## Chess Challenge

From the 2011 South African Olympiad:

In chess tournaments, players get 1 point for a win, 0.5 points for a draw and 0 points for a loss. In a recent round-robin tournament each pair of players met exactly once, and the top four scores were 4.5, 3.5, 3 and 1.5. What was the lowest score at the tournament?

Let n be the total number of players. Someone scored 4.5 points, so they must have played at least 5 games, and hence n is at least 6.

The total number of matches played is

$1 + 2 + \cdots + (n-1) = \frac{n(n-1)}{2}$.

This is also the total number of points awarded across the whole tournament. Thus the average score each player is $\frac{n-1}{2}$. Since the average score of the top four players is 3.125, the average score for the whole tournament will be at most 3.125. That is,

$\frac{n-1}{2} \leq 3.125$.

This requires n to be at most 7, and so the only possibilities are n = 6 or 7.

If n = 7, then 21 points are awarded in total, of which 8.5 go to the bottom three players. But the 4th highest total is 1.5, so at most 1.5 x 3 = 4.5 may be awarded to the bottom three players, which is impossible.

If n = 6, then 15 points are awarded in total, of which 2.5 go to the bottom two players. But the 4th highest total is 1.5, so it must be the case that 5th scores 1.5 and 6th scores 1. This scenario is indeed possible, as shown by the following list of results:

A draws with B and beats C, D, E, F;

B loses to C and beats D, E, F;

C loses to D and beats E and F;

D draws with E and loses to F;

E beats F.

Therefore there were 6 players, and the lowest score was 1.

Another great problem from South Africa, in particular very well crafted, to yield a unique solution.