BMO 2010 Shortlist

a) The total number of matches is $n(n-1)/2$. Let $w$ be the number of games ended with a victory and $e$ the number of games ended in a draw ($e\ge 1$ due to the last team). Thus, $w+e=n(n-1)/2$. If $r$ is the step of the arithmetical progression, we have that the total number of points in the final classification is $$\frac{(2+(n-1)r)n}{2}=3w+2e=w+2(w+e)=w+n(n-1),$$ or $$2w-2n=n(n-1)(r-2).$$ The case $r=0$ is obviously impossible (each team should have only 1 point). For $r=1$ we get $2w=3n-n^2$, a contradiction (for $n=3$ one has 3 games, $w=0$, all games ended in a draw, but the last team has only 1 point; the case $n\ge 3$ implies $w<0$). If $r\ge 3$, then $2w\ge 2n+n(n-1)=n(n+1)$ in contradiction with the fact that the total number of matches is $n(n-1)/2$. Thus, the only possible value is $r=2$. In this case $w=n$ and the number of points of each team in decreasing order is the sequence $2n-1,2n-3,2n-5,\dots ,1$. Denote by $w_i$ and by $e_i$ the number of victories and draws of the $i$-th classified team (in decreasing order). Note that $w_i+e_i\le n-1$. Considering the number of points obtained by the first three teams, that is $2n-1,2n-3,2n-5$, we get $2n-1=3w_1+e_1=2w_1+w_1+e_1\le 2w_1+n-1$, that is $w_1\ge n/2$. Analogously, $w_2\ge n/2-1$ and $w_3\ge n/2-2$. For $n\ge 7$ we obtain that $w_1+w_2+w_3\ge 3n/2-3>n$, a contradiction with the fact that the total number of victories is $n$. It implies that $n\le 6$ and it is impossible to have $n=12$.

b) It is clear that such a configuration does not exist for $n=3$ ($w=3,e=n(n-1)/2-w=0$). The case $n=4$ is possible, where the points $7,5,3,1$ in the final classification are realized by the following results of the matches of teams: T1-T2 (a draw), T1-T3 (T1 won), T1-T4 (T1 won), T2-T3 (T2 won), T2-T4 (a draw), T3-T4 (T3 won).

In this case we have to write 7 points of the third as a sum of at most five numbers of 1's, which is impossible. Therefore, the only possibility is $n=4$, the configuration being described above. ☐