# IMO 2007 Problem 3

In a mathematical competition some competitors are friends. Friendship is always mutual. Call a group of competitors a clique if each two of them are friends. The number of members of a clique is called its size. Given that, in this competition, the largest size of a clique is even, prove that the competitors can be arranged in two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.

=========================================

Proof:

Suppose not. Let graphs A and B be G and an empty graph, respectively. We move vertex one by one from A to B unless a move is invalid, meaning that after the move \(w(A)\)

Now we claim that w(A)=w(B)+1=d+1=|V(A)|. The first equalities are obvious. If there are more than d+1 vertices in A, then pick a maximum (d+1)-clique and move a vertex not in it to B.

Move a vertex v from A to B. Now w(A)=w(B)-1=d=|V(A)|. How many (d+1)-cliques in B does v belong to?

If v is in only one (d+1)-clique C, then by assumption moving every vertex in C back to A creates a (d+1)-clique in A, meaning that C, A, and v together is a clique of size 2d+1. Since the maximum clique in G has even size, it is at least of size 2d+2, a contradiction that the moving process above end up with maximum cliques of size d+1 and d in A and B, respectively.

If v is in two (d+1)-cliques C1 and C2, we arbitrarily pick vertices a and b from C1-C2 and C2-C1, respectively. Moving a or b to A does not increase w(A), but moving both does. So a and b are adjacent. Because a and b are arbitrarily chosen, the union of C1 and C2 is a clique bigger than C1 and C2, a contradiction. Notice here we do not need A to be a d-clique, but only has to contain a d-clique.

If v is in a set of k>=3 (d+1)-cliques C, we start a process of moving vertex one by one in C to A. A vertex u is moved to A if it does not belong to all cliques in C, and all cliques in C that contain u is removed from the set C right away. By the previous argument C could not be left with two (d+1)-cliques. Nor can moving a vertex not in all cliques in C results in an empty C. The process terminates when moving any remaining vertex from C to A leaves only one clique in C.

Now let's study C with k>1 (d+1)-cliques. Every vertex, except for the vertices common to all cliques in C, belongs to exactly k-1 cliques in C, meaning that any two vertices belong to at least one common clique and are therefore adjacent. So all vertices in C form a larger clique, contradiction. This concludes our proof.

Q.E.D.

=========================================

Proof:

Suppose not. Let graphs A and B be G and an empty graph, respectively. We move vertex one by one from A to B unless a move is invalid, meaning that after the move \(w(A)\)

Now we claim that w(A)=w(B)+1=d+1=|V(A)|. The first equalities are obvious. If there are more than d+1 vertices in A, then pick a maximum (d+1)-clique and move a vertex not in it to B.

Move a vertex v from A to B. Now w(A)=w(B)-1=d=|V(A)|. How many (d+1)-cliques in B does v belong to?

If v is in only one (d+1)-clique C, then by assumption moving every vertex in C back to A creates a (d+1)-clique in A, meaning that C, A, and v together is a clique of size 2d+1. Since the maximum clique in G has even size, it is at least of size 2d+2, a contradiction that the moving process above end up with maximum cliques of size d+1 and d in A and B, respectively.

If v is in two (d+1)-cliques C1 and C2, we arbitrarily pick vertices a and b from C1-C2 and C2-C1, respectively. Moving a or b to A does not increase w(A), but moving both does. So a and b are adjacent. Because a and b are arbitrarily chosen, the union of C1 and C2 is a clique bigger than C1 and C2, a contradiction. Notice here we do not need A to be a d-clique, but only has to contain a d-clique.

If v is in a set of k>=3 (d+1)-cliques C, we start a process of moving vertex one by one in C to A. A vertex u is moved to A if it does not belong to all cliques in C, and all cliques in C that contain u is removed from the set C right away. By the previous argument C could not be left with two (d+1)-cliques. Nor can moving a vertex not in all cliques in C results in an empty C. The process terminates when moving any remaining vertex from C to A leaves only one clique in C.

Now let's study C with k>1 (d+1)-cliques. Every vertex, except for the vertices common to all cliques in C, belongs to exactly k-1 cliques in C, meaning that any two vertices belong to at least one common clique and are therefore adjacent. So all vertices in C form a larger clique, contradiction. This concludes our proof.

Q.E.D.

post by Shen-Fu Tsai