According to the estimated number of participants who gave a correct solution, this was the hardest (!) problem from today's paper. So here is this great German killer - be warned! Given a circle k and three pairwisely distinct points A, B, C on this circle. Let h and g be the perpendiculars to the line BC at the points B and C. The perpendicular bisector of the segment AB meets the line h at a point F; the perpendicular bisector of the segment AC meets the line g at a point G. Prove that the product $BF\cdot CG$ is independent from the position of the point A, as long as the points B and C stay fixed. The actual problem behind the problem: Why on hell should the points B and C stay fixed? Darij

In triangle $ABC$ let $F$ denote the midpoint of side $BC$. Let the circle passing through point $A$ and tangent to side $BC$ at point $F$ intersect sides $AB$ and $AC$ at points $M$ and $N$, respectively. Let the line segments $CM$ and $BN$ intersect in point $X$. Let $P$ be the second point of intersection of the circumcircles of triangles $BMX$ and $CNX$. Prove that points $A, F$ and $P$ are collinear. Proposed by Imolay András, Budapest

The graphs of $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have $y$-intercepts of $2013$ and $2014$, respectively, and each graph has two positive integer $x$-intercepts. Find $h$.

There is a set with three elements: (2,3,5). It has got an interesting property: (2*3) mod 5=(2*5) mod 3=(3*5) mod 2. Prove that it is the only one set with such property.

Triangle $A_0B_0C_0$ is given in the plane. Consider all triangles $ABC$ such that: (i) The lines $AB,BC,CA$ pass through $C_0,A_0,B_0$, respectvely, (ii) The triangles $ABC$ and $A_0B_0C_0$ are similar. Find the possible positions of the circumcenter of triangle $ABC$.

Let $n$ be a positive integer. A \emph{pseudo-Gangnam Style} is a dance competition between players $A$ and $B$. At time $0$, both players face to the north. For every $k\ge 1$, at time $2k-1$, player $A$ can either choose to stay stationary, or turn $90^{\circ}$ clockwise, and player $B$ is forced to follow him; at time $2k$, player $B$ can either choose to stay stationary, or turn $90^{\circ}$ clockwise, and player $A$ is forced to follow him. After time $n$, the music stops and the competition is over. If the final position of both players is north or east, $A$ wins. If the final position of both players is south or west, $B$ wins. Determine who has a winning strategy when: (a) $n=2013^{2012}$ (b) $n=2013^{2013}$

For $x,y,z > 0$ and $xyz=1$, prove that \[\frac{x^{9}+y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}+\frac{x^{9}+z^{9}}{x^{6}+x^{3}z^{3}+z^{6}}+\frac{y^{9}+z^{9}}{y^{6}+y^{3}z^{3}+z^{6}}\geq 2\]

Let $ABC$ be an acute-angled triangle with $AC < AB$ and circumradius $R$. Furthermore, let $D$ be the foot ofthe altitude from $A$ on $BC$ and let $T$ denote the point on the line $AD$ such that $AT = 2R$ holds with $D$ lying between $A$ and $T$. Finally, let $S$ denote the mid-point of the arc $BC$ on the circumcircle that does not include $A$. Prove: $\angle AST = 90^\circ$. (Karl Czakler)

Bokos tribus have $2021$ closed chests, we know that every chest have some amount of rupias and some amount of diamonts. They are going to do a deal with Link, that consits that Link will stay with a amount of chests and Bokos with the rest. Before opening the chests, Link has to say the amount of chest that he will stay with. After this the chests open and Link has to choose the chests with the amount that he previously said. Link doesn´t want to make Bokos angry so he wants to say the smallest number of chest that he will stay with, but guaranteeing that he stay with at least with the half of diamonts, and at least the half of the rupias. What number does Link needs to say?

Consider 2009 cards which are lying in sequence on a table. Initially, all cards have their top face white and bottom face black. The cards are enumerated from 1 to 2009. Two players, Amir and Ercole, make alternating moves, with Amir starting. Each move consists of a player choosing a card with the number $k$ such that $k < 1969$ whose top face is white, and then this player turns all cards at positions $k,k+1,\ldots,k+40.$ The last player who can make a legal move wins. (a) Does the game necessarily end? (b) Does there exist a winning strategy for the starting player? [i]Also compare shortlist 2009, combinatorics problem C1.[/i]