There are 10 teams. Each will fought every other exactly once, each team will have exactly 9 matches.

Question

There are 10 teams in a competition. Each team will fought every other team exactly once, so each team will have exactly 9 matches. If a team wins, then it will get 3 points, if it loses he will get 0, but if a match is a draw then both will get only 1 point.

After the competition finish, the top 2 teams have collected the same total points. What is the maximum possible points that the team at the 3rd position can attain?

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attempt:

Let the teams be denoted as $x_{1},...,x_{10}$ ranked according to the index ($x_{1}$ and $x_{2}$ are the top two teams). It is clear that neither of them can get 27 points (the highest possible total points).

So one possibility is that $x_{1}, x_{2}$ both get total points of 25. This means that each won 8 times and draw one time. Now the draw must be from the match between $x_{1},x_{2}$, because if not then for example $x_{1}$ defeat $x_{2}$ it is impossible because these two were unbeaten.

Now, we have 8 teams left to analyze. With each team has hope for the highest total points which is 21. So the answer is 21?

Answer 1

You can exclude cases where the third-place team scores:

• more than $27$ points (not enough matches)
• $27$ points (must have beaten every other team, in which case they all got fewer than $27$ points and this team would have come first)
• $26$ points (no arithmetical possibility of this from nine matches)
• $25$ points (wins all but one and draws that, so one other team could also get at least $25$, but no more other teams could do so well)

The next score to consider is $24$ (eight wins and one loss). That is possible if the top three teams each beat the seven other teams, and also win one of the three matches between them. So the answer is $24$