## Task A-4.5.

In a chess tournament, boys and girls participated. Each competitor played one game with every other competitor, and no game ended in a draw. Determine the smallest possible number of competitors in the tournament if it is known that each girl won at least 21 boys and each boy won at least 12 girls.
## First Solution.

Let $m$ be the number of girls, and $n$ be the number of boys in the competition. We need to determine the minimum value of $m+n$. Since there are no conditions that need to be satisfied for matches between girls and matches between boys, in the solution, we only consider matches between girls and boys.

Among them, $mn$ matches have been played (each boy with each girl). The number of matches won by girls, according to the problem, is at least $21m$, while the number of matches won by boys is at least 12. From this, we conclude that

$$
m n \geqslant 21 m + 12 n
$$

This inequality can be written as $(m-12)(n-21) \geqslant 252$.

By applying the inequality between the arithmetic and geometric means, we get that

$$
(m-12) + (n-21) \geqslant 2 \sqrt{(m-12)(n-21)} \geqslant 2 \sqrt{252} > \sqrt{961} = 31
$$

Thus, $m+n > 64$. From this, we conclude that $m+n \geqslant 65$.

It remains to show that it is possible for there to be $m+n=65$ competitors at the tournament. We will construct an example for the case $m=30$ and $n=35$.

Divide the girls into six groups (call them $X_{1}, \ldots, X_{6}$), and the boys into seven groups (call them $Y_{1}, \ldots, Y_{7}$), each with 5 people.

For each pair of groups $X_{i}$ and $Y_{j}$, let the girl in position $k$ in group $X_{i}$ win the boys in positions $k, k+1$, and $k+2$ in group $Y_{j}$, considering positions cyclically (the sixth position is the first, the seventh is the second, the eighth is the third).

![](https://cdn.mathpix.com/cropped/2024_05_30_a630fb9dab34a5225850g-24.jpg?height=314&width=151&top_left_y=2256&top_left_x=881)

Then each girl has won $7 \cdot 3 = 21$ boys, and each boy has won $6 \cdot 2 = 12$ girls.

![](https://cdn.mathpix.com/cropped/2024_05_30_a630fb9dab34a5225850g-25.jpg?height=329&width=1119&top_left_y=338&top_left_x=400)

Note: From the inequality $(m-12)(n-21) \geqslant 252$, we cannot directly conclude that $m+n$ will be the smallest when equality holds. For example, $a b = 7 + 11$. Therefore, we need to use some form of the inequality between the arithmetic and geometric means or reasoning as in the second solution.

Note: The example from the solution can also be described in a different way. Divide the girls into five groups of 6 each (call them $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$), and the boys into five groups of 7 each (call them $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$). We have 5 groups of girls and 5 groups of boys (1 point).

Let all boys in group $B_{i}$ win the girls in groups $A_{i}$ and $A_{i+1}$, for $1 \leqslant i \leqslant 4$, and $B_{5}$ win the girls in $A_{5}$ and $A_{1}$. They lose to all other girls. (2 points)

This means that each boy has won girls from exactly two groups of girls, i.e., in exactly $2 \cdot 6 = 12$ matches. Each girl has won boys from exactly three of their groups, i.e., in exactly $3 \cdot 7 = 21$ matches. (1 point)