Blinkov A. A:

The teams held a football tournament in a round-robin format (each team played one match against every other team, with 3 points for a win, 1 point for a draw, and 0 points for a loss). It turned out that the sole winner scored less than $50 \%$ of the maximum possible points for one participant. What is the minimum number of teams that could have participated in the tournament?
Let's prove that there could not have been fewer than six teams. If, for example, there were five teams in the tournament, then they played $5 \cdot 4: 2=10$ matches and scored a total of at least 20 points. Therefore, the sole winner scored more than $20: 5=4$ points. However, according to the conditions, he scored no more than 5 points out of 12 possible. Thus, the winner scored exactly 5 points, and each of the other teams scored no more than 4. Therefore, the total points scored by all participants does not exceed $5+4 \cdot 4=21$. But the number of points scored by the winner means that he won at least once and lost at least once, which means that in this case, the total points scored by all teams cannot be less than 22. Contradiction.

The reasoning for a tournament with two, three, or four teams is similar.

Let's provide two possible examples for a tournament with six teams:

| Team | Total | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | **7** |  | 1 | 1 | 1 | 1 | 3 |
| 2 | **5** | 1 |  | 1 | 1 | 1 | 1 |
| 3 | **5** | 1 | 1 |  | 1 | 1 | 1 |
| 4 | **5** | 1 | 1 | 1 |  | 1 | 1 |
| 5 | **5** | 1 | 1 | 1 | 1 |  | 1 |
| 6 | **4** | 0 | 1 | 1 | 1 | 1 |  |


| Team | Total | **1** | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | **7** |  | 3 | 3 | 0 | 0 | 1 |
| 2 | **6** | 0 |  | 1 | 1 | 3 | 1 |
| 3 | **6** | 0 | 1 |  | 3 | 1 | 1 |
| 4 | **6** | 3 | 1 | 0 |  | 1 | 1 |
| 5 | **6** | 3 | 0 | 1 | 1 |  | 1 |
| 6 | **5** | 1 | 1 | 1 | 1 | 1 |  |

## Answer

6 teams.