Exercise 5. In a football tournament, each team plays exactly twice against each of the others. There are no draws; a win earns two points, and a loss earns none. It turns out that only one team won the tournament with 26 points, and there are two teams tied for last with 20 points each. Determine the number of teams, and provide an example of a tournament where such results occur.
Solution to Exercise 5 Let $n$ be the number of teams.

Each team wins a number of matches between 10 and 13. Since there are $n(n-1)$ matches, each team wins on average $n-1$ matches. The winning team and the two last teams tied win on average 11 matches, while the other teams win on average between 11 and 12 matches. Therefore, the average number of matches won per team is $\geqslant 11$ and $<12$. Since $n-1$ is an integer, we must have $n-1=11$, that is, $n=12$.

Let's show that such a tournament is possible. For the sake of clarity, let's say that the 12 teams belong to 12 different cities. Each pair of cities plays two matches, one in the city of one of the teams and the other in the city of the other team. We will say that a team plays at home if the match takes place in its city, and away otherwise.

Here is an example of a suitable tournament: when two of the teams $2, \ldots, 12$ meet, the team playing at home wins. In addition, team 1 wins all its matches against teams 2 and 3, and wins only at home against teams $4, \ldots, 12$.

Alternative solution for the first part. Let $a$ be the number of teams that win 11 matches and $b$ be the number of teams that win 12 matches. The total number of teams is $n=3+a+b$. On the other hand, the number of matches won is $n(n-1)=10 \times 2+11 a+12 b+13=33+11 a+12 b=11(3+a+b)+b=11 n+b$. We deduce that $b=n(n-1)-11 n=n(n-12)$. Since $0 \leqslant b<n$, we have $0 \leqslant n(n-12)<n$. Dividing by $n$, we get $0 \leqslant n-12<1$, hence $n=12$.