A complete graph of $n$ vertices is a set of $n$ vertices and those vertices are connected in pairs by edges. Suppose the graph has $n$ vertices $A_1, A_2, ..., A_n$, the cycle is a set of edges of the form $A_{i_1}A_{i_2}, A_{i_2}A_{i_3},..., A_{i_m}A_{i_1}$ with $i_1, i_2, ..., i_m \in {1, 2, ..., n}$ double one different. We call $m$ the length of this cycle. Find the smallest positive integer$ n$ such that for every way of coloring all edges of a complete graph of $n$ vertices, each edge filled with one of three different colors, there is always a cycle of even length with the same color. PS. The same problem with another wording [url=https://artofproblemsolving.com/community/c6h151391p852296]here [/url].

Let $k$ be a given integer, $3 < k \leq n$. Consider a graph $G$ with $n$ vertices satisfying the condition: for any two non-adjacent vertices $x$ and $y$ in graph $G$, the sum of their degrees must satisfy $d(x) + d(y) \geq k$. Please answer the following questions and prove your conclusions. (1) Suppose the length of the longest path in graph $G$ is $l$ satisfying the inequality $3 \leq l < k$, does graph $G$ necessarily contain a cycle of length $l+1$? (The length of a path or cycle refers to the number of edges that make up the path or cycle.) (2) For the case where $3 < k \leq n-1$ and graph $G$ is connected, can we determine that the length of the longest path in graph $G$, $l \geq k$? (3) For the case where $3 < k = n-1$, is it necessary for graph $G$ to have a path of length $n-1$ (i.e., a Hamiltonian path)?

In a company of people some pairs are enemies. A group of people is called [i]unsociable[/i] if the number of members in the group is odd and at least $3$, and it is possible to arrange all its members around a round table so that every two neighbors are enemies. Given that there are at most $2015$ unsociable groups, prove that it is possible to partition the company into $11$ parts so that no two enemies are in the same part. [i]Proposed by Russia[/i]

Here $G_{n}$ denotes a simple undirected graph with $n$ vertices, $K_{n}$ denotes the complete graph with $n$ vertices, $K_{n,m}$ the complete bipartite graph whose components have $m$ and $n$ vertices, and $C_{n}$ a circuit with $n$ vertices. The number of edges in the graph $G_{n}$ is denoted $e(G_{n})$. The edges of $K_{n}(n \geq 3)$ are colored with $n$ colors, and every color is used. Show that there is a triangle whose sides have different colors.

The following operation is allowed on a finite graph: Choose an arbitrary cycle of length 4 (if there is any), choose an arbitrary edge in that cycle, and delete it from the graph. For a fixed integer ${n\ge 4}$, find the least number of edges of a graph that can be obtained by repeated applications of this operation from the complete graph on $n$ vertices (where each pair of vertices are joined by an edge). [i]Proposed by Norman Do, Australia[/i]

A tournament is a directed graph for which every (unordered) pair of vertices has a single directed edge from one vertex to the other. Let us define a proper directed-edge-coloring to be an assignment of a color to every (directed) edge, so that for every pair of directed edges $\overrightarrow{uv}$ and $\overrightarrow{vw}$, those two edges are in different colors. Note that it is permissible for $\overrightarrow{uv}$ and $\overrightarrow{uw}$ to be the same color. The directed-edge-chromatic-number of a tournament is defined to be the minimum total number of colors that can be used in order to create a proper directed-edge-coloring. For each $n$, determine the minimum directed-edge-chromatic-number over all tournaments on $n$ vertices. [i]Proposed by Po-Shen Loh[/i]

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 or not. Our aim is to determine whether it is possible to travel from any town to any other by a sequence of roads. Prove that there is no algorithm which enables us to do so in less than $2016$ questions. (Proposed by Konstantin Knop)

There are $4n$ pebbles of weights $1, 2, 3, \dots, 4n.$ Each pebble is coloured in one of $n$ colours and there are four pebbles of each colour. Show that we can arrange the pebbles into two piles so that the following two conditions are both satisfied: [list] [*]The total weights of both piles are the same. [*] Each pile contains two pebbles of each colour. [/list] [i]Proposed by Milan Haiman, Hungary and Carl Schildkraut, USA[/i]

We say that a graph $G$ is [i]divisive[/i], if we can write a positive integer on each of its vertices such that all the integers are distinct, and any two of these integers divide each other if and only if there is an edge running between them in $G$. Which Platonic solids form a divisive graph? [img]https://cdn.artofproblemsolving.com/attachments/1/5/7c81439ee148ccda09c429556e0740865723e0.png[/img]

A complete graph is in a plane such that no three of its vertices are collinear. The edges of the graph, which are straight segments connecting the vertices, are colored with two colors. Prove that there is a non-self-intersecting spanning tree consisting of edges of the same color.

Each two of $n$ students, who attended an activity, have different ages. It is given that each student shook hands with at least one student, who did not shake hands with anyone younger than the other. Find all possible values of $n$.

