Find all pairs of integers $(x,y)$ satisfying the following equality $8x^2y^2 + x^2 + y^2 = 10xy$

$ABC$ is a triangle with an area of $1$ square meter. Given the point $D$ on $BC$, point $E$ on $CA$, point $F$ on $AB$, such that quadrilateral $AFDE$ is cyclic. Prove that the area of $DEF \le \frac{EF^2}{4 AD^2}$. [i](holmes)[/i]

At ARML, Santa is asked to give rubber duckies to $2013$ students, one for each student. The students are conveniently numbered $1,2,\cdots,2013$, and for any integers $1 \le m < n \le 2013$, students $m$ and $n$ are friends if and only if $0 \le n-2m \le 1$. Santa has only four different colors of duckies, but because he wants each student to feel special, he decides to give duckies of different colors to any two students who are either friends or who share a common friend. Let $N$ denote the number of ways in which he can select a color for each student. Find the remainder when $N$ is divided by $1000$. [i]Proposed by Lewis Chen[/i]

Triangle $ABC$ is inscribed in a unit circle $\omega$. Let $H$ be its orthocenter and $D$ be the foot of the perpendicular from $A$ to $BC$. Let $\triangle XY Z$ be the triangle formed by drawing the tangents to $\omega$ at $A, B, C$. If $\overline{AH} = \overline{HD}$ and the side lengths of $\triangle XY Z$ form an arithmetic sequence, the area of $\triangle ABC$ can be expressed in the form $\tfrac{p}{q}$ for relatively prime positive integers $p, q$. What is $p + q$?

A secant line splits a circle into two segments. Inside those segments, one draws two squares such that both squares has two corners on a secant line and two on the circumference. The ratio of the square's side lengths is $5:9$. Compute the ratio of the secant line versus circle radius.

Consider the numbers$$A= \frac{1}{4}\cdot \frac{3}{6}\cdot \frac{5}{8}\cdot ...\frac{595}{598}\cdot \frac{597}{600}$$and$$B= \frac{2}{5}\cdot \frac{4}{7}\cdot \frac{6}{9}\cdot ...\frac{596}{599}\cdot \frac{598}{601}$$. Prove that: (a) $A < B$, (b) $A < \frac{1}{5990}$

Seven cyclists follow one another, in a line, on a narrow way. By the end of the training, each cyclist complains about the style of driving of the one in front of him. They agree to rearrange themselves the next day, so that no cyclist would follow the same cyclist he follows the first day. How many such rearrangements are possible?

What is $\frac{2+4+6}{1+3+5}-\frac{1+3+5}{2+4+6}$? $\textbf{(A) }-1\qquad\textbf{(B) }\frac5{36}\qquad\textbf{(C) }\frac7{12}\qquad\textbf{(D) }\frac{49}{20}\qquad\textbf{(E) }\frac{43}3$

Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function with the property that for any $a,b \in \mathbb{R},$ $a<b,$ there are $c_1,c_2 \in [a,b],$ $c_1 \le c_2$ such that $f(c_1)= \min_{x \in [a,b]} f(x)$ and $f(c_2)= \max_{x \in [a,b]} f(x).$ Prove that $f$ is increasing.

A circle $\omega$ with radius $1$ is given. A collection $T$ of triangles is called [i]good[/i], if the following conditions hold: [list=1] [*] each triangle from $T$ is inscribed in $\omega$; [*] no two triangles from $T$ have a common interior point. [/list] Determine all positive real numbers $t$ such that, for each positive integer $n$, there exists a good collection of $n$ triangles, each of perimeter greater than $t$.

The infinite geometric series of positive reals $a_1, a_2, \dots$ satisfies \[ 1 = \sum_{n=1}^\infty a_n = -\frac{1}{2013} + \sum_{n=1}^{\infty} \text{GM}(a_1, a_2, \dots, a_n) = \frac{1}{N} + a_1 \] where $\text{GM}(x_1, x_2, \dots, x_k) = \sqrt[k]{x_1x_2\cdots x_k}$ denotes the geometric mean. Compute $N$. [i]Proposed by Aaron Lin[/i]

