Problem 9.2. At a party, 24 people gathered. A guest is considered an introvert if they have no more than three acquaintances among the other guests. It turned out that each guest has at least three acquaintances who are introverts. How many introverts could there have been at the party? Provide all possible answers and prove that there are no others.

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# Answer: 24.

Solution. Let $a$ be the number of pairs of introverts who know each other, and $b$ be the number of pairs of acquaintances where one of the pair is an introvert. Each person who came to the party is included in at least three pairs, since they know at least three other introverts, and pairs of introvert-introvert are counted twice. Therefore, $2a + b \geq 3 \cdot 24$.

On the other hand, each introvert can be included in no more than three pairs, so $2a + b$ is no more than three times the number of introverts. As a result, we conclude that three times the number of introverts is not less than three times the number of guests, which is only possible if all those who came to the party are introverts.

Another solution. From the condition, it follows that each introvert knows three other introverts and no one else; that is, there can be no acquaintances between introverts and extroverts (let's call the rest of the people extroverts). But then an extrovert cannot know three introverts, as required by the condition. Therefore, there are no extroverts at this party.

## Criteria

## Kurchatov School Olympiad in Mathematics - 2019 Final Stage.

Any correct solution to the problem is worth 7 points.

Solutions that unjustifiably use the fact that some cases are "optimal" ("best", "worst", etc.) are not credited.

In the absence of a correct solution, the following criterion is used:

## 16. There is a correct answer.