Show that $$1+\frac{n}{1!} + \frac{n^{2}}{2!} +\ldots+ \frac{n^{n}}{n!} > \frac{e^{n}}{2}$$ for every integer $n\geq 0.$

Let $n$ be a given positive integer. Sisyphus performs a sequence of turns on a board consisting of $n + 1$ squares in a row, numbered $0$ to $n$ from left to right. Initially, $n$ stones are put into square $0$, and the other squares are empty. At every turn, Sisyphus chooses any nonempty square, say with $k$ stones, takes one of these stones and moves it to the right by at most $k$ squares (the stone should say within the board). Sisyphus' aim is to move all $n$ stones to square $n$. Prove that Sisyphus cannot reach the aim in less than \[ \left \lceil \frac{n}{1} \right \rceil + \left \lceil \frac{n}{2} \right \rceil + \left \lceil \frac{n}{3} \right \rceil + \dots + \left \lceil \frac{n}{n} \right \rceil \] turns. (As usual, $\lceil x \rceil$ stands for the least integer not smaller than $x$. )

A schoolgirl forgot to write a multiplication sign between two $3$-digit numbers and wrote them as one number. This $6$-digit result proved to be $3$ times greater than the product (obtained by multiplication). Find these numbers. (A Kovaldzhi,

Let $BC=6$, $BX=3$, $CX=5$, and let $F$ be the midpoint of $\overline{BC}$. Let $\overline{AX}\perp\overline{BC}$ and $AF=\sqrt{247}$. If $AC$ is of the form $\sqrt{b}$ and $AB$ is of the form $\sqrt{c}$ where $b$ and $c$ are nonnegative integers, find $2c+3b$.

Let $ABCD$ be a square, and let $E$ and $F$ be points on sides $\overline{AB}$ and $\overline{CD}$, respectively, such that $AE:EB=AF:FD=2:1$. Let $G$ be the intersection of $\overline{AF}$ and $\overline{DE}$, and let $H$ be the intersection of $\overline{BF}$ and $\overline{CE}$. Find the ratio of the area of quadrilateral $EGFH$ to the area of square $ABCD$. [asy] size(5cm,0); draw((0,0)--(3,0)); draw((3,0)--(3,3)); draw((3,3)--(0,3)); draw((0,3)--(0,0)); draw((0,0)--(2,3)); draw((1,0)--(3,3)); draw((0,3)--(1,0)); draw((2,3)--(3,0)); label("$A$",(0,3),NW); label("$B$",(3,3),NE); label("$C$",(3,0),SE); label("$D$",(0,0),SW); label("$E$",(2,3),N); label("$F$",(1,0),S); label("$G$",(0.66666667,1),E); label("$H$",(2.33333333,2),W); [/asy]

An equilateral triangle is placed inside a larger equilateral triangle so that the region between them can be divided into three congruent trapezoids, as shown below. The side length of the inner triangle is $\tfrac23$ the side length of the larger triangle. What is the ratio of the area of one trapezoid to the area of the inner triangle? [asy] size(5cm); fill((0,0)--(2/3,1.155/3)--(4-(4-2)/3,1.155/3)--(4,0)--cycle,lightgray*0.5+mediumgray*0.5); draw((0,0)--(4,0)--(2,2*sqrt(3))--cycle); //center: 2,1.155 draw((2/3,1.155/3)--(4-(4-2)/3,1.155/3)--(2,2*sqrt(3)-0.7697)--cycle); dot((0,0)^^(4,0)^^(2,2*sqrt(3))^^(2/3,1.155/3)^^(4-(4-2)/3,1.155/3)^^(2,2*sqrt(3)-0.7697)); draw((0,0)--(2/3,1.155/3)); draw((4,0)--(4-(4-2)/3,1.155/3)); draw((2,2*sqrt(3))--(2,2*sqrt(3)-0.7697)); [/asy] $\textbf{(A) } 1:3\qquad\textbf{(B) } 3:8\qquad\textbf{(C) } 5:12\qquad\textbf{(D) } 7:16\qquad\textbf{(E) } 4:9$

The triangle $ABC$ with $AB < AC$ has an altitude $AD$. The points $E$ and $A$ lie on opposite sides of $BC$, with $E$ on the circumcircle of $ABC$. Furthermore, $AD = DE$ and $\angle ADO=\angle CDE$, where $O$ is the circumcentre of $ABC$. Determine $\angle BAC$.

Let $\mathbb R$ be the real numbers. Set $S = \{1, -1\}$ and define a function $\operatorname{sign} : \mathbb R \to S$ by \[ \operatorname{sign} (x) = \begin{cases} 1 & \text{if } x \ge 0; \\ -1 & \text{if } x < 0. \end{cases} \] Fix an odd integer $n$. Determine whether one can find $n^2+n$ real numbers $a_{ij}, b_i \in S$ (here $1 \le i, j \le n$) with the following property: Suppose we take any choice of $x_1, x_2, \dots, x_n \in S$ and consider the values \begin{align*} y_i &= \operatorname{sign} \left( \sum_{j=1}^n a_{ij} x_j \right), \quad \forall 1 \le i \le n; \\ z &= \operatorname{sign} \left( \sum_{i=1}^n y_i b_i \right) \end{align*} Then $z=x_1 x_2 \dots x_n$.

We have $2^m$ sheets of paper, with the number $1$ written on each of them. We perform the following operation. In every step we choose two distinct sheets; if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a + b$ on both sheets. Prove that after $m2^{m -1}$ steps, the sum of the numbers on all the sheets is at least $4^m$ . [i]Proposed by Abbas Mehrabian, Iran[/i]

Let $A{}$ be a point in the Cartesian plane. At each step, Ann tells Bob a number $0< a\leqslant 1$ and he then moves $A{}$ in one of the four cardinal directions, at his choice, by a distance of $a{}$. This process cotinues as long as Ann wishes. Amongst every $100$ consecutive moves, each of the four possible moves should have been made at least once. Ann's goal is to force Bob to eventually choose a point at a distance greater than $100$ from the initial position of $A{}$. Can Ann achieve her goal?

