A [i]windmill[/i] is a closed line segment of unit length with a distinguished endpoint, the [i]pivot[/i]. Let $S$ be a finite set of $n$ points such that the distance between any two points of $S$ is greater than $c$. A configuration of $n$ windmills is [i]admissible[/i] if no two windmills intersect and each point of $S$ is used exactly once as a pivot. An admissible configuration of windmills is initially given to Geoff in the plane. In one operation Geoff can rotate any windmill around its pivot, either clockwise or counterclockwise and by any amount, as long as no two windmills intersect during the process. Show that Geoff can reach any other admissible configuration in finitely many operations, where (i) $c = \sqrt 3$, (ii) $c = \sqrt 2$. [i]Proposed by Michael Ren[/i]

Fred and George play a game, as follows. Initially, $x = 1$. Each turn, they pick $r \in \{3,5,8,9\}$ uniformly at random and multiply $x$ by $r$. If $x+1$ is a multiple of 13, Fred wins; if $x+3$ is a multiple of 13, George wins; otherwise, they repeat. Determine the probability that Fred wins the game.

$X$ has $n$ elements. $F$ is a family of subsets of $X$ each with three elements, such that any two of the subsets have at most one element in common. Show that there is a subset of $X$ with at least $\sqrt{2n}$ members which does not contain any members of $F$.

An $n$-tuple $(a_1, a_2 . . . , a_n)$ is [i]occasionally periodic[/i] if there exist a non-negative integer $i$ and a positive integer $p$ satisfying $i + 2p \le n$ and $a_{i+j} = a_{i+j+p}$ for every $j = 1, 2, . . . , p$. Let $k$ be a positive integer. Find the least positive integer $n$ for which there exists an $n$-tuple $(a_1, a_2 . . . , a_n)$ with elements from the set $\{1, 2, . . . , k\}$, which is not occasionally periodic but whose arbitrary extension $(a_1, a_2, . . . , a_n, a_{n+1})$ is occasionally periodic for any $a_{n+1} \in \{1, 2, . . . , k\}$.

The Bank of Oslo issues two types of coin: aluminum (denoted A) and bronze (denoted B). Marianne has $n$ aluminum coins and $n$ bronze coins arranged in a row in some arbitrary initial order. A chain is any subsequence of consecutive coins of the same type. Given a fixed positive integer $k \leq 2n$, Gilberty repeatedly performs the following operation: he identifies the longest chain containing the $k^{th}$ coin from the left and moves all coins in that chain to the left end of the row. For example, if $n=4$ and $k=4$, the process starting from the ordering $AABBBABA$ would be $AABBBABA \to BBBAAABA \to AAABBBBA \to BBBBAAAA \to ...$ Find all pairs $(n,k)$ with $1 \leq k \leq 2n$ such that for every initial ordering, at some moment during the process, the leftmost $n$ coins will all be of the same type.

Sixteen wooden Cs are placed in a $4$-by-$4$ grid, all with the same orientation, and each is to be colored either red or blue. A quadrant operation on the grid consists of choosing one of the four two-by-two subgrids of Cs found at the corners of the grid and moving each C in the subgrid to the adjacent square in the subgrid that is $90$ degrees away in the clockwise direction, without changing the orientation of the C. Given that two colorings are the considered same if and only if one can be obtained from the other by a series of quadrant operations, determine the number of distinct colorings of the Cs. [img]https://cdn.artofproblemsolving.com/attachments/a/9/1e59dce4d33374960953c0c99343eef807a5d2.png[/img]

Given a $2025 \times 2025$ board and $k$ chips lying on the table. Initially, the board is empty. It is allowed to place a chip from the table on any free square $A$ if two conditions are met simultaneously: – the cell next to $A$ from below has a chip (or $A$ is on the bottom edge of the board); – the cell next to $A$ on the left has a chip (or $A$ is on the left edge of the board). In addition, it is allowed to remove any chip from the board and put it on the table. At what minimum $k$ can a chip appear in the upper-right corner of the board?

What is the largest number of knights that can be put on a $5 \times 5$ chess board so that each knight attacks exactly two other knights? (M Gorelov)

A frog is located on a unit square of an infinite grid oriented according to the cardinal directions. The frog makes moves consisting of jumping either one or two squares in the direction it is facing, and then turning according to the following rules: i) If the frog jumps one square, it then turns $90^\circ$ to the right; ii) If the frog jumps two squares, it then turns $90^\circ$ to the left. Is it possible for the frog to reach the square exactly $2024$ squares north of the initial square after some finite number of moves if it is initially facing: a) North; b) East?

For each sequence of $n$ zeros and $n$ units, we assign a number that is a number sections of the same digits in it. (For example, sequence $00111001$ has $4$ such sections $00, 111,00, 1$.) For a given $n$ we sum up all the numbers assigned to each such sequence. Prove that the sum total is equal to $(n+1){2n \choose n} $

a) A game master divides a group of $40$ players into four teams of ten. The players do not know what the teams are, however the master gives each player a card containing the names of two other players: one of them is a teammate and the other is not, but the master does not tell the player which is which. Can the master write the names on the cards in such a way that the players can determine the teams? (All of the players can work together to do so.) b) On the next occasion, the game master writes the names of $7$ teammates and $2$ opposing players on each card (possibly in a mixed up order). Now he wants to write the names in such a way that the players together cannot determine the four teams. Is it possible for him to achieve this? c) Can he write the names in such a way that the players together cannot determine the four teams, if now each card contains the names of $6$ teammates and $2$ opposing players (possibly in a mixed up order)?

