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

(Solution)
(Solution)
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<cmath>\sin^{2}{(\pi x)}+\sin^{2}{(\pi y)}>1</cmath>
 
<cmath>\sin^{2}{(\pi x)}+\sin^{2}{(\pi y)}>1</cmath>
<cmath>\sin^{2}{\pi x}>1-\sin^{2}{\pi y}=\cos^{2}{\pi y}</cmath>
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<cmath>\sin^{2}{(\pi x)}>1-\sin^{2}{(\pi y)}=\cos^{2}{(\pi y)}</cmath>
  
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^{2}{(\pi x)}=\cos^{2}{(\pi y)}</math> and <math>\sin^{2}{(\pi x)}=-\cos^{2}{(\pi y)}</math>
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<cmath>\sin{(\pi x)}=\cos{(\pi y)}=sin{(\frac{\pi}{2}\pm \pi y)}</cmath>

Revision as of 00:36, 8 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

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

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