2. In a chess tournament, everyone played against each other once. The winner won half of the games and drew the other half. It turned out that he scored 9 times fewer points than all the others combined. (1 point for a win, 0.5 for a draw, 0 for a loss.) How many chess players were there in the tournament?
# Answer: 15 chess players.

Solution. If the number of participants is $n$, then each played $n-1$ games. The winner won half of the games and scored $\frac{1}{2}(n-1)$ points. They drew the other half of the games and scored another $\frac{1}{4}(n-1)$ points. In total, the winner scored $\frac{1}{2}(n-1)+\frac{1}{4}(n-1)=\frac{3}{4}(n-1)$ points.

In the tournament, $\frac{1}{2} n(n-1)$ games were played. In each game, the players played for one point, so they scored a total of $\frac{1}{2} n(n-1)$ points. Therefore,

$$
9 \cdot \frac{3(n-1)}{4}=\frac{n(n-1)}{2}-\frac{3(n-1)}{4}
$$

from which $n=15$. Thus, in a tournament of 15 chess players, the winner played 14 games and scored $7+3.5=10.5$ points, while the others scored $\frac{1}{2} \cdot 15(15-1)-10.5=94.5$ points, which is 9 times more than the winner.

Criteria. Only the answer -0 points. Proven that for $n=15$ the tournament meets the condition, but not justified that there are no other solutions -2 points. The equation for the number of participants is correctly set up -5 points. Complete solution -7 points.