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

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== Problem ==
 
== 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 <math>\tfrac{1}{2}</math> and <math>\tfrac{2}{3}.</math> Armed with this information, what number should Carol choose to maximize her chance of winning?
+
Alice, Bob, and Carol play a game in which each of them chooses a real number between <math>0</math> and <math>1.</math> 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 <math>0</math> and <math>1,</math> and Bob announces that he will choose his number uniformly at random from all the numbers between <math>\tfrac{1}{2}</math> and <math>\tfrac{2}{3}.</math> Armed with this information, what number should Carol choose to maximize her chance of winning?
−
 
 
  
 
<math>
 
<math>
Line 12: Line 11:
 
</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.
 +
 
 +
Based on the value of <math>c,</math> we construct the following table:
 +
<cmath>\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}</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) ==
 +
 
 +
Note that Carol's number must lie in the interval <math>\left[\frac{1}{2}, \frac{2}{3}\right]</math> because it never needs to be less than <math>\frac{1}{2}</math> in order to be less than Bob's number, and it never needs to be greater than <math>\frac{2}{3}</math> 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 <math>\frac{1}{2}+x</math>, where <math>x \in \left[0, \frac{1}{6}\right]</math>. The probability of Case 1 can be expressed as <math>\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)</math>, and the probability of Case 2 can be expressed as <math>\frac{\frac{1}{2} - x}{1}\cdot\frac{x}{\frac{1}{6}}=\left(\frac{1}{2} - x\right)\left(6x\right)</math>.
 +
 
 +
Thus, the probability of Carol winning can be expressed as the sum of the probabilities of Cases 1 and 2: <math>P = \left(\frac{1}{2} + x\right)\left(1 - 6x\right) + \left(\frac{1}{2} - x\right)\left(6x\right)</math>, which simplifies to <math>P = \frac{1}{2} + x - 12x^2</math>. The maximum value of <math>P</math> is obtained through the value of <math>x</math> where the slope is <math>0</math>. We take the first derivative and get <math>1 - 24x</math>, which yields <math>0</math> at <math>x = \frac{1}{24}</math>. Hence, Carol should select <math>\frac{1}{2} + \frac{1}{24} = \boxed{\textbf{(B) }\frac{13}{24}}</math>.
 +
 
 +
Note that the same value of <math>x</math> can be obtained through the Vertex Formula, <math>x=-\frac{b}{2a}</math>, without using Calculus.
  
−
Plug in all the answer choices to get <math>\boxed{\textbf{(B)}.}</math>
+
== Solution 5 (Calculus) ==
  
−
==Solution 2==
+
It suffices to find the average (expected) value of <math>C=\frac{\left(A+B\right)}{2}</math> over the intervals <math>A \in \left[0,1\right]</math> and <math>B \in \left[\frac{1}{2},\frac{2}{3}\right]</math>. We do this by finding <math>\int_0^1 \int_\frac{1}{2}^\frac{2}{3}\frac{\left(A+B\right)}{2}\,dB\,dA</math> and divide by the area of the interval we're integrating over, namely <math>{\left(1-0\right)\left(\frac{2}{3}-\frac{1}{2}\right)}=\frac{1}{6}</math>. <math>\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}</math>. Dividing by <math>\frac{1}{6}</math> we get <math>\boxed{\textbf{(B) }\frac{13}{24}}</math>.
−
Let the value we want be <math>x</math>. The probability that Alice's number is less than Carol's number and Bob's number is greater than Carol's number is <math>x(\frac{2}{3}-x)</math>. Similarly, the probability that Bob's number is less than Carol's number and Alice's number is greater than Carol's number is <math>(x-\frac{1}{2})(1-x)</math>. Adding these together, the probability that Carol wins given a certain number <math>x</math> is <math>-2x^2+\frac{13}{6}x-\frac{1}{2}</math>. Using calculus or the fact that the extremum of a parabola occurs at <math>\frac{-b}{2a}</math>, the maximum value occurs at <math>x=\frac{13}{24}</math>, which is <math>\boxed{\textbf{(B)}.}</math>
 
  
−
== Solution 3 ==
+
~Joeythetoey
−
The expected value of Alice's number is <math>\frac{1}{2}</math> and the expected value of Bob's number is <math>\frac{7}{12}</math>. To maximize her chance of winning, Carol would choose number exactly in between the two expected values, giving:<math>\frac{6+7}{12*2}=\frac{13}{24}</math>. This is <math>\boxed{\textbf{(B)}}</math>. (Random_Guy)
 
  
−
EDIT: I believe this method is incorrect. Assume Bob can only choose <math>\frac{7}{12}</math> but Alice chooses from the same range as before. The answer, using the above method, remains <math>\frac{13}{24}</math> which is clearly wrong in this case. Correct me if I misunderstood the solution.
+
== Video Solution by Richard Rusczyk ==
−
(turnip123)
 
  
−
EDIT2: It would make sense for the answer to remain the same, given that Bob's expected value stays the same. Why should the answer change in your scenario? (KenV)
+
https://artofproblemsolving.com/videos/amc/2018amc12a/474
  
−
EDIT3: I realized my mistake. The method works in all situations where the expected value falls within both of their range. In my case, Carol's number would be less than Bob's number if Carol chooses any number from the range [0, 7/12). She would then want to maximize the chances of picking a number greater than Alice, which is achieved by picking the largest number possible from the range [0, 7/12), which is not 13/24.
+
~ dolphin7
  
−
== Solution 4 ==
+
== Video Solution (Meta-Solving Technique) ==
 +
https://youtu.be/GmUWIXXf_uk?t=926
  
−
Let’s call Alice’s number a, Bob’s number b, and Carol’s number c. Then, in order to maximize her chance of choosing a number that is in between a and b, she should choose c = (a+b)/2.
+
~ pi_is_3.14
  
−
We need to find the average value of (a+b)/2 over the region [0, 1] x[1/2, 2/3] in the a-b plane.
+
==See Also==
  
−
We can set up a double integral with bounds 0 to 1 for the outer integral and 1/2 to 2/3 for the inner integral with an integrand of (a+b)/2. We need to divide our answer by 1/6, the area of the region of interest. This should yield 13/24, B.
+
{{AMC12 box|year=2018|ab=A|num-b=23|num-a=25}}
 +
{{MAA Notice}}

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

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

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