Find all function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following equality holds \[ f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \] where $\left\lfloor a\right\rfloor $ is greatest integer not greater than $a.$ Proposed by Pierre Bornsztein, France

Let the real numbers $a,b,c,d$ satisfy the relations $a+b+c+d=6$ and $a^2+b^2+c^2+d^2=12.$ Prove that \[36 \leq 4 \left(a^3+b^3+c^3+d^3\right) - \left(a^4+b^4+c^4+d^4 \right) \leq 48.\] Proposed by Nazar Serdyuk, Ukraine

Let $x_1, \ldots , x_{100}$ be nonnegative real numbers such that $x_i + x_{i+1} + x_{i+2} \leq 1$ for all $i = 1, \ldots , 100$ (we put $x_{101 } = x_1, x_{102} = x_2).$ Find the maximal possible value of the sum $S = \sum^{100}_{i=1} x_i x_{i+2}.$ Proposed by Sergei Berlov, Ilya Bogdanov, Russia

A sequence $x_1, x_2, \ldots$ is defined by $x_1 = 1$ and $x_{2k}=-x_k, x_{2k-1} = (-1)^{k+1}x_k$ for all $k \geq 1.$ Prove that $\forall n \geq 1$ $x_1 + x_2 + \ldots + x_n \geq 0.$ Proposed by Gerhard Wöginger, Austria

Denote by $\mathbb{Q}^+$ the set of all positive rational numbers. Determine all functions $f : \mathbb{Q}^+ \mapsto \mathbb{Q}^+$ which satisfy the following equation for all $x, y \in \mathbb{Q}^+:$ \[f\left( f(x)^2y \right) = x^3 f(xy).\] Proposed by Thomas Huber, Switzerland

Suppose that $f$ and $g$ are two functions defined on the set of positive integers and taking positive integer values. Suppose also that the equations $f(g(n)) = f(n) + 1$ and $g(f(n)) = g(n) + 1$ hold for all positive integers. Prove that $f(n) = g(n)$ for all positive integer $n.$ Proposed by Alex Schreiber, Germany

Let $a_1, a_2, a_3, \ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that \[a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.\] Prove there exist positive integers $\ell \leq s$ and $N$, such that \[a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.\] Proposed by Morteza Saghafiyan, Iran

Given six positive numbers $a,b,c,d,e,f$ such that $a < b < c < d < e < f.$ Let $a+c+e=S$ and $b+d+f=T.$ Prove that \[2ST > \sqrt{3(S+T)\left(S(bd + bf + df) + T(ac + ae + ce) \right)}.\] Proposed by Sung Yun Kim, South Korea

In a concert, 20 singers will perform. For each singer, there is a (possibly empty) set of other singers such that he wishes to perform later than all the singers from that set. Can it happen that there are exactly 2010 orders of the singers such that all their wishes are satisfied? Proposed by Gerhard Wöginger, Austria

On some planet, there are $2^N$ countries $(N \geq 4).$ Each country has a flag $N$ units wide and one unit high composed of $N$ fields of size $1 \times 1,$ each field being either yellow or blue. No two countries have the same flag. We say that a set of $N$ flags is diverse if these flags can be arranged into an $N \times N$ square so that all $N$ fields on its main diagonal will have the same color. Determine the smallest positive integer $M$ such that among any $M$ distinct flags, there exist $N$ flags forming a diverse set. Proposed by Tonći Kokan, Croatia

2500 chess kings have to be placed on a $100 \times 100$ chessboard so that (i) no king can capture any other one (i.e. no two kings are placed in two squares sharing a common vertex); (ii) each row and each column contains exactly 25 kings. Find the number of such arrangements. (Two arrangements differing by rotation or symmetry are supposed to be different.) Proposed by Sergei Berlov, Russia

Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$ initially contains one coin. The following operations are allowed Type 1) Choose a non-empty box $B_j$, $1\leq j \leq 5$, remove one coin from $B_j$ and add two coins to $B_{j+1}$; Type 2) Choose a non-empty box $B_k$, $1\leq k \leq 4$, remove one coin from $B_k$ and swap the contents (maybe empty) of the boxes $B_{k+1}$ and $B_{k+2}$. Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ become empty, while box $B_6$ contains exactly $2010^{2010^{2010}}$ coins. Proposed by Hans Zantema, Netherlands

