2019 AMC 8 Problem 19
Attempt Problem 19 of the 2019 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
19.
In a tournament there are six teams that play each other twice. A team earns points for a win, point for a draw, and points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?
Answer: C
Small hint:
The top three should win all games against the bottom three
Big hint:
Among the top three, keep all points while splitting each pair evenly
Video solution:
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Written solution:
Each top team plays games against the bottom three teams, so it can earn at most points from those games.
Among the top three teams, there are pairs and hence games. Each game awards at most points total, so these games award at most points among the three teams. If their final scores are equal, each can therefore receive at most of those points.
Thus each top team has at most points. This bound is attainable: let every top team win all its games against the bottom teams, and for each pair of top teams let each team win one of their two games. Then each top team earns points.
Thus, the correct answer is C.
Problem 19 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
- 1994 AMC 8
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- 1998 AMC 8
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