Given an infinite positive integer sequence $\{x_i\}$ such that $$x_{n+2}=x_nx_{n+1}+1$$ Prove that for any positive integer $i$ there exists a positive integer $j$ such that $x_j^j$ is divisible by $x_i^i$. [i]Remark: Unfortunately, there was a mistake in the problem statement during the contest itself. In the last sentence, it should say "for any positive integer $i>1$ ..."[/i]

A rectangular board of 8 columns has squared numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result? [asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label("$2$",(1.5,8.2),N); label("$4$",(3.5,8.2),N); label("$5$",(4.5,8.2),N); label("$7$",(6.5,8.2),N); label("$8$",(7.5,8.2),N); label("$9$",(0.5,7.2),N); label("$11$",(2.5,7.2),N); label("$12$",(3.5,7.2),N); label("$13$",(4.5,7.2),N); label("$14$",(5.5,7.2),N); label("$16$",(7.5,7.2),N); [/asy] $\text{(A)}\ 36 \qquad \text{(B)}\ 64 \qquad \text{(C)}\ 78 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 120$

Compute the sum $\sum^{200}_{n=1}\frac{1}{n(n+1)(n+2)}$ .

The addition below is incorrect. The display can be made correct by changing one digit $d$, wherever it occurs, to another digit $e$. Find the sum of $d$ and $e$. \[\begin{tabular}{ccccccc}& 7 & 4 & 2 & 5 & 8 & 6 \\ + & 8 & 2 & 9 & 4 & 3 & 0\\ \hline 1 & 2 & 1 & 2 & 0 & 1 & 6 \end{tabular}\] $\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ \text{More than } 10$

We have a triangle with $ \angle BAC=\angle BCA. $ The point $ E $ is on the interior bisector of $ \angle ABC $ so that $ \angle EAB =\angle ACB. $ Let $ D $ be a point on $ BC $ such that $ B $ is on the segment $ CD $ (endpoints excluded) and $ BD=AB. $ Show that the midpoint of $ AC $ is on the line $ DE. $

Two incongruent triangles $ABC$ and $XYZ$ are called a pair of [i]pals[/i] if they satisfy the following conditions: (a) the two triangles have the same area; (b) let $M$ and $W$ be the respective midpoints of sides $BC$ and $YZ$. The two sets of lengths $\{AB, AM, AC\}$ and $\{XY, XW, XZ\}$ are identical $3$-element sets of pairwise relatively prime integers. Determine if there are infinitely many pairs of triangles that are pals of each other.

Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(2f(x)) = f(x - f(y)) + f(x) + y$$ for all $x, y \in \mathbb{R}$.

Let $ABC$ be an acute-angled triangle inscribed in a circle $k$. It is given that the tangent from $A$ to the circle meets the line $BC$ at point $P$. Let $M$ be the midpoint of the line segment $AP$ and $R$ be the second intersection point of the circle $k$ with the line $BM$. The line $PR$ meets again the circle $k$ at point $S$ different from $R$. Prove that the lines $AP$ and $CS$ are parallel.

Consider a sequence $x_1,x_2,\cdots x_{12}$ of real numbers such that $x_1=1$ and for $n=1,2,\dots,10$ let \[ x_{n+2}=\frac{(x_{n+1}+1)(x_{n+1}-1)}{x_n}. \] Suppose $x_n>0$ for $n=1,2,\dots,11$ and $x_{12}=0$. Then the value of $x_2$ can be written as $\frac{\sqrt{a}+\sqrt{b}}{c}$ for positive integers $a,b,c$ with $a>b$ and no square dividing $a$ or $b$. Find $100a+10b+c$. [i]Proposed by Michael Kural[/i]

Let n is a natural number,for which $\sqrt{1+12n^2}$ is a whole number.Prove that $2+2\sqrt{1+12n^2}$ is perfect square.

Nicu plays the Next game on the computer. Initially the number $S$ in the computer has the value $S = 0$. At each step Nicu chooses a certain number $a$ ($0 <a <1$) and enters it in computer. The computer arbitrarily either adds this number $a$ to the number $S$ or it subtracts from $S$ and displays on the screen the new result for $S$. After that Nicu does Next step. It is known that among any $100$ consecutive operations the computer the at least once apply the assembly. Give an arbitrary number $M> 0$. Show that there is a strategy for Nicu that will always allow him after a finite number of steps to get a result $S> M$. [hide=original wording]Nicu joacă la calculator următorul joc. Iniţial numărul S din calculator are valoarea S = 0. La fiecare pas Nicu alege un număr oarecare a (0 < a < 1) şi îl introduce în calculator. Calculatorul, în mod arbitrar, sau adună acest număr a la numărul S sau îl scade din S şi afişează pe ecran rezultatul nou pentru S. După aceasta Nicu face următorul pas. Se ştie că printre oricare 100 de operaţii consecutive calculatorul cel puţin o dată aplică adunarea. Fie dat un număr arbitrar M > 0. Să se arate că există o strategie pentru Nicu care oricând îi va permite lui după un număr finit de paşi să obţină un rezulat S > M.[/hide]

The polynomial $P(x) = x^3 - 6x - 2$ has three real roots, $\alpha$, $\beta$, and $\gamma$. Depending on the assignment of the roots, there exist two different quadratics $Q$ such that the graph of $y=Q(x)$ pass through the points $(\alpha,\beta)$, $(\beta,\gamma)$, and $(\gamma,\alpha)$. What is the larger of the two values of $Q(1)$?

