AMC10 2012 B

AMC10 2012 B · Q15

AMC10 2012 B · Q15. It mainly tests Counting & probability misc, Invariants.

In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end of the tournament?

在一个有6支队伍的循环赛中，每支队伍与其他每支队伍各进行一场比赛，每场比赛有一支队伍胜、一支队伍负。锦标赛结束时，按胜场数对队伍排名。最多有多少支队伍可能并列胜场数最多？

(A) 2 2

(B) 3 3

(D) 5 5

(E) 6 6

Answer

Correct choice: (D)

正确答案：（D）

Solution

A total of 15 games are played, so all 6 teams could not be\ntied for the most wins as this would require $15/6 = 2.5$ wins per team. However,\nit is possible for 5 teams to be tied, each with 3 wins and 2 losses. One such\noutcome can be constructed by labeling 5 of the teams A, B, C, D, and E, and\nplacing these labels at distinct points on a circle. If each of these teams beat\nthe 2 labeled teams clockwise from its respective labeled point, as well as the\nremaining unlabeled team, all 5 would tie with 3 wins and 2 losses.

总共进行15场比赛，所以6支队伍不可能并列最多胜场，因为这需要每队 $15/6 = 2.5$ 场胜利。然而，5支队伍可以并列，每支3胜2负。可以这样构造结果：将5支队伍标记为A、B、C、D和E，放置在圆周上的不同点。如果每支队伍击败其顺时针方向的两个标记队伍以及剩余的无标记队伍，则这5支队伍都以3胜2负并列。

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