers N for which A can winindependently of the moves of B

Problem I-3

We are given a cyclic quadrilateral ABCD with a point E on the diagonal AC such that

AD = AE and CB = CE Let M be the center of the circumcircle k of the triangle BDE.The circle k intersects the line AC in the points E and F Prove that the lines F M , AD, and

BC meet at one point

Problem I-4

Find all positive integers n which satisfy the following two conditions:

(i) n has at least four different positive divisors;

(ii) for any divisors a and b of n satisfying 1 < a < b < n, the number b − a divides n

Time: 5 hours

Time for questions: 45 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

OL YMPIAD

Problem T-3

In each vertex of a regular n-gon there is a fortress At the same moment each fortress shoots

at one of the two nearest fortresses and hits it The result of the shooting is the set of the hitfortresses; we do not distinguish whether a fortress was hit once or twice Let P (n) be thenumber of possible results of the shooting Prove that for every positive integer k ≥ 3, P (k)and P (k + 1) are relatively prime

Problem T-4

Let n be a positive integer A square ABCD is partitioned into n2 unit squares Each of them

is divided into two triangles by the diagonal parallel to BD Some of the vertices of the unitsquares are colored red in such a way that each of these 2n2 triangles contains at least one redvertex Find the least number of red vertices

Problem T-5.

The incircle of the triangle ABC touches the sides BC, CA, and AB in the points D, E, and

F , respectively Let K be the point symmetric to D with respect to the incenter The lines

DE and F K intersect at S Prove that AS is parallel to BC

2 0 0

| {z }n

2 0 0

| {z }n1

Prove that an/3 is always the sum of two positive perfect cubes but never the sum of twoperfect squares

Problem T-8

We are given a positive integer n which is not a power of 2 Show that there exists a positiveinteger m with the following two properties:

(i) m is the product of two consecutive positive integers;

(ii) the decimal representation of m consists of two identical blocks of n digits

Time: 5 hours

Time for questions: 45 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

re-4(a1+ a2+ a3+ a4) of the four new integers is equal to the number a In a step we neously replace all the integers on the board in the above way After 30 steps we end up with

simulta-n = 430 integers b1, b2, , bn on the board Prove that

Let n ≥ 3 be an integer John and Mary play the following game: First John labels the sides

of a regular n-gon with the numbers 1, 2, , n in whatever order he wants, using each numberexactly once Then Mary divides this n-gon into triangles by drawing n − 3 diagonals which donot intersect each other inside the n-gon All these diagonals are labeled with number 1 Intoeach of the triangles the product of the numbers on its sides is written Let S be the sum ofthose n − 2 products

Determine the value of S if Mary wants the number S to be as small as possible and Johnwants S to be as large as possible and if they both make the best possible choices

Problem I-3

In a plane the circles K1 and K2 with centers I1 and I2, respectively, intersect in two points Aand B Assume that ∠I1AI2 is obtuse The tangent to K1 in A intersects K2 again in C andthe tangent to K2 in A intersects K1 again in D Let K3 be the circumcircle of the triangleBCD Let E be the midpoint of that arc CD of K3 that contains B The lines AC and ADintersect K3 again in K and L, respectively Prove that the line AE is perpendicular to KL

Problem I-4

Let k and m, with k > m, be positive integers such that the number km(k2− m2) is divisible

by k3− m3 Prove that (k − m)3 > 3km

Time: 5 hours

Time for questions: 45 min

Each problem is worth 8 points

The order of the problems does not depend on their diculty

Let ABCDE be a convex pentagon with all ve sides equal in length The diagonals AD and

EC meet in S with ∠ASE = 60◦ Prove that ABCDE has a pair of parallel sides

We call a positive integer n amazing if there exist positive integers a, b, c such that the equality

n = (b, c)(a, bc) + (c, a)(b, ca) + (a, b)(c, ab)holds Prove that there exist 2011 consecutive positive integers which are amazing

(By (m, n) we denote the greatest common divisor of positive integers m and n.)

Time: 5 hours

Time for questions: 45 min

Each problem is worth 8 points

The order of the problems does not depend on their diculty

Problem I-2.

Let N be a positive integer A set S ⊆ {1, 2, , N } is called allowed if it does not contain three distinct elements a, b, c such that a divides b and b divides c Determine the largest possible number of elements in an allowed set S.

2 + an+ an−1for all positive integers n.

Determine all prime numbers p for which there exists a positive integer m such that p divides the number am− 1.

Time: 5 hours

Time for questions: 45 min

Each problem is worth 8 points.

The order of the problems does not depend on their difficulty.

of such words containing an odd number of blocks M E and an odd number of blocks M O Prove that a > b.

