Let $n$ be a given positive integer. Find the smallest positive integer $k$ for which the following statement is true: for any given simple connected graph $G$ and minimal cuts $V_1, V_2,\ldots, V_n$, at most $k$ vertices can be chosen with the property that picking any two of the chosen vertices there exists an integer $1\le i\le n$ such that $V_i$ separates the two vertices. A partition of the vertices of $G$ into two disjoint non-empty sets is called a [i]minimal cut[/i] if the number of edges crossing the partition is minimal. [i]Proposed by András Imolay, Budapest[/i]

Prove that if $n$ is large enough, then for each coloring of the subsets of the set $\{1,2,...,n\}$ with $1391$ colors, two non-empty disjoint subsets $A$ and $B$ exist such that $A$, $B$ and $A\cup B$ are of the same color.

In a family album, there are ten photos. On each of them, three people are pictured: in the middle stands a man, to the right of him stands his brother, and to the left of him stands his son. What is the least possible total number of people pictured, if all ten of the people standing in the middle of the ten pictures are different.

In a graph with $2n-1$ vertices throwing out any vertex the remaining graph has a complete subgraph with $n$ vertices. Prove that the initial graph has a complete subgraph with $n+1$ vertices.

Is it possible for any finite group of people to choose a positive integer $N$ and assign a positive integer to each person in the group such that the product of two persons' number is divisible by $N$ if and only if they are friends?

Prove that there is a constant $c>0$ such that for any $n>3$ there exists a planar graph $G$ with $n$ vertices such that every straight-edged plane embedding of $G$ has a pair of edges with ratio of lengths at least $cn$.

$2017$ engineers attend a conference. Any two engineers if they converse, converse with each other in either Chinese or English. No two engineers converse with each other more than once. It is known that within any four engineers, there was an even number of conversations and furthermore within this even number of conversations: i) At least one conversation is in Chinese. ii) Either no conversations are in English or the number of English conversations is at least that of Chinese conversations. Show that there exists $673$ engineers such that any two of them conversed with each other in Chinese.

Let $n>1$ be an integer. Given a simple graph $G$ on $n$ vertices $v_1, v_2, \dots, v_n$ we let $k(G)$ be the minimal value of $k$ for which there exist $n$ $k$-dimensional rectangular boxes $R_1, R_2, \dots, R_n$ in a $k$-dimensional coordinate system with edges parallel to the axes, so that for each $1\leq i<j\leq n$, $R_i$ and $R_j$ intersect if and only if there is an edge between $v_i$ and $v_j$ in $G$. Define $M$ to be the maximal value of $k(G)$ over all graphs on $n$ vertices. Calculate $M$ as a function of $n$.

There are $n \geq 3$ islands in a city. Initially, the ferry company offers some routes between some pairs of islands so that it is impossible to divide the islands into two groups such that no two islands in different groups are connected by a ferry route. After each year, the ferry company will close a ferry route between some two islands $X$ and $Y$. At the same time, in order to maintain its service, the company will open new routes according to the following rule: for any island which is connected to a ferry route to exactly one of $X$ and $Y$, a new route between this island and the other of $X$ and $Y$ is added. Suppose at any moment, if we partition all islands into two nonempty groups in any way, then it is known that the ferry company will close a certain route connecting two islands from the two groups after some years. Prove that after some years there will be an island which is connected to all other islands by ferry routes.

$110$ teams participate in a volleyball tournament. Every team has played every other team exactly once (there are no ties in volleyball). Turns out that in any set of $55$ teams, there is one which has lost to no more than $4$ of the remaining $54$ teams. Prove that in the entire tournament, there is a team that has lost to no more than $4$ of the remaining $109$ teams.

Let $n \geq 3$ be an integer. Two players play a game on an empty graph with $n + 1$ vertices, consisting of the vertices of a regular n-gon and its center. They alternately select a vertex of the n-gon and draw an edge (that has not been drawn) to an adjacent vertex on the n-gon or to the center of the n-gon. The player who first makes the graph connected wins. Between the player who goes first and the player who goes second, who has a winning strategy? [i]Note: an empty graph is a graph with no edges.[/i]

In a tournament, every two of the $n$ players played exactly one match with each other (no draws). Prove that it is possible either (i) to partition the league in two groups $A$ and $B$ such that everybody in $A$ defeated everybody in $B$; or (ii) to arrange all the players in a chain $x_1, x_2, . . . , x_n, x_1$ in such a way that each player defeated his successor.

In Sikinia, there are $2024$ cities. Between some of them there are flight connections, which can be used in either direction. No city has a direct flight to all $2023$ other cities. It is known, however, that there is a positive integer $n$ with the following property: For any $n$ cities in Sikinia, there is another city which is directly connected to all these cities. Determine the largest possible value of $n$.

Given the graph $G$ and cycle $C$ in it, we can perform the following operation: add another vertex $v$ to the graph, connect it to all vertices in $C$ and erase all the edges from $C$. Prove that we cannot perform the operation indefinitely on a given graph.

