XXIX Rioplatense Mathematical Olympiad

Let $d$ be the number of draws in the tournament. Then the total number of points awarded equals $2d+3(12-d)=36-d$, because every draw awards 2 points in total and every non-draw awards 3 points in total. Since teams $A$, $B$, $C$ combined have 24 points, the score of team $D$ equals $12-d$.

If $D$ has less than 3 points, then there are at least 10 draws in the tournament. Since $A$, $B$, $C$ play only 6 matches among themselves, at least 4 draws involve team $D$, but then the score of team $D$ is at least 4, a contradiction. Hence $D$ has at least 3 points.

On the other hand, notice that each of the teams $A$, $B$ and $C$ needs at least 2 draws to have a final score of 8 (because 8 has a remainder of 2 upon division by 3). This implies that there are at least 3 draws in the tournament, which in turn means that $12-d\le 9$.

3 points 4 points 6 points 7 points 8 points 9 points

It only remains to see if the score of team D can be equal to 5. Suppose $12-d=5$, then there are exactly 7 draws in the tournament. Since both 8 and 5 have a remainder of 2 upon division by 3, we know that each team must have 2 or 5 draws. Since each draw involves two teams, if for each team we count its number of draws, these four numbers will add up to $2\cdot 7=14$. This is only possible if two teams have 5 draws and the other two teams have 2 draws. Let $X$, $Y$ be the teams with 5 draws and $Z$, $W$ be the teams with 2 draws. Since $X$, $Y$ play each other only twice, each of them has at least 3 draws with a team from $Z$, $W$. But this means that $Z$, $W$ combined have at least 6 draws, which is a contradiction because we stated that each of them has 2 draws. This proves that the final score of team D cannot be 5.

In conclusion, the only possible scores of team D are 3, 4, 6, 7, 8, 9.