Problem T-4.

Let p > 2 be a prime number For any permutation π = (π(1), π(2), , π(p)) of the set

S = {1, 2, , p}, let f (π) denote the number of multiples of p among the following p numbers:

π(1), π(1) + π(2), , π(1) + π(2) + · · · + π(p).

Determine the average value of f (π) taken over all permutations π of S.

Let ABCD be a convex quadrilateral with no pair of parallel sides, such that ^ABC =

^CDA Assume that the intersections of the pairs of neighbouring angle bisectors of ABCD form a convex quadrilateral EF GH Let K be the intersection of the diagonals of EF GH Prove that the lines AB and CD intersect on the circumcircle of the triangle BKD.

Time for questions: 45 min

Each problem is worth 8 points.

The order of the problems does not depend on their difficulty.

Find all triples (a, b, c) for which equality holds.

Problem I-2 Let n be a positive integer On a board consisting of 4n × 4n squares,

exactly 4n tokens are placed so that each row and each column contains one token In a

step, a token is moved horizontally or vertically to a neighbouring square Several tokensmay occupy the same square at the same time The tokens are to be moved to occupy allthe squares of one of the two diagonals

Determine the smallest number k(n) such that for any initial situation, we can do it

in at most k(n) steps.

Problem I-3 Let ABC be an isosceles triangle with AC = BC Let N be a point inside

the triangle such that 2∠ANB = 180◦ +∠ACB Let D be the intersection of the line

BN and the line parallel to AN that passes through C Let P be the intersection of the

angle bisectors of the angles CAN and ABN

Show that the lines DP and AN are perpendicular.

Problem I-4 Let a and b be positive integers Prove that there exist positive integers x

and y such that

Time for questions: 60 min

Each problem is worth 8 points.

The order of the problems does not depend on their difficulty.

Problem T-2 Let x, y, z, w ∈ R \ {0} such that x + y 6= 0, z + w 6= 0, and xy + zw ≥ 0.

Prove the inequality

2 ≥

x

z +

z x

 −1

w +

w y

! −1

.

Problem T-3 There are n ≥ 2 houses on the northern side of a street Going from

the west to the east, the houses are numbered from 1 to n The number of each house

is shown on a plate One day the inhabitants of the street make fun of the postman byshuffling their number plates in the following way: for each pair of neighbouring houses,the current number plates are swapped exactly once during the day

How many different sequences of number plates are possible at the end of the day?

Problem T-4 Consider finitely many points in the plane with no three points on a line.

All these points can be coloured red or green such that any triangle with vertices of thesame colour contains at least one point of the other colour in its interior

What is the maximal possible number of points with this property?

Problem T-5 Let ABC be an acute triangle Construct a triangle P QR such that

AB = 2P Q, BC = 2QR, CA = 2RP , and the lines P Q, QR, and RP pass through the

points A, B, and C, respectively (All six points A, B, C, P , Q, and R are distinct.)

Problem T-6 Let K be a point inside an acute triangle ABC, such that BC is a common

tangent of the circumcircles of AKB and AKC Let D be the intersection of the lines CK and AB, and let E be the intersection of the lines BK and AC Let F be the intersection

of the line BC and the perpendicular bisector of the segment DE The circumcircle of

ABC and the circle k with centre F and radius F D intersect at points P and Q.

Prove that the segment P Q is a diameter of k.

Problem T-7 The numbers from 1 to 20132 are written row by row into a table consisting

of 2013 × 2013 cells Afterwards, all columns and all rows containing at least one of the

perfect squares 1, 4, 9, , 20132 are simultaneously deleted

How many cells remain?

Problem T-8 The expression

±  ±  ±  ±  ±  ± 

is written on the blackboard Two players, A and B, play a game, taking turns Player

A takes the first turn In each turn, the player on turn replaces a symbol  by a positive

integer After all the symbols  are replaced, player A replaces each of the signs ± by either + or −, independently of each other Player A wins if the value of the expression

on the blackboard is not divisible by any of the numbers 11, 12, , 18 Otherwise, player

B wins.

Determine which player has a winning strategy

Time: 5 hours

Time for questions: 60 min

Each problem is worth 8 points.

The order of the problems does not depend on their difficulty.

Individual Competition

Sept 20, 2014 English version

Problem I–1

Determine all functions f : R ÝÑ R such that

xf pyq ` f `xfpyq˘ ´ xf`fpyq˘ ´ fpxyq “ 2x ` fpyq ´ fpx ` yq holds for all x, y P R.

