Let $n$ be a positive integer. We start with $n$ piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. Find (in terms of $n$) the smallest number of nonempty piles that one can obtain by performing a finite sequence of moves of this form.

The cirumcentre of the cyclic quadrilateral $ABCD$ is $O$. The second intersection point of the circles $ABO$ and $CDO$, other than $O$, is $P$, which lies in the interior of the triangle $DAO$. Choose a point $Q$ on the extension of $OP$ beyond $P$, and a point $R$ on the extension of $OP$ beyond $O$. Prove that $\angle QAP=\angle OBR$ if and only if $\angle PDQ=\angle RCO$.

Kobar and Borah are playing on a whiteboard with the following rules: They start with two distinct positive integers on the board. On each step, beginning with Kobar, each player takes turns changing the numbers on the board, either from $P$ and $Q$ to $2P-Q$ and $2Q-P$, or from $P$ and $Q$ to $5P-4Q$ and $5Q-4P$. The game ends if a player writes an integer that is not positive. That player is declared to lose, and the opponent is declared the winner. At the beginning of the game, the two numbers on the board are $2024$ and $A$. If it is known that Kobar does not lose on his first move, determine the largest possible value of $A$ so that Borah can win this game.

The three roots of $P(x) = x^3 - 2x^2 - x + 1$ are $a>b>c \in \mathbb{R}$. Find the value of $a^2b+b^2c+c^2a$. :D

We are given a natural number $k$. Let us consider the following game on an infinite onedimensional board. At the start of the game, we distrubute $n$ coins on the fields of the given board (one field can have multiple coins on itself). After that, we have two choices for the following moves: $(i)$ We choose two nonempty fields next to each other, and we transfer all the coins from one of the fields to the other. $(ii)$ We choose a field with at least $2$ coins on it, and we transfer one coin from the chosen field to the $k-\mathrm{th}$ field on the left , and one coin from the chosen field to the $k-\mathrm{th}$ field on the right. $\mathbf{(a)}$ If $n\leq k+1$, prove that we can play only finitely many moves. $\mathbf{(b)}$ For which values of $k$ we can choose a natural number $n$ and distribute $n$ coins on the given board such that we can play infinitely many moves.

Let $f_n\ (n=1,\ 2,\ \cdots)$ be a linear transformation expressed by a matrix $\left( \begin{array}{cc} 1-n & 1 \\ -n(n+1) & n+2 \end{array} \right)$ on the $xy$ plane. Answer the following questions: (1) Prove that there exists 2 lines passing through the origin $O(0,\ 0)$ such that all points of the lines are mapped to the same lines, then find the equation of the lines. (2) Find the area $S_n$ of the figure enclosed by the lines obtained in (1) and the curve $y=x^2$. (3) Find $\sum_{n=1}^{\infty} \frac{1}{S_n-\frac 16}.$ [i]2011 Tokyo Institute of Technlogy entrance exam, Problem 1[/i]

Let $n \ge 3$ be a fixed integer. A game is played by $n$ players sitting in a circle. Initially, each player draws three cards from a shuffled deck of $3n$ cards numbered $1, 2, \dots, 3n$. Then, on each turn, every player simultaneously passes the smallest-numbered card in their hand one place clockwise and the largest-numbered card in their hand one place counterclockwise, while keeping the middle card. Let $T_r$ denote the configuration after $r$ turns (so $T_0$ is the initial configuration). Show that $T_r$ is eventually periodic with period $n$, and find the smallest integer $m$ for which, regardless of the initial configuration, $T_m=T_{m+n}$. [i]Proposed by Carl Schildkraut and Colin Tang[/i]

