Question

一个锦标赛中有六支队伍，每两队互相比赛两次。胜一场得 $3$ 分，平一场得 $1$ 分，负一场得 $0$ 分。所有比赛结束后，前三名队伍的总分相同。前三名每队可能得到的最大总分是多少？

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?

$22$

$23$

$24$

$26$

$30$

答案：C

知识点：最优化极端原理

难度评级：1610

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文字解答：

每支前三名队伍与后三名队伍比赛 $6$ 场，所以从这些比赛中最多得到 $6 \cdot 3 = 18$ 分。

前三名队伍之间有 $3$ 对，因此共有 $6$ 场比赛。每场比赛总共最多产生 $3$ 分，所以这六场比赛给三支队伍的总分至多为 $18$ 分。若三队最终同分，每队从中至多得到 $18/3=6$ 分。

因此每支前三名队伍最多得到 $18+6=24$ 分。这个上界可以达到：让每支前三名队伍赢下所有对后三名队伍的比赛，并让每对前三名队伍在两场交锋中各赢一场。这样每支前三名队伍都得到 $18+6=24$ 分。

所以正确答案是 C。

Each top team plays $6$ games against the bottom three teams, so it can earn at most $6 \cdot 3 = 18$ points from those games.

Among the top three teams, there are $3$ pairs and hence $6$ games. Each game awards at most $3$ points total, so these games award at most $18$ points among the three teams. If their final scores are equal, each can therefore receive at most $18/3=6$ of those points.

Thus each top team has at most $18+6=24$ 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 $18+6=24$ points.

Thus, the correct answer is C.

2019 AMC 8 第 19 题

答案：C

难度评级：1610

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