Difference between revisions of "2018 AMC 12A Problems/Problem 24"
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1-c & \mathrm{if} \ \frac23<c<1 | 1-c & \mathrm{if} \ \frac23<c<1 | ||
\end{cases}.</cmath> | \end{cases}.</cmath> | ||
− | Note that | + | Note that <math>P(c)</math> is continuous in the interval <math>(0,1),</math> increasing in the interval <math>\left(0,\frac12\right),</math> increasing and then decreasing in the interval <math>\left(\frac12,\frac23\right),</math> and decreasing in the interval <math>\left(\frac23,1\right).</math> Therefore, the maximum point of <math>P(c)</math> occurs in the interval <math>\left[\frac12,\frac23\right],</math> namely at <math>c=-\frac{13}{2\cdot(-12)}=\boxed{\textbf{(B) }\frac{13}{24}}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 22:57, 12 September 2021
Contents
Problem
Alice, Bob, and Carol play a game in which each of them chooses a real number between and
The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between
and
and Bob announces that he will choose his number uniformly at random from all the numbers between
and
Armed with this information, what number should Carol choose to maximize her chance of winning?
Solution 1 (Expected Values)
The expected value of Alice's number is and the expected value of Bob's number is
To maximize her chance of winning, Carol should choose the midpoint between these two expected values. So, the answer is
Alternatively, once we recognize that the answer lies in the interval we should choose
since no other answer choices lie in this interval.
~Random_Guy (Solution)
~MRENTHUSIASM (Revision)
Solution 2 (Piecewise Function)
Let and
be the numbers that Alice, Bob, and Carol choose, respectively.
Based on the value of we construct the following table:
Let
be Carol's probability of winning when she chooses
We write
as a piecewise function:
Note that
is continuous in the interval
increasing in the interval
increasing and then decreasing in the interval
and decreasing in the interval
Therefore, the maximum point of
occurs in the interval
namely at
~MRENTHUSIASM
Solution 3 (Answer Choices)
Let and
be the numbers that Alice, Bob, and Carol choose, respectively.
From the answer choices, we construct the following table:
Therefore, Carol should choose
to maximize her chance of winning.
~MRENTHUSIASM
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2018amc12a/474
~ dolphin7
Video Solution (Meta-Solving Technique)
https://youtu.be/GmUWIXXf_uk?t=926
~ pi_is_3.14
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.