Difference between revisions of "2018 AMC 12A Problems/Problem 24"

Latest revision as of 01:49, 10 October 2024

Problem

Alice, Bob, and Carol play a game in which each of them chooses a real number between $0$ and $1.$ 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 $0$ and $1,$ and Bob announces that he will choose his number uniformly at random from all the numbers between $\tfrac{1}{2}$ and $\tfrac{2}{3}.$ Armed with this information, what number should Carol choose to maximize her chance of winning?

$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }\frac{13}{24} \qquad \textbf{(C) }\frac{7}{12} \qquad \textbf{(D) }\frac{5}{8} \qquad \textbf{(E) }\frac{2}{3}\qquad$

Solution 1 (Expected Values)

The expected value of Alice's number is $\frac12\left(0+1\right)=\frac12,$ and the expected value of Bob's number is $\frac12\left(\frac12+\frac23\right)=\frac{7}{12}.$ To maximize her chance of winning, Carol should choose the midpoint between these two expected values. So, the answer is $\frac12\left(\frac12+\frac{7}{12}\right)=\boxed{\textbf{(B) }\frac{13}{24}}.$

Alternatively, once we recognize that the answer lies in the interval $\left(\frac12,\frac{7}{12}\right),$ we should choose $\textbf{(B)}$ since no other answer choices lie in this interval.

~Random_Guy ~MRENTHUSIASM

Solution 2 (Piecewise Function)

Let $a,b,$ and $c$ be the numbers that Alice, Bob, and Carol choose, respectively.

Based on the value of $c,$ we construct the following table: $\begin{array}{c|c|c} & & \\ [-2ex] \textbf{Case} & \textbf{Conditions for }\boldsymbol{a}\textbf{ and }\boldsymbol{b} & \textbf{Carol's Probability of Winning} \\ [0.5ex] \hline & & \\ [-1.5ex] 0<c<\frac12 & 0<a<c \text{ and } \frac12<b<\frac23 & \hspace{1.25mm}\frac{c}{1}\cdot\frac{1/6}{1/6}=c \\ [1.5ex] \frac12\leq c\leq\frac23 & \left(0<a<c \text{ and } c<b<\frac23\right) \text{ or } \left(c<a<1 \text{ and } \frac12<b<c\right) & \hspace{1.25mm}\frac{c}{1}\cdot\frac{2/3-c}{1/6}+\frac{1-c}{1}\cdot\frac{c-1/2}{1/6}=-12c^2+13c-3 \\ [1.5ex] \frac23<c<1 & c<a<1 \text{ and } \frac12<b<\frac23 & \hspace{4.375mm}\frac{1-c}{1}\cdot\frac{1/6}{1/6}=1-c \\ [1.5ex] \end{array}$ Let $P(c)$ be Carol's probability of winning when she chooses $c.$ We write $P(c)$ as a piecewise function: $P(c) = \begin{cases} c & \mathrm{if} \ 0<c<\frac12 \\ -12c^2+13c-3 & \mathrm{if} \ \frac12\leq c\leq\frac23 \\ 1-c & \mathrm{if} \ \frac23<c<1 \end{cases}.$ Note that $P(c)$ is continuous in the interval $(0,1),$ increasing in the interval $\left(0,\frac12\right),$ increasing and then decreasing in the interval $\left(\frac12,\frac23\right),$ and decreasing in the interval $\left(\frac23,1\right).$ The graph of $y=P(c)$ 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 $P(c)$ occurs in the interval $\left[\frac12,\frac23\right],$ namely at $c=-\frac{13}{2\cdot(-12)}=\boxed{\textbf{(B) }\frac{13}{24}}.$

~MRENTHUSIASM

Solution 3 (Answer Choices)

Let $a,b,$ and $c$ be the numbers that Alice, Bob, and Carol choose, respectively.

