Difference between revisions of "2019 AMC 8 Problems/Problem 19"

Revision as of 19:10, 17 October 2020

Solution 3

We can name the top three teams as $A, B,$ and $C$ . We can see that $A=B=C$ , 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: $AB, BC,$ and $AC$ come twice. In order to even out the scores and get the maximum score, we can say that in match $AB, A$ and $B$ each win once out of the two games that they play. We can say the same thing for $AC$ and $BC$ . This tells us that each team $A, B,$ and $C$ win and lose twice. This gives each team a total of 3 + 3 + 0 + 0 = 6 points. Now, we need to include the other three teams. We can label these teams as $D, E,$ and $F$ . We can write down every match that $A, B,$ or $C$ plays in that we haven't counted yet: $AD, AD, AE, AE, AF, AF, BD, BD, BE, BE, BF, BF, CD, CD, CE, CE, CF,$ and $CF$ . We can say $A, B,$ and $C$ win each of these in order to obtain the maximum score that $A, B,$ and $C$ can have. If $A, B,$ and $C$ win all six of their matches, $A, B,$ and $C$ will have a score of $18$ . $18 + 6$ results in a maximum score of $\boxed{24}$ . This tells us that the correct answer choice is $\boxed{C}$ .

~Champion1234

Video Solutions

Associated Video - https://youtu.be/s0O3_uXZrOI

Video Solution - https://youtu.be/Lw8fSbX_8FU (Also explains problems 11-20)

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.

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-==Solution 1==
-After fully understanding the problem, we immediately know that the three top teams, say team <math>A</math>, team <math>B</math>, and team <math>C</math>, must beat the other three teams <math>D</math>, <math>E</math>, <math>F</math>. Therefore, <math>A</math>,<math>B</math>,<math>C</math> must each obtain <math>(3+3+3)=9</math> points. However, they play against each team twice, for a total of <math>18</math> points against <math>D</math>, <math>E</math>, and <math>F</math>. For games between <math>A</math>, <math>B</math>, <math>C</math>, we have 2 cases. In both cases, there is an equality of points between <math>A</math>, <math>B</math>, and <math>C</math>.
-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\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>.
-~A1337h4x0r
---------------------------
-Note that case 2 can be easily seen to be better as follows. Let <math>x_A</math> be the number of points <math>A</math> gets, <math>x_B</math> be the number of points <math>B</math> gets, and <math>x_C</math> be the number of points <math>C</math> gets. Since <math>x_A = x_B = x_C</math>, to maximize <math>x_A</math>, we can just maximize <math>x_A + x_B + x_C</math>. But in each match, if one team wins then the total sum increases by <math>3</math> points, whereas if they tie, the total sum increases by <math>2</math> points. So it is best if there are the fewest ties possible.
 ==Solution 3==

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Difference between revisions of "2019 AMC 8 Problems/Problem 19"

Revision as of 19:10, 17 October 2020

Solution 3

Video Solutions

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