In a tournament, any two players play against each other. Each player gets one point for a win, 1/2 for a draw, and 0 points for a loss. Let $S$ be the set of the 10 lowest scores. We know that each player obtained half of their score playing against players from $S$.

a) What is the sum of the scores of the players in $S$?

b) Determine how many participants are in the tournament.

Note: Each player plays only once against each opponent.
Solution

a) The players of $S$, in matches played only among themselves, scored $\frac{10(10-1)}{2}=45$ points. Since the points they scored playing among themselves correspond to half of the points each one scored in the tournament, we can conclude that the sum of the points of the players of $S$ is $45+45=90$.

b) Let $n$ be the number of players in the tournament and $T$ be the set of the other $n-10$ players. Playing only among themselves, the players of $T$ scored $\frac{(n-10)(n-11)}{2}$ points. Since they scored half of their points playing against players of $S$, we can conclude that they scored the other half playing among themselves. Consequently, the sum of the scores of all the players of $T$ is $(n-10)(n-11)$. Since the sum of the scores of all the players corresponds to the total number of games, we have

$$
\begin{aligned}
90+(n-10)(n-11) & =\frac{n(n-1)}{2} \\
n^{2}-41 n+400 & =0
\end{aligned}
$$

The roots of this quadratic equation are 16 and 25. If $n=16$, the $16-10=6$ best players who scored, in total, $(16-10)(16-11)=30$ points. Since each player among the 6 best scored at least as many points as the 10 worst, the average score in $T$ is at least the average score of $S$, that is,

$$
5=\frac{30}{6} \geq \frac{90}{10}=9
$$

This is an absurdity. Therefore, we are left with $n=25$. It remains to show that it is possible to meet the tournament requirements with 25 players.

Consider the division of the 25 people into three groups:

$$
\begin{aligned}
& G_{1}=\left\{J_{1}, J_{2}, J_{3}, J_{4}, J_{5}, J_{6}\right\} \\
& G_{2}=\left\{J_{7}, J_{8}, J_{9}, J_{10}, J_{11}, J_{12}, J_{13}, J_{14}, J_{15}\right\} \\
& G_{3}=\left\{J_{16}, J_{17}, J_{18}, J_{19}, J_{20}, J_{21}, J_{22}, J_{23}, J_{24}, J_{25}\right\}
\end{aligned}
$$

All games between elements of $G_{2} \cup G_{3}$ will end in draws. We will now define the outcome of the matches between the elements of $G_{1}$ and the other competitors:

i) The elements of $G_{1}$ will win all the matches against elements of $G_{3}$.

ii) All the games between two elements of $G_{1}$ will end in draws.

iii) Each element of $G_{1}$ will win exactly six elements of $G_{2}$ and draw with the other 3 remaining. For this, representing a win by means of an arrow $(\rightarrow)$, we will establish:

$$
\begin{aligned}
J_{1} & \rightarrow J_{10}, J_{11}, J_{12}, J_{13}, J_{14}, J_{15} \\
J_{2} & \rightarrow J_{10}, J_{11}, J_{12}, J_{13}, J_{14}, J_{15} \\
J_{3} & \rightarrow J_{7}, J_{8}, J_{9}, J_{13}, J_{14}, J_{15} \\
J_{4} & \rightarrow J_{7}, J_{8}, J_{9}, J_{13}, J_{14}, J_{15} \\
J_{5} & \rightarrow J_{7}, J_{8}, J_{9}, J_{10}, J_{11}, J_{12} \\
J_{6} & \rightarrow J_{7}, J_{8}, J_{9}, J_{10}, J_{11}, J_{12}
\end{aligned}
$$

With this distribution, the players of $G_{1}, G_{2}$, and $G_{3}$ will score 20, 10, and 9 points, respectively. In addition, each element of these three groups will have scored half of their points against the elements of $G_{3}$.