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Difference between revisions of "2020 AMC 12B Problems/Problem 25"

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From the region graph, notice that in order to maximize <math>P(a)</math>, <math>a\geq\tfrac{1}{2}</math>. We can solve the rest with geometric probability.
 
From the region graph, notice that in order to maximize <math>P(a)</math>, <math>a\geq\tfrac{1}{2}</math>. We can solve the rest with geometric probability.
  
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When <math>a\geq\tfrac{1}{2}, P(a)</math> consists of a triangle with area <math>\tfrac{1}{4}</math> and a trapezoid with bases <math>1</math> and <math>2-2a</math> and height <math>a-\frac{1}{2}</math>. Finally, to calculate <math>P(a)</math>, we divide this area by <math>a</math>, so <cmath>P(a)=\frac{1}{a}\left(\frac{1}{4}+\frac{(a-\frac{1}{2})(3-2a)}{2}\right)</cmath>
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Instead of maximizing <math>P(a)</math>, we minimize <math>Q(a)=1-P(a)</math>. <math>Q(a)</math> consists of two squares (each broken into two triangles), one of area <math>\tfrac{1}{4}</math> and another of area <math>(a-\tfrac 12)^2</math>. To calculate <math>Q(a)</math>, we divide this area by <math>a</math>, so <cmath>Q(a) = \frac{1}{a}\left(\frac{1}{4}+(a-\tfrac 12)^2\right) = \frac{1}{a}\left(\frac{1}{2}+a^2-a\right)= \left(a+\frac 1{2a}-1\right).</cmath>
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After expanding out, we get <math>P(a)=2-a-\frac{1}{2a}</math>. In order to maximize this expression, we must minimize <math>a+\frac{1}{2a}</math>.
 
  
 
By AM-GM, <math>a+\frac{1}{2a}\geq 2\sqrt{\frac{a}{2a}}=\sqrt{2}</math>, which we can achieve by setting <math>a=\frac{\sqrt{2}}{2}</math>.
 
By AM-GM, <math>a+\frac{1}{2a}\geq 2\sqrt{\frac{a}{2a}}=\sqrt{2}</math>, which we can achieve by setting <math>a=\frac{\sqrt{2}}{2}</math>.
  
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Therefore, the maximum value of <math>P(a)</math> is <math>P\left(\frac{\sqrt{2}}{2}\right)=\boxed{\textbf{(B)}\ 2 - \sqrt{2}}</math>
+
Therefore, the maximum value of <math>P(a)</math> is <math>1-\min(Q(a))</math>, which is <math>1-(\sqrt{2}-1) =\boxed{\textbf{(B)}\ 2 - \sqrt{2}}</math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 13:52, 13 July 2022

Problem

For each real number $a$ with $0 \leq a \leq 1$, let numbers $x$ and $y$ be chosen independently at random from the intervals $[0, a]$ and $[0, 1]$, respectively, and let $P(a)$ be the probability that

\[\sin^2{(\pi x)} + \sin^2{(\pi y)} > 1\] What is the maximum value of $P(a)?$

$\textbf{(A)}\ \frac{7}{12} \qquad\textbf{(B)}\ 2 - \sqrt{2} \qquad\textbf{(C)}\ \frac{1+\sqrt{2}}{4} \qquad\textbf{(D)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(E)}\ \frac{5}{8}$

Solution

Let's start first by manipulating the given inequality.

\[\sin^{2}{(\pi x)}+\sin^{2}{(\pi y)}>1\] \[\sin^{2}{(\pi x)}>1-\sin^{2}{(\pi y)}=\cos^{2}{(\pi y)}\]

Let's consider the boundary cases: $\sin{(\pi x)}=\cos{(\pi y)}$ and $\sin{(\pi x)}=-\cos{(\pi y)}$

\[\sin{(\pi x)}=\cos{(\pi y)}=\sin{(\frac{\pi}{2}\pm \pi y)}\]

Solving the first case gives us \[y=\tfrac{1}{2}-x \quad \textrm{and} \quad y=x-\tfrac{1}{2}.\] Solving the second case gives us \[y=x+\tfrac{1}{2}\quad \textrm{and} \quad y=\tfrac{3}{2}-x.\] If we graph these equations in $[0,1]\times[0,1]$, we get a rhombus shape. [asy] defaultpen(fontsize(9)+0.8); size(200); pen p=fontsize(10); pair A,B,C,D,A1,A2,B1,B2,C1,C2,D1,D2,I,L; A=MP("(0,0)",origin,down+left,p); B=MP("(1,0)",right,down+right,p); C=MP("(1,1)",right+up,up+right,p); D=MP("(0,1)",up,up+left,p); A1=MP("",extension(A,B,(0.5,0),(0,0.5)),2*down,p); dot(A1); A2=MP("",extension(A,D,(0.5,0),(0,0.5)),2*left,p); dot(A2); B1=MP("",extension(B,C,(0.5,0),(0,-0.5)),2*right,p); dot(B1); B2=MP("",extension(C,D,(0.5,1),(0,0.5)),2*up,p); dot(B2); real a=0.7; draw(A1--B1--B2--A2--cycle, gray+0.6); draw(a*right--a*right+up, royalblue); draw(A1--B2, royalblue+dashed); draw(A--B--C--D--A, black+1.2); dot("$(a,0)$",(a,0),down); dot("$(a,1)$",(a,1),up); [/asy] Testing points in each section tells us that the inside of the rhombus satisfies the inequality in the problem statement.

From the region graph, notice that in order to maximize $P(a)$, $a\geq\tfrac{1}{2}$. We can solve the rest with geometric probability.

Instead of maximizing $P(a)$, we minimize $Q(a)=1-P(a)$. $Q(a)$ consists of two squares (each broken into two triangles), one of area $\tfrac{1}{4}$ and another of area $(a-\tfrac 12)^2$. To calculate $Q(a)$, we divide this area by $a$, so \[Q(a) = \frac{1}{a}\left(\frac{1}{4}+(a-\tfrac 12)^2\right) = \frac{1}{a}\left(\frac{1}{2}+a^2-a\right)= \left(a+\frac 1{2a}-1\right).\]

By AM-GM, $a+\frac{1}{2a}\geq 2\sqrt{\frac{a}{2a}}=\sqrt{2}$, which we can achieve by setting $a=\frac{\sqrt{2}}{2}$.

Therefore, the maximum value of $P(a)$ is $1-\min(Q(a))$, which is $1-(\sqrt{2}-1) =\boxed{\textbf{(B)}\ 2 - \sqrt{2}}$

Video Solution

On The Spot STEM: https://www.youtube.com/watch?v=5goLUdObBrY

See Also

2020 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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