Problem 19 · 2019 AMC 8
Hard
Logic & Word Problems
caseworksum-constraint
In a tournament there are six teams that play each other twice. A team earns 3 points for a win, 1 point for a draw, and 0 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?
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Answer: C — 24 points each.
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Hint 1 of 2
To push the top three as high as possible, give them every point that's "free" — let all three crush the bottom three in every game. The only points in question are the ones they fight over among themselves.
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Hint 2 of 2
A win-then-loss split between two teams yields 3 + 0 = 3 points total, but a draw-draw yields only 1 + 1 = 2. So to maximize and keep the trio tied, settle their head-to-heads with decisive wins, not draws.
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Approach: sweep the bottom, split decisively among the tied top
- Each top team plays the bottom three twice each = 6 games. Winning all of them gives 6 × 3 = 18 points — the maximum any team can grab from outside the trio.
- Within the top three, the three pairs play twice each. To keep all three level, give each pair a 1-win, 1-loss split. A win is worth 3 and a loss 0 (better than two draws at 1 each), and the splits cancel out so the trio stays tied.
- Each top team sits in 2 of those pairs and wins one game in each: +3 + 3 = 6 more.
- Maximum each = 18 + 6 = 24.
- Why this transfers: in "maximize a tied group" problems, hand the group all the outside wins, then realize that 3-for-a-win beats splitting points via draws — symmetry keeps the tie while decisive results keep the total high.
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