In a triangle $ABC$ with a right angle at $C$, the angle bisector $AL$ (where $L$ is on segment $BC$) intersects the altitude $CH$ at point $K$. The bisector of angle $BCH$ intersects segment $AB$ at point $M$. Prove that $CK=ML$

Does there exist a positive integer $n$ such that all its digits (in the decimal system) are greather than 5, while all the digits of $n^2$ are less than 5?

Let $n>1$ be a given integer. The Mint issues coins of $n$ different values $a_1, a_2, ..., a_n$, where each $a_i$ is a positive integer (the number of coins of each value is unlimited). A set of values $\{a_1, a_2,..., a_n\}$ is called lucky, if the sum $a_1+ a_2+...+ a_n$ can be collected in a unique way (namely, by taking one coin of each value). (a) Prove that there exists a lucky set of values $\{a_1, a_2, ..., a_n\}$ with $$a_1+ a_2+...+ a_n < n \cdot 2^n.$$(b) Prove that every lucky set of values $\{a_1, a_2,..., a_n\}$ satisfies $$a_1+ a_2+...+ a_n >n \cdot 2^{n-1}.$$ Proposed by Ilya Bogdanov

Given three positive real numbers $a,b,c$ such that following holds $a^2=b^2+bc$, $b^2=c^2+ac$ Prove that $\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$.

There is an empty table with $2^{100}$ rows and $100$ columns. Alice and Eva take turns filling the empty cells of the first row of the table, Alice plays first. In each move, Alice chooses an empty cell and puts a cross in it; Eva in each move chooses an empty cell and puts a zero. When no empty cells remain in the first row, the players move on to the second row, and so on (in each new row Alice plays first). The game ends when all the rows are filled. Alice wants to make as many different rows in the table as possible, while Eva wants to make as few as possible. How many different rows will be there in the table if both follow their best strategies? Proposed by Denis Afrizonov

In convex pentagon $ABCDE$ points $A_1$, $B_1$, $C_1$, $D_1$, $E_1$ are intersections of pairs of diagonals $(BD, CE)$, $(CE, DA)$, $(DA, EB)$, $(EB, AC)$ and $(AC, BD)$ respectively. Prove that if four of quadrilaterals $AB_{1}A_{1}B$, $BC_{1}B_{1}C$, $CD_{1}C_{1}D$, $DE_{1}D_{1}E$ and $EA_{1}E_{1}A$ are cyclic then the fifth one is also cyclic.