Difference between revisions of "2019 AMC 8 Problems/Problem 19"
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<math>\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30</math> | <math>\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30</math> | ||
− | ==Solution 1== | + | ==Solution 1== to another. This gives equality, as each team wins once and loses once as well. For a win, we have <math>3</math> points, so a team gets <math>3\times2=6</math> points if they each win a game and lose a game. This case brings a total of <math>18+6=24</math> points. |
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Therefore, we use Case 2 since it brings the greater amount of points, or <math>\boxed {\textbf {(C) }24}</math>. | Therefore, we use Case 2 since it brings the greater amount of points, or <math>\boxed {\textbf {(C) }24}</math>. |
Revision as of 19:27, 16 January 2023
Contents
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== 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 .
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
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.