$n \geq 4$ players participated in a tennis tournament. Any two players have played exactly one game, and there was no tie game. We call a company of four players $bad$ if one player was defeated by the other three players, and each of these three players won a game and lost another game among themselves. Suppose that there is no bad company in this tournament. Let $w_i$ and $l_i$ be respectively the number of wins and losses of the $i$-th player. Prove that \[\sum^n_{i=1} \left(w_i - l_i\right)^3 \geq 0.\] Proposed by Sung Yun Kim, South Korea

Given a positive integer $k$ and other two integers $b > w > 1.$ There are two strings of pearls, a string of $b$ black pearls and a string of $w$ white pearls. The length of a string is the number of pearls on it. One cuts these strings in some steps by the following rules. In each step: (i) The strings are ordered by their lengths in a non-increasing order. If there are some strings of equal lengths, then the white ones precede the black ones. Then $k$ first ones (if they consist of more than one pearl) are chosen; if there are less than $k$ strings longer than 1, then one chooses all of them. (ii) Next, one cuts each chosen string into two parts differing in length by at most one. (For instance, if there are strings of $5, 4, 4, 2$ black pearls, strings of $8, 4, 3$ white pearls and $k = 4,$ then the strings of 8 white, 5 black, 4 white and 4 black pearls are cut into the parts $(4,4), (3,2), (2,2)$ and $(2,2)$ respectively.) The process stops immediately after the step when a first isolated white pearl appears. Prove that at this stage, there will still exist a string of at least two black pearls. Proposed by Bill Sands, Thao Do, Canada

Let $P_1, \ldots , P_s$ be arithmetic progressions of integers, the following conditions being satisfied: (i) each integer belongs to at least one of them; (ii) each progression contains a number which does not belong to other progressions. Denote by $n$ the least common multiple of the ratios of these progressions; let $n=p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ its prime factorization. Prove that \[s \geq 1 + \sum^k_{i=1} \alpha_i (p_i - 1).\] Proposed by Dierk Schleicher, Germany

Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$ Proposed by Christopher Bradley, United Kingdom

Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$. Proposed by Marcin E. Kuczma, Poland

Let $A_1A_2 \ldots A_n$ be a convex polygon. Point $P$ inside this polygon is chosen so that its projections $P_1, \ldots , P_n$ onto lines $A_1A_2, \ldots , A_nA_1$ respectively lie on the sides of the polygon. Prove that for arbitrary points $X_1, \ldots , X_n$ on sides $A_1A_2, \ldots , A_nA_1$ respectively, \[\max \left\{ \frac{X_1X_2}{P_1P_2}, \ldots, \frac{X_nX_1}{P_nP_1} \right\} \geq 1.\] Proposed by Nairi Sedrakyan, Armenia

Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$. Proposed by Tai Wai Ming and Wang Chongli, Hong Kong

Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$ Proposed by Nazar Serdyuk, Ukraine

The vertices $X, Y , Z$ of an equilateral triangle $XYZ$ lie respectively on the sides $BC, CA, AB$ of an acute-angled triangle $ABC.$ Prove that the incenter of triangle $ABC$ lies inside triangle $XYZ.$ Proposed by Nikolay Beluhov, Bulgaria

