Given is a grid 11x11 with 121 cells. Four of them are colored in black, the rest are white. We have to cut a completely white rectangle (it could be a square and the rectangle must have its sides parralel to the lines of the grid), so that this rectangle has maximal possible area. What largest area of this rectangle we can guarantee? (We can cut this rectangle for every placement of the black squares)

$(MON 2)$ Given reals $x_0, x_1, \alpha, \beta$, find an expression for the solution of the system \[x_{n+2} -\alpha x_{n+1} -\beta x_n = 0, \qquad n= 0, 1, 2, \ldots\]

Quadrangle $ABCD$ is inscribed in a circle. The perpendicular from the vertex $C$ on the bisector of $\angle ABD$ intersects the line $AB$ at the point $C_1$. The perpendicular from the vertex $B$ on the bisector of $\angle ACD$ intersects the line $CD$ at the point $B_1$. Prove that $B_1C_1 \parallel AD$.

Determine the number of all numbers which are represented as $x^2+y^2$ with $x, y \in \{1, 2, 3, \ldots, 1000\}$ and which are divisible by 121.

Let $m,n$ be naturals satisfying $n \geq m \geq 2$ and let $S$ be a set consisting of $n$ naturals. Prove that $S$ has at least $2^{n-m+1}$ distinct subsets, each whose sum is divisible by $m$. (The zero set counts as a subset).

let $ABC$ be a triangle and $I$ be its incentre. let the incircle of $ABC$ touch $BC$ at $D$. let incircle of triangle $ABD$ touch $AB$ at $E$ and incircle of triangle $ACD$ touch $AC$ at $F$. prove that $B,E,I,F$ are concyclic.

In a 10 by 10 grid of dots, what is the maximum number of lines that can be drawn connecting two dots on the grid so that no two lines are parallel? [i]Proposed by Aaron Lin[/i]

An alphabet consists of $n$ letters. What is the maximal length of a word if we know that any two consecutive letters $a,b$ of the word are different and that the word cannot be reduced to a word of the kind $abab$ with $a\neq b$ by removing letters.

Find all second degree polynomials $P(x)$ such that for all $a \in\mathbb{R} , a \geq 1$, then $P(a^2+a) \geq a.P(a+1)$

Find the remainder of $$\prod_{n = 2}^{99} (1 - n^2 + n^4)(1 - 2n^2 + n^4)$$ when divided by $101$.

Let there be two circles. Find all points $M$ such that there exist two points, one on each circle such that $M$ is their midpoint.

Call a positive integer $n$ $k$[i]-pretty[/i] if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$. For example, $18$ is $6$[i]-pretty[/i]. Let $S$ be the sum of positive integers less than $2019$ that are $20$[i]-pretty[/i]. Find $\tfrac{S}{20}$.

Prove that $p(x)=1+x+x^2 +\ldots+x^n$ is reducible over $\mathbb{F}_{2}$ in case $n+1$ is composite. If $n+1$ is prime, is $p(x)$ irreducible over $\mathbb{F}_{2}$ ?

Triangle $ABC$ has sidelengths $AB=1$, $BC=\sqrt{3}$, and $AC=2$. Points $D,E$, and $F$ are chosen on $AB, BC$, and $AC$ respectively, such that $\angle EDF = \angle DFA = 90^{\circ}$. Given that the maximum possible value of $[DEF]^2$ can be expressed as $\frac{a}{b}$ for positive integers $a, b$ with $\gcd (a, b) = 1$, find $a + b$. (Here $[DEF]$ denotes the area of triangle $DEF$.) [i]Proposed by Vismay Sharan[/i]

Over all real numbers $x$ and $y$ such that $$x^3=3x+y \qquad \text{and} \qquad y^3=3y+x,$$ compute the sum of all possible values of $x^2+y^2.$

$$\sin 30^o + \sin 45^o + \sin 60^o + \sin 90^o + \cos 120^o + \cos 135^o + \cos 150^o + \cos 180^o = ?$$

In a word of more than $10$ letters, any two consecutive letters are different. Prove that one can change places of two consecutive letters so that the resulting word is not [i]periodic[/i], that is, cannot be divided into equal subwords.

