In how many ways can the integers from $1$ to $n$ be ordered subject to the condition that, except for the first integer on the left, every integer differs by $1$ from some integer to the left of it?

Let $ABCD$ be a trapezoid such that $|AC|=8$, $|BD|=6$, and $AD \parallel BC$. Let $P$ and $S$ be the midpoints of $[AD]$ and $[BC]$, respectively. If $|PS|=5$, find the area of the trapezoid $ABCD$.

Let $P(z)= z^n + c_1 z^{n-1} + c_2 z^{n-2} + \cdots + c_n$ be a polynomial in the complex variable $z$, with real coefficients $c_k$. Suppose that $|P(i)| < 1$. Prove that there exist real numbers $a$ and $b$ such that $P(a + bi) = 0$ and $(a^2 + b^2 + 1)^2 < 4 b^2 + 1$.

For non negative reals $a,b$ we know that $a^2+a+b^2\ge a^4+a^3+b^4$. Prove that $$\frac{1-a^4}{a^2}\ge \frac{b^2-1}{b}$$

Let $n \geq 2$ be a positive integer. In a mathematics competition, there are $n+1$ students, with one of them being a hacker. The competition is conducted as follows: each receives the same problem with an open-ended answer, has 5 minutes to give their own answer, after which all answers are submitted simultaneously, the correct answer is announced, then they receive a new problem, and so on. The hacker cheats by using spy cameras to see the answers of the other participants. A correct answer gives 1 point, while a wrong answer gives -1 point to everyone except the hacker; for him, it's 0 points because he managed to hack the scoring system. Prove that regardless of the total number of problems, if at some point the hacker is ahead of the second-place contestant by at least $2^{n-2} + 1$ points, then he has a strategy to ensure he will be the sole winner by the end of the competition.

Prove that for any four positive real numbers $ a$, $ b$, $ c$, $ d$ the inequality \[ \frac {(a \minus{} b)(a \minus{} c)}{a \plus{} b \plus{} c} \plus{} \frac {(b \minus{} c)(b \minus{} d)}{b \plus{} c \plus{} d} \plus{} \frac {(c \minus{} d)(c \minus{} a)}{c \plus{} d \plus{} a} \plus{} \frac {(d \minus{} a)(d \minus{} b)}{d \plus{} a \plus{} b}\ge 0\] holds. Determine all cases of equality. [i]Author: Darij Grinberg (Problem Proposal), Christian Reiher (Solution), Germany[/i]

Some positive integer $n$ is written on the board. Andrey and Petya is playing the following game. Andrey finds all the different representations of the number n as a product of powers of prime numbers (values degrees greater than 1), in which each factor is greater than all previous or equal to the previous one. Petya finds all different representations of the number $n$ as a product of integers greater than $1$, in which each factor is divisible by all the previous factors. The one who finds more performances wins, if everyone finds the same number of representations, the game ends in a draw. Find all positive integers $n$ for which the game will end in a draw. Note. The representation of the number $n$ as a product is also considered a representation consisting of a single factor $n$.

Let $R$ be the set of points $(x, y)$ such that $x$ and $y$ are positive, $x + y$ is at most 2013, and \[ \lceil x \rceil \lfloor y \rfloor = \lfloor x \rfloor \lceil y \rceil. \] Compute the area of set $R$. Recall that $\lfloor a \rfloor$ is the greatest integer that is less than or equal to $a$, and $\lceil a \rceil$ is the least integer that is greater than or equal to $a$.

Five girls have a little store that opens from Monday through Friday. Since two people are always enough for taking care of it, they decide to do a work plan for the week, specifying who will work each day, and fulfilling the following conditions: a) Each girl will work exactly two days a week b) The 5 assigned couples for the week must be different In how many ways can the girls do the work plan?

Prove The Following identity: $ \sum_{j \equal{} 0}^n \left ({3n \plus{} 2 \minus{} j \choose j}2^j \minus{} {3n \plus{} 1 \minus{} j \choose j \minus{} 1}2^{j \minus{} 1}\right ) \equal{} 2^{3n}$. The Second term on left hand side is to be regarded zero for j=0.

Let $ABC$ be an equilateral triangle and $D$ a point on side $AC$. Let $E$ be a point on $BC$ such that $DE \perp BC$, $F$ on $AB$ such that $EF \perp AB$, and $G$ on $AC$ such that $FG \perp AC$. Lines $FG$ and $DE$ intersect in $P$. If $M$ is the midpoint of $BC$, show that $BP$ bisects $AM$.

Let $ G$ be a graph with $ 2n$ vertexes and $ 2n(n\minus{}1)$ edges.If we color some edge to red,then vertexes,which are connected by this edge,must be colored to red too. But not necessary that all edges from the red vertex are red. Prove that it is possible to color some vertexes and edges in $ G$,such that all red vertexes has exactly $ n$ red edges.

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]

Let $\{x_n\}_{n=1}^\infty$ be a sequence such that $x_1=25$, $x_n=\operatorname{arctan}(x_{n-1})$. Prove that this sequence has a limit and find it.

A point $T$ is chosen inside a triangle $ABC$. Let $A_1$, $B_1$, and $C_1$ be the reflections of $T$ in $BC$, $CA$, and $AB$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_1B_1C_1$. The lines $A_1T$, $B_1T$, and $C_1T$ meet $\Omega$ again at $A_2$, $B_2$, and $C_2$, respectively. Prove that the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent on $\Omega$. [i]Proposed by Mongolia[/i]

In Hidro planet were living $2008^2$ hydras some time ago. One of them had 1 tentacle, other 2 and so on to the last with $2008^2$ tentacles. Let $t(H)$ be the number of tentacles of hydra $H$. The pairing of $H_1$ and $H_2$ (where $t(H_1) < t(H_2)$) is a new hydra with $t(H_2)-t(H_1)+8$ tentacles, in case of $t(H_1)=t(H_2)$ they die. An expedition found in Hidro the last hydra with 23 tentacles. That could be true ?