Difference between revisions of "2018 AMC 12A Problems/Problem 24"
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</math> | </math> | ||
− | == Solution 1 ( | + | == Solution 1 (Expected Values) == |
+ | The expected value of Alice's number is <math>\frac12\left(0+1\right)=\frac12,</math> and the expected value of Bob's number is <math>\frac12\left(\frac12+\frac23\right)=\frac{7}{12}.</math> To maximize her chance of winning, Carol should choose the midpoint between these two expected values. So, the answer is <math>\frac12\left(\frac12+\frac{7}{12}\right)=\boxed{\textbf{(B) }\frac{13}{24}}.</math> | ||
+ | |||
+ | Alternatively, once we recognize that the answer lies in the interval <math>\left(\frac12,\frac{7}{12}\right),</math> we should choose <math>\textbf{(B)}</math> since no other answer choices lie in this interval. | ||
+ | |||
+ | ~Random_Guy ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2 (Piecewise Function)== | ||
Let <math>a,b,</math> and <math>c</math> be the numbers that Alice, Bob, and Carol choose, respectively. | Let <math>a,b,</math> and <math>c</math> be the numbers that Alice, Bob, and Carol choose, respectively. | ||
− | + | Based on the value of <math>c,</math> we construct the following table: | |
<cmath>\begin{array}{c|c|c} | <cmath>\begin{array}{c|c|c} | ||
& & \\ [-2ex] | & & \\ [-2ex] | ||
− | \ | + | \textbf{Case} & \textbf{Conditions for }\boldsymbol{a}\textbf{ and }\boldsymbol{b} & \textbf{Carol's Probability of Winning} \\ [0.5ex] |
\hline | \hline | ||
& & \\ [-1.5ex] | & & \\ [-1.5ex] | ||
− | + | 0<c<\frac12 & 0<a<c \text{ and } \frac12<b<\frac23 & \hspace{1.25mm}\frac{c}{1}\cdot\frac{1/6}{1/6}=c \\ [1.5ex] | |
− | + | \frac12\leq c\leq\frac23 & \left(0<a<c \text{ and } c<b<\frac23\right) \text{ or } \left(c<a<1 \text{ and } \frac12<b<c\right) & \hspace{1.25mm}\frac{c}{1}\cdot\frac{2/3-c}{1/6}+\frac{1-c}{1}\cdot\frac{c-1/2}{1/6}=-12c^2+13c-3 \\ [1.5ex] | |
− | \ | + | \frac23<c<1 & c<a<1 \text{ and } \frac12<b<\frac23 & \hspace{4.375mm}\frac{1-c}{1}\cdot\frac{1/6}{1/6}=1-c \\ [1.5ex] |
− | |||
− | \frac23 & c<a<1 \text{ and } \frac12<b< | ||
\end{array}</cmath> | \end{array}</cmath> | ||
− | + | Let <math>P(c)</math> be Carol's probability of winning when she chooses <math>c.</math> We write <math>P(c)</math> as a piecewise function: | |
+ | <cmath>P(c) = \begin{cases} | ||
+ | c & \mathrm{if} \ 0<c<\frac12 \\ | ||
+ | -12c^2+13c-3 & \mathrm{if} \ \frac12\leq c\leq\frac23 \\ | ||
+ | 1-c & \mathrm{if} \ \frac23<c<1 | ||
+ | \end{cases}.</cmath> | ||
+ | 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> The graph of <math>y=P(c)</math> is shown below. | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(200); | ||
+ | |||
+ | real f(real x) { return x; } | ||
+ | real g(real x) { return -12x^2+13x-3; } | ||
+ | real h(real x) { return 1-x; } | ||
+ | |||
+ | draw((1/2,0)--(1/2,1.25),dashed); | ||
+ | draw((2/3,0)--(2/3,1.25),dashed); | ||
+ | draw(graph(f,0,1/2),red); | ||
+ | draw(graph(g,1/2,2/3),red); | ||
+ | draw(graph(h,2/3,1),red); | ||
+ | |||
+ | real xMin = -0.25; | ||
+ | real xMax = 1.25; | ||
+ | real yMin = -0.25; | ||
+ | real yMax = 1.25; | ||
+ | |||
+ | draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); | ||
+ | draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); | ||
+ | label("$c$",(xMax,0),(2,0)); | ||
+ | label("$y$",(0,yMax),(0,2)); | ||
− | + | pair A[]; | |
+ | A[0] = (0,0); | ||
+ | A[1] = (1/2,1/2); | ||
+ | A[2] = (2/3,1/3); | ||
+ | A[3] = (1,0); | ||
− | + | dot(A[1],red+linewidth(3.5)); | |
+ | dot(A[2],red+linewidth(3.5)); | ||
− | + | label("$0$",A[0],(-1.5,-1.5)); | |
+ | label("$\frac12$",(1/2,0),(0,-1.5)); | ||
+ | label("$\frac23$",(2/3,0),(0,-1.5)); | ||
+ | label("$1$",A[3],(0,-1.5)); | ||
+ | label("$1$",(0,1),(-1.5,0)); | ||
− | + | draw((1/2,-0.02)--(1/2,0.02),linewidth(1)); | |
+ | draw((2/3,-0.02)--(2/3,0.02),linewidth(1)); | ||
+ | draw((1,-0.02)--(1,0.02),linewidth(1)); | ||