Problem I–2

We consider dissections of regular n-gons into n ´ 2 triangles by n ´ 3 diagonals which do not intersect inside the n-gon A bicoloured triangulation is such a dissection of an n-gon in which

each triangle is coloured black or white and any two triangles which share an edge have different

colours We call a positive integer n ě 4 triangulable if every regular n-gon has a bicoloured triangulation such that for each vertex A of the n-gon the number of black triangles of which A

is a vertex is greater than the number of white triangles of which A is a vertex.

Find all triangulable numbers

Problem I–3

Let ABC be a triangle with AB ă AC and incentre I Let E be the point on the side AC such that AE “ AB Let G be the point on the line EI such that >IBG “ >CBA and such that E and G lie on opposite sides of I.

Prove that the line AI, the perpendicular to AE at E, and the bisector of the angle >BGI are

Remark The double factorial n!! is defined to be the product of all even positive integers up to n

if n is even and the product of all odd positive integers up to n if n is odd So e.g 0!! “ 1,

4!! “ 2 ¨ 4 “ 8, and 7!! “ 1 ¨ 3 ¨ 5 ¨ 7 “ 105.

Time: 5 hours

Time for questions: 60 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

1 / 1

Team Competition Sept 21, 2014 English version

Determine all functions f : R ÝÑ R such that

xf pxyq ` xyf pxq ě f px2qf pyq ` x2y

holds for all x, y P R.

Problem T–3

Let K and L be positive integers On a board consisting of 2K ˆ 2L unit squares an ant starts

in the lower left corner square and walks to the upper right corner square In each step it goeshorizontally or vertically to a neighbouring square It never visits a square twice At the endsome squares may remain unvisited

In some cases the collection of all unvisited squares forms a single rectangle In such cases, we

call this rectangle memorable.

Determine the number of different memorable rectangles

Remark Rectangles are different unless they consist of exactly the same squares.

Problem T–4

In Happy City there are 2014 citizens called A1, A2, , A2014 Each of them is either happy or

unhappy at any moment in time The mood of any citizen A changes (from being unhappy to

being happy or vice versa) if and only if some other happy citizen smiles at A On Monday morning there were N happy citizens in the city.

The following happened on Monday during the day: citizen A1 smiled at citizen A2, then A2smiled at A3, etc., and, finally, A2013 smiled at A2014 Nobody smiled at anyone else apart fromthis Exactly the same repeated on Tuesday, Wednesday and Thursday There were exactly 2000happy citizens on Thursday evening

Determine the largest possible value of N

Time: 5 hours

Time for questions: 60 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

1 / 2

Team Competition Sept 21, 2014 English version

Problem T–5

Let ABC be a triangle with AB ă AC Its incircle with centre I touches the sides BC, CA, and

AB in the points D, E, and F respectively The angle bisector AI intersects the lines DE and

DF in the points X and Y respectively Let Z be the foot of the altitude through A with respect

to BC.

Prove that D is the incentre of the triangle XY Z.

Problem T–6

Let the incircle k of the triangle ABC touch its side BC at D Let the line AD intersect k at

L ‰ D and denote the excentre of ABC opposite to A by K Let M and N be the midpoints of

an integer whenever k ě 1 and a1, , a k P A are distinct.

Given a positive integer n, determine the least possible sum of the elements of a meanly n-element

Time for questions: 60 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

2 / 2

Individual Competition 27th August, 2015 English version

Problem I–1

Find all surjective functions f : N Ñ N such that for all positive integers a and b, exactly one of

the following equations is true:

f (a) = f (b),

f (a + b) = mintf (a), f (b)u.

Remarks: N denotes the set of all positive integers A function f : X Ñ Y is said to be surjective

if for every y P Y there exists x P X such that f (x) = y.

Problem I–2

Let n ě 3 be an integer An inner diagonal of a simple n-gon is a diagonal that is contained in the n-gon Denote by D(P ) the number of all inner diagonals of a simple n-gon P and by D(n) the least possible value of D(Q), where Q is a simple n-gon Prove that no two inner diagonals

of P intersect (except possibly at a common endpoint) if and only if D(P ) = D(n).

Remark: A simple n-gon is a non-self-intersecting polygon with n vertices A polygon is not

Find all pairs of positive integers (m, n) for which there exist relatively prime integers a and b

greater than 1 such that

a m + b m

a n + b n

is an integer

Time: 5 hours

Time for questions: 60 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

1 / 1

Team Competition 28th August, 2015 English version

There are n students standing in line in positions 1 to n While the teacher looks away, some

students change their positions When the teacher looks back, they are standing in line again If

a student who was initially in position i is now in position j, we say the student moved for |i ´ j|

steps Determine the maximal sum of steps of all students that they can achieve

Problem T–4

Let N be a positive integer In each of the N2 unit squares of an N ˆ N board, one of the two diagonals is drawn The drawn diagonals divide the N ˆ N board into K regions For each N , determine the smallest and the largest possible values of K.

