Difference between revisions of "2019 AMC 8 Problems/Problem 19"
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− | ==Problem | + | ==Problem== |
− | In a tournament there are six | + | In a tournament there are six teams that play each other twice. A team earns <math>3</math> points for a win, <math>1</math> point for a draw, and <math>0</math> 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? |
<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> | ||
− | ==Solution 1== | + | ==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 <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 {\textbf {(C) }24}</math>. | ||
+ | |||
+ | -------------------------- | ||
+ | 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== | ||
+ | |||
+ | |||
+ | 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> | ||
+ | <!-- Edited by Lvluo --> | ||
+ | |||
+ | == Solution 3 == | ||
+ | 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> | ||
+ | or <math>\boxed{\textbf{(C) }24}</math> | ||
+ | |||
+ | == Solution 4== | ||
+ | Let the top <math>3</math> teams be <math>A, B, C</math> and the bottom teams to be <math>D, E, F</math>. There are <math>2</math> types of games the winners (<math>A, B, C</math>) can play: Winners vs losers and Winner vs Winner. | ||
+ | |||
+ | Case 1: Winner vs Loser | ||
+ | |||
+ | For A, it would be <math>AD, AE, AF</math>. We want to maximize this or make <math>A</math> win in all <math>6</math> (<math>3\times2</math>) games. Therefore, A would get <math>6\times3</math> points or <math>18</math>. | ||
+ | |||
+ | Case 2: Winner vs Winner | ||
+ | |||
+ | There are <math>_{3}P_{2}=\frac{3!}{1!}=6</math> games they can play against each other. Since we want each <math>3</math> to have an equal amount. Each team should win <math>2</math> games and get <math>6</math> points. In total, <math>18+6=\boxed{\textbf{(C) }24}</math> | ||
+ | ~RandomMathGuy500 | ||
+ | |||
+ | ==Video Solution by Math-X== | ||
+ | https://youtu.be/IgpayYB48C4?si=ZFTK7CQBPH6nrE9h&t=5702 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution== | ||
+ | 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 == | ||
+ | |||
+ | 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(1 to 25 Full Solution)== | ||
+ | https://youtu.be/Xm4ZGND9WoY | ||
+ | |||
+ | ~Hayabusa1 | ||
==See Also== | ==See Also== |
Revision as of 22:47, 27 December 2024
Contents
[hide]Problem
In a tournament there are six teams 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
This isn't finished
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 .
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
We can name the top three teams as ,
, and
. We can see that (respective scores of)
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:
,
, and
come twice. In order to even out the scores and get the maximum score, we can say that in match
,
and
each win once out of the two games that they play. We can say the same thing for
and
. This tells us that each team
,
, and
win and lose twice. This gives each team a total of
points. Now, we need to include the other three teams. We can label these teams as
,
, and
. We can write down every match that
or
plays in that we haven't counted yet:
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
, and
. We can say
,
, and
win each of these in order to obtain the maximum score that
,
, and
can have. If
,
, and
win all six of their matches,
,
, and
will have a score of
.
results in a maximum score of
Solution 3
To start, we calculate how many games each team plays. Each team can play against people twice, so there are
games that each team plays. So the answer is
which is
But wait... if we want
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
, and
plays
and
twice so in order to maximize the games being played, we can split it
between the
games
plays against
or
. We find that we just subtract
games or
points. Therefore the answer is
,
or
Solution 4
Let the top teams be
and the bottom teams to be
. There are
types of games the winners (
) can play: Winners vs losers and Winner vs Winner.
Case 1: Winner vs Loser
For A, it would be . We want to maximize this or make
win in all
(
) games. Therefore, A would get
points or
.
Case 2: Winner vs Winner
There are games they can play against each other. Since we want each
to have an equal amount. Each team should win
games and get
points. In total,
~RandomMathGuy500
Video Solution by Math-X
https://youtu.be/IgpayYB48C4?si=ZFTK7CQBPH6nrE9h&t=5702
~Math-X
Video Solution
~Education, the Study of Everything
Video Solutions
Associated Video - https://youtu.be/s0O3_uXZrOI
-Happpytwin
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=4616
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=k_AuB_bzidc&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=20
Video Solution
~savannahsolver
Video Solution by The Power of Logic(1 to 25 Full Solution)
~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.