Let $a_1, a_2, \cdots, a_{2n}$ be $2n$ elements of $\{1, 2, 3, \cdots, 2n-1\}$ ($n>3$) with the sum $a_1+a_2+\cdots+a_{2n}=4n$. Prove that exist some numbers $a_i$ with the sum is $2n$.

Given two natural numbers $a < b$, Xavier and Ze play the following game. First, Xavier writes $a$ consecutive numbers of his choice; then, repeat some of them, also of his choice, until he has $b$ numbers, with the condition that the sum of the $b$ numbers written is an even number. Ze wins the game if he manages to separate the numbers into two groups with the same amount. Otherwise, Xavier wins. For example, for $a = 4$ and $b = 7$, if Xavier wrote the numbers $3,4,5,6,3,3,4$, Ze could win, separating these numbers into groups $3,3 ,4,4$ and $3,5,6$. For what values of $a$ and $b$ can Xavier guarantee victory?

Given a $101\times 200$ sheet of graph paper, we start moving from a corner square in the direction of the square’s diagonal (not the sheet’s diagonal) to the border of the sheet, then change direction obeying the laws of light’s reflection. Will we ever reach a corner square? [img]https://cdn.artofproblemsolving.com/attachments/b/8/4ec2f4583f406feda004c7fb4f11a424c9b9ae.png[/img]

[b]O[/b]n a blackboard a stranger writes the values of $s_7(n)^2$ for $n=0,1,...,7^{20}-1$, where $s_7(n)$ denotes the sum of digits of $n$ in base $7$. Compute the average value of all the numbers on the board.

There are $n$ bus stops placed around the circular lake. Each bus stop is connected by a road to the two adjacent stops (we call a [i]segment [/i] the entire road between two stops). Determine the number of bus routes that start and end in the fixed bus stop A, pass through each bus stop at least once and travel through exactly $n+1$ [i]segments[/i].

There are There are $64$ towns in a country, and some pairs of towns are connected by roads but we do not know these pairs. We may choose any pair of towns and find out whether they are connected by a road. Our aim is to determine whether it is possible to travel between any two towns using roads. Prove that there is no algorithm which would enable us to do this in less than $2016$ questions. but we do not know these pairs. We may choose any pair of towns and find out whether they are connected by a road. Our aim is to determine whether it is possible to travel between any two towns using roads. Prove that there is no algorithm which would enable us to do this in less than $2016$ questions.

Ten square cardboards of $3$ centimeters on a side are cut by a line, as indicated in the figure. After the cuts, there are $20$ pieces: $10$ triangles and $10$ trapezoids. Assemble a square that uses all $20$ pieces without overlaps or gaps. [img]https://cdn.artofproblemsolving.com/attachments/7/9/ec2242cca617305b02eef7a5409e6a6b482d66.gif[/img]

Let $X$ be a set containing $n$ elements. Find the number of ordered triples $(A,B, C)$ of subsets of $X$ such that $A$ is a subset of $B$ and $B$ is a proper subset of $C$.

The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point \(A\). At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path \(AJABCHCHIJA\), which has \(10\) steps. Let \(n\) be the number of paths with \(15\) steps that begin and end at point \(A\). Find the remainder when \(n\) is divided by \(1000\). [asy] unitsize(32); draw(unitcircle); draw(scale(2) * unitcircle); for(int d = 90; d < 360 + 90; d += 72){ draw(2 * dir(d) -- dir(d)); } real s = 4; dot(1 * dir( 90), linewidth(s)); dot(1 * dir(162), linewidth(s)); dot(1 * dir(234), linewidth(s)); dot(1 * dir(306), linewidth(s)); dot(1 * dir(378), linewidth(s)); dot(2 * dir(378), linewidth(s)); dot(2 * dir(306), linewidth(s)); dot(2 * dir(234), linewidth(s)); dot(2 * dir(162), linewidth(s)); dot(2 * dir( 90), linewidth(s)); defaultpen(fontsize(10pt)); real r = 0.05; label("$A$", (1-r) * dir( 90), -dir( 90)); label("$B$", (1-r) * dir(162), -dir(162)); label("$C$", (1-r) * dir(234), -dir(234)); label("$D$", (1-r) * dir(306), -dir(306)); label("$E$", (1-r) * dir(378), -dir(378)); label("$F$", (2+r) * dir(378), dir(378)); label("$G$", (2+r) * dir(306), dir(306)); label("$H$", (2+r) * dir(234), dir(234)); label("$I$", (2+r) * dir(162), dir(162)); label("$J$", (2+r) * dir( 90), dir( 90)); [/asy]

There are $N$ cities in a country. Any two of them are connected either by a road or by an airway. A tourist wants to visit every city exactly once and return to the city at which he started the trip. Prove that he can choose a starting city and make a path, changing means of transportation at most once.

There is a $8*8$ board and the numbers $1,2,3,4,...,63,64$. In all the unit squares of the board, these numbers are places such that only $1$ numbers goes to only one unit square. Prove that there is atleast $4$ $2*2$ squares such that the sum of the numbers in $2*2$ is greater than $100$.

Given an integer $n \ge 2$ and a regular $2n$-polygon at each vertex of which sitting on an ant. At some points in time, each ant creeps into one of two adjacent peaks (some peaks may have several ants at a time). Through $k$ such operations, it turned out to be an arbitrary line connecting two different ones the vertices of a polygon with ants do not pass through its center. For given $n$ find the lowest possible value of $k$.

Is it true that in any $2$-connected graph with a countably infinite number of vertices it's always possible to find a trail that is infinite in one direction? [i]Submitted by Balázs Bursics and Anett Kocsis, Budapest[/i]

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]