1

2

34

Time for questions: 60 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

1 / 2

Team Competition 28th August, 2015 English version

Problem T–5

Let ABC be an acute triangle with AB ą AC Prove that there exists a point D with the following property: whenever two distinct points X and Y lie in the interior of ABC such that the points B, C, X, and Y lie on a circle and

=AXB ´ =ACB = =CY A ´ =CBA

holds, the line XY passes through D.

Problem T–6

Let I be the incentre of triangle ABC with AB ą AC and let the line AI intersect the side BC

at D Suppose that point P lies on the segment BC and satisfies P I = P D Further, let J be the point obtained by reflecting I over the perpendicular bisector of BC, and let Q be the other intersection of the circumcircles of the triangles ABC and AP D Prove that =BAQ = =CAJ.

Time for questions: 60 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

2 / 2

V ¨ OCKLABRUCK

AUSTRIA 2016

Individual Competition

24 August 2016 English version

Problem I–1

Let n ě 2 be an integer and x1, x2, , x n be real numbers satisfying

(a) x j ą ´1 for j “ 1, 2, , n and

Show that an infinite sequence of moves cannot exist

Problem I–3

Let ABC be an acute-angled triangle with >BAC ą 45˝ and with circumcentre O The point P lies in its interior such that the points A, P , O, B lie on a circle and BP is perpendicular to CP The point Q lies on the segment BP such that AQ is parallel to P O.

Prove that >QCB “ >P CO.

Problem I–4

Find all functions f : N Ñ N such that f paq ` f pbq divides 2pa ` b ´ 1q for all a, b P N.

Remark: N “ t1, 2, 3, u denotes the set of positive integers.

Time: 5 hours

Time for questions: 60 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

1 / 1

V ¨ OCKLABRUCK

AUSTRIA 2016

Team Competition

25 August 2016 English version

A house is said to be blocked from sunlight if there are three houses on the plots immediately to

its east, west and south

What is the maximum number of houses that can simultaneously exist, such that none of them

is blocked from sunlight?

Remark: By definition, houses on the east, west and south borders are never blocked from light

sun-Problem T–4

A class of high school students wrote a test Every question was graded as either 1 point for acorrect answer or 0 points otherwise It is known that each question was answered correctly by

at least one student and the students did not all achieve the same total score

Prove that there was a question on the test with the following property: The students whoanswered the question correctly got a higher average test score than those who did not

Time: 5 hours

Time for questions: 60 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

1 / 2

V ¨ OCKLABRUCK

AUSTRIA 2016

Team Competition

25 August 2016 English version

Problem T–5

Let ABC be an acute-angled triangle with AB ‰ AC, and let O be its circumcentre The line

AO intersects the circumcircle ω of ABC a second time in point D, and the line BC in point E.

The circumcircle of CDE intersects the line CA a second time in point P The line P E intersects the line AB in point Q The line through O parallel to P E intersects the altitude of the triangle

ABC that passes through A in point F

Prove that F P “ F Q.

Problem T–6

Let ABC be a triangle with AB ‰ AC The points K, L, M are the midpoints of the sides BC,

CA, AB, respectively The inscribed circle of ABC with centre I touches the side BC at point

D The line g, which passes through the midpoint of segment ID and is perpendicular to IK,

intersects the line LM at point P

Prove that >P IA “ 90˝

Problem T–7

A positive integer n is called a Mozartian number if the numbers 1, 2, , n together contain an

even number of each digit (in base 10)

Prove:

(a) All Mozartian numbers are even

(b) There are infinitely many Mozartian numbers

Problem T–8

We consider the equation a2` b2` c2` n “ abc, where a, b, c are positive integers.

Prove:

(a) There are no solutions pa, b, cq for n “ 2017.

(b) For n “ 2016, a must be divisible by 3 for every solution pa, b, cq.

(c) The equation has infinitely many solutions pa, b, cq for n “ 2016.

Time: 5 hours

Time for questions: 60 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

2 / 2

Individual Competition

Aug 23, 2017 English version

Problem I–1

Determine all functions f : R Ñ R satisfying

f px2 f pxqfpyqq  xfpx yq for all real numbers x and y.

Problem I–2

Let n ¥ 3 be an integer A labelling of the n vertices, the n sides and the interior of a regular

n-gon by 2n 1 distinct integers is called memorable if the following conditions hold:

(a) Each side has a label that is the arithmetic mean of the labels of its endpoints

(b) The interior of the n-gon has a label that is the arithmetic mean of the labels of all the

vertices

Determine all integers n ¥ 3 for which there exists a memorable labelling of a regular n-gon consisting of 2n 1 consecutive integers

Problem I–3

Let ABCDE be a convex pentagon Let P be the intersection of the lines CE and BD Assume

that=P AD  =ACB and =CAP  =EDA Prove that the circumcentres of the triangles ABC and ADE are collinear with P

Problem I–4

Determine the smallest possible value of

|2m
181n |, where m and n are positive integers.