From the answer choices, we construct the following table: $\begin{array}{c|c|c} & & \\ [-2ex] \boldsymbol{c} & \textbf{Conditions for }\boldsymbol{a}\textbf{ and }\boldsymbol{b} & \textbf{Carol's Probability of Winning} \\ [0.5ex] \hline & & \\ [-1.5ex] \frac12 & 0<a<\frac12 \text{ and } \frac12<b<\frac23 & \hspace{23.375mm}\frac{1/2}{1}\cdot\frac{1/6}{1/6}=\frac12 \\ [1.5ex] \frac{13}{24} & \left(0<a<\frac{13}{24} \text{ and } \frac{13}{24}<b<\frac23\right) \text{ or } \left(\frac{13}{24}<a<1 \text{ and } \frac12<b<\frac{13}{24}\right) & \frac{13/24}{1}\cdot\frac{1/8}{1/6}+\frac{11/24}{1}\cdot\frac{1/24}{1/6}=\frac{25}{48} \\ [1.5ex] \frac{7}{12} & \left(0<a<\frac{7}{12} \text{ and } \frac{7}{12}<b<\frac23\right) \text{ or } \left(\frac{7}{12}<a<1 \text{ and } \frac12<b<\frac{7}{12}\right) & \frac{7/12}{1}\cdot\frac{1/12}{1/6}+\frac{5/12}{1}\cdot\frac{1/12}{1/6}=\frac12 \\ [1.5ex] \frac58 & \left(0<a<\frac58 \text{ and } \frac58<b<\frac23\right) \text{ or } \left(\frac58<a<1 \text{ and } \frac12<b<\frac58\right) & \hspace{5.625mm}\frac{5/8}{1}\cdot\frac{1/24}{1/6}+\frac{3/8}{1}\cdot\frac{1/8}{1/6}=\frac{7}{16} \\ [1.5ex] \frac23 & \frac23<a<1 \text{ and } \frac12<b<\frac23 & \hspace{23.25mm}\frac{1/3}{1}\cdot\frac{1/6}{1/6}=\frac13 \\ [1.5ex] \end{array}$ Therefore, Carol should choose $\boxed{\textbf{(B) }\frac{13}{24}}$ to maximize her chance of winning.

~MRENTHUSIASM

Solution 4 (Calculus)

Note that Carol's number must lie in the interval $\left[\frac{1}{2}, \frac{2}{3}\right]$ because it never needs to be less than $\frac{1}{2}$ in order to be less than Bob's number, and it never needs to be greater than $\frac{2}{3}$ in order to be greater than Bob's number. Going past either value will only decrease the probability of being on the correct side of Alice's number.

There are two cases of winning:

Case 1: Alice chooses a number that is smaller than Carol's, and Bob chooses a number that is bigger.

Case 2: Alice chooses a number that is bigger than Carol's, and Bob chooses a number that is smaller.

Let Carol's number be $\frac{1}{2}+x$ , where $x \in \left[0, \frac{1}{6}\right]$ . The probability of Case 1 can be expressed as $\frac{\frac{1}{2} + x}{1}\cdot\frac{\frac{1}{6} - x}{\frac{1}{6}}=\left(\frac{1}{2} + x\right)\left(1 - 6x\right)$ , and the probability of Case 2 can be expressed as $\frac{\frac{1}{2} - x}{1}\cdot\frac{x}{\frac{1}{6}}=\left(\frac{1}{2} - x\right)\left(6x\right)$ .

Thus, the probability of Carol winning can be expressed as the sum of the probabilities of Cases 1 and 2: $P = \left(\frac{1}{2} + x\right)\left(1 - 6x\right) + \left(\frac{1}{2} - x\right)\left(6x\right)$ , which simplifies to $P = \frac{1}{2} + x - 12x^2$ . The maximum value of $P$ is obtained through the value of $x$ where the slope is $0$ . We take the first derivative and get $1 - 24x$ , which yields $0$ at $x = \frac{1}{24}$ . Hence, Carol should select $\frac{1}{2} + \frac{1}{24} = \boxed{\textbf{(B) }\frac{13}{24}}$ .

Note that the same value of $x$ can be obtained through the Vertex Formula, $x=-\frac{b}{2a}$ , without using Calculus.

Solution 5 (Calculus)

It suffices to find the average (expected) value of $C=\frac{\left(A+B\right)}{2}$ over the intervals $A \in \left[0,1\right]$ and $B \in \left[\frac{1}{2},\frac{2}{3}\right]$ . We do this by finding $\int_0^1 \int_\frac{1}{2}^\frac{2}{3}\frac{\left(A+B\right)}{2}\,dB\,dA$ and divide by the area of the interval we're integrating over, namely ${\left(1-0\right)\left(\frac{2}{3}-\frac{1}{2}\right)}=\frac{1}{6}$ . $\int_0^1 \left[\frac{AB}{2}+\frac{B^2}{4}\right]_{B=\frac{1}{2}}^\frac{2}{3}\,dA = \left[\frac{A^2}{24}+\frac{7A}{144}\right]_{A=0}^1=\frac{13}{144}$ . Dividing by $\frac{1}{6}$ we get $\boxed{\textbf{(B) }\frac{13}{24}}$ .

~Joeythetoey

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

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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