Determine the prime numbers $p, q, r$ with the property that: $p(p-7) + q (q-7) = r (r-7)$.

Let $\gamma$ be the incircle in the triangle $A_0A_1A_2$. For all $i\in\{0,1,2\}$ we make the following constructions (all indices are considered modulo 3): $\gamma_i$ is the circle tangent to $\gamma$ which passes through the points $A_{i+1}$ and $A_{i+2}$; $T_i$ is the point of tangency between $\gamma_i$ and $\gamma$; finally, the common tangent in $T_i$ of $\gamma_i$ and $\gamma$ intersects the line $A_{i+1}A_{i+2}$ in the point $P_i$. Prove that a) the points $P_0$, $P_1$ and $P_2$ are collinear; b) the lines $A_0T_0$, $A_1T_1$ and $A_2T_2$ are concurrent.

The figure below shows a $9\times7$ arrangement of $2\times2$ squares. Alternate squares of the grid are split into two triangles with one of the triangles shaded. Find the area of the shaded region. [asy] size(5cm); defaultpen(linewidth(.6)); fill((0,1)--(1,1)--(1,0)--cycle^^(0,3)--(1,3)--(1,2)--cycle^^(1,2)--(2,2)--(2,1)--cycle^^(2,1)--(3,1)--(3,0)--cycle,rgb(.76,.76,.76)); fill((0,5)--(1,5)--(1,4)--cycle^^(1,4)--(2,4)--(2,3)--cycle^^(2,3)--(3,3)--(3,2)--cycle^^(3,2)--(4,2)--(4,1)--cycle^^(4,1)--(5,1)--(5,0)--cycle,rgb(.76,.76,.76)); fill((0,7)--(1,7)--(1,6)--cycle^^(1,6)--(2,6)--(2,5)--cycle^^(2,5)--(3,5)--(3,4)--cycle^^(3,4)--(4,4)--(4,3)--cycle^^(4,3)--(5,3)--(5,2)--cycle^^(5,2)--(6,2)--(6,1)--cycle^^(6,1)--(7,1)--(7,0)--cycle,rgb(.76,.76,.76)); fill((2,7)--(3,7)--(3,6)--cycle^^(3,6)--(4,6)--(4,5)--cycle^^(4,5)--(5,5)--(5,4)--cycle^^(5,4)--(6,4)--(6,3)--cycle^^(6,3)--(7,3)--(7,2)--cycle^^(7,2)--(8,2)--(8,1)--cycle^^(8,1)--(9,1)--(9,0)--cycle,rgb(.76,.76,.76)); fill((4,7)--(5,7)--(5,6)--cycle^^(5,6)--(6,6)--(6,5)--cycle^^(6,5)--(7,5)--(7,4)--cycle^^(7,4)--(8,4)--(8,3)--cycle^^(8,3)--(9,3)--(9,2)--cycle,rgb(.76,.76,.76)); fill((6,7)--(7,7)--(7,6)--cycle^^(7,6)--(8,6)--(8,5)--cycle^^(8,5)--(9,5)--(9,4)--cycle,rgb(.76,.76,.76)); fill((8,7)--(9,7)--(9,6)--cycle,rgb(.76,.76,.76)); draw((0,0)--(0,7)^^(1,0)--(1,7)^^(2,0)--(2,7)^^(3,0)--(3,7)^^(4,0)--(4,7)^^(5,0)--(5,7)^^(6,0)--(6,7)^^(7,0)--(7,7)^^(8,0)--(8,7)^^(9,0)--(9,7)); draw((0,0)--(9,0)^^(0,1)--(9,1)^^(0,2)--(9,2)^^(0,3)--(9,3)^^(0,4)--(9,4)^^(0,5)--(9,5)^^(0,6)--(9,6)^^(0,7)--(9,7)); draw((0,1)--(1,0)^^(0,3)--(3,0)^^(0,5)--(5,0)^^(0,7)--(7,0)^^(2,7)--(9,0)^^(4,7)--(9,2)^^(6,7)--(9,4)^^(8,7)--(9,6)); [/asy]

In the game of Guess the Card, two players each have a $\frac{1}{2}$ chance of winning and there is exactly one winner. Sixteen competitors stand in a circle, numbered $1,2,\dots,16$ clockwise. They participate in an $4$-round single-elimination tournament of Guess the Card. Each round, the referee randomly chooses one of the remaining players, and the players pair off going clockwise, starting from the chosen one; each pair then plays Guess the Card and the losers leave the circle. If the probability that players $1$ and $9$ face each other in the last round is $\frac{m}{n}$ where $m,n$ are positive integers, find $100m+n$. [i]Proposed by Evan Chen[/i]