Time: 5 hours

Time for questions: 60 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

1 / 1

Team Competition Aug 24, 2017 English version

Problem T–1

Determine all pairs of polynomialspP, Qq with real coefficients satisfying

P px Qpyqq  Qpx P pyqq for all real numbers x and y.

There is a lamp on each cell of a 2017 2017 square board Each lamp is either on or off A lamp

is called bad if it has an even number of neighbours that are on What is the smallest possible

number of bad lamps on such a board?

(Two lamps are neighbours if their respective cells share a side.)

Problem T–4

Let n ¥ 3 be an integer A sequence P1, P2, , P n of distinct points in the plane is called good

if no three of them are collinear, the polyline P1P2 P n is non-self-intersecting and the triangle

P i P i 1 P i 2 is oriented counterclockwise for every i  1, 2, , n
2.

For every integer n ¥ 3 determine the greatest possible integer k with the following property: there exist n distinct points A1, A2, , A n in the plane for which there are k distinct permutations

σ : t1, 2, , nu Ñ t1, 2, , nu such that A σp1q, A σp2q, , A σ pnq is good.

(A polyline P1P2 P n consists of the segments P1P2, P2P3, , P n
1P n.)

Time: 5 hours

Time for questions: 60 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

1 / 2

Team Competition Aug 24, 2017 English version

Problem T–5

Let ABC be an acute-angled triangle with AB ¡ AC and circumcircle Γ Let M be the midpoint

of the shorter arc BC of Γ, and let D be the intersection of the rays AC and BM Let E  C

be the intersection of the internal bisector of the angle ACB and the circumcircle of the triangle

BDC Let us assume that E is inside the triangle ABC and there is an intersection N of the line

DE and the circle Γ such that E is the midpoint of the segment DN

Show that N is the midpoint of the segment I B I C , where I B and I C are the excentres of ABC opposite to B and C, respectively.

Problem T–6

Let ABC be an acute-angled triangle with AB  AC, circumcentre O and circumcircle Γ Let the tangents to Γ through B and C meet each other at D, and let the line AO intersect BC at

E Denote the midpoint of BC by M and let AM meet Γ again at N  A Finally, let F  A be

a point on Γ such that A, M , E and F are concyclic Prove that F N bisects the segment M D.

Problem T–7

Determine all integers n ¥ 2 such that there exists a permutation x0, x1, , x n
1 of the numbers

0, 1, , n
1 with the property that the n numbers

x0, x0 x1, , x0 x1 x n
1

are pairwise distinct modulo n.

Problem T–8

For an integer n ¥ 3 we define the sequence α1, α2, , α k as the sequence of exponents in the

prime factor decomposition of n!  p α1

1 p α2

2 p α k

k , where p1   p2       p k are primes

Determine all integers n ¥ 3 for which α1, α2, , α k is a geometric progression

Time: 5 hours

Time for questions: 60 min

Each problem is worth 8 points

The order of the problems does not depend on their difficulty

2 / 2

Individual Problems

and Solutions

2(n2−n+ 2) Since the sequence an = 1

2(n2−n+ 2) is as required(transform ai+ al > aj + ak to i2+ l2 > j2+ k2 and substitute i = d − y, l = d + y,

j = d − x, k = d + x, where 0 6 x < y), the smallest value of a2008 is 2,015,029

I–2 Consider a chessboard n × n where n > 1 is a positive integer We select thecenters of 2n − 2 squares How many selections are there such that no two selectedcenters lie on a line parallel to one of the diagonals of the chessboard?

Solution By a k-diagonal we mean any chessboard diagonal formed by k squares,where 1 6 k 6 n Since the number of stones is 2n−2, while the number of chessboard

1

diagonals in one direction is 2n−1 and two of them, which are 1-diagonals, must not beoccupied by stones simultaneously, we can conclude that each k-diagonal with k > 1contains exactly 1 stone and that exactly two of the 4 corner squares (1-diagonals)are occupied (and lie on the same border side) Let us call two different directions ofdiagonals as A and B.