Three circular arcs $\gamma_1, \gamma_2,$ and $\gamma_3$ connect the points $A$ and $C.$ These arcs lie in the same half-plane defined by line $AC$ in such a way that arc $\gamma_2$ lies between the arcs $\gamma_1$ and $\gamma_3.$ Point $B$ lies on the segment $AC.$ Let $h_1, h_2$, and $h_3$ be three rays starting at $B,$ lying in the same half-plane, $h_2$ being between $h_1$ and $h_3.$ For $i, j = 1, 2, 3,$ denote by $V_{ij}$ the point of intersection of $h_i$ and $\gamma_j$ (see the Figure below). Denote by $\widehat{V_{ij}V_{kj}}\widehat{V_{kl}V_{il}}$ the curved quadrilateral, whose sides are the segments $V_{ij}V_{il},$ $V_{kj}V_{kl}$ and arcs $V_{ij}V_{kj}$ and $V_{il}V_{kl}.$ We say that this quadrilateral is $circumscribed$ if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}, \widehat{V_{12}V_{22}}\widehat{V_{23}V_{13}},\widehat{V_{21}V_{31}}\widehat{V_{32}V_{22}}$ are circumscribed, then the curved quadrilateral $\widehat{V_{22}V_{32}}\widehat{V_{33}V_{23}}$ is circumscribed, too. Proposed by Géza Kós, Hungary [asy][asy] pathpen=black; size(400); pair A=(0,0), B=(4,0), C=(10,0); draw(L(A,C,0.3)); MP("A",A); MP("B",B); MP("C",C); pair X=(5,-7); path G1=D(arc(X,C,A)); pair Y=(5,7), Z=(9,6); draw(Z--B--Y); struct T {pair C;real r;}; T f(pair X, pair B, pair Y, pair Z) { pair S=unit(Y-B)+unit(Z-B); real s=abs(sin(angle((Y-B)/(Z-B))/2)); real t=10, r=abs(X-A); pair Q; for(int k=0;k<30;++k) { Q=B+t*S; t-=(abs(X-Q)-r)/abs(S)-s*t; } T T=new T; T.C=Q; T.r=s*t*abs(S); return T; } void g(pair Q, real r) { real t=0; for(int k=0;k<30;++k) { X=(5,t); t+=(abs(X-Q)+r-abs(X-A)); } } pair Z1=(1.07,6); draw(B--Z1); T T=f(X,B,Y,Z1); draw(CR(T.C,T.r)); T T=f(X,B,Y,Z); draw(CR(T.C,T.r)); g(T.C,T.r); path G2=D(arc(X,C,A)); T T=f(X,B,Y,Z1); draw(CR(T.C,T.r)); T=f(X,B,Y,Z); draw(CR(T.C,T.r)); g(T.C,T.r); path G3=D(arc(X,C,A)); pen p=black+fontsize(8); MC("\gamma_1",G1,0.85,p); MC("\gamma_2",G2,0.85,NNW,p); MC("\gamma_3",G3,0.85,WNW,p); MC("h_1",B--Z1,0.95,E,p); MC("h_2",B--Y,0.95,E,p); MC("h_3",B--Z,0.95,E,p); path[] G={G1,G2,G3}; path[] H={B--Z1,B--Y,B--Z}; pair[][] al={{S+SSW,S+SSW,3*S},{SE,NE,NW},{2*SSE,2*SSE,2*E}}; for(int i=0;i<3;++i) for(int j=0;j<3;++j) MP("V_{"+string(i+1)+string(j+1)+"}",IP(H[i],G[j]),al[i][j],fontsize(8));[/asy][/asy]

Find the least positive integer $n$ for which there exists a set $\{s_1, s_2, \ldots , s_n\}$ consisting of $n$ distinct positive integers such that \[ \left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) = \frac{51}{2010}.\] Proposed by Daniel Brown, Canada

Find all pairs $(m,n)$ of nonnegative integers for which \[m^2 + 2 \cdot 3^n = m\left(2^{n+1} - 1\right).\] Proposed by Angelo Di Pasquale, Australia

Find the smallest number $n$ such that there exist polynomials $f_1, f_2, \ldots , f_n$ with rational coefficients satisfying \[x^2+7 = f_1\left(x\right)^2 + f_2\left(x\right)^2 + \ldots + f_n\left(x\right)^2.\] Proposed by Mariusz Skałba, Poland

Let $a, b$ be integers, and let $P(x) = ax^3+bx.$ For any positive integer $n$ we say that the pair $(a,b)$ is $n$-good if $n | P(m)-P(k)$ implies $n | m - k$ for all integers $m, k.$ We say that $(a,b)$ is $very \ good$ if $(a,b)$ is $n$-good for infinitely many positive integers $n.$ (a) Find a pair $(a,b)$ which is 51-good, but not very good. (b) Show that all 2010-good pairs are very good. Proposed by Okan Tekman, Turkey

Find all functions $g:\mathbb{N}\rightarrow\mathbb{N}$ such that \[\left(g(m)+n\right)\left(g(n)+m\right)\] is a perfect square for all $m,n\in\mathbb{N}.$ Proposed by Gabriel Carroll, USA

The rows and columns of a $2^n \times 2^n$ table are numbered from $0$ to $2^{n}-1.$ The cells of the table have been coloured with the following property being satisfied: for each $0 \leq i,j \leq 2^n - 1,$ the $j$-th cell in the $i$-th row and the $(i+j)$-th cell in the $j$-th row have the same colour. (The indices of the cells in a row are considered modulo $2^n$.) Prove that the maximal possible number of colours is $2^n$. Proposed by Hossein Dabirian, Sepehr Ghazi-nezami, Iran