2019 AMC 8 Problems/Problem 19
Contents
- 1 Problem 19
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Video Solution by Math-X (First understand the problem!!!)
- 6 Video Solution (HOW TO THINK CREATIVELY!!!)
- 7 Video Solutions
- 8 Video Solution by OmegaLearn
- 9 Video Solution
- 10 Video Solution
- 11 Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
- 12 See Also
Problem 19
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?
Solution 1
This isn't finished 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 .
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.
Solution 2
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 3
To start, we calculate how many games each team plays. Each team can play against people 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 player 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
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/IgpayYB48C4?si=ZFTK7CQBPH6nrE9h&t=5702
~Math-X
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solutions
Associated Video - https://youtu.be/s0O3_uXZrOI
-Happpytwin
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=4616
~ pi_is_3.14
Video Solution
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=k_AuB_bzidc&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=20
Video Solution
~savannahsolver
Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~Hayabusa1
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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