2019 AMC 8 Problems/Problem 19

Problem 19

In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ 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?

$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$

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 2

(1st match(3) + 2nd match(1)) * number of teams(6) = 24, $\boxed{C}$ .

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 that 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 $\boxed{C}$ .

~MRLUIGI

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 $\boxed{24}$ .

~Champion1234

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2019 AMC 8 Problems/Problem 19

Contents

Problem 19

Solution 1

Solution 2

Solution 3

See Also