Describe all convex, inscriptible polygons which have the property that however we choose three distinct vertexes of of one of them, those vertexes form an isosceles triangle. [i]Gheorghe Iurea[/i]

Let $ \mathcal{F}$ be a family of hexagons $ H$ satisfying the following properties: i) $ H$ has parallel opposite sides. ii) Any 3 vertices of $ H$ can be covered with a strip of width 1. Determine the least $ \ell\in\mathbb{R}$ such that every hexagon belonging to $ \mathcal{F}$ can be covered with a strip of width $ \ell$. Note: A strip is the area bounded by two parallel lines separated by a distance $ \ell$. The lines belong to the strip, too.

Let $ f(x)$ be a polynomial and $ C$ be a real number. Find the $ f(x)$ and $ C$ such that $ \int_0^x f(y)dy\plus{}\int_0^1 (x\plus{}y)^2f(y)dy\equal{}x^2\plus{}C$.

$m,n$ are positive intergers and $x,y,z$ positive real numbers such that $0 \leq x,y,z \leq 1$. Let $m+n=p$. Prove that: $0 \leq x^p+y^p+z^p-x^m*y^n-y^m*z^n-z^m*x^n \leq 1$

Solve the system of equations: \[ x+y+z=a \] \[ x^2+y^2+z^2=b^2 \] \[ xy=z^2 \] where $a$ and $b$ are constants. Give the conditions that $a$ and $b$ must satisfy so that $x,y,z$ are distinct positive numbers.

In the non-isosceles $\vartriangle ABC$, the interior bisectors of vertices $B$ and $C$ are drawn, which cut the sides $AC$ and $AB$ at $E$ and $F$ respectively.The line $EF$ cuts the extension of side $BC$ at $T$. In the side$ BC$ a point D is located, so that $\frac{DB}{DC} = \frac{TB}{TC}$. Prove that $AT$ is the exterior bisector of angle $A$.

$a_n$ sequence is a arithmetic sequence with all terms be positive integers. (for $a_n$ non-constant sequence) Let $p_n$ is greatest prime divisor of $a_n$. Prove that $$(\frac{a_n}{p_n})$$ sequence is infinity. [hide]Note: If we find a $M>0$ constant such that $x_n \leq M$ for all $n \in {\mathbb N}$'s, $(x_n)$ sequence is non-infinite, but we can't find $M$, $(x_n)$ sequence is infinity [/hide]

Given an acute scalene triangle $ABC$ inscribed in circle $(O)$. Let $H$ be its orthocenter and $M$ be the midpoint of $BC$. Let $D$ lie on the opposite rays of $HA$ so that $BC=2DM$. Let $D'$ be the reflection of $D$ through line $BC$ and $X$ be the intersection of $AO$ and $MD$. a) Show that $AM$ bisects $D'X$. b) Similarly, we define the points $E,F$ like $D$ and $Y,Z$ like $X$. Let $S$ be the intersection of tangent lines from $B,C$ with respect to $(O)$. Let $G$ be the projection of the midpoint of $AS$ to the line $AO$. Show that there exists a point with the same power to all the circles $(BEY),(CFZ),(SGO)$ and $(O)$.

A $2022 \times 2022$ squareboard was divided into $L$ and $Z$ tetrominoes. Each tetromino consists of four squares, which can be rotated or flipped. Determine the least number of $Z$-tetrominoes necessary to cover the $2022 \times 2022$ squareboard.

Let $ABC$ be a triangle with $AB < AC$. Let $D$ be the intersection point of the internal bisector of angle $BAC$ and the circumcircle of $ABC$. Let $Z$ be the intersection point of the perpendicular bisector of $AC$ with the external bisector of angle $\angle{BAC}$. Prove that the midpoint of the segment $AB$ lies on the circumcircle of triangle $ADZ$. [i]Olimpiada de Matemáticas, Nicaragua[/i]

In Shvambrania there are $N$ towns, every two of which are connected by a road. These roads do not intersect. If necessary, some of them pass over or under others via bridges. An evil magician establishes one-way rules along the roads in such a way that if someone goes out of a certain town he is unable to come back. Prove that (a) It is possible to establish such rules. (b) There exists a town from which it is possible to reach any other town, and there exists a town from which it is not possible to go out. (c) There is one and only one route passing through all towns. (d) The magician can realise his intention in $N!$ ways. (LM Koganov, Moscow) PS. (a),(b),(c) for Juniors, (a),(b),(d) for Seniors

A positive integer $N$ is a [i]triple-double[/i] if there exists non-negative integers $a$, $b$, $c$ such that $2^a + 2^b + 2^c = N$. How many three-digit numbers are triple-doubles? [i]Proposed by Giacomo Rizzo[/i]

Let $M$ be the midpoint of the base $AC$ of an isosceles $\triangle ABC$. Points $E$ and $F$ on the sides $AB$ and $BC$ respectively are chosen so that $AE \neq CF$ and $\angle FMC = \angle MEF = \alpha$. Determine $\angle AEM$. [i](6 points) [/i] [i]Maxim Prasolov[/i]

