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55th IMO Team Selection Test

Bulgaria counting and probability

Problem

There are citizens in a town every two of which are either friends or enemies. For some positive integer each citizen has at most enemies and there exists a citizen having exactly enemies. A group is called friendly if any two members of the group are friends. It is known that a friendly group with more than members does not exist and all citizens can be partitioned into two friendly groups having members each. Prove that the number of friendly groups having members is not greater than .
Solution
Let and be the two friendly groups. For arbitrary group from denote by the group of all people from each of which is an enemy of at least one person from . If then is a friendly group having more than members, a contradiction.

Therefore the sets satisfy the Hall's condition for system of distinct representatives. Hence, we may assume that and are enemies for all .

This means that every friendly group of members include one person from every pair for .

Suppose the enemies of are . For a friendly group such that we have . Hence, . From every of the remaining pairs , we have to choose one of and to be an element of . Therefore there are at most such group.

For a friendly group with members such that we have . Since from each of the remaining pairs , we have to choose one of and to be an element of we find that there are at most such groups.

Therefore, the total number of friendly groups of members is at most .

Techniques

Matchings, Marriage Lemma, Tutte's theorem