Question

For each real number $a$ with $0 \le a \le 1,$ 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 $\sin^2(\pi x) + \sin^2(\pi y) \gt 1.$ What is the maximum value of $P(a)?$

$\dfrac{7}{12}$

$2 - \sqrt{2}$

$\dfrac{1 + \sqrt{2}}{4}$

$\dfrac{\sqrt{5} - 1}{2}$

$\dfrac58$

Answer: B

Concepts:geometric probability optimization calculus

Difficulty rating: 2540

Solution:

Since $\sin^2(\pi y) = 1 - \cos^2(\pi y),$ the condition is $|\sin \pi x| \gt |\cos \pi y|.$ For fixed $x,$ the probability over $y \in [0, 1]$ is $g(x) = 2x$ for $0 \le x \le \tfrac12$ and $g(x) = 2 - 2x$ for $\tfrac12 \le x \le 1.$

Then $P(a) = \tfrac1a \int_0^a g(x)\,dx.$ For $a \le \tfrac12,$ $P(a) = a,$ increasing to $\tfrac12.$ For $a \ge \tfrac12,$ $\begin{gathered} P(a) = \frac{2a - a^2 - \tfrac12}{a} \\ {}= 2 - a - \frac{1}{2a}. \end{gathered}$

Setting the derivative to zero gives $2a^2 = 1,$ so $a = \tfrac{1}{\sqrt2},$ and $P\!\left(\tfrac{1}{\sqrt2}\right) = 2 - \sqrt2.$

Thus, the correct answer is B.

2020 AMC 12B Problem 25

Answer: B

Difficulty rating: 2540

Solution:

Problem 25 in Other Years