Now let us consider the set P of all the pairs (s, f ), for which the stone s lies onthe same diagonal as the unoccupied (“free”) square f There are exactly n2

−2n + 2free squares on the chessboard, two of them are corner, hence for each of the n2

−2nfree squares f which lie on two k-diagonals with k > 1, we have (s, f ) ∈ P for exactlytwo stones s Thus the total number p of the pairs in P is given by the formula

p = 2(n2

−2n) + 2 = 2n2

−4n + 2,where +2 stands for the two free corner squares

If a stone s lies on the intersection of a k1-diagonal and a k2-diagonal with

k1, k2 > 1, then the number of pairs (s, f ) ∈ P with this s equals k1 + k2−2 Thesame holds also for the two other stones with {k1, k2} = {1, n} Obviously, for anystone we have k1+ k2 > n + 1 with equality iff the stone lies on a border square Thusfor each stone s, the number of pairs (s, f ) ∈ P is at least n − 1, and therefore

in the other direction Hence there are exactly 2n possibilities how to distribute thestones on the chessboard as required

Comment The proof of the fact that all the stones must lie on some of the bordersquares from the preceding solution can be presented in the following algebraic formwithout counting the pairs (s, f )

Consider the chessboard n × n as the grid {0, 1, , m} × {0, 1, , m} with

m = n − 1, in which the occupied squares are represented by points (ai, bi) with

i= 1, 2, , 2m Since ai−biare 2m distinct integers from {−m, −m+1, , m−1, m}and the boundary values ±m are not reached simultaneously, the values of |ai−bi|(in nondecreasing order) are the numbers

0, 1, 1, 2, 2, , m − 1, m − 1, mwhose sum equals m2 Thus we have

i=2mXi=1

and, similarly,

i=2mXi=1

|ai+ bi−m|= m2

2

Summing (1) and (2) and taking into account the identity

|a − b|+ |a + b − m| = max(|2a − m|, |2b − m|) for any a, b, m ∈R,

we obtain the equality

i=2mXi=1

Since |2ai−m| 6 m and |2bi−m| 6 m for each i, the following inequality

i=2mXi=1max(|2ai−m|, |2bi−m|) 6 2m · m = 2m2

holds and hence the equality (3) implies that

max(|2ai−m|, |2bi−m|) = m for any i = 1, 2, , 2m

This means that any (ai, bi) is a boundary point of the grid and the proof is complete.Another solution (Bernd Mulansky, Germany) The first paragraph is identical withthat from the first solution It follows that each satisfactory distribution of the 2n − 2stones can be derived as a result of the following procedure in n steps:

⊲ Step 1: One stone is placed on one of the two 1-diagonals of direction A

⊲ Step k (where 2 6 k 6 n − 1): Two stones are placed, each on one of the twok-diagonals of direction A

⊲ Step n: One stone is placed on the n-diagonal of direction A

Notice that for each m = 1, 2, , n − 1 the following conclusion clearly holds:after m steps of our procedure, well-done in the sense that no two stones were placed

on the same diagonal (of direction B), all the 2m−1 longest k-diagonals of direction B(those with k > n + 1 − m) are occupied by stones Consequently, if in addition

m < n −1, in the next step m + 1 the two stones must be placed on the bordersquares of the two (m + 1)-diagonals of direction A (their other squares lie on theoccupied diagonals of direction B) and there are exactly two ways in which this can

be well-done Analogously for the case m + 1 = n Thus we have two possibilities ineach of the n steps and the number of all satisfactory distributions equals 2n

Another solution (Pavol Novotný, Slovakia) Let us colour the chessboard squares asusual, with the black square in the left upper corner It is easy to show that n − 1stones must be placed on the black squares (let us call them black stones), analogouslyfor the n − 1 white stones The number sn of all satisfactory stone distributions onthe chessboard n ×n is equal to the product bn·wn, where bn and wn are the numbers

of satisfactory distributions of black and white stones, respectively We have w1 = 1,

w2 = w3 = 2, b1 = 1, b2 = 2 and b3 = 4 Easy arguments show that for each n > 3,

wn = 2wn−2,1 and bn = 2wn−1,2 hence sn = bnwn = 4wn−2wn−1 = 2bn−1wn−1 =2sn−1 and the result sn= 2n follows

1 Remove two white 2-diagonals of one direction and two white ( n − 1)-diagonals of the other direction; the remaining white squares form the same diagonals as white squares of the chessboard (n − 2) × (n − 2).

2 Remove one black n-diagonal; the remaining black squares form the same diagonals as the white squares of the chessboard ( n − 1) × (n − 1).

3

I–3 Let ABC be an isosceles triangle with |AC| = |BC| Its incircle touches ABand BC at D and E, respectively A line (different from AE) passes through A andintersects the incircle at F and G The lines EF and EG intersect the line AB at Kand L, respectively Prove that |DK| = |DL|.

