In chess tournaments, usually ${ }^{2}$, each win is worth 1 point, each draw, 0.5 points, and each loss, zero points. In the "Magistral Championship" chess tournament, only

International Masters - IMs and Grandmasters - GMs participated. The number of GMs was ten times the number of IMs. Each player played only once against all opponents, and thus, if $n$ was the number of players, then $\frac{n(n-1)}{2}$ games took place. The total points of all GMs was 4.5 times the total points of all IMs. Therefore, the questions are:

a) How many International Masters participated in this competition?

b) How many Grandmasters participated in this championship?

c) How many games were played in the tournament?
Solution

a) Let $i$ and $g$ be the numbers of IMs and GMs, respectively. Denote the sums of scores in each category by $I$ and $G$ and the total number of players by $n$. From the statement, $g=10 i$, $G=4.5 I$ and $n=i+g=i+10 i=11 i$. Now, the total number of games (and points contested 3) is

$$
\begin{aligned}
\frac{n(n-1)}{2} & =\frac{11 i \cdot(11 i-1)}{2} \\
I+G & =I+4.5 I=5.5 I
\end{aligned}
$$

Moreover, we know that each chess player will play $(11 i-1)$ games, which is also the maximum number of points a player can achieve. Therefore, we can write that

$$
\begin{aligned}
\frac{11 i \cdot(11 i-1)}{2} & =5.5 I \\
11 i \cdot(11 i-1) & =11 I \\
i \cdot(11 i-1) & =I
\end{aligned}
$$

The equation $I=i \cdot(11 i-1)$ can be interpreted as if all the IMs had achieved the maximum score, since $I$ is the total sum of their points, $(11 i-1)$ is the number of games (and the maximum score) for each IM, and $i$ is the number of IMs. However, this could not have happened, because with two or more IMs in the tournament, they would face each other, resulting in a draw or loss for one of them. Therefore, there was only 1 IM in the competition.[^0]b) Since $g=10 i$, then $g=10 \cdot 1=10$. Ten GMs participated in the tournament.

c) And there were $\frac{11(11-1)}{2}=55$ games in the championship.