Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2020 AMC 12B Problems/Problem 25

Revision as of 13:16, 13 July 2022 by Shihan (talk | contribs) (→‎Solution)

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.

When $a\geq\tfrac{1}{2}, P(a)$ consists of a triangle with area $\tfrac{1}{4}$ 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 $P(a)=2-a-\frac{1}{2a}$. 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)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_25&oldid=175891"