Olympiad Maths Prep

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Problem 998

AIME late
Combinatorics Difficulty 6.0 Prove it

1. A resident of a house is called sociable if they are acquainted with at least 10 other residents of the same house (If Petrov is acquainted with Ivanov, then Ivanov is acquainted with Petrov). Prove that in any house there are either two sociable residents who are acquainted with each other or two unsociable residents who are not acquainted with each other.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

# Solution

1) A sociable person has at least 10 acquaintances.

If among them there is a sociable one, then we have found a pair of sociable acquaintances.

2) If there are no sociable people among them, then they are all unsociable. In this case, we will prove that there are two of them who are not acquainted with each other.

Assume the opposite: they are all acquainted with each other, and since they are acquainted with the sociable person we initially considered, they are all sociable. This is a contradiction, therefore, there are two unsociable residents who are not acquainted with each other.

## Criteria

The idea of the proof is correct, but the proof is not completed or contains unjustified conclusions (for example, in the second point, it is stated but not proven that there are two unsociable residents who are not acquainted) - 4 points.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.