57. In a football tournament, 30 teams participated. At the end of the tournament, it turned out that among any three teams, there were two that had scored the same number of points in the three matches within this trio of teams (3 points are awarded for a win, 1 for a draw, and 0 for a loss). What is the minimum number of draws that could be in such a tournament?
Solution. Let us choose an arbitrary team $A$ and divide all teams into three groups: teams that won against $A$, teams that lost to $A$, and teams that drew with $A$. To the last group, we will add team $A$ itself. Note that any two teams from the same group played a draw - this easily follows from the condition. Let the groups contain $a, b$, and $c$ teams. Then the number of draws is not less than

$$
\frac{a(a-1)}{2}+\frac{b(b-1)}{2}+\frac{c(c-1)}{2}=\frac{1}{2}\left(a^{2}+b^{2}+c^{2}-30\right).
$$

The sum of the numbers $a, b$, and $c$ is 30, so the sum of the squares is minimal when all numbers are equal. From this, it follows that the number of draws is not less than 135. An example is obtained when the described groups indeed contain exactly 10 teams each, and the teams in the groups win against each other in a cycle.