Let $ p \geq 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1,2,\ldots, p-1 \}$ that was not chosen before by either of the two players and then chooses an element $a_i$ from the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices have been chosen .Then the following number is computed: $$M=a_0+a_110+a_210^2+\cdots+a_{p-1}10^{p-1}= \sum_{i=0}^{p-1}a_i.10^i$$. The goal of Eduardo is to make $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy. [i]Proposed by Amine Natik, Morocco[/i]

Square $ABCD$ has side length $4$. Points $E$ and $F$ are the midpoints of sides $AB$ and $CD$, respectively. Eight $1$ by $2$ rectangles are placed inside the square so that no two of the eight rectangles overlap (see diagram). If the arrangement of eight rectangles is chosen randomly, then there are relatively prime positive integers $m$ and $n$ so that $\tfrac{m}{n}$ is the probability that none of the rectangles crosses the line segment $EF$ (as in the arrangement on the right). Find $m + n$. [asy] size(200); defaultpen(linewidth(0.8)+fontsize(10pt)); real r = 7; path square=origin--(4,0)--(4,4)--(0,4)--cycle; draw(square^^shift((r,0))*square,linewidth(1)); draw((1,4)--(1,0)^^(3,4)--(3,0)^^(0,2)--(1,2)^^(1,3)--(3,3)^^(1,1)--(3,1)^^(2,3)--(2,1)^^(3,2)--(4,2)); draw(shift((r,0))*((2,4)--(2,0)^^(0,2)--(4,2)^^(0,1)--(4,1)^^(0,3)--(2,3)^^(3,4)--(3,2))); label("A",(4,4),NE); label("A",(4+r,4),NE); label("B",(0,4),NW); label("B",(r,4),NW); label("C",(0,0),SW); label("C",(r,0),SW); label("D",(4,0),SE); label("D",(4+r,0),SE); label("E",(2,4),N); label("E",(2+r,4),N); label("F",(2,0),S); label("F",(2+r,0),S); [/asy]

(a) Prove that $x^4+3x^3+6x^2+9x+12$ cannot be expressed as product of two polynomials of degree 2 with integers coefficients. (b) $2n+1$ segments are marked on a line. Each of these segments intersects at least $n$ other segments. Prove that one of these segments intersects all other segments.

Given a set of 102 elements. Is it possible to choose 102 17-element subsets so that the intersection of any two subsets contains no more than 3 elements?

Prove that for all $\displaystyle a,b \in \left( 0 ,\frac{\pi}{4} \right)$ and $\displaystyle n \in \mathbb N^\ast$ we have \[ \frac{\sin^n a + \sin^n b}{\left( \sin a + \sin b \right)^n} \geq \frac{\sin^n 2a + \sin^n 2b}{\left( \sin 2a + \sin 2b \right)^n} . \]

Determine for what $n\ge 3$ integer numbers, it is possible to find positive integer numbers $a_1 < a_2 < ...< a_n$ such $\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}=1$ and $a_1 a_2\cdot\cdot\cdot a_n$ is a perfect square.

Let the roots of $x^2-3x+1=0$ be $r$ and $s$. Then the expression $r^2+s^2$ is: $ \textbf{(A)}\ \text{a positive integer} \qquad\textbf{(B)}\ \text{a positive fraction greater than 1}\qquad\textbf{(C)}\ \text{a positive fraction less than 1}$ $\textbf{(D)}\ \text{an irrational number}\qquad\textbf{(E)}\ \text{an imaginary number}$

Define the sequence of positive integers $f_n$ by $f_0 = 1, f_1 = 1, f_{n+2} = f_{n+1} + f_n$. Show that $f_n =\frac{ (a^{n+1} - b^{n+1})}{\sqrt5}$, where $a, b$ are real numbers such that $a + b = 1, ab = -1$ and $a > b$.

Let \(\triangle ABC\) have side lengths \(AB=30\), \(BC=32\), and \(AC=34\). Point \(X\) lies in the interior of \(\overline{BC}\), and points \(I_1\) and \(I_2\) are the incenters of \(\triangle ABX\) and \(\triangle ACX\), respectively. Find the minimum possible area of \(\triangle AI_1I_2\) as \( X\) varies along \(\overline{BC}\).

Fow which real values of $m$ does the polynomial $x^3+1995x^2-1994x+m$ have all three roots integers?

Give an isosceles triangle $ABC$ at $A$. Draw ray $Cx$ being perpendicular to $CA, BE$ perpendicular to $Cx$ ($E \in Cx$).Let $M$ be the midpoint of $BE$, and $D$ be the intersection point of $AM$ and $Cx$. Prove that $BD \perp BC$.