+ | draw((-0.02,1)--(0.02,1),linewidth(1)); | ||
+ | </asy> | ||
+ | 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 | |
− | + | == Solution 3 (Answer Choices)== | |
+ | Let <math>a,b,</math> and <math>c</math> be the numbers that Alice, Bob, and Carol choose, respectively. | ||
− | == | + | From the answer choices, we construct the following table: |
+ | <cmath>\begin{array}{c|c|c} | ||
+ | & & \\ [-2ex] | ||
+ | \boldsymbol{c} & \textbf{Conditions for }\boldsymbol{a}\textbf{ and }\boldsymbol{b} & \textbf{Carol's Probability of Winning} \\ [0.5ex] | ||
+ | \hline | ||
+ | & & \\ [-1.5ex] | ||
+ | \frac12 & 0<a<\frac12 \text{ and } \frac12<b<\frac23 & \hspace{23.375mm}\frac{1/2}{1}\cdot\frac{1/6}{1/6}=\frac12 \\ [1.5ex] | ||
+ | \frac{13}{24} & \left(0<a<\frac{13}{24} \text{ and } \frac{13}{24}<b<\frac23\right) \text{ or } \left(\frac{13}{24}<a<1 \text{ and } \frac12<b<\frac{13}{24}\right) & \frac{13/24}{1}\cdot\frac{1/8}{1/6}+\frac{11/24}{1}\cdot\frac{1/24}{1/6}=\frac{25}{48} \\ [1.5ex] | ||
+ | \frac{7}{12} & \left(0<a<\frac{7}{12} \text{ and } \frac{7}{12}<b<\frac23\right) \text{ or } \left(\frac{7}{12}<a<1 \text{ and } \frac12<b<\frac{7}{12}\right) & \frac{7/12}{1}\cdot\frac{1/12}{1/6}+\frac{5/12}{1}\cdot\frac{1/12}{1/6}=\frac12 \\ [1.5ex] | ||
+ | \frac58 & \left(0<a<\frac58 \text{ and } \frac58<b<\frac23\right) \text{ or } \left(\frac58<a<1 \text{ and } \frac12<b<\frac58\right) & \hspace{5.625mm}\frac{5/8}{1}\cdot\frac{1/24}{1/6}+\frac{3/8}{1}\cdot\frac{1/8}{1/6}=\frac{7}{16} \\ [1.5ex] | ||
+ | \frac23 & \frac23<a<1 \text{ and } \frac12<b<\frac23 & \hspace{23.25mm}\frac{1/3}{1}\cdot\frac{1/6}{1/6}=\frac13 \\ [1.5ex] | ||
+ | \end{array}</cmath> | ||
+ | Therefore, Carol should choose <math>\boxed{\textbf{(B) }\frac{13}{24}}</math> to maximize her chance of winning. | ||
− | + | ~MRENTHUSIASM | |
− | + | == Solution 4 (Calculus) == | |
+ | There are two cases of winning: | ||
− | + | Case 1: Alice choose a number that is smaller than Carol's, and Bob choose a number that is bigger. | |
− | + | Case 2: Alice choose a number that is bigger, and Bob choose a number that is smaller. | |
− | + | Let Carol's number be <math>\frac{1}{2}+x</math>, then the probability of Case 1 can be expressed by <math>\frac{1/2 + x}{1}\cdot\frac{1/3 - x }{1/3}</math>, and the probability of Case 2 can be expressed by <math>\frac{1/3 + 1/6 - x}{1}\cdot\frac{x}{1/3}</math>. | |
− | + | Then the probability of Carol winning can be expressed by the sum of the probability of Case 1 and the probability of Case 2: <math>3\left(\frac{1}{2} + x\right)\left(\frac{1}{3} - x\right)+ 3x\left(\frac{1}{2} - x\right)</math>, which simplify to <math>\frac{1}{2} + x - 6x^2</math>. Taking the first derivative, we get <math>1 - 12x</math>, which yield a maximum point at <math>x = \frac{1}{12}</math>. Hence, Carol should select <math>\frac{1}{2} + \frac{1}{12} = \boxed{\textbf{(B) }\frac{13}{24}}</math>. | |
− | |||
== Video Solution by Richard Rusczyk == | == Video Solution by Richard Rusczyk == |
Revision as of 23:18, 6 August 2024
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 ~MRENTHUSIASM
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
The graph of
is shown below.
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
Solution 4 (Calculus)
There are two cases of winning:
Case 1: Alice choose a number that is smaller than Carol's, and Bob choose a number that is bigger.
Case 2: Alice choose a number that is bigger, and Bob choose a number that is smaller.
Let Carol's number be , then the probability of Case 1 can be expressed by
, and the probability of Case 2 can be expressed by
.
Then the probability of Carol winning can be expressed by the sum of the probability of Case 1 and the probability of Case 2: , which simplify to
. Taking the first derivative, we get
, which yield a maximum point at
. Hence, Carol should select
.
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.