Art of Problem Solving

2020 AMC 12B Problems/Problem 25: Difference between revisions

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=Solution=
==Problem 25==


Let's start by manipulating the given inequality.
For each real number <math>a</math> with <math>0 \leq a \leq 1</math>, let numbers <math>x</math> and <math>y</math> be chosen independently at random from the intervals <math>[0, a]</math> and <math>[0, 1]</math>, respectively, and let <math>P(a)</math> be the probability that


<cmath></cmath>
<cmath>\sin^2{(\pi x)} + \sin^2{(\pi y)} > 1</cmath>
What is the maximum value of <math>P(a)?</math>
 
<math>\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}</math>
 
[[2020 AMC 12B Problems/Problem 25|Solution]]

Revision as of 23:29, 7 February 2020

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