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

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

(→‎Solution)
(→‎Solution)
Line 23: Line 23:
 
From the region graph, notice that in order to maximize <math>P(a)</math>, <math>a\geq\frac{1}{2}</math>. We can solve the rest with geometric probability.
 
From the region graph, notice that in order to maximize <math>P(a)</math>, <math>a\geq\frac{1}{2}</math>. We can solve the rest with geometric probability.
  
−
When <math>a\geq\frac{1}{2}, P(a)</math> consists of a triangle with area <math>\frac{1}{4}</math> and a trapezoid with bases <math>1</math> and <math>2-2a</math> and height <math>a-\frac{1}{2}</math>. Finally, to calculate <math>P(a)</math>, we divide this area by <math>a</math>, so <cmath>P(a)=\frac{1}{a}\left(\frac{1}{4}+\frac{(a-\frac{1}{2})(2-2a)}{2}\right)</cmath>
+
When <math>a\geq\frac{1}{2}, P(a)</math> consists of a triangle with area <math>\frac{1}{4}</math> and a trapezoid with bases <math>1</math> and <math>2-2a</math> and height <math>a-\frac{1}{2}</math>. Finally, to calculate <math>P(a)</math>, we divide this area by <math>a</math>, so <cmath>P(a)=\frac{1}{a}\left(\frac{1}{4}+\frac{(a-\frac{1}{2})(3-2a)}{2}\right)</cmath>
 +
 
 +
After expanding out, we get <math>P(a)=\frac{-4a^{2}+8a-2}{4a}=2-a-\frac{1}{2a}</math>. In order to maximize this expression, we must minimize <math>a+\frac{1}{2a}</math>.
 +
 
 +
By AM-GM, <math>a+\frac{1}{2a}\geq 2\sqrt{\frac{a}{2a}}=\sqrt{2}</math>, which we can achieve by setting <math>x=\frac{\sqrt{2}}{2}</math>.
 +
 
 +
Therefore, the maximum value of <math>P(a)</math> is <math>P(\frac{\sqrt{2}}{2})=\boxed{\textbf{(B)}2-\sqrt{2}}</math>

Revision as of 23:49, 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

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 $x=\frac{\sqrt{2}}{2}$.

Therefore, the maximum value of $P(a)$ is $P(\frac{\sqrt{2}}{2})=\boxed{\textbf{(B)}2-\sqrt{2}}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_25&oldid=117310"