Question

在一个六支队伍的循环赛中，每支队伍都与其他每支队伍比赛一场，每场比赛都有一胜一负。比赛结束后，队伍按胜场数排名。最终并列第一的队伍最多可以有多少支？

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?

$2$

$3$

$4$

$5$

$6$

答案：D

知识点：图论极端原理

难度评级：1540

解答：

总比赛数，即总胜场数，为 $\binom 62 = 15$ 。

若 $6$ 支队伍全都并列，每队应有 $2.5$ 胜，不可能。

若有 $5$ 支队伍并列第一，这 $5$ 支队伍可以各有 $3$ 胜。

把这 $5$ 支队伍编号为 $1$ 到 $5$ ，并指定第 $6$ 支队伍输掉所有比赛。对每个 $1\le x \le 5$ ，让该队击败队伍 $x+1 \mod 5$ 、队伍 $x+2 \mod 5$ ，以及队伍 $6$ 。

所以正确答案是 D。

They would have to share $\binom 62 = 15$ wins.

This means we cannot have a $6$ way tie as that would be $2.5$ wins per team.

If we had a $5$ way tie, each team could have $3$ wins, which is possible if one team loses all of its games, and out of the $5$ winning teams, they each split their games.

If we label the teams from $1$ to $5,$ and designate team $6$ to lose all their games. To get a $5$ way tie, we could have each team $1\le x \le 5$ beat team $x+1 \mod 5$ and team $x+2 \mod 5,$ as well as team $6.$

Thus, the correct answer is D .

2012 AMC 10B 第 15 题

答案：D

难度评级：1540

解答：

其他年份的第 15 题