AMC 8 2019 Problem 19
AMC8 2019 年第 19 题
基本计数原理★★★☆☆
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?
在一场循环赛中，一共有 6 支队伍，每两支队伍之间都要比赛 2 次。某个队伍赢一次得 3 分，平局得 1 分，输了则得 0 分。当所有比赛都结束后，结果发现排名前三的队伍得分相同。问排名前三的队伍每支队最多可能得多少分？
- A.22
- B.23
- C.24
- D.26
- E.30
正确答案 · Correct Answer
C
解析 · Solution
Solution 1
Each team wins once and loses once to get the highest points. For a win, we have 3 points, so a team gets 3×2=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 points This brings a total of 18+6=24 points.
We can name the top three teams as , , and . We can see that (respective scores of) A=B=C because these teams have the same points. If we look at the matches that involve the top three teams, we see that there are some duplicates: , , and come twice. In order to even out the scores and get the maximum score, we can say that in match , and each win once out of the two games that they play. We can say the same thing for and . This tells us that each team , , and win and lose twice. This gives each team a total of 3 + 3 + 0 + 0 = 6 points. Now, we need to include the other three teams. We can label these teams as , , and . We can write down every match that $A, B,$ or plays in that we haven't counted yet: , , , , , , , , , , , , , , , , , and . We can say , , and win each of these in order to obtain the maximum score that , , and can have. If , , and win all six of their matches, , , and will have a score of 18. 18 + 6 results in a maximum score of 24
To start, we calculate how many games each team plays. Each team can play against 5 other teams twice, so there are 10 games that each team plays. So the answer is which is 30 But wait... if we want 3 teams to have the same amount of points, there can't possibly be a team who wins all their games. Let the top three teams be A,B, and . plays and twice, so in order to maximize the games being played, we can split it 50-50 between the 4 games plays against or . We find that we just subtract 2 games or 6 points. Therefore the answer is 30-6, 24
Let the top 3 teams be A, B, C and the bottom teams be D, E, F. There are 2 types of games the winners ( A, B, C ) can play: Winner vs loser and Winner vs Winner. Case 1: Winner vs Loser. For A, it would be $AD, AE, AF$. We want to maximize this, making win in all 6 ( 3×2 ) games. Therefore, A would get 6×3 points, or 18. Case 2: Winner vs Winner. There are games they can play against each other. Since we want each of the 3 to have an equal amount, each team should win 2 games and get 6 points. In total, 18+6=24.
Name the teams A, B, C, D, E, and F. As we know that the top three teams, A, B, and C, tied, we can just focus on one, A. When A plays B or C, we want them to both win one and lose one, maximising the points for both teams. When A plays D, E, or F, we want A to win both times. This adds up to 3+3 (top teams) +6+6+6 (other teams), which equals 24 points. This solution is by far the fastest and probably most used
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