a) Let $a\in \mathbb{R}$ and $f \colon \mathbb{R} \to \mathbb{R}$ be a continuous function for which there exists an antiderivative $F \colon \mathbb{R} \to \mathbb{R} $, such that $F(x)+a\cdot f(x) \geq 0$, for any $x \in \mathbb{R}$, and$ \lim_{|x| \to \infty} \frac{F(x)}{e^{|\alpha \cdot x|}}=0$ holds for any $\alpha \in \mathbb{R}^*$. Prove that $F(x) \geq 0$ for all $x \in \mathbb{R}$. b) Let $n\geq 2$ be a positive integer, $g \in \mathbb{R}[X]$, $g = X^n + a_1X^{n-1}+ \dots + a_{n-1}X+a_n$ be a polynomial with all of its roots being real, and $f \colon \mathbb{R} \to \mathbb{R}$ a polynomial function such that $f(x)+a_1\cdot f'(x)+a_2\cdot f^{(2)}(x)+\dots+a_n\cdot f^{(n)}(x) \geq 0$ for any $x \in \mathbb{R}$. Prove that $f(x) \geq 0$ for all $x \in \mathbb{R}$.

Prove that the six sides of any tetrahedron can be the sides of a convex hexagon.

Does there exist a natural number $n$, for which $n.2^{2^{2014}}-81-n$ is a perfect square?

The table below gives the percent of students in each grade at Annville and Cleona elementary schools: \[\begin{tabular}{rccccccc} & \textbf{\underline{K}} & \textbf{\underline{1}} & \textbf{\underline{2}} & \textbf{\underline{3}} & \textbf{\underline{4}} & \textbf{\underline{5}} & \textbf{\underline{6}} \\ \textbf{Annville:} & 16\% & 15\% & 15\% & 14\% & 13\% & 16\% & 11\% \\ \textbf{Cleona:} & 12\% & 15\% & 14\% & 13\% & 15\% & 14\% & 17\% \end{tabular}\] Annville has 100 students and Cleona has 200 students. In the two schools combined, what percent of the students are in grade 6? $\text{(A)}\ 12\% \qquad \text{(B)}\ 13\% \qquad \text{(C)}\ 14\% \qquad \text{(D)}\ 15\% \qquad \text{(E)}\ 28\%$

Opposite sides of a convex hexagon are parallel. Let's call the "height" of such a hexagon a segment with ends on straight lines containing opposite sides and perpendicular to them. Prove that a circle can be circumscribed around this hexagon if and only if its "heights" can be parallelly moved so that they form a triangle. (A. Zaslavsky)

Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that$$\frac{a^2-b^2}{a+bc}+\frac{b^2-c^2}{b+ca}+\frac{c^2-a^2}{c+ab}\leq a+b+c-3.$$

Given a regular $n$-gon $A_{1}A_{2}...A_{n}$ (with $n\geq 3$) in a plane. How many triangles of the kind $A_{i}A_{j}A_{k}$ are obtuse ?

Express the sum of $99$ terms$$\frac{1\cdot 4}{2\cdot 5}+\frac{2\cdot 7}{5\cdot 8}+\ldots +\frac{k(3k+1 )}{(3k-1)(3k+2)}+\ldots +\frac{99\cdot 298}{296\cdot 299}$$ as an irreducible fraction.

Compute \[\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}}}\]

$(SWE 6)$ Prove that there are infinitely many positive integers that cannot be expressed as the sum of squares of three positive integers.

Sir Alex plays the following game on a row of 9 cells. Initially, all cells are empty. In each move, Sir Alex is allowed to perform exactly one of the following two operations: [list=1] [*] Choose any number of the form $2^j$, where $j$ is a non-negative integer, and put it into an empty cell. [*] Choose two (not necessarily adjacent) cells with the same number in them; denote that number by $2^j$. Replace the number in one of the cells with $2^{j+1}$ and erase the number in the other cell. [/list] At the end of the game, one cell contains $2^n$, where $n$ is a given positive integer, while the other cells are empty. Determine the maximum number of moves that Sir Alex could have made, in terms of $n$. [i]Proposed by Warut Suksompong, Thailand[/i]

Compute how many permutations of the numbers $1,2,\dots,8$ have no adjacent numbers that sum to $9$.

Alice asserts that after her recent visit to Addis-Ababa she now has spent the New Year inside every possible hemisphere of Earth except one. What is the minimal number of places where Alice has spent the New Year? Note: we consider places of spending the New Year to be points on the sphere. A point on the border of a hemisphere does not lie inside the hemisphere. Ilya Dumansky, Roman Krutovsky

$10$ students take the Analysis Round. The average score was a $3$ and the high score was a $7$. If no one got a $0$, what is the maximum number of students that could have achieved the high score?

