Problem:
In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was more than the number of games won by right-handed players (there were no ties and no ambidextrous players). What is the total number of games played?
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Solution:
Let denote the number of left-handed players, and let denote the number of righthanded players. There were games between two left-handed players, games between\
two right-handed players, and games involving one left-handed and one right-handed player. Let denote the number of games won by right-handed players against left-handed players. Then righthanded players won a total of games, and left-handed players won games. Because left-handed players won more games than right-handed players,
Using and simplifying this equation gives . If , there are no real solutions for this quadratic equation. Therefore , giving and . The total number of games played is
Note that the right-handed players won a total of 15 games (all among themselves), and the left-handed players won a total of 21 games ( 3 games among themselves and all 18 games against right-handed players), and 21 is indeed more than 15.
The problems on this page are the property of the MAA's American Mathematics Competitions