On square $ABCD,$ points $E,F,G,$ and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34.$ Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P,$ and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$. [asy] size(200); defaultpen(linewidth(0.8)+fontsize(10.6)); pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(120))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", (H+E)/2,fontsize(15)); label("$x$", (E+F)/2,fontsize(15)); label("$y$", (G+F)/2,fontsize(15)); label("$z$", (H+G)/2,fontsize(15)); label("$w:x:y:z=269:275:405:411$",(sqrt(850)/2,-4.5),fontsize(11)); [/asy]

On the plane are three straight lines $\ell_1, \ell_2,\ell_3$, forming a triangle, and the point $O$ is marked, the center of the circumscribed circle of this triangle. For an arbitrary point X of the plane, we denote by $X_i$ the point symmetric to the point X with respect to the line $\ell_i, i = 1,2,3$. a) Prove that for an arbitrary point $M$ the straight lines connecting the midpoints of the segments $O_1O_2$ and $M_1M_2, O_2O_3$ and $M_2M_3, O_3O_1$ and $M_3M_1$ intersect at one point, b) where can this intersection point lie?

Consider an arc of a planar curve; let the radius of curvature at any point of the arc be a differentiable function of the arc length and its derivative be everywhere different from zero; moreover, let the total curvature be less than $ \frac{\pi}{2}$. Let $ P_1,P_2,P_3,P_4,P_5$ and $ P_6$ be any points on this arc, subject to the only condition that the radius of curvature at $ P_k$ is greater than at $ P_j$ if $ j<k$. Prove that the radius of the circle passing through the points $ P_1,P_3$ and $ P_5$ is less than the radius of the circle through $ P_2,P_4$ and $ P_6$

Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$, such that $$f(xy+f(x^2))=xf(x+y)$$ for all reals $x, y$.

Points $A', B', C'$ lie on sides $BC, CA, AB$ of triangle $ABC.$ for a point $X$ one has $\angle AXB =\angle A'C'B' + \angle ACB$ and $\angle BXC = \angle B'A'C' +\angle BAC.$ Prove that the quadrilateral $XA'BC'$ is cyclic.

In a building there are 6018 desks in 2007 rooms, and in every room there is at least one desk. Every room can be cleared dividing the desks in the oher rooms such that in every room is the same number of desks. Find out what methods can be used for dividing the desks initially.

Which of the following triangles cannot exist? $\textbf{(A)}\ \text{An acute isosceles triangle}$ $\textbf{(B)}\ \text{An isosceles right triangle}$ $\textbf{(C)}\ \text{An obtuse right triangle}$ $\textbf{(D)}\ \text{A scalene right triangle}$ $\textbf{(E)}\ \text{A scalene obtuse triangle}$

Prove that there doesn't exist any positive integer $n$ such that $2n^2+1,3n^2+1$ and $6n^2+1$ are perfect squares.

Let $a$ be a real number greater than $1$ such that $\frac{20a}{a^2+1} = \sqrt{2}$. Find $\frac{14a}{a^2 - 1}$.

For $n$ an odd positive integer, the unit squares of an $n\times n$ chessboard are coloured alternately black and white, with the four corners coloured black. A [i]tromino[/i] is an $L$-shape formed by three connected unit squares. $(a)$ For which values of $n$ is it possible to cover all the black squares with non-overlapping trominoes lying entirely on the chessboard? $(b)$ When it is possible, find the minimum number of trominoes needed.

Let $a_1>0$ and for $n \ge 1$ define \[a_{n+1}=a_n+\frac{1}{a_1+a_2+\dots+a_n}.\] Prove that \[\lim_{n \to \infty} \frac{a_n^2}{\ln n}=2.\]

An equilateral triangle is drawn with a side of length $ a$. A new equilateral triangle is formed by joining the midpoints of the sides of the first one. Then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. The limit of the sum of the perimeters of all the triangles thus drawn is: $ \textbf{(A)}\ \text{Infinite} \qquad\textbf{(B)}\ 5\frac {1}{4}a \qquad\textbf{(C)}\ 2a \qquad\textbf{(D)}\ 6a \qquad\textbf{(E)}\ 4\frac {1}{2}a$

Given an isosceles triangle $ABC$ with a vertex at the point $B$. Based on $AC$, an arbitrary point $D $ is selected, different from the vertices $A$ and $C $. On the line $AC $ select the point $E $ outside the segment $AC$, for which $AE = CD$. Prove that the perimeter $\Delta BDE$ is larger than the perimeter $\Delta ABC$.

Prove that there exists a positive integer $k$ such that $k\cdot2^n+1$ is composite for every integer $n$.