What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than $2019$? $\textbf{(A) } 11 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 22 \qquad\textbf{(D) } 23 \qquad\textbf{(E) } 27$

For the NEMO, Kevin needs to compute the product \[ 9 \times 99 \times 999 \times \cdots \times 999999999. \] Kevin takes exactly $ab$ seconds to multiply an $a$-digit integer by a $b$-digit integer. Compute the minimum number of seconds necessary for Kevin to evaluate the expression together by performing eight such multiplications. [i]Proposed by Evan Chen[/i]

Let $a,b,c>0$ be real numbers. Prove that $$\frac{a+b}{\sqrt{b+c}}+\frac{b+c}{\sqrt{c+a}}+\frac{c+a}{\sqrt{a+b}}\geq \sqrt{2a}+ \sqrt{2b}+ \sqrt{2c}$$ Б. Кайрат (Казахстан), А. Храбров

let x,y are positive and $ \in R$ that : $ x\plus{}2y\equal{}1$.prove that : \[ \frac{1}{x}\plus{}\frac{2}{y} \geq \frac{25}{1\plus{}48xy^2}\]

Determine the number of integers $a$ with $1\leq a\leq 1007$ and the property that both $a$ and $a+1$ are quadratic residues mod $1009$.

Given $26$ different positive integers , in any six numbers of the $26$ integers , there are at least two numbers , one can be devided by another. Then prove : There exists six numbers , one of them can be devided by the other five numbers .

Alice creates the graphs $y=|x-a|$ and $y=c-|x-b|$ , where $a,b,c\in\mathbb{R^+}$. She observes that these two graphs and $x$ axis divides the positive side of the plane ($x,y>0$) into two triangles and a quadrilateral. Find the ratio of sums of two triangles' areas to the area of quadrilateral. [hide=There might be a translation error] In the original statement,it says $XOY$ plane,instead of positive side of the plane. I think these 2 are the same,but I might be wrong [/hide]

The polygon enclosed by the solid lines in the figure consists of $ 4$ congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing? [asy]unitsize(10mm); defaultpen(fontsize(10pt)); pen finedashed=linetype("4 4"); filldraw((1,1)--(2,1)--(2,2)--(4,2)--(4,3)--(1,3)--cycle,grey,black+linewidth(.8pt)); draw((0,1)--(0,3)--(1,3)--(1,4)--(4,4)--(4,3)-- (5,3)--(5,2)--(4,2)--(4,1)--(2,1)--(2,0)--(1,0)--(1,1)--cycle,finedashed); draw((0,2)--(2,2)--(2,4),finedashed); draw((3,1)--(3,4),finedashed); label("$1$",(1.5,0.5)); draw(circle((1.5,0.5),.17)); label("$2$",(2.5,1.5)); draw(circle((2.5,1.5),.17)); label("$3$",(3.5,1.5)); draw(circle((3.5,1.5),.17)); label("$4$",(4.5,2.5)); draw(circle((4.5,2.5),.17)); label("$5$",(3.5,3.5)); draw(circle((3.5,3.5),.17)); label("$6$",(2.5,3.5)); draw(circle((2.5,3.5),.17)); label("$7$",(1.5,3.5)); draw(circle((1.5,3.5),.17)); label("$8$",(0.5,2.5)); draw(circle((0.5,2.5),.17)); label("$9$",(0.5,1.5)); draw(circle((0.5,1.5),.17));[/asy] $ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

A sequence $\{a_n\}_{n\geq 1}$ is defined by a recurrence relation $$a_1 = 1,\quad a_{n+1} = \log \frac{e^{a_n}-1}{a_n}$$ And a sequence $\{b_n\}_{n\geq 1}$ is defined as $b_n = \prod\limits_{i=1}^n a_i$. Evaluate an infinite series $\sum\limits_{n=1}^\infty b_n$.

Find all functions $f: (0, +\infty)\cap\mathbb{Q}\to (0, +\infty)\cap\mathbb{Q}$ satisfying thefollowing conditions: [list=1] [*] $f(ax) \leq (f(x))^a$, for every $x\in (0, +\infty)\cap\mathbb{Q}$ and $a \in (0, 1)\cap\mathbb{Q}$ [*] $f(x+y) \leq f(x)f(y)$, for every $x,y\in (0, +\infty)\cap\mathbb{Q}$ [/list]

Harold has $3$ red checkers and $3$ black checkers. Find the number of distinct ways that Harold can place these checkers in stacks. Two ways of stacking checkers are the same if each stack of the rst way matches a corresponding stack in the second way in both size and color arrangement. So, for example, the $3$ stack arrangement $RBR, BR, B$ is distinct from $RBR, RB, B$, but the $4$ stack arrangement $RB, BR, B, R$ is the same as $B, BR, R, RB$.