In a football tournament there are $8$ teams, each of which plays exacly one match against every other team. If a team $A$ defeats team $B$, then $A$ is awarded $3$ points and $B$ gets $0$ points. If they end up in a tie, they receive $1$ point each. It turned out that in this tournament, whenever a match ended up in a tie, the two teams involved did not finish with the same final score. Find the maximum number of ties that could have happened in such a tournament.

Let $ABC$ be a triangle with $AB = AC$. The incircle touches $BC$, $AC$ and $AB$ at $D$, $E$ and $F$ respectively. Let $P$ be a point on the arc $\overarc{EF}$ that does not contain $D$. Let $Q$ be the second point of intersection of $BP$ and the incircle of $ABC$. The lines $EP$ and $EQ$ meet the line $BC$ at $M$ and $N$, respectively. Prove that the four points $P, F, B, M$ lie on a circle and $\frac{EM}{EN} = \frac{BF}{BP}$.

Find all functions $f: \mathbb R \rightarrow \mathbb R$ such that \[f(x+xy+f(y)) = \left(f(x)+\frac{1}{2}\right) \left(f(y)+\frac{1}{2}\right)\]holds for all real numbers $x,y$.

Two players, $A$ and $B$, play a game on a board which is a rhombus of side $n$ and angles of $60^{\circ}$ and $120^{\circ}$, divided into $2n^2$ equilateral triangles, as shown in the diagram for $n=4$. $A$ uses a red token and $B$ uses a blue token, which are initially placed in cells containing opposite corners of the board (the $60^{\circ}$ ones). In turns, players move their token to a neighboring cell (sharing a side with the previous one). To win the game, a player must either place his token on the cell containing the other player's token, or get to the opposite corner to the one where he started. If $A$ starts the game, determine which player has a winning strategy.

Let $p$ and $q$ be prime numbers. The sequence $(x_n)$ is defined by $x_1 = 1$, $x_2 = p$ and $x_{n+1} = px_n - qx_{n-1}$ for all $n \geq 2$. Given that there is some $k$ such that $x_{3k} = -3$, find $p$ and $q$.

Suppose $a_1, a_2, ..., a_r$ are integers with $a_i \geq 2$ for all $i$ such that $a_1 + a_2 + ... + a_r = 2010$. Prove that the set $\{1,2,3,...,2010\}$ can be partitioned in $r$ subsets $A_1, A_2, ..., A_r$ each with $a_1, a_2, ..., a_r$ elements respectively, such that the sum of the numbers on each subset is divisible by $2011$. Decide whether this property still holds if we replace $2010$ by $2011$ and $2011$ by $2012$ (that is, if the set to be partitioned is $\{1,2,3,...,2011\}$).