Determine all integers solutions of the equation $x + x^3 = 5y^2$.

Many distinct points are marked in the plane. A student draws all the segments determined by those points, and then draws a line [i]r[/i] that does not pass through any of the marked points, but cuts exactly $60$ drawn segments. How many segments were not cut by [i]r[/i]? Give all possibilites.

Let $f:S\to\mathbb{R}$ be the function from the set of all right triangles into the set of real numbers, defined by $f(\Delta ABC)=\frac{h}{r}$, where $h$ is the height with respect to the hypotenuse and $r$ is the inscribed circle's radius. Find the image, $Im(f)$, of the function.

Let $ABC$ be an equilateral triangle, $M$ be a point inside it, and $A',B',C'$ be the intersections of $AM,\; BM,\; CM$ with the sides of $ABC$. If $A'',\; B'',\; C''$ are the midpoints of $BC$, $CA$, $AB$, show that there is a triangle with sides $A'A''$, $B'B''$ and $C'C''$. [i]Laurențiu Panaitopol[/i]

Let $f:(0;\infty) -- (0;\infty)$ such that $f(x^y)=(f(x))^{f(y)}$. Prove $f(xy)=f(x)f(y)$ and $f(x+y)=f(x)+f(y)$ for all positive real $x,y$.

$ABC$ is a triangle, $G$ its centroid and $A',B',C'$ the midpoints of its sides $BC,CA,AB$, respectively. Prove that if the quadrilateral $AC'GB'$ is cyclic then $AB \cdot CC' = AC \cdot BB'$:

How many positive integers less than 2005 are relatively prime to 1001?

Cauchy the magician has a new card trick. He takes a standard deck(which consists of 52 cards with 13 denominations in each 4 suits) and let Schwartz to shuffle randomly. Schwartz is told to take $ m $ cards not more than $ \frac{1}{3} $ form the top of the deck. Then, Cauchy take $ 18 $ cards one by one from the top of the remaining deck and show it to Schwartz with the second card is placed in front of the first card (from Schwartz view) and so on. He ask Schwartz to memorize the $ m-th $ card when showing the cards. Let it be $ C_1 $. After that, Cauchy places the $ 18 $ cards and the $ m $ cards on the bottom of the deck with the $ m $ cards are placed lower than the $ 18 $ cards. Now, Cauchy distributes and flip the cards on the table from the top of the deck while shouting the numbers $ 10 $ until $ 1 $ with the following operation: a) When a card flipped has the number which is same as the number shouted by Cauchy, stop the distribution and continue with another set.\\ b) When $ 10 $ cards are flipped and none of the cards flipped has the number which is same as the number shouted by Cauchy, take a card from the top of the deck and place it on top of the set with backside(the site which has no value) facing up. Then continue with another set.\\ Cauchy stops when 3 sets of cards are placed. Then, he adds up all the numbers on top of each sets of cards( backside is consider $ 0 $ ). Let $ k $ be the sum. He placed another $ k $ cards to the table from the top of the remaining deck. Finally, he shows the first card on top of the remaining deck to Schwartz. Let it be $ C_2 $. Show that $ C_1 = C_2 $.

A cyclic $n$-gon is divided by non-intersecting (inside the $n$-gon) diagonals to $n-2$ triangles. Each of these triangles is similar to at least one of the remaining ones. For what $n$ this is possible?

In a tetrahedron we know that sum of angles of all vertices is $180^\circ.$ (e.g. for vertex $A$, we have $\angle BAC + \angle CAD + \angle DAB=180^\circ.$) Prove that faces of this tetrahedron are four congruent triangles.

A linear binomial $l(z) = Az + B$ with complex coefficients $A$ and $B$ is given. It is known that the maximal value of $|l(z)|$ on the segment $-1 \leq x \leq 1$ $(y = 0)$ of the real line in the complex plane $z = x + iy$ is equal to $M.$ Prove that for every $z$ \[|l(z)| \leq M \rho,\] where $\rho$ is the sum of distances from the point $P=z$ to the points $Q_1: z = 1$ and $Q_3: z = -1.$

In the given figure hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$. What is the area of $\triangle KBC$? $\textbf{(A) }6\sqrt{2}\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }9\sqrt{2}\qquad\textbf{(E) }32$ [asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80)); label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250)); label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("$G$",(-8-6*sqrt(2),-4),W); label("$32$",(-10,-8),N); label("$18$",(0,6*sqrt(2)+2),N); [/asy]

Let $(x_1,x_2,\ldots)$ be a sequence of positive real numbers satisfying ${\displaystyle \sum_{n=1}^{\infty}\frac{x_n}{2n-1}=1}$. Prove that $$ \displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{k} \frac{x_n}{k^2} \le2. $$ (Proposed by Gerhard J. Woeginger, The Netherlands)

Does there exist an irreducible two variable polynomial $f(x,y)\in \mathbb{Q}[x,y]$ such that it has only four roots $(0,1),(1,0),(0,-1),(-1,0)$ on the unit circle.

The positive integers $ A$, $ B$, $ A \minus{} B$, and $ A \plus{} B$ are all prime numbers. The sum of these four primes is $ \textbf{(A)}\ \text{even} \qquad \textbf{(B)}\ \text{divisible by }3 \qquad \textbf{(C)}\ \text{divisible by }5 \qquad \textbf{(D)}\ \text{divisible by }7 \\ \textbf{(E)}\ \text{prime}$