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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>
  
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==Solution 1==
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==Solution 1==  
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This isn't finished
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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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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>.
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Therefore, we use Case 2  since it brings the greater amount of points, or <math>\boxed {\textbf {(C) }24}</math>.
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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.
 
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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.
 
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Therefore, we use Case2 since it brings the greater amount of points, or <math>\boxed{24}</math>, so the answer is <math>\boxed{C}</math>.
 
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~A1337h4x0r
 
  
 
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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.
+
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 2==
 
==Solution 2==
  
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(1st match(3) + 2nd match(1)) * number of teams(6) = 24, <math>\boxed{C}</math>.
 
  
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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 <math>\boxed{C}</math>.
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We can name the top three teams as <math>A</math>, <math>B</math>, and <math>C</math>. We can see that (respective scores of) <math>A=B=C</math> 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: <math>AB</math>, <math>BC</math>, and <math>AC</math> come twice. In order to even out the scores and get the maximum score, we can say that in match <math>AB</math>, <math>A</math> and <math>B</math> each win once out of the two games that they play. We can say the same thing for <math>AC</math> and <math>BC</math>. This tells us that each team <math>A</math>, <math>B</math>, and <math>C</math> win and lose twice. This gives each team a total of <math>3 + 3 + 0 + 0 = 6</math> points. Now, we need to include the other three teams. We can label these teams as <math>D</math>, <math>E</math>, and <math>F</math>. We can write down every match that <math>A, B,</math> or <math>C</math> plays in that we haven't counted yet: <math>AD</math>, <math>AD</math>, <math>AE</math>, <math>AE</math>, <math>AF</math>, <math>AF</math>, <math>BD</math>, <math>BD</math>, <math>BE</math>, <math>BE</math>, <math>BF</math>, <math>BF</math>, <math>CD</math>, <math>CD</math>, <math>CE</math>, <math>CE</math>, <math>CF</math>, and <math>CF</math>. We can say <math>A</math>, <math>B</math>, and <math>C</math> win each of these in order to obtain the maximum score that <math>A</math>, <math>B</math>, and <math>C</math> can have. If <math>A</math>, <math>B</math>, and <math>C</math> win all six of their matches, <math>A</math>, <math>B</math>, and <math>C</math> will have a score of <math>18</math>. <math>18 + 6</math> results in a maximum score of <math>\boxed{\textbf{(C) }24}</math>
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<!-- Edited by Lvluo -->
  
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~MRLUIGI
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== Solution 3 ==
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To start, we calculate how many games each team plays. Each team can play against <math>5</math> people twice, so there are <math>10</math> games that each team plays. So the answer is <math>10\cdot 3</math> which is <math>30!</math> But wait... if we want <math>3</math> teams to have the same amount of points, there can't possibly be a player who wins all their games. Let the top three teams be <math>A,B</math>, and <math>C.</math> <math>A</math> plays <math>B</math> and <math>C</math> twice so in order to maximize the games being played, we can split it <math>50-50</math> between the <math>4</math> games <math>A</math> plays against <math>B</math> or <math>C</math>. We find that we just subtract <math>2</math> games or <math>6</math> points. Therefore the answer is <math>30-6</math>, <math>24</math>
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or <math>\boxed{\textbf{(C) }24}</math>
  
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==Video Solution by Math-X (First understand the problem!!!)==
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https://youtu.be/IgpayYB48C4?si=ZFTK7CQBPH6nrE9h&t=5702
  
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==Solution 3==
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~Math-X
  
  
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We can name the top three teams as <math>A, B,</math> and <math>C</math>. We can see that <math>A=B=C</math>, 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: <math>AB, BC,</math> and <math>AC</math> come twice. In order to even out the scores and get the maximum score, we can say that in match <math>AB, A</math> and <math>B</math> each win once out of the two games that they play. We can say the same thing for <math>AC</math> and <math>BC</math>. This tells us that each team <math>A, B,</math> and <math>C</math> 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 <math>D, E,</math> and <math>F</math>. We can write down every match that <math>A, B,</math> or <math>C</math> plays in that we haven't counted yet: <math>AD, AD, AE, AE, AF, AF, BD, BD, BE, BE, BF, BF, CD, CD, CE, CE, CF,</math> and <math>CF</math>. We can say <math>A, B,</math> and <math>C</math> win each of these in order to obtain the maximum score that <math>A, B,</math> and <math>C</math> can have. If <math>A, B,</math> and <math>C</math> win all six of their matches, <math>A, B,</math> and <math>C</math> will have a score of <math>18</math>. <math>18 + 6</math> results in a maximum score of <math>\boxed{24}</math>. This tells us that the correct answer choice is <math>\boxed{C}</math>.
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==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/sPdg92Alud4
  
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~Champion1234
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~Education, the Study of Everything
  
  
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-Happpytwin
 
-Happpytwin
 +
 +
== Video Solution by OmegaLearn==
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https://youtu.be/HISL2-N5NVg?t=4616
 +
 +
~ pi_is_3.14
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 +
== Video Solution ==
 +
 +
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=k_AuB_bzidc&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=20
 +
 +
==Video Solution==
 +
https://youtu.be/d-JoEwIOlKQ
 +
 +
~savannahsolver
 +
 +
==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)==
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https://youtu.be/Xm4ZGND9WoY
 +
 +
~Hayabusa1
  
 
==See Also==
 
==See Also==

Latest revision as of 11:49, 30 December 2023

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

This isn't finished 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 $\boxed {\textbf {(C) }24}$.

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

We can name the top three teams as $A$, $B$, and $C$. We can see that (respective scores of) $A=B=C$ 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 $\boxed{\textbf{(C) }24}$

Solution 3

To start, we calculate how many games each team plays. Each team can play against $5$ people twice, so there are $10$ games that each team plays. So the answer is $10\cdot 3$ which is $30!$ But wait... if we want $3$ teams to have the same amount of points, there can't possibly be a player who wins all their games. Let the top three teams be $A,B$, and $C.$ $A$ plays $B$ and $C$ twice so in order to maximize the games being played, we can split it $50-50$ between the $4$ games $A$ plays against $B$ or $C$. We find that we just subtract $2$ games or $6$ points. Therefore the answer is $30-6$, $24$ or $\boxed{\textbf{(C) }24}$

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/IgpayYB48C4?si=ZFTK7CQBPH6nrE9h&t=5702

~Math-X


Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/sPdg92Alud4

~Education, the Study of Everything


Video Solutions

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

https://youtu.be/hM4sHJSMNDs

-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

https://youtu.be/d-JoEwIOlKQ

~savannahsolver

Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

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. AMC logo.png

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