A sequence of natural numbers $a_1,a_2,...$ satisfies $a_1 = 1, a_{n+2} = 2a_{n+1} - a_n +2$ for $n \in N$. Prove that for every natural $n$ there exists a natural $m$ such that $a_na_{n+1} = a_m$.

The paper is written on consecutive integers $1$ through $n$. Then are deleted all numbers ending in $4$ and $9$ and the rest alternating between $-$ and $+$. Finally, an opening parenthesis is added after each character and at the end of the expression the corresponding number of parentheses: $1 - (2 + 3 - (5 + 6 - (7 + 8 - (10 +...))))$. Find all numbers $n$ such that the value of this expression is $13$.

Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m.$

Let $n$ be a given positive integer. Sisyphus performs a sequence of turns on a board consisting of $n + 1$ squares in a row, numbered $0$ to $n$ from left to right. Initially, $n$ stones are put into square $0$, and the other squares are empty. At every turn, Sisyphus chooses any nonempty square, say with $k$ stones, takes one of these stones and moves it to the right by at most $k$ squares (the stone should say within the board). Sisyphus' aim is to move all $n$ stones to square $n$. Prove that Sisyphus cannot reach the aim in less than \[ \left \lceil \frac{n}{1} \right \rceil + \left \lceil \frac{n}{2} \right \rceil + \left \lceil \frac{n}{3} \right \rceil + \dots + \left \lceil \frac{n}{n} \right \rceil \] turns. (As usual, $\lceil x \rceil$ stands for the least integer not smaller than $x$. )

A ring is of the shape of a hoola-hoop of negligible thickness. A ring of radius $R$ carries a current $I$. Prove that the magnetic field at a given point in the plane of the ring at a distance $a$ from the center, due to the magnetic field of the ring, is \[ B = \dfrac {\mu_0}{2\pi} \cdot IR \cdot \displaystyle\int_{0}^{\pi} \dfrac {R - a \cos \theta}{\sqrt{\left( a^2 + R^2 - 2aR \cos \theta \right)^3}} \, \mathrm{d}\theta. \] [i]Problem proposed by Ahaan Rungta[/i]

Find all functions $f:\mathbb{Q}\rightarrow\mathbb{R}$ such that \[ f(x)^2-f(y)^2=f(x+y)\cdot f(x-y), \] for all $x,y\in \mathbb{Q}$. [i]Proposed by Portugal.[/i]

A trapezoid is divided into seven strips of equal width as shown. What fraction of the trapezoid’s area is shaded?

Given a real number $t\ge3$, suppose a polynomial $f\in\mathbb R[x]$ satisfies $$\left|f(k)-t^k\right|<1,\enspace\forall k=0,1,\ldots,n.$$Prove that $\deg f\ge n$.

Let $a_1$ be any positive integer. For all $i$, write $5^{2020}$ times $a_i$ in base $10$, replace each digit with its remainder when divided by $2$, read off the result in binary, and call that $a_{i+1}$. Prove that $a_N = a_{N+2^{2020}}$ for all sufficiently large $N$.

Let $x$ be a real number between 0 and $\tfrac{\pi}{2}$ for which the function $3\sin^2 x + 8\sin x \cos x + 9\cos^2 x$ obtains its maximum value, $M$. Find the value of $M + 100\cos^2x$.

Prove that the infinite series $ 1\minus{}\frac{1}{x(x\plus{}1)}\minus{}\frac{x\minus{}1}{2!x^2(2x\plus{}1)}\minus{}\frac{(x\minus{}1)(2x\minus{}1)}{3!(x^3(3x\plus{}1))}\minus{}\frac{(x\minus{}1)(2x\minus{}1)(3x\minus{}1)}{4!x^4(4x\plus{}1)}\minus{}\cdots$ is convergent for every positive $ x$. Denoting its sum by $ F(x)$, find $ \lim_{x\to \plus{}0}F(x)$ and $ \lim_{x\to \infty}F(x)$.

Given a line $p$ and a triangle $\Delta$ in the plane, construct an equilateral triangle one of whose vertices lies on the line $p$, while the other two halve the perimeter of $\Delta.$

Solve the system of equations $$\begin{cases} x+y+z+t=6 \\ \sqrt{1-x^2}+\sqrt{4-y^2}+\sqrt{9-z^2}+\sqrt{16-t^2}=8 \end{cases}$$

The vertices of a convex $1991$-gon are enumerated with integers from $1$ to $1991$. Each side and diagonal of the $1991$-gon is colored either red or blue. Prove that, for an arbitrary renumeration of vertices, one can find integers $k$ and $l$ such that the segment connecting the vertices numbered $k$ and $l$ before the renumeration has the same color as the segment connecting the vertices numbered $k$ and $l$ after the renumeration.