The natural numbers $x_1, x_2, \ldots , x_n$ are such that all their $2^n$ partial sums are distinct. Prove that: $$ {x_1}^2 + {x_2}^2 + \ldots + {x_n}^2 \geq \frac{4^n – 1}{3}. $$

A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring? [asy] size(200); defaultpen(linewidth(3)); real[] inrad = {40,34,28,21}; real[] outrad = {55,49,37,30}; real[] center; path[][] quad = new path[4][4]; center[0] = 0; for(int i=0;i<=3;i=i+1) { if(i != 0) { center[i] = center[i-1] - inrad[i-1] - inrad[i]+3.5; } quad[0][i] = arc((0,center[i]),inrad[i],0,90)--arc((0,center[i]),outrad[i],90,0)--cycle; quad[1][i] = arc((0,center[i]),inrad[i],90,180)--arc((0,center[i]),outrad[i],180,90)--cycle; quad[2][i] = arc((0,center[i]),inrad[i],180,270)--arc((0,center[i]),outrad[i],270,180)--cycle; quad[3][i] = arc((0,center[i]),inrad[i],270,360)--arc((0,center[i]),outrad[i],360,270)--cycle; draw(circle((0,center[i]),inrad[i])^^circle((0,center[i]),outrad[i])); } void fillring(int i,int j) { if ((j % 2) == 0) { fill(quad[i][j],white); } else { filldraw(quad[i][j],black); } } for(int i=0;i<=3;i=i+1) { for(int j=0;j<=3;j=j+1) { fillring(((2-i) % 4),j); } } for(int k=0;k<=2;k=k+1) { filldraw(circle((0,-228 - 25 * k),3),black); } real r = 130, s = -90; draw((0,57)--(r,57)^^(0,-57)--(r,-57),linewidth(0.7)); draw((2*r/3,56)--(2*r/3,-56),linewidth(0.7),Arrows(size=3)); label("$20$",(2*r/3,-10),E); draw((0,39)--(s,39)^^(0,-39)--(s,-39),linewidth(0.7)); draw((9*s/10,38)--(9*s/10,-38),linewidth(0.7),Arrows(size=3)); label("$18$",(9*s/10,0),W); [/asy] $ \textbf{(A) } 171\qquad \textbf{(B) } 173\qquad \textbf{(C) } 182\qquad \textbf{(D) } 188\qquad \textbf{(E) } 210$

Eight tiles are located on an $8\times 8$ board in such a way that no pair of them are in the same row or in the same column. Prove that, among the distances between each pair of tiles, we can find two of them that are equal (the distance between two tiles is the distance between the centers of the squares in which they are located).

Let $A\in \mathcal{M}_n(\mathbb{R}^*)$. If $A\cdot\ ^t A=I_n$, prove that: a)$|\text{Tr}(A)|\le n$; b)If $n$ is odd, then $\det(A^2-I_n)=0$.

Let $f$, $g$, $h : [a, b] \to \mathbb{R}$, three integrable functions such that:$$\int \limits_a^b fgdx=\int \limits_a^bghdx=\int \limits_a^bhfdx=\int \limits_a^bg^2dx\int \limits_a^bh^2dx=1$$Then$$\int \limits_a^bg^2dx=\int \limits_a^bh^2dx=1$$

Paul and Sara are playing a game with integers on a whiteboard, with Paul going first. When it is Paul’s turn, he can pick any two integers on the board and replace them with their product; when it is Sara’s turn, she can pick any two integers on the board and replace them with their sum. Play continues until exactly one integer remains on the board. Paul wins if that integer is odd, and Sara wins if it is even. Initially, there are $2021$ integers on the board, each one sampled uniformly at random from the set $\{0, 1, 2, 3, . . . , 2021\}$. Assuming both players play optimally, the probability that Paul wins is $m/n$ , where $m, n$ are positive integers and $gcd(m, n) = 1$. Find the remainder when $m + n$ is divided by $1000$.

For an integer $m\geq 4,$ let $T_{m}$ denote the number of sequences $a_{1},\dots,a_{m}$ such that the following conditions hold: (1) For all $i=1,2,\dots,m$ we have $a_{i}\in \{1,2,3,4\}$ (2) $a_{1} = a_{m} = 1$ and $a_{2}\neq 1$ (3) For all $i=3,4\cdots, m, a_{i}\neq a_{i-1}, a_{i}\neq a_{i-2}.$ Prove that there exists a geometric sequence of positive integers $\{g_{n}\}$ such that for $n\geq 4$ we have that \[ g_{n} - 2\sqrt{g_{n}} < T_{n} < g_{n} + 2\sqrt{g_{n}}.\]

Show that there are infinitely many positive integers $c$, such that the following equations both have solutions in positive integers: $(x^2 - c)(y^2 -c) = z^2 -c$ and $(x^2 + c)(y^2 - c) = z^2 - c$.