Difference between revisions of "2019 AMC 8 Problems/Problem 19"
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Case 1: A team ties the two other teams. For a tie, we have 1 point, so we have <math>(1+1)*2=4</math> points (they play twice). Therefore, this case brings a total of <math>4+18=22</math> points. | Case 1: A team ties the two other teams. For a tie, we have 1 point, so we have <math>(1+1)*2=4</math> points (they play twice). Therefore, this case brings a total of <math>4+18=22</math> 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 <math>3</math> points, so a team gets <math>3 | + | 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 <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. |
Therefore, we use Case 2 since it brings the greater amount of points, or <math>\boxed{24}</math>, so the answer is <math>\boxed{C}</math>. | Therefore, we use Case 2 since it brings the greater amount of points, or <math>\boxed{24}</math>, so the answer is <math>\boxed{C}</math>. | ||
| − | ~A1337h4x0r | + | ~A1337h4x0r & ViratKohli2018 |
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Revision as of 20:44, 13 December 2019
Contents
Problem 19
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?
Solution 1
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 
.
~A1337h4x0r & ViratKohli2018
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
bruh: (1st match(3) + 2nd match(1)) * number of teams(6) = 24, .
Explanation: So after reading the problem we see that there are 6 teams and each team versus each other twice. This means one of the two matches has to be a win, so 3 points so far. Now if we say that the team won again and make it 6 points, that would mean that team would be dominating the leader-board and the problem says that all the top 3 people have the same score. So bruh means the maximum amount of points we could get is 1 so that each team gets the same amount of matches won & drawn so that adds up to 4. 4 * the number of teams(6) = 24 so the answer is .
bruh bruh
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. 
