Prove that $\arcsin x+\arccos x=\frac{\pi}{2}$, where $x\in[-1,1]$.

Prove that $$\int \limits_0^{\infty} e^{3t}\frac{4e^{4t}(3t - 1) + 2e^{2t}(15t - 17) + 18(1 - t)}{\left(1 + e^{4t} - e^{2t}\right)^2}=12\sum \limits_{k=0}^{\infty}\frac{(-1)^k}{(2k + 1)^2}-10$$

Moor owns $3$ shirts, one each of black, red, and green. Moor also owns $3$ pairs of pants, one each of white, red, and green. Being stylish, he decides to wear an outfit consisting of one shirt and one pair of pants that are different colors. How many combinations of shirts and pants can Moor choose?

A circle cuts each side of a rhombus twice thus dividing each side into three segments. Let us go around the perimeter of the rhombus clockwise beginning at a vertex and paint these segments successively in red, white and blue. Prove that the sum of lengths of the blue segments equals that of the red ones. (V Proizvolov)

Is there a triangle with sides of integer lengths such that the length of the shortest side is $ 2007$ and that the largest angle is twice the smallest?

Let $n$ be a positive integer. Tasty and Stacy are given a circular necklace with $3n$ sapphire beads and $3n$ turquoise beads, such that no three consecutive beads have the same color. They play a cooperative game where they alternate turns removing three consecutive beads, subject to the following conditions: [list] [*]Tasty must remove three consecutive beads which are turquoise, sapphire, and turquoise, in that order, on each of his turns. [*]Stacy must remove three consecutive beads which are sapphire, turquoise, and sapphire, in that order, on each of her turns. [/list] They win if all the beads are removed in $2n$ turns. Prove that if they can win with Tasty going first, they can also win with Stacy going first. [i]Yannick Yao[/i]

Let $a, b, c$ be positive real numbers such that $abc = 8$. Prove that $$\frac{a-2}{a+1}+\frac{b-2}{b+1}+\frac{c-2}{c+1} \le 0$$

Let $A_1A_2 \dots A_n$ a cyclic $n$-agon with center $O$ and $P$, $Q$ being two isogonal conjugates of it (i.e, $\angle PA_{i+1}A_i = \angle QA_{i+1}A_{i+2}$ for all $i$). Let $P_i$ be the circumcenter of $\triangle PA_iA_{i+1}$ and $Q_i$ the circumcenter of $\triangle QA_iA_{i+1}$ for all $i$. Prove that: $a) ~P_1P_2 \dots P_n$ and $Q_1Q_2 \dots Q_n$ are cyclic, with centers $O_P$ and $O_Q$, respectively. $b)~O, O_P$ and $O_Q$ are collinears. $c)~O_PO_Q \mid \mid PQ.$ Remark: indices are taken modulo $n$.

Let $n=3k$ be a positive integer (with $k\geq 2$). An equilateral triangle is divided in $n^2$ unit equilateral triangles with sides parallel to the initial, forming a grid. We will call "trapezoid" the trapezoid which is formed by three equilateral triangles (one base is equal to one and the other is equal to two). We colour the points of the grid with three colours (red, blue and green) such that each two neighboring points have different colour. Finally, the colour of a "trapezoid" will be the colour of the midpoint of its big base. Find the number of all "trapezoids" in the grid (not necessarily disjoint) and determine the number of red, blue and green "trapezoids".

Suppose that $x$, $y$, $z$ are real numbers satisfying \[x+y+z=12,\qquad\text{and}\qquad x^2+y^2+z^2=54.\] Prove that:[list](a) Each of the numbers $xy$, $yz$, $zx$ is at least $9$, but at most $25$. (b) One of the numbers $x$, $y$, $z$ is at most $3$, and another one is at least $5$.[/list]

Find the number of ordered pairs $(x,y)$, where $x$ and $y$ are both integers between $1$ and $9$, inclusive, such that the product $x\times y$ ends in the digit $5$. [i]Proposed by Andrew Wen[/i]

In a polygon, every external angle is one sixth of its corresponding internal angle. How many sides does the polygon have?

Let $N$ denote the number of ordered 2011-tuples of positive integers $(a_1,a_2,\ldots,a_{2011})$ with $1\le a_1,a_2,\ldots,a_{2011} \le 2011^2$ such that there exists a polynomial $f$ of degree $4019$ satisfying the following three properties: [list] [*] $f(n)$ is an integer for every integer $n$; [*] $2011^2 \mid f(i) - a_i$ for $i=1,2,\ldots,2011$; [*] $2011^2 \mid f(n+2011) - f(n)$ for every integer $n$. [/list] Find the remainder when $N$ is divided by $1000$. [i]Victor Wang[/i]

Prove that for every positive integer $m$ we can find a finite set $S$ of points in the plane, such that given any point $A$ of $S$, there are exactly $m$ points in $S$ at unit distance from $A$.