Jean made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is $8$ feet high and the other peak is $12$ feet high. Each peak forms a $90^\circ$ angle, and the straight sides of the mountains form $45^\circ$ with the ground. The artwork has an area of $183$ square feet. The sides of the mountains meet at an intersection point near the center of the artwork, $h$ feet above the ground. What is the value of $h$? [asy] unitsize(.3cm); filldraw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,gray(0.7),linewidth(1)); draw((-1,0)--(-1,8),linewidth(.75)); draw((-1.4,0)--(-.6,0),linewidth(.75)); draw((-1.4,8)--(-.6,8),linewidth(.75)); label("$8$",(-1,4),W); label("$12$",(31,6),E); draw((-1,8)--(8,8),dashed); draw((31,0)--(31,12),linewidth(.75)); draw((30.6,0)--(31.4,0),linewidth(.75)); draw((30.6,12)--(31.4,12),linewidth(.75)); draw((31,12)--(18,12),dashed); label("$45^{\circ}$",(.75,0),NE,fontsize(10pt)); label("$45^{\circ}$",(29.25,0),NW,fontsize(10pt)); draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle); draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle); draw((11,5)--(11,0),dashed); label("$h$",(11,2.5),E); [/asy] $\textbf{(A)}~4 \qquad \textbf{(B)}~5 \qquad \textbf{(C)}~4 \sqrt{2} \qquad \textbf{(D)}~6 \qquad \textbf{(E)}~5 \sqrt{2}$

Let $ABC$ be an isosceles triangle with base $AC$. Points $D$ and $E$ are chosen on the sides $AC$ and $BC$, respectively, such that $CD = DE$. Let $H, J,$ and $K$ be the midpoints of $DE, AE,$ and $BD$, respectively. The circumcircle of triangle $DHK$ intersects $AD$ at point $F$, whereas the circumcircle of triangle $HEJ$ intersects $BE$ at $G$. The line through $K$ parallel to $AC$ intersects $AB$ at $I$. Let $IH \cap GF =$ {$M$}. Prove that $J, M,$ and $K$ are collinear points.

Let $x_1,x_2,\ldots,x_n$ be arbitrary real numbers. Prove the inequality \[ \frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots + \frac{x_n}{1 + x_1^2 + \cdots + x_n^2} < \sqrt{n}. \]

In a group of $kn$ persons, each person knows more than $(k-1)n$ others ($k,n$ are positive integers). Prove that one can choose $k+1$ persons from this group so that each chosen person knows all the others chosen. Note: If a person $A$ knows $B$, then $B$ knows $A$.

In the coordinate plane, define $M = \{(a, b),a,b \in Z\}$. A transformation $S$, which is defined on $M$, sends $(a,b)$ to $(a + b, b)$. Transformation $T$, also defined on $M$, sends $(a, b)$ to $(-b, a)$. Prove that for all $(a, b) \in M$, we can use $S,T$ denitely to map it to $(g,0)$.

All the positive divisors of a positive integer $n$ are stored into an increasing array. Mary is writing a programme which decides for an arbitrarily chosen divisor $d > 1$ whether it is a prime. Let $n$ have $k$ divisors not greater than $d$. Mary claims that it suffices to check divisibility of $d$ by the first $\left\lceil\frac{k}{2}\right\rceil$ divisors of $n$: $d$ is prime if and only if none of them but $1$ divides $d$. Is Mary right?

Let $ABC$ be a triangle. For a point $P$ in its interior, we draw the threee lines through $P$ parallel to the sides of the triangle. This partitions $ABC$ in three triangles and three quadrilaterals. Let $V_A$ be the area of the quadrilateral which has $A$ as one vertex. Let $D_A$ be the area of the triangle which has a part of $BC$ as one of its sides. Define $V_B, D_B$ and $V_C, D_C$ similarly. Determine all possible values of $\frac{D_A}{V_A}+\frac{D_B}{V_B}+\frac{D_C}{V_C}$, as $P$ varies in the interior of the triangle.

Six points are marked inside the $3\times 4$ rectangle. Prove that there is a pair of marked points with the distance between them not greater than $\sqrt5$.

$10$ people attended a party. For every $3$ people, there exist at least $2$ people who don’t know each other. Prove that there exist $4$ people who don’t know each other.

Compute: \[\lim_{x\to 0}\text{ }\dfrac{x^2}{1-\cos(x)}\]

The nine digits from 1 to 9 are to be placed around a circle so that the average of any three consecutive digits is a multiple of 3. Is this possible? Justify your answer.

In a "Hanger Party", the guests are initially dressed. In certain moments, the host chooses a guest, and the chosen guest and all his friends will wear its respective clothes if they are naked, and undress it if they are dressed. It is possible that, in some moment, the guests are naked, independent of their mutual friendships? (Suppose friendship is reciprocal.)