Alice has got a circular key ring with $n$ keys, $n\ge3$. When she takes it out of her pocket, she does not know whether it got rotated and/or flipped. The only way she can distinguish the keys is by coloring them (a color is assigned to each key). What is the minimum number of colors needed?

What is the value of \[ \log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27? \] $\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10 $

Let $f^1(x)=x^3-3x$. Let $f^n(x)=f(f^{n-1}(x))$. Let $\mathcal{R}$ be the set of roots of $\tfrac{f^{2022}(x)}{x}$. If \[\sum_{r\in\mathcal{R}}\frac{1}{r^2}=\frac{a^b-c}{d}\] for positive integers $a,b,c,d$, where $b$ is as large as possible and $c$ and $d$ are relatively prime, find $a+b+c+d$.

At a certain beach if it is at least $80 ^\circ F$ and sunny, then the beach will be crowded. On June 10 the beach was not crowded. What can be said about the weather conditions on June 10? $(A)$ The temperature was cooler than $80 ^\circ F$ and it was not sunny. $(B)$ The temperature was cooler than $80 ^\circ F$ or it was not sunny. $(C)$ If the temperature was at least $80 ^\circ F$, then it was sunny. $(D)$ If the temperature was cooler than $80 ^\circ F$, then it was sunny. $(E)$ If the temperature was cooler than $80 ^\circ F$, then it was not sunny.

A set of different positive integers is called meaningful if for any finite nonempty subset the corresponding arithmetic and geometric means are both integers. $a)$ Does there exist a meaningful set which consists of $2019$ numbers? $b)$ Does there exist an infinite meaningful set? Note: The geometric mean of the non-negative numbers $a_1, a_2,\cdots, $ $a_n$ is defined as $\sqrt[n]{a_1a_2\cdots a_n} .$

Andriyko has rectangle desk and a lot of stripes that lie parallel to sides of the desk. For every pair of stripes we can say that first of them is under second one. In desired configuration for every four stripes such that two of them are parallel to one side of the desk and two others are parallel to other side, one of them is under two other stripes that lie perpendicular to it. Prove that Andriyko can put stripes one by one such way that every next stripe lie upper than previous and get desired configuration. [i]Proposed by Denys Smirnov[/i]

Consider a function $ f:\mathbb{R}\longrightarrow\mathbb{R} $ that admits primitives. Prove that: $ \text{(i)} $ Every term (function) of the sequence functions $ \left( h_n\right)_{n\ge 2}:\mathbb{R}\longrightarrow\mathbb{R} $ defined, for any natural number $ n $ as $ h_n(x)=x^nf\left( x^3 \right) , $ is primitivable. $ \text{(ii)} $ The function $ \phi :\mathbb{R}\longrightarrow\mathbb{R} $ defined as $$ \phi (x) =\left\{ \begin{matrix} e^{-1/x^2} f(x),& \quad x\neq 0 \\ 0,& \quad x=0 \end{matrix} \right. $$ is primitivable. [i]Cristinel Mortici[/i]

Let $\mathbb{Q}^{+}$ denote the set of positive rational numbers. Find, with proof, all functions $f:\mathbb{Q}^+ \to \mathbb{Q}^+$ such that, for all positive rational numbers $x$ and $y,$ we have \[f(x)=f(x+y)+f(x+x^2f(y)).\]

How many whole numbers between 99 and 999 contain exactly one 0? $\text{(A)}\ 72 \qquad \text{(B)}\ 90 \qquad \text{(C)}\ 144 \qquad \text{(D)}\ 162 \qquad \text{(E)}\ 180$

Which is larger: $2^{\sqrt{2}}$; $2^{1+\frac{1}{\sqrt{2}}}$ and 3 ?

Let $N,K,L$ be points on the sides $\overline{AB}, \overline{BC}, \overline{CA}$ respectively. Suppose $AL=BK$ and $\overline{CN}$ is the internal bisector of angle $ACB$. Let $P$ be the intersection of lines $\overline{AK}$ and $\overline{BL}$ and let $I,J$ be the incenters of triangles $APL$ and $BPK$ respectively. Let $Q$ be the intersection of lines $\overline{IJ}$ and $\overline{CN}$. Prove that $IP=JQ$.

Determine the last two digits of the product of the squares of all positive odd integers less than $2014$.

The diagram below shows two parallel rows with seven points in the upper row and nine points in the lower row. The points in each row are spaced one unit apart, and the two rows are two units apart. How many trapezoids which are not parallelograms have vertices in this set of $16$ points and have area of at least six square units? [asy] import graph; size(7cm); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; dot((-2,4),linewidth(6pt) + dotstyle); dot((-1,4),linewidth(6pt) + dotstyle); dot((0,4),linewidth(6pt) + dotstyle); dot((1,4),linewidth(6pt) + dotstyle); dot((2,4),linewidth(6pt) + dotstyle); dot((3,4),linewidth(6pt) + dotstyle); dot((4,4),linewidth(6pt) + dotstyle); dot((-3,2),linewidth(6pt) + dotstyle); dot((-2,2),linewidth(6pt) + dotstyle); dot((-1,2),linewidth(6pt) + dotstyle); dot((0,2),linewidth(6pt) + dotstyle); dot((1,2),linewidth(6pt) + dotstyle); dot((2,2),linewidth(6pt) + dotstyle); dot((3,2),linewidth(6pt) + dotstyle); dot((4,2),linewidth(6pt) + dotstyle); dot((5,2),linewidth(6pt) + dotstyle); [/asy]