There are $n$ cities in a country, where $n>1$. There are railroads connecting some of the cities so that you can travel between any two cities through a series of railroads (railroads run in both direction.) In addition, in this country, it is impossible to travel from a city, through a series of distinct cities, and return back to the original city. We define the [b]degree[/b] of a city as the number of cities directly connected to it by a single segment of railroad. For a city $A$ that is directly connected to $x$ cities, with $y$ of those cities having a smaller degree than city $A$, the [b]significance[/b] of city $A$ is defined as $\frac{y}{x}$. Find the smallest positive real number $t$ so that, for any $n>1$, the sum of the significance of all cities is less than $tn$, no matter how the railroads are paved. [i]Proposed by houkai[/i]

Find a countable family of natural solutions to $ \frac{1}{a} +\frac{1}{b} +\frac{1}{ab}=\frac{1}{c} . $

Let $\Gamma$ be the circumcircle of isosceles triangle $ABC$ with vertex $C$. An arbitrary point $M$ is chosen on the segment $BC$ and point $N$ lies on the ray $AM$ with $M$ between $A,N$ such that $AN=AC$. The circumcircle of $CMN$ cuts $\Gamma$ in $P$ other than $C$ and $AB,CP$ intersect at $Q$. Prove that $\angle BMQ = \angle QMN.$

In $\triangle ABC$ with $AB=10$, $AC=13$, and $\measuredangle ABC = 30^\circ$, $M$ is the midpoint of $\overline{BC}$ and the circle with diameter $\overline{AM}$ meets $\overline{CB}$ and $\overline{CA}$ again at $D$ and $E$, respectively. The area of $\triangle DEM$ can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m, n$. Compute $100m + n$. [i]Based on a proposal by Matthew Babbitt[/i]

Find all primes $ p,q $ such that $ \alpha^{3pq} -\alpha \equiv 0 \pmod {3pq} $ for all integers $ \alpha $.

Given that $4^{x_1} = 5, 5^{x_2} = 6, \dots , 2047^{x_{2044}} = 2048$, compute the product $x_1 \dots x_{2044}$.

Each square of a $7 \times 8$ board is painted black or white, in such a way that each $3 \times 3$ subboard has at least two black squares that are neighboring. What is the least number of black squares that can be on the entire board? Clarification: Two squares are [i]neighbors [/i] if they have a common side.

Nine copies of a certain pamphlet cost less than $ \$10.00$ while ten copies of the same pamphlet (at the same price) cost more than $ \$11.00$. How much does one copy of this pamphlet cost? \[ \textbf{(A)}\ \$1.07 \qquad \textbf{(B)}\ \$1.08 \qquad \textbf{(C)}\ \$1.09 \qquad \textbf{(D)}\ \$1.10 \qquad \textbf{(E)}\ \$1.11 \]

Let ${ABC}$ be an acute angled triangle, let ${O}$ be its circumcentre, and let ${D,E,F}$ be points on the sides ${BC,CA,AB}$, respectively. The circle ${(c_1)}$ of radius ${FA}$, centered at ${F}$, crosses the segment ${OA}$ at ${A'}$ and the circumcircle ${(c)}$ of the triangle ${ABC}$again at ${K}$. Similarly, the circle ${(c_2)}$ of radius $DB$, centered at $D$, crosses the segment $\left( OB \right)$ at ${B}'$ and the circle ${(c)}$ again at ${L}$. Finally, the circle ${(c_3)}$ of radius $EC$, centered at $E$, crosses the segment $\left( OC \right)$at ${C}'$ and the circle ${(c)}$ again at ${M}$. Prove that the quadrilaterals $BKF{A}',CLD{B}'$ and $AME{C}'$ are all cyclic, and their circumcircles share a common point. Evangelos Psychas (Greece)

The game of rock-scissors is played just like rock-paper-scissors, except that neither player is allowed to play paper. You play against a poorly-designed computer program that plays rock with $50\%$ probability and scissors with $50\%$ probability. If you play optimally against the computer, find the probability that after $8$ games you have won at least $4$. [i]In the game of rock-paper-scissors, two players each choose one of rock, paper, or scissors to play. Rock beats scissors, scissors beats paper, and paper beats rock. If the players play the same thing, the match is considered a draw.[/i]

Let $AD$ be the internal angle bisector of triangle $ABC$. The incircles of triangles $ABC$ and $ACD$ touch each other externally. Prove that $\angle ABC > 120^{\circ}$. (Recall that the incircle of a triangle is a circle inside the triangle that is tangent to its three sides.) [i]Proposed by Volodymyr Brayman (Ukraine)[/i]