Let $n\geq 2$ be a positive integer. We call a [i]vertex[/i] every point in the coordinate plane, whose $x$ and $y$ coordinates both are from the set $\{1,2,3,...,n\}$. We call a segment between two vertices an [i]edge[/i], if its length if $1$. We've colored some edges red, such that between any two vertices, there is a unique path of red edges (a path may contain each edge at most once). The red edge $f$ is [i]vital[/i] for an edge $e$, if the path of red edges connecting the two endpoints of $e$ contain $f$. Prove that there is a red edge, such that it is vital for at least $n$ edges.

A $2010\times 2010$ board is divided into corner-shaped figures of three cells. Prove that it is possible to mark one cell in each figure such that each row and each column will have the same number of marked cells. [i]I. Bogdanov & O. Podlipsky[/i]

Given a simple, connected graph with $n$ vertices and $m$ edges. Prove that one can find at least $m$ ways separating the set of vertices into two parts, such that the induced subgraphs on both parts are connected.

We have a complete graph with $n$ vertices. We have to color the vertices and the edges in a way such that: no two edges pointing to the same vertice are of the same color; a vertice and an edge pointing him are coloured in a different way. What is the minimum number of colors we need?

Can the set of lattice points $\{(x, y) \mid x, y \in \mathbb{Z}, 1 \le x, y \le 252, x \neq y\}$ be colored using 10 distinct colors such that for all $a \neq b$, $b \neq c$, the colors of $(a, b)$ and $(b, c)$ are distinct?

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]

At an IMOTC party, all people have pairwise distinct ages. Some pairs of people are friends and friendship is mutual. Call a person [i]junior[/i] if they are younger than all their friends, and [i]senior[/i] if they are older than all their friends. A person with no friends is both [i]junior[/i] and [i]senior[/i]. A sequence of pairwise distinct people $A_1, \dots, A_m$ is called [i]photogenic[/i] if: 1. $A_1$ is [i]junior[/i], 2. $A_m$ is [i]senior[/i], and 3. $A_i$ and $A_{i+1}$ are friends, and $A_{i+1}$ is older than $A_i$ for all $1 \leq i \leq m-1$. Let $k$ be a positive integer such that for every [i]photogenic[/i] sequence $A_1, \dots, A_m$, $m$ is not divisible by $k$. Prove that the people at the party can be partitioned into $k$ groups so that no two people in the same group are friends. [i]Proposed by Shantanu Nene[/i]

Misha came to country with $n$ cities, and every $2$ cities are connected by the road. Misha want visit some cities, but he doesn`t visit one city two time. Every time, when Misha goes from city $A$ to city $B$, president of country destroy $k$ roads from city $B$(president can`t destroy road, where Misha goes). What maximal number of cities Misha can visit, no matter how president does?

Let $n>1$ be a given integer. Let $G$ be a finite simple graph with the property that each of its edges is contained in at most $n$ cycles. Prove that the chromatic number of the graph is at most $n+1$.

Here $G_{n}$ denotes a simple undirected graph with $n$ vertices, $K_{n}$ denotes the complete graph with $n$ vertices, $K_{n,m}$ the complete bipartite graph whose components have $m$ and $n$ vertices, and $C_{n}$ a circuit with $n$ vertices. The number of edges in the graph $G_{n}$ is denoted $e(G_{n})$. If $n \geq 5$ and $e(G_{n}) \geq \frac{n^{2}}{4}+2$, prove that $G_{n}$ contains two triangles that share exactly one vertex.

Armadillo and Badger are playing a game. Armadillo chooses a nonempty tree (a simple acyclic graph) and places apples at some of its vertices (there may be several apples on a single vertex). First, Badger picks a vertex $v_0$ and eats all its apples. Next, Armadillo and Badger take turns alternatingly, with Armadillo starting. On the $i$-th turn the animal whose turn it is picks a vertex $v_i$ adjacent to $v_{i-1}$ that hasn't been picked before and eats all its apples. The game ends when there is no such vertex $v_i$. Armadillo's goal is to have eaten more apples than Badger once the game ends. Can she fulfill her wish?

A group consist of n tourists. Among every 3 of them there are 2 which are not familiar. For every partition of the tourists in 2 buses you can find 2 tourists that are in the same bus and are familiar with each other. Prove that is a tourist familiar to at most $\displaystyle \frac 2{5}n$ tourists.

Let \(L_3 = \{3\}\), \(L_n = \{3, 4, \ldots, h\}\) (where \(h > 3\)). For any given integer \(n \geq 3\), consider a graph \(G\) with \(n\) vertices that contains a Hamiltonian cycle \(C\) and has more than \(\frac{n^2}{4}\) edges. For which lengths \(l \in L_n\) must the graph \(G\) necessarily contain a cycle of length \(l\)?

In a group of nine persons it is not possible to choose four persons such that every one knows the three others. Prove that this group of nine persons can be partitioned into four groups such that nobody knows anyone from his or her group.