Let $ABC$ be a triangle with $AB < AC$. Let $D$ and $E$ be on the lines $CA$ and $BA$, respectively, such that $CD = AB$, $BE = AC$, and $A$, $D$ and $E$ lie on the same side of $BC$. Let $I$ be the incenter of triangle $ABC$, and let $H$ be the orthocenter of triangle $BCI$. Show that $D$, $E$, and $H$ are collinear.

$\indent$ For a positive integer $n$, let $S_n$ be the set of bijective functions from $\{1,2,\dots ,n\}$ to itself. For a pair of positive integers $(a,b)$ such that $1 \leq a <b \leq n$, and for a permutation $\sigma \in S_n$, we say the pair $(a,b)$ is [i][u]expanding[/u][/i] for $\sigma$ if $|\sigma (a)- \sigma(b)| \geq |a-b|$ $\indent$ [b](a)[/b] Is it true that for all integers $n > 1$, there exists $\sigma \in S_n$ so that the number of pairs $(a,b)$ that are expanding for permutation $\sigma$ is less than $1000n\sqrt n$ ? $\indent$ [b](b)[/b] Does there exist a positive integer $n>1$ and a permutation $\sigma \in S_n$ so that the number of pairs $(a,b)$ that are expanding for the permutation $\sigma$ is less than $\frac{n\sqrt n}{1000}$?

$$Problem 1 $$The set S = {1,2,3,...,2006} is partitioned into two disjoint subsets A and B such that: (i) 13 ∈ A; (ii) if a ∈ A, b ∈ B, a+b ∈ S, then a+b ∈ B; (iii) if a ∈ A, b ∈ B, ab ∈ S, then ab ∈ A. Determine the number of elements of A

For the positive integer $k$ we denote by the $a_n$ , the $k$ from the left digit in the decimal notation of the number $2^n$ ($a_n = 0$ if in the notation of the number $2^n$ less than the digits). Consider the infinite decimal fraction $a = \overline{0, a_1a_2a_3...}$. Prove that the number $a$ is irrational.

You are given $25$ pieces of cheese of different weights. Is it always possible to cut one of the pieces into two parts and put the $26$ pieces in two packets so that $\bullet$ each packet contains $13$ pieces; $\bullet$ the total weights of the two packets are equal; $\bullet$ the two parts of the piece which has been cut are in different packets? (VL Dolnikov)

Prove that there are infinitely many positive $n$ that for all prime divisors $p$ of $n^2 + 3, \exists 0 \leq k \leq \sqrt{n}$ and $p \mid k^2+3$

Let $x$ and $y$ are real numbers such that $$\begin{cases} x + y = 2 \\ x^3 + y^3 = 3\end{cases} $$ What is $x^2+y^2$?

In a half-plane, bounded by a line \(r\), equilateral triangles \(S_1, S_2, \ldots, S_n\) are placed, each with one side parallel to \(r\), and their opposite vertex is the point of the triangle farthest from \(r\). For each triangle \(S_i\), let \(T_i\) be its medial triangle. Let \(S\) be the region covered by triangles \(S_1, S_2, \ldots, S_n\), and let \(T\) be the region covered by triangles \(T_1, T_2, \ldots, T_n\). Prove that \[\text{area}(S) \leq 4 \cdot \text{area}(T).\]

Show that for any natural number $n$, the number $A = [\frac{n + 3}{4}] + [ \frac{n + 5}{4} ] + [\frac{n}{2} ] +n^2 + 3n + 3$ is a perfect square. ($[x]$ denotes the integer part of the real number x.)

$n$ people (with names $1,2,\dots,n$) are around a table. Some of them are friends. At each step 2 friend can change their place. Find a necessary and sufficient condition for friendship relation between them that with these steps we can always reach to all of posiible permutations.

Let $ABCD$ be a parallelogram with $\angle BAD < 90^{\circ}$. A circle tangent to sides $\overline{DA}$, $\overline{AB}$, and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ$, as shown. Suppose that $AP = 3$, $PQ = 9$, and $QC = 16$. Then the area of $ABCD$ can be expressed in the form $m\sqrt n$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$. [asy] defaultpen(linewidth(0.6)+fontsize(11)); size(8cm); pair A,B,C,D,P,Q; A=(0,0); label("$A$", A, SW); B=(6,15); label("$B$", B, NW); C=(30,15); label("$C$", C, NE); D=(24,0); label("$D$", D, SE); P=(5.2,2.6); label("$P$", (5.8,2.6), N); Q=(18.3,9.1); label("$Q$", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy]