Given a convex quadrilateral $ABCD$ with $AB = BC $and $\angle ABD = \angle BCD = 90$.Let point $E$ be the intersection of diagonals $AC$ and $BD$. Point $F$ lies on the side $AD$ such that $\frac{AF}{F D}=\frac{CE}{EA}$.. Circle $\omega$ with diameter $DF$ and the circumcircle of triangle $ABF$ intersect for the second time at point $K$. Point $L$ is the second intersection of $EF$ and $\omega$. Prove that the line $KL$ passes through the midpoint of $CE$. [i]Proposed by Mahdi Etesamifard and Amir Parsa Hosseini - Iran[/i]

The chords $AB$ and $CD$ of a sphere intersect at $X. A, C$ and $X$ are equidistant from a point $Y$ on the sphere. Show that $BD$ and $XY$ are perpendicular.

Prove that $a^3 + b^3 + c^3 + abc +a^{3}b^{2}c^{-1}+a^{3}c^{2}b^{-1}+b^{3}a^{2}c^{-1}+b^{3}c^{2}a^{-1}+c^{3}a^{2}b^{-1}+c^{3}b^{2}a^{-1}+a^{5}b^{3}c^{-3}+ abc^{14} + a^{5}c^{3}b^{-3}+b^{5}a^{3}c^{-3}+b^{5}c^{3}a^{-3}+c^{5}a^{3}b^{-3}+c^{5}b^{3}a^{-3}+a^{6}b^{1}c^{-1}+a^{6}c^{1}b^{-1}+b^{6}a^{1}c^{-1}+b^{6}c^{1}a^{-1}+c^{6}a^{1}b^{-1}+c^{6}b^{1}a^{-1}+ a^{6}b^{4}c^{-3}+a^{6}c^{4}b^{-3}+b^{6}a^{4}c^{-3}+b^{6}c^{4}a^{-3}+c^{6}a^{4}b^{-3}+c^{6}b^{4}a^{-3}+a^{7}b^{2}c^{-1}+a^{7}c^{2}b^{-1}+b^{7}a^{2}c^{-1}+b^{7}c^{2}a^{-1}+c^{7}a^{2}b^{-1}+ abc + a^{14}bc + c^{7}b^{2}a^{-1}+a^{4}b^{1}c^{4}+a^{4}c^{1}b^{4}+b^{4}a^{1}c^{4}+b^{4}c^{1}a^{4}+c^{4}a^{1}b^{4}+c^{4}b^{1}a^{4}+a^{6}c^{4}+a^{6}b^{4}+b^{6}c^{4}+b^{6}a^{4}+c^{6}b^{4}+c^{6}a^{4}+a^{9}b^{6}c^{-4}+a^{9}c^{6}b^{-4}+ ab^{14}c + b^{9}a^{6}c^{-4}+b^{9}c^{6}a^{-4}+c^{9}a^{6}b^{-4}+ abc + c^{9}b^{6}a^{-4}+a^{12}b^{1}c^{-1}+a^{12}c^{1}b^{-1}+b^{12}a^{1}c^{-1}+b^{12}c^{1}a^{-1}+c^{12}a^{1}b^{-1}+ c^5 b^5 a^5 - c^5 b^5 a^2 + 3 c^5 b^5 - c^5 b^2 a^5 + c^5 b^2 a^2 - 3 c^5 b^2 + 3 c^5 a^5 - 3 c^5 a^2 + 9 c^5 - c^2 b^5 a^5 + c^2 b^5 a^2 - 3 c^2 b^5 + c^2 b^2 a^5 - c^2 b^2 a^2 + 3 c^2 b^2 - 3 c^2 a^5 + 3 c^2 a^2 - 9 c^2 + 3 b^5 a^5 - 3 b^5 a^2 + 9 b^5 - 3 b^2 a^5 + 3 b^2 a^2 - 9 b^2 + 9 a^5 - 9 a^2 + 27 + c^{12}b^{1}a^{-1}+a^{13}b^{9}c^{-9}+a^{13}c^{9}b^{-9}+b^{13}a^{9}c^{-9}+b^{13}c^{9}a^{-9}+c^{13}a^{9}b^{-9}+c^{13}b^{9}a^{-9}+a^{12}b^{11}c^{-9}+a^{12}c^{11}b^{-9}+b^{12}a^{11}c^{-9}+b^{12}c^{11}a^{-9}+c^{12}a^{11}b^{-9}+c^{12}b^{11}a^{-9}+a^{8}b^{7}+a^{8}c^{7}+b^{8}a^{7}+b^{8}c^{7}+c^{8}a^{7}+c^{8}b^{7} + a^{16} + b^{16} + c^{16} + a^{16} + b^{16} + c^{16} + a^{16} + b^{16} + c^{16}\ge c^3 + 3 c^2 a + 3 c b^2 + 6 c b a + b^3 + 3 b^2 a + a^3 + a^{1}c^{2}+a^{1}b^{2}+4b^{1}c^{2}+4b^{1}a^{2}+c^{1}b^{2}+4c^{1}a^{2}+a^{1}c^{3}+a^{1}b^{3}+b^{1}c^{3}+b^{1}a^{3}+c^{1}b^{3}+c^{1}a^{3}+a^{3}b^{2}+a^{3}c^{2}+b^{3}a^{2}+b^{3}c^{2}+c^{3}a^{2}+c^{3}b^{2}+a^{5}c^{1}+a^{5}b^{1}+b^{5}c^{1}+b^{5}a^{1}+c^{5}b^{1}+c^{5}a^{1}+a^{2}b^{1}c^{4}+a^{2}c^{1}b^{4}+b^{2}a^{1}c^{4}+b^{2}c^{1}a^{4}+c^{2}a^{1}b^{4}+c^{2}b^{1}a^{4}+a^{1}c^{7}+a^{1}b^{7}+b^{1}c^{7}+b^{1}a^{7}+c^{1}b^{7}+c^{1}a^{7}+a^{1}c^{8}+a^{1}b^{8}+b^{1}c^{8}+b^{1}a^{8}+c^{1}b^{8}+c^{1}a^{8}+a^{5}b^{1}c^{4}+a^{5}c^{1}b^{4}+b^{5}a^{1}c^{4}+b^{5}c^{1}a^{4}+c^{5}a^{1}b^{4}+c^{5}b^{1}a^{4}+a^{2}b^{1}c^{8}+a^{2}c^{1}b^{8}+b^{2}a^{1}c^{8}+b^{2}c^{1}a^{8}+c^{2}a^{1}b^{8}+c^{2}b^{1}a^{8}+a^{1}c^{11}+a^{1}b^{11}+b^{1}c^{11}+b^{1}a^{11}+c^{1}b^{11}+c^{1}a^{11}+a^{6}b^{2}c^{5}+a^{6}c^{2}b^{5}+b^{6}a^{2}c^{5}+b^{6}c^{2}a^{5}+c^{6}a^{2}b^{5}+c^{6}b^{2}a^{5}+a^{3}b^{2}c^{9}+a^{3}c^{2}b^{9}+b^{3}a^{2}c^{9}+b^{3}c^{2}a^{9}+c^{3}a^{2}b^{9}+c^{3}b^{2}a^{9}+a^{3}b^{1}c^{11}+a^{3}c^{1}b^{11}+b^{3}a^{1}c^{11}+b^{3}c^{1}a^{11}+c^{3}a^{1}b^{11}+c^{3}b^{1}a^{11} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15}+c^{2}a^{1}b^{4}+c^{2}b^{1}a^{4}+a^{1}c^{7}+a^{1}b^{7}+b^{1}c^{7}+b^{1}a^{7}+c^{1}b^{7}+c^{1}a^{7}+a^{1}c^{8}+a^{1}b^{8}+b^{1}c^{8}+b^{1}a^{8}+c^{1}b^{8}+c^{1}a^{8}+a^{5}b^{1}c^{4}+a^{5}c^{1}b^{4}+b^{5}a^{1}c^{4}+b^{5}c^{1}a^{4}+c^{5}a^{1}b^{4}+c^{5}b^{1}a^{4}+a^{2}b^{1}c^{8}+a^{2}c^{1}b^{8}+b^{2}a^{1}c^{8}+b^{2}c^{1}a^{8}+c^{2}a^{1}b^{8}+c^{2}b^{1}a^{8}+a^{1}c^{11}+a^{1}b^{11}+b^{1}c^{11}+b^{1}a^{11}+c^{1}b^{11}+c^{1}a^{11}+a^{6}b^{2}c^{5}+a^{6}c^{2}b^{5}+b^{6}a^{2}c^{5}+b^{6}c^{2}a^{5}+c^{6}a^{2}b^{5}+c^{6}b^{2}a^{5}+a^{3}b^{2}c^{9}+a^{3}c^{2}b^{9}+b^{3}a^{2}c^{9}+b^{3}c^{2}a^{9}+c^{3}a^{2}b^{9}+c^{3}b^{2}a^{9}+a^{3}b^{1}c^{11}+a^{3}c^{1}b^{11}+b^{3}a^{1}c^{11}+b^{3}c^{1}a^{11}+c^{3}a^{1}b^{11}+c^{3}b^{1}a^{11} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15}$ for all $a,b,c\in\mathbb R^+$. [i]Proposed by Henry Jiang and C++[/i]

Do there exist distinct non-zero digits $a, b$ and $c$ such that the two-digit number $\overline{ab}$ is divisible by $c$, the number $\overline{bc}$ is divisible by $a$ and $\overline{ca}$, is divisible by $b$?

A positive integer $N$ has base-eleven representation $\underline{a}\,\underline{b}\,\underline{c}$ and base-eight representation $\underline{1}\,\underline{b}\,\underline{c}\,\underline{a}$, where $a$, $b$, and $c$ represent (not necessarily distinct) digits. Find the least such $N$ expressed in base ten.

Consider triangle $ABC$ with $AB<AC<BC$, inscribed in triangle $\Gamma_1$ and the circles $\Gamma_2 (B,AC)$ and $\Gamma_2 (C,AB)$. A common point of circle $\Gamma_2$ and $\Gamma_3$ is point $E$, a common point of circle $\Gamma_1$ and $\Gamma_3$ is point $F$ and a common point of circle $\Gamma_1$ and $\Gamma_2$ is point $G$, where the points $E,F,G$ lie on the same semiplane defined by line $BC$, that point $A$ doesn't lie in. Prove that circumcenter of triangle $EFG$ lies on circle $\Gamma_1$. Note: By notation $\Gamma (K,R)$, we mean random circle $\Gamma$ has center $K$ and radius $R$.

Find the number of pairs $(m,n)$ of integers with $-2014\le m,n\le 2014$ such that $x^3+y^3 = m + 3nxy$ has infinitely many integer solutions $(x,y)$. [i]Proposed by Victor Wang[/i]

There are $n$ blocks placed on the unit squares of a $n \times n$ chessboard such that there is exactly one block in each row and each column. Find the maximum value $k$, in terms of $n$, such that however the blocks are arranged, we can place $k$ rooks on the board without any two of them threatening each other. (Two rooks are not threatening each other if there is a block lying between them.)

We know that $201$ and $9$ give the same remainder when divided by $24$. What is the smallest positive integer $k$ such that $201+k$ and $9+k$ give the same remainder when divided by $24$? [i]2020 CCA Math Bonanza Lightning Round #1.1[/i]

Several non-intersecting diagonals divide a convex polygon into triangles. At each vertex of the polygon the number of triangles adjacent to it is written. Is it possible to reconstruct all the diagonals using these numbers if the diagonals are erased?

Find all monotone functions $ f: \mathbb{R} \to \mathbb{R} $ satisfying the equation $$ f(4x)-f(3x) = 2x \ \ \text{ for } \ \ x \in \mathbb{R}.$$

If $x, y, z$ are positive real numbers with sum $12$, prove that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+ 3 \ge \sqrt{x} +\sqrt{y }+\sqrt{z}$. When equality is valid?

Let $A_1$, $B_1$ and $C_1$ be the midpoints of sides $BC$, $CA$ and $AB$ of triangle $ABC$, respectively. Points $B_2$ and $C_2$ are the midpoints of segments $BA_1$ and $CA_1$ respectively. Point $B_3$ is symmetric to $C_1$ wrt $B$, and $C_3$ is symmetric to $B_1$ wrt $C$. Prove that one of common points of circles $BB_2B_3$ and $CC_2C_3$ lies on the circumcircle of triangle $ABC$.

Find all positive integers $n$ such that $7^n+147$ is a perfect square.

We call a positive integer [i]alternating[/i] if every two consecutive digits in its decimal representation are of different parity. Find all positive integers $n$ such that $n$ has a multiple which is alternating.

Solve $2|3-2x|-|x-2|=x$ , $x\in\mathbb{R}$ .

Is there an integer coefficients polynomial $P(x)$ satisfying \[ \begin{cases} P(1+\sqrt[3]{2})=1+\sqrt[3]{2}\\ P(1+\sqrt{5})=2+3\sqrt{5}\end{cases} \]

Let $k\ge t\ge 2$ positive integers. For integers $n\ge k$ let $p_n$ be the probability that if we choose $k$ from the first $n$ positive integers randomly, any $t$ of the $k$ chosen integers have greatest common divisor $1$. Let qn be the probability that if we choose $k-t+1$ from the first $n$ positive integers the product is not divisible by a perfect $t^{th}$ power that is greater then $1$. Prove that sequences $p_n$ and $q_n$ converge to the same value.

A game is played on the board shown. In this game, a player can move three places in any direction (up, down, right or left) and then can move two places in a direction perpendicular to the first move. If a player starts at $S$, which position on the board ($P, Q, R, T$, or $W$) cannot be reached through any sequence of moves? \[ \begin{tabular}{|c|c|c|c|c|}\hline & & P & & \\ \hline & Q & & R &\\ \hline & & T & & \\ \hline S & & & & W\\ \hline\end{tabular} \] $\textbf{(A)}\ P \qquad \textbf{(B)}\ Q \qquad \textbf{(C)}\ R \qquad \textbf{(D)}\ T \qquad \textbf{(E)}\ W$

[b]p1.[/b] Consider a triangular grid: nodes of the grid are painted black and white. At a single step you are allowed to change colors of all nodes situated on any straight line (with the slope $0^o$ ,$60^o$, or $120^o$ ) going through the nodes of the grid. Can you transform the combination in the left picture into the one in the right picture in a finite number of steps? [img]https://cdn.artofproblemsolving.com/attachments/3/a/957b199149269ce1d0f66b91a1ac0737cf3f89.png[/img] [b]p2.[/b] Find $x$ satisfying $\sqrt{x\sqrt{x \sqrt{x ...}}} = \sqrt{2022}$ where it is an infinite expression on the left side. [b]p3.[/b] $179$ glasses are placed upside down on a table. You are allowed to do the following moves. An integer number $k$ is fixed. In one move you are allowed to turn any $k$ glasses . (a) Is it possible in a finite number of moves to turn all $179$ glasses into “bottom-down” positions if $k=3$? (b) Is it possible to do it if $k=4$? [b]p4.[/b] An interval of length $1$ is drawn on a paper. Using a compass and a simple ruler construct an interval of length $\sqrt{93}$. [b]p5.[/b] Show that $5^{2n+1} + 3^{n+2} 2^{n-1} $ is divisible by $19$ for any positive integer $n$. [b]p6.[/b] Solve the system $$\begin{cases} \dfrac{xy}{x+y}=1-z \\ \dfrac{yz}{y+z}=2-x \\ \dfrac{xz}{x+z}=2-y \end{cases}$$ PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

If $a,b,c$ are non-zero real numbers such that \[\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a},\] and \[x=\frac{(a+b)(b+c)(c+a)}{abc},\] and $x<0$, then $x$ equals $\textbf{(A) }-1\qquad\textbf{(B) }-2\qquad\textbf{(C) }-4\qquad\textbf{(D) }-6\qquad \textbf{(E) }-8$

An equilateral triangle whose side is $ 2$ is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides. If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is: $ \textbf{(A)}\ \frac{\sqrt{6}}{2} \qquad \textbf{(B)}\ \sqrt{2} \qquad \textbf{(C)}\ 2\plus{}\sqrt{2} \qquad \textbf{(D)}\ \frac{2\plus{}\sqrt{2}}{2} \qquad \textbf{(E)}\ \frac{2\sqrt{3}\minus{}\sqrt{6}}{2}$

The cells of the $200 \times 200$ table are painted black and white so that there are $404$ more black cells than white ones. Prove that there is a $2 \times 2$ square in which the number of white cells is odd.

In a right triangle $ABC$ ($\angle C = 90^o$) it is known that $AC = 4$ cm, $BC = 3$ cm. The points $A_1, B_1$ and $C_1$ are such that $AA_1 \parallel BC$, $BB_1\parallel A_1C$, $CC_1\parallel A_1B_1$, $A_1B_1C_1= 90^o$, $A_1B_1= 1$ cm. Find $B_1C_1$.

Calculate $\int \frac{x^{3}}{(x-1)^{3}(x-2)}\ dx$