Solution In view of symmetry, suppose that AF < AG, and, in addition, that G is

on the smaller arc DE (for the other case see the last two sentences below)

If the incircle touches AC at J, then6 CAB =6 CJ E =6 J DE =6 J F E (Fig 1),hence AJF K is a cyclic quadrilateral Thus 6 AJ K = 6 AF K = 6 EF G = 6 LEB,which implies that AJK and BEL are congruent triangles Since K and L are innerpoints of the segment AB, AK = BL means that DK = DL

If G is on the larger arc DE (between E and J), then K, A, B, L is the order ofthese collinear points and the cyclic quadrilateral is AKJF The rest of the proof isthe same

Another solution (Tomáš Pavlík, Czech Republic) Let us denote X the intersection

of line AF with side BC of the given triangle (Fig 2) The power of the point X withregard to the incircle of ABC gives XE2 = XF · XG which means that

AL · BE

LB · GA = KB · F A

AK · BE4

which gives

AK · AL · BE2

KB · LB · F A · GA = 1,hence

AK ·(AB ± BL) = AK · AL = KB · LB = (AB ± AK) · BL

which results into AK = BL in both cases This is equivalent to the wanted equality

k −4 = ±2k with any nonnegative integer k

On the other hand, if k − 4 has got an odd divisor p > 1, then we can easilyfind a multiple of p of the form 4n + 1 (for example, the number p2 or simply one ofthe numbers p, 3p) For any number 4n + 1 being a multiple of p, the above identityimplies that p | kn + 1, hence 4n + 1 and kn + 1 are not relatively prime

Answer: k = 4 ± 2k, where k = 0, 1, 2,

5

Team Problems

and Solutions

T–1 Let R denote the set of real numbers Find all functions f :R →R such that

xf(x + xy) = xf (x) + f (x2

)f (y)for all x, y ∈R

Let us distinguish the cases f (−1) = 0 and f (−1) 6= 0

The case f(−1) = 0 It follows from (1) that f (x) = 0 for all x 6= 0 As wealready know, f (0) = 0 Thus we get the zero function f (x) = 0, which is obviously

a solution

1

The case f(−1) 6= 0 Setting x = −1 in (1) yields f (1) = 1 Using this in (1)with x = 1, we get f (−1) = −1 and hence (1) can be transformed to

Assume that f (a) = 0 for some a 6= 0 Then f (a2) = 0 by (2) and hence (0) with

x = a implies that af (a + ay) = 0, i.e f (a + ay) = 0 Since y is arbitrary here, weget f (−1) = 0, which is not the case Therefore, for any x 6= 0 we have f (x) 6= 0, andhence f (x2) 6= 0 as well Thus (4) leads to the conclusion that f (x − 1) = x − 1 forany x 6= 0, i.e f (x) = x for any x 6= −1 Since we already know that f (−1) = −1,

we get the identity function f (x) = x, which is obviously a solution

T–2 Let n > 2 be an integer There are n positive integers written on a blackboard

In each step we choose two of the numbers on the blackboard and replace each ofthem by their sum Determine all values of n for which it is always possible to get

n identical integers in a finite number of steps

Solution Starting from the n-tuple (2, 2, 1, 1, , 1) with any n > 3, we get always

an n-tuple in which the number of maximal values is even Hence no odd n > 3 is asrequired

Let us show by induction that any even n > 2 is satisfactory, which is obvious

if n = 2 For an even n > 4, by the induction hypothesis, we can transform anyinitial n-tuple to (a, a, , a, b, b) If a 6= b, we apply repeatedly some of the followingseries of steps, which always lead to an n-tuple of type (a, , a

| {z }k

, b, , b

| {z }n−k) in which the

number k may differ from the initial value k = n − 2 (remaining to be even):

series α: (a, , a

| {z }k

, b, , b

| {z }n−k

series β: (a, , a

| {z }k

, b, , b

| {z }n−k

) → (a, , a

| {z }k

, b, , b

| {z }n−k

series γ2 (if k > n − k): (a, , a

| {z }k

, b, , b

| {z }n−k

) → (a, , a

| {z }2k−n

, b, , b

| {z }n−k) with

a 6= b, let us apply

2

⊲ series α if P (a) < P (b),

⊲ series β if P (a) > P (b),

⊲ series γ1 or γ2 if P (a) = P (b) (and hence N (a) 6= N (b))

Using the series α and β, the numbers N (a), N (b) do not change, while the series

γ1 and γ2 cause the changes exactly one of them, namely

N(b) → N(a) + N (b)

2m , or N (b) → N(a) + N (b)

2m respectively,where m = P (N (a) + N (b)) > 1 and hence

N(a) + N (b)

2m 6 N(a) + N (b)

