2020 AMC 12B Problems/Problem 25

Problem 25

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: $\sin^{2}{(\pi x)}=\cos^{2}{(\pi y)}$ and $\sin^{2}{(\pi x)}=-\cos^{2}{(\pi y)}$

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

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

-Solution by Qqqwerw

Video Solution

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

See Also

2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS