3. Pete, Sasha, and Misha are playing tennis in a knockout format. A knockout format means that in each match, two players compete while the third waits. The loser of the match gives up their place to the third player and becomes the waiting player in the next match. Pete played a total of 12 matches, Sasha - 7 matches, Misha - 11 matches. How many times did Pete win against Sasha?
Answer: 4

Solution. First, let's find the total number of games played. Petya, Pasha, and Misha participated in a total of $12+7+11=30$ games. Since each game involves two participants, the number of games is half of this: $30 / 2=15$.

Thus, Petya did not participate in $15-12=3$ games, Pasha in $15-7=8$ games, and Misha in $15-11=4$ games.

Notice now that in a round-robin tournament, one player cannot skip two consecutive games. Since Pasha did not participate in 8 out of 15 games, it means he did not participate in the very first game and then skipped every second game. This means Pasha lost all his games.

Therefore, the number of Petya's wins over Pasha is equal to the number of games in which Petya and Pasha met. This number is equal to the number of games in which Misha did not participate, which is 4, as found earlier.

Criteria.

$(-$.$) Correct answer for a specific case, or it is proven that 15 games were played.

$(-/+)$ Correct answer for a specific case + it is proven that 15 games were played

(+/2) It is proven that Pasha lost all his games, and the correct answer for a specific example, without sufficient justification.

$(+/-)$

(+.) The problem is solved correctly with one flaw: there is no strict proof that 15 games were played.

$(+)$ The answer is correct and strictly justified.

## Mathematics - solutions