2020 AMC 12B Problem 25

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

The condition is

|\sin \pi x| \gt |\cos \pi y|;

for fixed

x,

the probability over

y

is a simple triangular function

Express

P(a) = \tfrac1a \int_0^a g(x)\,dx

where

g

rises then falls, then maximize over

a

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.

Problem 25 in other years