In a short league, five teams face each other against each other. In the end, neither pair of teams shares a score. The question is how far the fourth-placed team can go.
The tournament of ties
Riddle statement
Five teams play a round-robin league, in a single round.
Score:
- victory: 3 points;
- tie: 1 point for each team;
- defeat: 0 points.
In the end, the five teams finish with different scores.
What is the highest possible score for the fourth place team?
Show solution
Solution
Answer: the highest possible score for the fourth-place finisher is 5 points.
Let's call the final scores, from highest to lowest:
We want to maximize $s_4$.
Suppose the fourth-place finisher could have at least 6 points. As the five scores are different, the three teams above it would have to reach at least 7, 8 and 9 points respectively.
That means that the first four would add at least:
Now, in a five-team league There are
matches, and each match distributes a maximum of 3 points. The maximum total points for the tournament is, therefore:
For the first four to already have 30, the fifth team would have to have 0 points, and all matches would have to end with a victory — without any ties. But if there are no ties, each score is a multiple of 3. Scores 8 and 7 would be impossible.
It has been proven that $s_4 \le 5$.
Now it remains to be seen that 5 is achievable. A possible classification is:
The following table of results is made by:
Team A beats B, C and D; ties with E. Ends with 10 points.
Team B beats C and D; tie with E; loses to A. Ends with 7 points.
Team C beats D and E; loses to A and B. Ends with 6 points.
Team E beats D; tie with A and B; loses to C. Finishes with 5 points.
Team D loses all of its games. It ends with 0 points.
The five scores are different and the fourth-placed team (team E) has 5 points.
Therefore, the highest possible score for the fourth-placed team is 5 points.