A set of $n$ points in Euclidean 3-dimensional space, no four of which are coplanar, is partitioned into two subsets $\mathcal{A}$ and $\mathcal{B}$. An $\mathcal{AB}$-tree is a configuration of $n-1$ segments, each of which has an endpoint in $\mathcal{A}$ and an endpoint in $\mathcal{B}$, and such that no segments form a closed polyline. An $\mathcal{AB}$-tree is transformed into another as follows: choose three distinct segments $A_1B_1$, $B_1A_2$, and $A_2B_2$ in the $\mathcal{AB}$-tree such that $A_1$ is in $\mathcal{A}$ and $|A_1B_1|+|A_2B_2|>|A_1B_2|+|A_2B_1|$, and remove the segment $A_1B_1$ to replace it by the segment $A_1B_2$. Given any $\mathcal{AB}$-tree, prove that every sequence of successive transformations comes to an end (no further transformation is possible) after finitely many steps.

On the table lay a pencil, sharpened at one end. The student can rotate the pencil around one of its ends at $45^{\circ}$ clockwise or counterclockwise. Can the student, after a few turns of the pencil, go back to the starting position so that the sharpened end and the not sharpened are reversed?

Let $n$ be a positive integer. Given a sequence $\varepsilon_1$, $\dots$, $\varepsilon_{n - 1}$ with $\varepsilon_i = 0$ or $\varepsilon_i = 1$ for each $i = 1$, $\dots$, $n - 1$, the sequences $a_0$, $\dots$, $a_n$ and $b_0$, $\dots$, $b_n$ are constructed by the following rules: \[a_0 = b_0 = 1, \quad a_1 = b_1 = 7,\] \[\begin{array}{lll} a_{i+1} = \begin{cases} 2a_{i-1} + 3a_i, \\ 3a_{i-1} + a_i, \end{cases} & \begin{array}{l} \text{if } \varepsilon_i = 0, \\ \text{if } \varepsilon_i = 1, \end{array} & \text{for each } i = 1, \dots, n - 1, \\[15pt] b_{i+1}= \begin{cases} 2b_{i-1} + 3b_i, \\ 3b_{i-1} + b_i, \end{cases} & \begin{array}{l} \text{if } \varepsilon_{n-i} = 0, \\ \text{if } \varepsilon_{n-i} = 1, \end{array} & \text{for each } i = 1, \dots, n - 1. \end{array}\] Prove that $a_n = b_n$. [i]Proposed by Ilya Bogdanov, Russia[/i]

There are $n$ sheep and a wolf in sheep's clothing . Some of the sheep are friends (friendship is mutual). The goal of the wolf is to eat all the sheep. First, the wolf chooses some sheep to make friend's with. In each of the following days, the wolf eats one of its friends. Whenever the wolf eats a sheep $A$: (a) If a friend of $A$ is originally a friend of the wolf, it un-friends the wolf. (b) If a friend of $A$ is originally not a friend of the wolf, it becomes a friend of the wolf. Repeat the procedure until the wolf has no friend left. Find the largest integer $m$ in terms of $n$ satisfying the following: There exists an initial friendsheep structure such that the wolf has $m$ different ways of choosing initial sheep to become friends, so that the wolf has a way to eat all of the sheep.

Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$, $\omega_{B}$ to $BC$ and $BA$, $\omega_{C}$ to $CA$ and $CB$, and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$. If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

There are $n+1$ containers arranged in a circle. One container has $n$ stones, the others are empty. A move is to choose two containers $A$ and $B$, take a stone from $A$ and put it in one of the containers adjacent to $B$, and to take a stone from $B$ and put it in one of the containers adjacent to $A$. We can take $A = B$. For which $n$ is it possible by series of moves to end up with one stone in each container except that which originally held $n$ stones.

George plays the following game: At every step he can replace a triple of integers $(x,y,z)$ which is written on the blackboard, with any of the following triples: (i) $(x,z,y)$ (ii) $(-x,y,z)$ (iii) $(x+y,y,2x+y+z)$ (iv) $(x-y,y,y+z-2x)$ Initially, the triple $(1,1,1)$ is written on the blackboard. Determine whether George can, with a sequence of allowed steps, end up at the triple $(2021,2019,2023)$, fully justifying your answer.