2 <max(N (a), N (b))(recall that N (a) 6= N (b)) Consequently, throughout our procedure, the value ofmax(N (a), N (b)) is a nonincreasing variable, and hence constant after a finite numbers

of series From this moment, we must still have either N (a) > N(b), or N(a) 6 N(b).This excludes either series γ1, or series γ2from future applications, in which, therefore,all possible changes of the parameter k are either k → 2k, or (n−k) → 2(n−k) Sincethis can repeat only r times, where 2r

6 n, at the end we always we get an n-tuple(a, , a, b, , b) for which (if a 6= b) the continuation of our procedure reduces only

to the series α and β Applying now either α, or β exactly |P (a) − P (b)| times, we get

an n-tuple (a′, , a′, b′, , b′) with P (a′) = P (b′) Since γ1, γ2are already excluded,

we have a′ = b′, which completes the induction proof

Another solution(German team, adapted) We show without induction on n that anyeven n = 2k is satisfactory At the beginning in the initial 2k-tuple (a1, , a2k) wereplace every pair (a2i−1, a2i) (for i = 1, , k) by the pair (a2i−1+ a2i, a2i−1+ a2i).From now on, we shall have always identical numbers on the (2i − 1)th and (2i)thposition Hence because of brevity we shall work with k-tuples (x, y, z, ) instead

of 2k-tuples (x, x, y, y, z, z, ) We are allowed to do the following transformations

on the k-tuples:

⊲ choose two of the numbers x, y and replace each of them by their sum (this sponds with two steps ( , x, x, , y, y, ) → ( , x +y, x, , x +y, y, ) →( , x + y, x + y, , x + y, x + y, ) performed on the 2k-tuple);

corre-⊲ choose one number x and multiply it by 2 (this corresponds with one step( , x, x, ) → ( , x + x, x + x, );

⊲ divide all numbers by 2 (this obviously does not affect anything; formally wecould remember how many times we have performed this dividing and multiplyall the numbers by the proper power of two at the end)

Our aim is to obtain k identical numbers We reach it by iterating the followingalgorithm:

1 While there are at least two odd numbers, find the minimum and the maximumodd number and replace each of them by their (even) sum

2 If there is one odd number left after finishing the first step, multiply it by two

3 Divide all numbers by 2

Clearly, after each iteration, the maximum number among all k numbers eitherdecreases or does not change As this maximum is permanently a positive integer, af-ter a finite number of iterations, it fixes at the value M and does not change anymore.From now on, look at the number N of M ’s in the k-tuple

3

Obviously M is odd (otherwise it would decrease in the third step in the nextiteration) If N < k, then there is at least one number m with m < M If m isodd, after the next iteration N decreases As it is impossible to increase N in theiterations, it must be constant after a finite number of steps and there must be onlyeven m with m < M But every even m is divided by 2 in each iteration and aftersome iterations some odd number less than M must appear So there are no numbersless than M , which completes the proof.

T–3 An acute-angled triangle ABC is given Let E be a point such that B and Elie on different sides of the line AC, and let D be an interior point of the segment

AE Suppose that 6 ADB = 6 CDE, 6 BAD= 6 ECD and 6 ACB =6 EBA Provethat B, C and E are collinear

Solution Condition6 ADB =6 CDEmotivates us to reflect B over AE to B′(Fig 3).Then C, D and B′ are collinear and 6 EAB′ = 6 EAB = 6 ECD = 6 ECB′, so

B′ACE is a cyclic quadrilateral This implies that 6 ECA = π − 6 EB′A = π −

6 EBA = π −6 ACB, hence 6 ECA+6 ACB = π and thus B, C, E are collinear

B′

CD

E

Fig 3

Comment With the same success, we can reflect C over AE to get a point C′collinearwith B, D and such that ABEC′ is cyclic, hence 6 ECA =6 EC′A = π −6 EBA=

π −6 ACB, i.e 6 ECA+6 ACB = π again

Because of the proved collinearity of B, C, E, the condition 6 ACB = 6 EBAimplies that AB = AC, while the condition 6 BAD = 6 ECD implies that ABCDmust be a cyclic quadrilateral The last fact serves as a good motivation for an-other solution Before we present it, let us note that the situations described in thestatement of the problem do exist and all of them are of the following form:

ABC is an isosceles triangle with AB = AC, points B, C, E are collinear (C isbetween B and E) and AE cuts the circumcircle of ABC at D

Another solution Suppose that B, C, E are not collinear The line through B, which

is parallel to CE, meets the lines CD and AD at C′ and E′, respectively Since

6 E′C′D =6 ECD = 6 BAD, the quadrilateral ABC′D is cyclic (Fig 4) Denote its

4