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Combinatorics in Tournament | AIME I, 1985 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on combinatorics in Tournament.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1985 based on Combinatorics in Tournament.

Combinatorics in Tournament- AIME I, 1985


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 player 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?

  • is 107
  • is 25
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Combinatorics

Algebra

Check the Answer


Answer: is 25.

AIME I, 1985, Question 14

Elementary Algebra by Hall and Knight

Try with Hints


Let there be n+10 players

Case I from n players not in weakest 10, \({n \choose 2}\) games played and \({n \choose 2}\) points earned

Case II n players also earned \({n \choose 2}\) points against weakest 10

Case III now weakest 10 played among themselves \({10 \choose 2}\)=45 games and 45 points earned

Case IV 10 players also earned 45 points against stronger n

So total points earned= 2[\({n \choose 2}\)+45]=\(n^{2}-n+90\)

case V 1 point earned per game \({n+10 \choose 2}\)=\(\frac{(n+10)(n+9)}{2}\) games and \(\frac{(n+10)(n+9)}{2}\) points earned

So \(n^{2}-n+90=\frac{(n+10)(n+9)}{2}\)

or, \(n^{2}-21n+90=0\)

or, n=6, n=15 here taking n>10,

or, n=15 or, n+10=25.

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