$n\ge 3$ guests met at a party. Some of them know each other but there is no quartet of different guests $a, b, c, d$ such that in pairs $\lbrace a, b \rbrace, \lbrace b, c \rbrace, \lbrace c, d \rbrace, \lbrace d, a \rbrace$ guests know each other but in pairs $\lbrace a, c \rbrace, \lbrace b, d \rbrace$ guests don't know each other. We say a nonempty set of guests $X$ is an [i]ingroup[/i], when guests from $X$ know each other pairwise and there are no guests not from $X$ knowing all guests from $X$. Prove that there are at most $\frac{n(n-1)}{2}$ different ingroups at that party.

There are $ n\ge 1 $ ordered bulbs controlled by $ n $ ordered switches such that the $ k\text{-th} $ switch controls the $ k\text{-th} $ bulb and also the $ j\text{-th} $ bulb if and only if the $ j\text{-th} $ switch controls the $ k\text{-th} $ bulb, for any $ 1\le k,j\le n. $ If all bulbs are off, show that it can be chosen some switches such that, if pushed simmultaneously, the bulbs turn all on.

A hilly island has $2023$ lookouts. It is known that each of them is in line of sight with at least $42$ of the other lookouts. For any two distinct lookouts $X$ and $Y$ there is a positive integer $n$ and lookouts $A_1,A_2,\dots,A_{n+1}$ such that $A_1=X$ and $A_{n+1}=Y$ and $A_1$ is in line of sight with $A_2$, $A_2$ with $A_3$, $\dots$ and $A_n$ with $A_{n+1}$. The smallest such number $n$ is called the [i]viewing distance[/i] of $X$ and $Y$. Determine the largest possible viewing distance that can exist between two lookouts under these conditions.

On a table there are 2004 boxes, and in each box a ball lies. I know that some the balls are white and that the number of white balls is even. In each case I may point to two arbitrary boxes and ask whether in the box contains at least a white ball lies. After which minimum number of questions I can indicate two boxes for sure, in which white balls lie?

In a football tournament $2n$ teams play in a round. Every round consists of $n$ pairs of teams that haven’t played with each other yet. Every round’s schedule is determined before the round is held. Find the minimal positive integer $k$ such that the following situation is possible: after $k$ rounds it’s impossible to schedule the next round.

Let $M$ be a positive integer. At a party with 120 people, 30 wear red hats, 40 wear blue hats, and 50 wear green hats. Before the party begins, $M$ pairs of people are friends. (Friendship is mutual.) Suppose also that no two friends wear the same colored hat to the party. During the party, $X$ and $Y$ can become friends if and only if the following two conditions hold: [list] [*] There exists a person $Z$ such that $X$ and $Y$ are both friends with $Z$. (The friendship(s) between $Z,X$ and $Z,Y$ could have been formed during the party.) [*] $X$ and $Y$ are not wearing the same colored hat. [/list] Suppose the party lasts long enough so that all possible friendships are formed. Let $M_1$ be the largest value of $M$ such that regardless of which $M$ pairs of people are friends before the party, there will always be at least one pair of people $X$ and $Y$ with different colored hats who are not friends after the party. Let $M_2$ be the smallest value of $M$ such that regardless of which $M$ pairs of people are friends before the party, every pair of people $X$ and $Y$ with different colored hats are friends after the party. Find $M_1+M_2$. [hide="Clarifications"] [list] [*] The definition of $M_2$ should read, ``Let $M_2$ be the [i]smallest[/i] value of $M$ such that...''. An earlier version of the test read ``largest value of $M$''.[/list][/hide] [i]Victor Wang[/i]

Let $\mathfrak F$ be the family of all $k$-element subsets of the set $\{1, 2, \ldots, 2k + 1\}$. Prove that there exists a bijective function $f :\mathfrak F \to \mathfrak F$ such that for every $A \in \mathfrak F$, the sets $A$ and $f(A)$ are disjoint.

A directed graph has each vertex with outdegree 2. Prove that it is possible to split the vertices into 3 sets so that for each vertex $v$, $v$ is not simultaneously in the same set with both of the vertices that it points to. [i]David Yang.[/i] [hide="Stronger Version"]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=492100]here[/url].[/hide]

The kingdom of Anisotropy consists of $n$ cities. For every two cities there exists exactly one direct one-way road between them. We say that a [i]path from $X$ to $Y$[/i] is a sequence of roads such that one can move from $X$ to $Y$ along this sequence without returning to an already visited city. A collection of paths is called [i]diverse[/i] if no road belongs to two or more paths in the collection. Let $A$ and $B$ be two distinct cities in Anisotropy. Let $N_{AB}$ denote the maximal number of paths in a diverse collection of paths from $A$ to $B$. Similarly, let $N_{BA}$ denote the maximal number of paths in a diverse collection of paths from $B$ to $A$. Prove that the equality $N_{AB} = N_{BA}$ holds if and only if the number of roads going out from $A$ is the same as the number of roads going out from $B$. [i]Proposed by Warut Suksompong, Thailand[/i]

On a $16 \times 16$ torus as shown all $512$ edges are colored red or blue. A coloring is [i]good [/i]if every vertex is an endpoint of an even number of red edges. A move consists of switching the color of each of the $4$ edges of an arbitrary cell. What is the largest number of good colorings so that none of them can be converted to another by a sequence of moves?

For a positive integer \( n \geq 6 \), find the smallest integer \( S(n) \) such that any graph with \( n \) vertices and at least \( S(n) \) edges must contain at least two disjoint cycles (cycles with no common vertices).