The tournament of ties

1. If the fourth has a lot of points, the first three must have even more.
2. The maximum total of tournament points occurs when there are no ties.
3. After finding an upper limit, it remains to construct a classification that reaches it.

Answer: the highest possible score for the fourth place finisher is 5 points. Let's call the final scores, from highest to lowest: $$
s_1>s_2>s_3>s_4>s_5.

$$ We want to maximize $s_4$. Let's assume that the fourth placed team could have at least 6 points. Since the five scores are different, the three teams above him would have to have at least 7, 8 and 9 points, in some order. Therefore, the first four would add up to at least: $$

9+8+7+6=30.

$$ But in a five-team league there are $$

\binom{5}{2}=10

$$ matches, and each match distributes a maximum of 3 points. So the maximum tournament point total is: $$

10\cdot3=30.
$$ That would require the five teams to add exactly 30 points, without any ties, and the first four scores to be exactly 9, 8, 7 and 6. But if there are no ties, each team only adds points from victories, so all the scores must be multiples of 3. Scores 8 and 7 would be impossible. Therefore, the fourth placed team cannot reach 6 points. So $$s_4\le 5.$$ Now it remains to be seen that 5 can be achieved. A possible classification is: $$10,7,6,5,0.$$ It can be obtained, for example, with these results: - Team A: beat B, C and D; ties with E. Ends with 10.

• Team B: beat C and D; tie with E; loses with A. Ends with 7.
• Team C: beat D and E; loses with A and B. Ends with 6.
• Team E: beat D; tie with A and B; loses with C. Ends with 5.
• Team D: loses all its games. Ends with 0. Thus the scores are different and the fourth classified has 5 points. Therefore, the highest possible score for the fourth place finisher is 5 points.