AMC8 2019

AMC8 2019 · Q19

AMC8 2019 · Q19. It mainly tests Casework, Optimization (basic).

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?

锦标赛中有六支队伍，每两队交手两次。胜一场得 3 分，平局得 1 分，负一场得 0 分。所有比赛结束后，前三名队伍的总积分相同。前三名队伍每队可能的最大总积分是多少？

(A) 22 22

(B) 23 23

(D) 26 26

(E) 30 30

Answer

Correct choice: (C)

正确答案：（C）

Solution

Each team wins once and loses once to get the highest points. For a win, we have $3$ points, so a team gets $3\times2=6$ points if they each win a game and lose a game. Against the other 3 teams, there are 6 games. If each team wins each of those six games, they get $3\cdot6=18$ points This brings a total of $18+6=\boxed {\textbf {(C) }24}$ points. Note that this can be easily seen to be the best case as follows. Let $x_A$ be the number of points $A$ gets, $x_B$ be the number of points $B$ gets, and $x_C$ be the number of points $C$ gets. Since $x_A = x_B = x_C$, to maximize $x_A$, we can just maximize $x_A + x_B + x_C$. But in each match, if one team wins then the total sum increases by $3$ points, whereas if they tie, the total sum increases by $2$ points. So, it is best if there are the fewest ties possible.

每队与前三队之间各胜一场各负一场，以获得最高积分。对另一队胜一场得 3 分，所以每队得 3×2=6 分。与其他 3 队有 6 场比赛。如果每队赢下这些比赛，得 3·6=18 分，总计 18+6=\boxed{\textbf{(C) }24} 分。这可以通过以下方式看出是最佳情况：设 x_A、B、C 为三队的积分，且相等。要最大化 x_A，即最大化 x_A + x_B + x_C。每场比赛一方胜则总积分增 3 分，平局增 2 分，所以平局越少越好。

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