2019 AMC 8
Complete problem set with solutions and individual problem pages
In a tournament there are six teams that play each other twice. A team earns points for a win, point for a draw, and 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?
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Solution 1
We can name the top three teams as , , and . We can see that (respective scores of) 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 points. Now, we need to include the other three teams. We can label these teams as , , and . We can write down every match that 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 . results in a maximum score of
Solution 2
To start, we calculate how many games each team plays. Each team can play against other teams twice, so there are games that each team plays. So the answer is which is But wait if we want 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 , and plays and twice, so in order to maximize the games being played, we can split it between the games plays against or . We find that we just subtract games or points. Therefore the answer is , or
