2019 AMC 8 Problems/Problem 19

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Solution 1

After fully understanding the problem, we immediately know that the three top teams, say team $A$, team $B$, and team $C$, must beat the other three teams $D$, $E$, $F$. Therefore, $A$,$B$,$C$ must each obtain $(3+3+3)=9$ points. However, they play against each team twice, for a total of $18$ points against $D$, $E$, and $F$. For games between $A$, $B$, $C$, we have 2 cases. In both cases, there is an equality of points between $A$, $B$, and $C$.

Case 1: A team ties the two other teams. For a tie, we have 1 point, so we have $(1+1)*2=4$ points (they play twice). Therefore, this case brings a total of $4+18=22$ 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 $3$ points, so a team gets $3\times2=6$ points if they each win a game and lose a game. This case brings a total of $18+6=24$ points.

Therefore, we use Case 2 since it brings the greater amount of points, or $\boxed{24}$, so the answer is $\boxed{C}$.

~A1337h4x0r


Note that case 2 can be easily seen to be better as follows. Let $x_A$ be the number of points $A$ gets, $x_B$ be the number of points $B$ gets, and $x_C$ be the number of points $C$ gets. Since $x_A = x_B = x_C$, to maximize $x_A$, we can just maximize $x_A + x_B + x_C$. But in each match, if one team wins then the total sum increases by $3$ points, whereas if they tie, the total sum increases by $2$ points. So it is best if there are the fewest ties possible.

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)

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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

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