Let $(a_n)^{+\infty}_{n=1}$ be a sequence defined recursively as follows: $a_1=1$ and $$a_{n+1}=1 + \sum\limits_{k=1}^{n}ka_k$$For every $n > 1$, prove that $\sqrt[n]{a_n} < \frac {n+1}{2}$.

In the City of Islands there are $2021$ islands connected by bridges. For any given pair of islands $A$ and $B$, one can go from island $A$ to island $B$ using the bridges. Moreover, for any four islands $A_1, A_2, A_3$ and $A_4$: if there is a bridge from $A_i$ to $A_{i+1}$ for each $i \in \left \{ 1,2,3 \right \}$, then there is a bridge between $A_{j}$ and $A_{k}$ for some $j,k \in \left \{ 1,2,3,4 \right \}$ with $|j-k|=2$. Show that there is at least one island which is connected to any other island by a bridge.

Let $ABCD$ be a trapezoid with $AD \parallel BC$ and $\angle BCD < \angle ABC < 90^\circ$. Let $E$ be the intersection point of the diagonals $AC$ and $BD$. The circumcircle $\omega$ of $\triangle BEC$ intersects the segment $CD$ at $X$. The lines $AX$ and $BC$ intersect at $Y$, while the lines $BX$ and $AD$ intersect at $Z$. Prove that the line $EZ$ is tangent to $\omega$ iff the line $BE$ is tangent to the circumcircle of $\triangle BXY$.

For a fixed positive integer $n \geq 3$ we are given a $n$ $\times$ $n$ board with all unit squares initially white. We define a floating plus as a $5$-tuple $(M,L,R,A,B)$ of unit squares such that $L$ is in the same row and left of $M$, $R$ is in the same row and right of $M$, $A$ is in the same column and above $M$ and $B$ is in the same column and below $M$. It is possible for $M$ to form a floating plus with unit squares that are not next to it. Find the largest positive integer $k$ (depending on $n$) such that we can color some $k$ squares black in such a way that there is no black colored floating plus. Proposed by Nikola Velov

Let $(x_{n})_{n=1}^{+\infty}$ be a sequence defined recursively with $x_{n+1} = x_{n}(x_{n}-2)$ and $x_{1} = \frac{7}{2}$. Let $x_{2021} = \frac{a}{b}$, where $a,b \in \mathbb{N}$ are coprime. Show that if $p$ is a prime divisor of $a$, then either $3|p-1$ or $p=3$. Proposed by Nikola Velov