Let $a_1,a_2,\ldots,a_k$ be real numbers and $a_1+a_2+\ldots+a_k=0$. Prove that \[ \max_{1\leq i \leq k} a_i^2 \leq \frac{k}{3} \left( (a_1-a_2)^2+(a_2-a_3)^2+\cdots +(a_{k-1}-a_k)^2\right). \]

For positive integers $a_1,a_2 ,\ldots,a_{2006}$ such that $\frac{a_1}{a_2},\frac{a_2}{a_3},\ldots,\frac{a_{2005}}{a_{2006}}$ are pairwise distinct, find the minimum possible amount of distinct positive integers in the set$\{a_1,a_2,...,a_{2006}\}$.

Positive integers $k, m, n$ satisfy $mn=k^2+k+3$, prove that at least one of the equations $x^2+11y^2=4m$ and $x^2+11y^2=4n$ has an odd solution.

In a right angled-triangle $ABC$, $\angle{ACB} = 90^o$. Its incircle $O$ meets $BC$, $AC$, $AB$ at $D$,$E$,$F$ respectively. $AD$ cuts $O$ at $P$. If $\angle{BPC} = 90^o$, prove $AE + AP = PD$.

Let $\{a_n\}$ be a sequence such that: $a_1 = \frac{1}{2}$, $a_{k+1}=-a_k+\frac{1}{2-a_k}$ for all $k = 1, 2,\ldots$. Prove that \[ \left(\frac{n}{2(a_1+a_2+\cdots+a_n)}-1\right)^n \leq \left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^n\left(\frac{1}{a_1}-1\right)\left(\frac{1}{a_2}-1\right)\cdots \left(\frac{1}{a_n}-1\right). \]

Suppose $X$ is a set with $|X| = 56$. Find the minimum value of $n$, so that for any 15 subsets of $X$, if the cardinality of the union of any 7 of them is greater or equal to $n$, then there exists 3 of them whose intersection is nonempty.

Click for solution I think it's 41. First, let us prove $n=41$ is good. Take $A_{1},A_{2},...,A_{15} \subseteq [56]$ such that $|A_{i_{1}} \cup ... \cup A_{i_{7}}| \geq 41$. Suppose by way of contradiction that every element of $X$ occurs in at most 2 sets. If there is some $x \in X$ which belongs to a single $A_{i}$, you can add this $x$ to some other $A_{j}$ without affecting the property that the union of any 7 is at least 41, so suppose WLOG that every element of $X$ occurs in exactly 2 sets. Take a graph $G$ with vertices $1,2,...,15$ in which vertices $i$ and $j$ are connected iff $A_{i} \cap A_{j} \neq \emptyset$. It is clear that $G$ has exactly 56 edges. We know that for any 7 vertices, the number of edges which have at least one endpoint among these 7 vertices is at least 41, hence for any 8 vertices, the number of edges which have both endpoints among them is at most 56-41=15. Consider some vertice $v$ whose degree is $\deg v$. By our last observation, the number of edges among these 14 points is \[ 56-\deg v \leq 15 \frac{{14 \choose 2}}{{8 \choose 2}} = 48.75 \Rightarrow \deg v \geq 8, \forall v \in V. \] Hence $112 = 2|E| \geq 8 \times 15=120$, a contradiction. To create a counterexample for $n=40$, just select some sets $A_{i}$ such that our graph $G$ is a $K_{8,7}$. I hope it's correct.