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

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<cmath>\sin^{2}{(\pi x)}>1-\sin^{2}{(\pi y)}=\cos^{2}{(\pi y)}</cmath>
 
<cmath>\sin^{2}{(\pi x)}>1-\sin^{2}{(\pi y)}=\cos^{2}{(\pi y)}</cmath>
  
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Let's consider the boundary cases: <math>\sin^{2}{(\pi x)}=\cos^{2}{(\pi y)}</math> and <math>\sin^{2}{(\pi x)}=-\cos^{2}{(\pi y)}</math>
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Let's consider the boundary cases: <math>\sin{(\pi x)}=\cos{(\pi y)}</math> and <math>\sin{(\pi x)}=-\cos{(\pi y)}</math>
  
 
<cmath>\sin{(\pi x)}=\cos{(\pi y)}=\sin{(\frac{\pi}{2}\pm \pi y)}</cmath>
 
<cmath>\sin{(\pi x)}=\cos{(\pi y)}=\sin{(\frac{\pi}{2}\pm \pi y)}</cmath>

Revision as of 01:21, 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, we get $y=\frac{1}{2}-x$ and $y=x-\frac{1}{2}$. Solving the second case gives us $y=x+\frac{1}{2}$ and $y=\frac{3}{2}-x$. If we graph these equations in $[0,1]\times[0,1]$, we get a rhombus shape. 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 $P(a)$, $a\geq\frac{1}{2}$. We can solve the rest with geometric probability.

When $a\geq\frac{1}{2}, P(a)$ consists of a triangle with area $\frac{1}{4}$ and a trapezoid with bases $1$ and $2-2a$ and height $a-\frac{1}{2}$. Finally, to calculate $P(a)$, we divide this area by $a$, so \[P(a)=\frac{1}{a}\left(\frac{1}{4}+\frac{(a-\frac{1}{2})(3-2a)}{2}\right)\]

After expanding out, we get $P(a)=\frac{-4a^{2}+8a-2}{4a}=2-a-\frac{1}{2a}$. In order to maximize this expression, we must minimize $a+\frac{1}{2a}$.

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 $P\left(\frac{\sqrt{2}}{2}\right)=\boxed{\textbf{(B)}\ 2 - \sqrt{2}}$

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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