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Difference between revisions of "2018 AMC 12A Problems/Problem 24"
(→‎Solution 4 (Calculus))
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</math>
 
</math>
  
−
== Solution 1 (Answer Choices)==
+
== Solution 1 (Expected Values) ==
−
Let <math>a,b,</math> and <math>c</math> be the numbers that Alice, Bob, and Carol choose, respectively.
+
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>
  
−
From the answer choices, we construct the following table:
+
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.
−
<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
+
~Random_Guy ~MRENTHUSIASM
  
 
==Solution 2 (Piecewise Function)==
 
==Solution 2 (Piecewise Function)==
Line 39: Line 27:
 
\hline
 
\hline
 
& & \\ [-1.5ex]
 
& & \\ [-1.5ex]
−
0<c\leq\frac12 & 0<a<c \text{ and } \frac12<b<\frac23 & \hspace{1.25mm}\frac{c}{1}\cdot\frac{1/6}{1/6}=c \\ [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<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]
+
\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<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>
 
\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:
 
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}
 
<cmath>P(c) = \begin{cases}
−
c & \mathrm{if} \ 0<c\leq\frac12 \\
+
c & \mathrm{if} \ 0<c<\frac12 \\
−
-12c^2+13c-3 & \mathrm{if} \ \frac12<c\leq\frac23 \\
+
-12c^2+13c-3 & \mathrm{if} \ \frac12\leq c\leq\frac23 \\
 
1-c & \mathrm{if} \ \frac23<c<1
 
1-c & \mathrm{if} \ \frac23<c<1
 
\end{cases}.</cmath>
 
\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);
  
−
<b>WILL FINISH BY THE MORNING. NO EDIT PLEASE.</b>
+
real f(real x) { return x; }
 +
real g(real x) { return -12x^2+13x-3; }
 +
real h(real x) { return 1-x; }
  
−
== Solution 3 (5-second way) ==
+
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);
  
−
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)
+
real xMin = -0.25;
 +
real xMax = 1.25;
 +
real yMin = -0.25;
 +
real yMax = 1.25;
  
−
Even faster, once you know it's between <math>\frac{1}{2}</math> and <math>\frac{7}{12},</math> the answer is <math>B</math> because no other option is in this interval.
+
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));
  
−
== Solution 4 ==
+
pair A[];
 +
A[0] = (0,0);
 +
A[1] = (1/2,1/2);
 +
A[2] = (2/3,1/3);
 +
A[3] = (1,0);
  
−
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.
+
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
  
−
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.
+
== Solution 4 (Calculus) ==
 +
There are two cases of winning:
  
−
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.
+
Case 1: Alice choose a number that is smaller than Carol's, and Bob choose a number that is bigger.  
  
−
== Solution 5 ==
+
Case 2: Alice choose a number that is bigger, and Bob choose a number that is smaller.
  
−
Have Carol pick a point <math>x</math> on the real line. The probability that Alice's number is below is thus <math>x</math>, and the probability that Alice's number is above is <math>1 - x</math>. Now, Carol must pick a point between <math>\frac{1}{2}</math> and <math>\frac{2}{3}</math>, exclusive.  If she does not, then it is impossible for Carol's point to be between both Alice and Bob's point. Now what is the probability that Bob's point is below Carol's? To make it simple, scale Bob's number to the normal number line by subtracting <math>\frac{1}{2}</math> and multiplying by <math>6</math>. So thus the chance Bob's number is below is <math>6(x - \frac{1}{2})</math> and above is <math>1 - 6(x-\frac{1}{2})</math>.
+
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>.  
  
−
Now we want to maximixe the probability of (Alice number below <math>x</math>) AND (Bob number above <math>x</math>) + (Alice number above <math>x</math>) AND (Bob number below <math>x</math>). This gives the formula <math>x(1 - 6(x-\frac{1}{2})) + (1 -x)6(x-\frac{1}{2}) = -3 + 13 x - 12 x^2</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>.
−
Then proceed using the process of completing the square or averaging the roots to get that the maximum of this is <math>\frac{13}{24}</math>.
 
  
 
== Video Solution by Richard Rusczyk ==
 
== Video Solution by Richard Rusczyk ==

Revision as of 11:39, 20 August 2023

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)

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 $\frac{1}{2}+x$, then the probability of Case 1 can be expressed by $\frac{1/2 + x}{1}\cdot\frac{1/3 - x }{1/3}$, and the probability of Case 2 can be expressed by $\frac{1/3 + 1/6 - x}{1}\cdot\frac{x}{1/3}$.

Then the probability of Carol winning can be expressed by the sum of the probability of Case 1 and the probability of Case 2: $3\left(\frac{1}{2} + x\right)\left(\frac{1}{3} - x\right)+ 3x\left(\frac{1}{2} - x\right)$, which simplify to $\frac{1}{2} + x - 6x^2$. Taking the first derivative, we get $1 - 12x$, which yield a maximum point at $x = \frac{1}{12}$. Hence, Carol should select $\frac{1}{2} + \frac{1}{12} = \boxed{\textbf{(B) }\frac{13}{24}}$.

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
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All AMC 12 Problems and Solutions

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