Let $P$ units be the increase in the circumference of a circle resulting from an increase in $\pi$ units in the diameter. Then $P$ equals: $\textbf{(A)}\ \dfrac{1}{\pi} \qquad \textbf{(B)}\ \pi \qquad \textbf{(C)}\ \dfrac{\pi^2}{2} \qquad \textbf{(D)}\ \pi^2 \qquad \textbf{(E)}\ 2\pi $

Denote $f(m) =\sum_{k=1}^m (-1)^k cos \frac{k\pi}{2 m + 1}$ For which positive integers $m$ is $f(m)$ rational?

A $m \times n$ table is filled with signs $"+"$ and $"-"$. A table is called irreducible if one cannot reduce it to the table filled with $"+"$, applying the following operations (as many times as one wishes). $a)$ It is allowed to flip all the signs in a row or in a column. Prove that an irreducible table contains an irreducible $2\times 2$ sub table. $b)$ It is allowed to flip all the signs in a row or in a column or on a diagonal (corner cells are diagonals of length $1$). Prove that an irreducible table contains an irreducible $4\times 4$ sub table.

Let $ABC$ be an acute-angled triangle with $AB<AC<BC$ and $c(O,R)$ the circumscribed circle. Let $D, E$ be points in the small arcs $AC, AB$ respectively. Let $K$ be the intersection point of $BD,CE$ and $N$ the second common point of the circumscribed circles of the triangles $BKE$ and $CKD$. Prove that $A, K, N$ are collinear if and only if $K$ belongs to the symmedian of $ABC$ passing from $A$.

Let $ x_{1},x_{2},\cdots,x_{m},y_{1},y_{2},\cdots,y_{n}$ be positive real numbers. Denote by $ X \equal{} \sum_{i \equal{} 1}^{m}x,Y \equal{} \sum_{j \equal{} 1}^{n}y.$ Prove that $ 2XY\sum_{i \equal{} 1}^{m}\sum_{j \equal{} 1}^{n}|x_{i} \minus{} y_{j}|\ge X^2\sum_{j \equal{} 1}^{n}\sum_{l \equal{} 1}^{n}|y_{i} \minus{} y_{l}| \plus{} Y^2\sum_{i \equal{} 1}^{m}\sum_{k \equal{} 1}^{m}|x_{i} \minus{} x_{k}|$

[b]a)[/b] Is it possible to find a function $f:\mathbb N^2\to\mathbb N$ such that for every function $g:\mathbb N\to\mathbb N$ and positive integer $M$ there exists $n\in\mathbb N$ such that set $\left\{k\in \mathbb N : f(n,k)=g(k)\right\}$ has at least $M$ elements? [b]b)[/b] Is it possible to find a function $f:\mathbb N^2\to\mathbb N$ such that for every function $g:\mathbb N\to\mathbb N$ there exists $n\in \mathbb N$ such that set $\left\{k\in\mathbb N : f(n,k)=g(k)\right\}$ has an infinite number of elements?

Let $a_1, a_2, \dots$ be a sequence of real numbers satisfying $a_{i+j} \leq a_i+a_j$ for all $i,j=1,2,\dots$. Prove that \[ a_1 + \frac{a_2}{2} + \frac{a_3}{3} + \cdots + \frac{a_n}{n} \geq a_n \] for each positive integer $n$.

Three sidelines of on acute-angled triangle are drawn on the plane. Fyodor wants to draw the altitudes of this triangle using a ruler and a compass. Ivan obstructs him using an eraser. For each move Fyodor may draw one line through two markeed points or one circle centered at a marked point and passing through another marked point. After this Fyodor may mark an arbitrary number of points (the common points of drawn lines, arbitrary points on the drawn lines or arbitrary points on the plane). For each move Ivan erases at most three of marked point. (Fyodor may not use the erased points in his constructions but he may mark them for the second time). They move by turns, Fydors begins. Initially no points are marked. Can Fyodor draw the altitudes?

In the beginnig, each unit square of a $m\times n$ board is colored white. We are supposed to color all the squares such that one of two adjacent squares having a common side is black and the other is white. At each move, we choose a $2 \times 2$ square, and we color each of $4$ squares inversely such that if the square is black then it is colored white or vice versa. For which of the following ordered pair $(m, n)$, can the board be colored in this way? $ \textbf{a)}\ (3,3) \qquad\textbf{b)}\ (2,6) \qquad\textbf{c)}\ (4,8) \qquad\textbf{d)}\ (5,5) \qquad\textbf{e)}\ \text{None of above} $

You flip a fair coin which results in heads ($\text{H}$) or tails ($\text{T}$) with equal probability. What is the probability that you see the consecutive sequence $\text{THH}$ before the sequence $\text{HHH}$?