In a chess tournament, eight players participated, and everyone played against each other once. Everyone scored a different number of points in the tournament. The second-place finisher accumulated as many points as the last four placed players combined. What was the result of the game between the 3rd and 5th place finishers?
Four people can be chosen in pairs in as many ways as the number of segments determined by four points, say the four vertices of a square, in a plane. The latter is equal to six, as the six segments are the four sides and the two diagonals. The last four competitors in the competition thus shared six points, the sum of their points being at least six. The first-place competitor can earn the most points by beating everyone, in which case they have seven points.

According to this, the second-place competitor has at least six and at most six and a half points. (Fractions of points smaller than a half do not occur in the competition.) However, six and a half points could only be achieved if they scored at least half a point from the first, in which case the latter could have at most $61 / 2$, so they would either not be first, or the points of the first and second place would be the same. Neither is possible, so the second has six points, and this is also the sum of the points of the last four. They thus only scored points from each other, specifically the fifth did not score points from the third, so the third-place competitor beat the fifth-place competitor.

Gábor Tóth (Budapest, Fazekas M. Gymnasium, 2nd grade)