Let $O$ be the circumcenter of the triangle $ABC, A'$ be a point symmetric of $A$ wrt line $BC, X$ is an arbitrary point on the ray $AA'$ ($X \ne A$). Angle bisector of angle $BAC$ intersects the circumcircle of triangle $ABC$ at point $D$ ($D \ne A$). Let $M$ be the midpoint of the segment $DX$. A line passing through point $O$ parallel to $AD$, intersects $DX$ at point $N$. Prove that angles $BAM$ and $CAN$ angles are equal.

In a football tournament, $n$ teams play one round ($n \vdots 2$). In each round should play $n / 2$ pairs of teams that have not yet played. Schedule of each round takes place before its holding. For which smallest natural $k$ such that the following situation is possible: after $k$ tours, making a schedule of $k + 1$ rounds already is not possible, i.e. these $n$ teams cannot be divided into $n / 2$ pairs, in each of which there are teams that have not played in the previous $k$ rounds. PS. The 3 vertical dots notation in the first row, I do not know what it means.

Find all triples $(p, x, y)$ consisting of a prime number $p$ and two positive integers $x$ and $y$ such that $x^{p -1} + y$ and $x + y^ {p -1}$ are both powers of $p$. Proposed by Belgium

A prime number $p> 3$ is given. Prove that integers less than $p$, it is possible to divide them into two non-empty sets such that the sum of the numbers in the first set will be congruent modulo p to the product of the numbers in the second set.

For a sequence $x_1,x_2,\ldots,x_n$ of real numbers, we define its $\textit{price}$ as \[\max_{1\le i\le n}|x_1+\cdots +x_i|.\] Given $n$ real numbers, Dave and George want to arrange them into a sequence with a low price. Diligent Dave checks all possible ways and finds the minimum possible price $D$. Greedy George, on the other hand, chooses $x_1$ such that $|x_1 |$ is as small as possible; among the remaining numbers, he chooses $x_2$ such that $|x_1 + x_2 |$ is as small as possible, and so on. Thus, in the $i$-th step he chooses $x_i$ among the remaining numbers so as to minimise the value of $|x_1 + x_2 + \cdots x_i |$. In each step, if several numbers provide the same value, George chooses one at random. Finally he gets a sequence with price $G$. Find the least possible constant $c$ such that for every positive integer $n$, for every collection of $n$ real numbers, and for every possible sequence that George might obtain, the resulting values satisfy the inequality $G\le cD$. Proposed by Georgia

Given an acute triangle $ABC, H$ is the foot of the altitude drawn from the point $A$ on the line $BC, P$ and $K \ne H$ are arbitrary points on the segments $AH$ and$ BC$ respectively. Segments $AC$ and $BP$ intersect at point $B_1$, lines $AB$ and $CP$ at point $C_1$. Let $X$ and $Y$ be the projections of point $H$ on the lines $KB_1$ and $KC_1$, respectively. Prove that points $A, P, X$ and $Y$ lie on one circle.

Let $A$ and $B$ be two sets of real numbers. Suppose that the elements of the set $AB = \{ab: a\in A, b\in B\}$ form a finite arithmetic progression. Prove that one of these sets contains no more than three elements

Find all functions $f: R \to R$ such that $f(x)f(yf(x)-1)=x^2f(y)-f(x)$ for all real $x ,y$

The set $M$ consists of $n$ points on the plane and satisfies the conditions: $\bullet$ there are $7$ points in the set $M$, which are vertices of a convex heptagon, $\bullet$ for arbitrary five points with $M$, which are vertices of a convex pentagon, there is a point that also belongs to $M$ and lies inside this pentagon. Find the smallest possible value that $n$ can take .

Determine all pairs $(x, y)$ of positive integers such that \[\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1.\] Proposed by Titu Andreescu, USA

Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$. (Here we always assume that an angle bisector is a ray.) Proposed by Sergey Berlov, Russia

For a given natural $n$, we consider the set $A\subset \{1,2, ..., n\}$, which consists of at least $\left[\frac{n+1}{2}\right]$ items. Prove that for $n \ge 2015$ the set $A$ contains a three-element arithmetic sequence.