Difference between revisions of "2019 AMC 8 Problems/Problem 19"
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Revision as of 21:32, 24 November 2019
In a tournament there are six team 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?
After fully understanding the problem, we immediately know that the three top teams, say team , team , and team , must beat the other three teams , , . Therefore, ,, must each obtain points. However, they play against each team twice, for a total of points against , , and . For games between , , , we have 2 cases. In both cases, there is an equality of points between , , and .
Case 1: A team ties the two other teams. For a tie, we have 1 point, so we have points (they play twice). Therefore, this case brings a total of points.
Case 2: A team beats one team while losing to another. This gives equality, as each team wins once and loses once as well. For a win, we have points, so a team gets points if they each win a game and lose a game. This case brings a total of points.
Therefore, we use Case 2 since it brings the greater amount of points, or , so the answer is .
Note that case 2 can be easily seen to be better as follows. Let be the number of points gets, be the number of points gets, and be the number of points gets. Since , to maximize , we can just maximize . But in each match, if one team wins then the total sum increases by points, whereas if they tie, the total sum increases by points. So it is best if there are the fewest ties possible.
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