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1985 AIME Problems/Problem 14

Revision as of 23:05, 18 October 2018 by Gamemaster402 (talk | contribs) (→‎Solution 2)

Problem

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 $\frac{1}{2}$ 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/his points against the other nine of the ten). What was the total number of players in the tournament?

Solution 1

Let us suppose for convenience that there were $n + 10$ players overall. Among the $n$ players not in the weakest 10 there were $n \choose 2$ games played and thus $n \choose 2$ points earned. By the givens, this means that these $n$ players also earned $n \choose 2$ points against our weakest 10. Now, the 10 weakest players playing amongst themselves played ${10 \choose 2} = 45$ games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger $n$ players. Since every point earned falls into one of these categories, It follows that the total number of points earned was $2{n \choose 2} + 90 = n^2 - n + 90$. However, there was one point earned per game, and there were a total of ${n + 10 \choose 2} = \frac{(n + 10)(n + 9)}{2}$ games played and thus $\frac{(n + 10)(n + 9)}{2}$ points earned. So we have $n^2 -n + 90 = \frac{(n + 10)(n + 9)}{2}$ so $2n^2 - 2n + 180 = n^2 + 19n + 90$ and $n^2 -21n + 90 = 0$ and $n = 6$ or $n = 15$. Now, note that the top $n$ players got $n(n - 1)$ points in total (by our previous calculation) for an average of $n - 1$, while the bottom 10 got 90 points total, for an average of 9. Thus we must have $n > 10$, so $n = 15$ and the answer is $15 + 10 = \boxed{25}$.

Solution 2

Suppose that there are $n$ players participating in the tournament. We break this up into a group of the weakest ten, and the other $n-10$ people. Note that the $10$ players who played each other generated a total of $\dbinom{10}{2} = 45$ points playing each other. Thus, they earned $45$ playing the $n-10$ other people. Thus, the $n-10$ people earned a total of $10(n-10)-45 = 10n-145$ points playing vs. this group of 10 people, and also earned a total of $10n-145$ playing against themselves. Since each match gives a total of one point, we must have that $\dbinom{n-10}{2}=10n-45$. Expanding and simplifying gives us $n^2-41+400=0$. Thus, $n=16$ or $n=25$. Note however that if $n=16$, then the strongest $16$ people get a total of $16*10-145=15$ playing against the weakest $10$ who gained $45$ points vs them, which is a contradiction since it must be larger. Thus, $n=\boxed{25}$.

Solution by GameMaster402

See also

1985 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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