I'm not sure this is a complete solution as I assume that (i) each pair of teams plays exactly once, and (ii) there was a winner in the match between the 4th and 8th placed teams (i.e. no draw). Anyway, here goes: First, some facts: - There are a total of $\binom{12}{2} = 66$. Since each match contributes exactly 2 points, the total scores of all the teams is $66 \times 2 = 132$. - The maximum possible score for any team is $11 \times 2 = 22$ (i.e. all wins). Hence, the maximum possible score for the 2nd placed team is $21$. Divide the teams into 2 groups: Group A consisting of the 1st to 7th place teams and Group B consisting of the 8th-12th place teams (i.e. the 5th lowest ranked teams). There are $\binom{5}{2} = 10$ games where both teams are from Group B. Hence, the sum of the scores for the teams in Group B is at least $10 \times 2 = 20$. This means that the 2nd placed team must have scored either 20 or 21. This in turn implies that for matches where one team comes from Group A and one team comes from Group B, all of these matches ended up with the team from Group A winning except possibly for one game which ended in a draw. Hence, assuming that there was a clear winner in the match between 4th & 8th place, the winner was the 4th placed team.

The assumption that there's a clear winner between 4th and 8th place teams is not necessary. If the second place scores 21 points, then the first place scores 22 points. Thus the first place defeats everyone, including the second place. But 21 points is only possible as 10 wins and 1 draw, so the second place cannot have lost, contradiction. So the second place scores 20 points; thus any match between Group A and Group B gives a win for the team in Group A.