11.6. In a competition held in a single round-robin format, each team plays against every other team exactly once. At the end of a single round-robin football tournament among 16 teams, it turned out that the team "Joker" won against all the teams that, in the final standings, are placed higher than "Joker" (by points scored), but lost to all those that are placed lower. Moreover, no team scored the same number of points as "Joker." What is the highest place "Joker" could have in the tournament? Justify your answer. (It is known that in football, 3 points are awarded for a win, 1 for a draw, and 0 for a loss.)
# Solution:

Method 1. We will call the teams that outperformed the "Joker" in the tournament table strong, and the teams that lagged behind - weak. Note that the position of the "Joker" will not change if we replay all matches where a strong team met a weak team and award the victory to the strong team. Therefore, we can assume (and we will do so) that all strong teams have won against all weak teams.

Moreover, we will assume that all weak teams played to a draw with each other. This is also permissible, as in this case, the total number of points scored by the weak teams will be the smallest, as well as the number of points scored by the best of the weak teams.

Suppose the "Joker" took 5th place (the situation would be even worse if it were 4th or higher). Then we have 4 strong teams and 11 weak teams. The "Joker" scored only 12 points in total, while each weak team scored 3 points for a win against the "Joker" and 10 for draws with other weak teams, totaling 13 points, which is more than 12. Contradiction. The "Joker" could have taken 6th place: indeed, in this case, it scored 15 points, each of the 5 strong teams won against 10 weak teams, which is already 30 points, clearly more than 15, and each weak team scored \(3 + 9 = 12\) points, which is less than 15. Thus, the highest place the "Joker" could have taken is 6th. Method 2. Let the "Joker" team take place \(n (1 \leqslant n \leqslant 16)\). Then it won \(n-1\) matches and scored \(3(n-1)\) points. Each of the \(16-n\) weak teams scored at least one point less, i.e., no more than \(3n-4\) points, in total, the weak teams scored no more than \((16-n)(3n-4)\) points. But these teams, first, all won against the "Joker," and second, played

\[
\frac{(16-n)(15-n)}{2} \text{ matches, }
\]

in each of which they scored a total of 2 or 3 points. Thus, they scored a total of no less than \(3(16-n) + (16-n)(15-n)\) points. We have the inequality

\[
(16-n)(3n-4) > 3(16-n) + (16-n)(15-n)
\]

from which \(n > 5.5 > 5\). Considering that \(n\) is a natural number, we get that \(n \geqslant 6\). It remains to show that the "Joker" can take 6th place. Indeed, let the "Joker" win 5 matches (and score 15 points), all the teams it defeated (they will be strong) won against all the teams it lost to (they will be weak), and all the teams it lost to played to a draw with each other. Then each strong team scored at least 30 points (won against 10 weak teams) and stands higher in the table than the "Joker." Each weak team scored 12 points (3 for a win against the "Joker" and 9 for 9 draws with other weak teams) and stands lower in the table than the "Joker." All conditions of the problem are met, the example is constructed.

Answer: 6th place.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| Proved that the "Joker" could not take a place higher than 6th, but there is no justification (or correct example) that it could take 6th place. | 5 points |
| Correct example of a tournament in which the "Joker" took 6th place (with an incorrect proof of the impossibility of taking a higher place) | 2 points |
| Obtained an inequality between the number of points of the "Joker" team and the number of points of all teams that lagged behind it. No further progress. | 2 points |
| Correct answer without justification, as well as an incorrect answer, even if it is supported by examples of tournaments | 0 points |