Question

In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac12$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest-scoring players earned half of her or his points against the other nine of the ten.) What was the total number of players in the tournament?

Answer

Answer: 25

Concepts:double counting graph theory quadratic

Difficulty rating: 2720

Small Hint:

Sum the scores of the ten lowest-scoring players and count their internal games

Big Hint:

Let $m$ be the number of other players and double-count points from games across the two groups

Solution:

The games among the ten lowest players contribute $\binom{10}{2}=45$ total points. These are half of those ten players’ combined score, so their combined score is $90.$ Hence they earned $45$ points in games against the other $m$ players.

The other players therefore earned $10m-45$ points against the lowest ten. By the condition, this is half their combined score, which is $\binom{m+10}{2}-90.$ Thus $2(10m-45)=\binom{m+10}{2}-90,$ giving $(m-6)(m-15)=0.$ If $m=6,$ the lowest ten average $9$ points while the other six average only $5,$ impossible for the latter group to rank above them. Hence $m=15,$ and the total number of players is $10+15=25.$

1985 AIME Problem 14

Answer: 25

Difficulty rating: 2720

Small Hint:

Big Hint:

Solution:

Problem 14 in Other Years