Difference between revisions of "1985 AIME Problems/Problem 14"
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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 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? | 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 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 == | == Solution == | ||
| − | Let us suppose for convenience that there were <math>n + 10</math> players over all. Among the <math>n</math> players not in the weakest 10 there were <math>n \choose 2</math> games played and thus <math>n \choose 2</math> points earned. By the givens, this means that these <math>n</math> players also earned <math>n \choose 2</math> points against our weakest 10. Now, the 10 weakest players playing amongst themselves played <math>{10 \choose 2} = 45</math> games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger <math>n</math> players. Since every point earned falls into one of these categories, It follows that the total number of points earned was <math>2{n \choose 2} + 90 = n^2 - n + 90</math>. However, there was one point earned per game, and there were a total of <math>{n + 10 \choose 2} = \frac{(n + 10)(n + 9)}{2}</math> games played and thus <math>\frac{(n + 10)(n + 9)}{2}</math> points earned. So we have <math>n^2 -n + 90 = \frac{(n + 10)(n + 9)}{2}</math> so <math>2n^2 - 2n + 180 = n^2 + 19n + 90</math> and <math>n^2 -21n + 90 = 0</math> and <math>n = 6</math> or <math>n = 15</math>. Now, note that the top <math>n</math> players got <math>n(n - 1)</math> points in total (by our previous calculation) for an average of <math>n - 1</math>, while the bottom 10 got 90 points total, for an average of 9. Thus we must have <math>n > 10</math>, so <math>n = 15</math> and the answer is <math>15 + 10 = 025</math> | + | Let us suppose for convenience that there were <math>n + 10</math> players over all. Among the <math>n</math> players not in the weakest 10 there were <math>n \choose 2</math> games played and thus <math>n \choose 2</math> points earned. By the givens, this means that these <math>n</math> players also earned <math>n \choose 2</math> points against our weakest 10. Now, the 10 weakest players playing amongst themselves played <math>{10 \choose 2} = 45</math> games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger <math>n</math> players. Since every point earned falls into one of these categories, It follows that the total number of points earned was <math>2{n \choose 2} + 90 = n^2 - n + 90</math>. However, there was one point earned per game, and there were a total of <math>{n + 10 \choose 2} = \frac{(n + 10)(n + 9)}{2}</math> games played and thus <math>\frac{(n + 10)(n + 9)}{2}</math> points earned. So we have <math>n^2 -n + 90 = \frac{(n + 10)(n + 9)}{2}</math> so <math>2n^2 - 2n + 180 = n^2 + 19n + 90</math> and <math>n^2 -21n + 90 = 0</math> and <math>n = 6</math> or <math>n = 15</math>. Now, note that the top <math>n</math> players got <math>n(n - 1)</math> points in total (by our previous calculation) for an average of <math>n - 1</math>, while the bottom 10 got 90 points total, for an average of 9. Thus we must have <math>n > 10</math>, so <math>n = 15</math> and the answer is <math>15 + 10 = \boxed{025}</math>. |
== See also == | == See also == | ||
Revision as of 22:16, 31 December 2013
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 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
Let us suppose for convenience that there were 
players over all. Among the 
players not in the weakest 10 there were 
games played and thus points earned. By the givens, this means that these players also earned points against our weakest 10. Now, the 10 weakest players playing amongst themselves played 
games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger players. Since every point earned falls into one of these categories, It follows that the total number of points earned was 
. However, there was one point earned per game, and there were a total of 
games played and thus 
points earned. So we have 
so 
and 
and 
or 
. Now, note that the top players got 
points in total (by our previous calculation) for an average of 
, while the bottom 10 got 90 points total, for an average of 9. Thus we must have 
, so and the answer is 
.
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 | ||
