# 2020 AMC 12B Problems/Problem 25

## 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 1

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: $\sin{(\pi x)}=\pm \cos{(\pi y)}$.

$$\sin{(\pi x)}=\pm \cos{(\pi y)}=\sin{(\tfrac 12 {\pi}\pm \pi y)}$$

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.

Instead of maximizing $P(a)$, we minimize $Q(a)=1-P(a)$. $Q(a)$ consists of two squares (each broken into two triangles), one of area $\tfrac{1}{4}$ and another of area $(a-\tfrac 12)^2$. To calculate $Q(a)$, we divide this area by $a$, so $$Q(a) = \frac{1}{a}\left(\frac{1}{4}+(a-\tfrac 12)^2\right) = \frac{1}{a}\left(\frac{1}{2}+a^2-a\right)= a+\frac 1{2a}-1.$$

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 $1-\min(Q(a))$, which is $1-(\sqrt{2}-1) =\boxed{\textbf{(B)}\ 2 - \sqrt{2}}$

## Solution 2 (Trigonometry Identities)

$$\sin^2{(\pi x)} + \sin^2{(\pi y)} > 1, \quad \sin^2{(\pi x)} - (1 - \sin^2{(\pi y)} ) > 0, \quad \sin^2{(\pi x)} - \cos^2{(\pi y)} > 0$$

$$( \sin{(\pi x)} + \cos{(\pi y)} )(\sin{(\pi x)} - \cos{(\pi y)})> 0, \quad \left( \sin{(\pi x)} + \sin{( \frac{\pi}{2} - \pi y)} \right) \left( \sin{(\pi x)} - \sin{(\frac{\pi}{2} - \pi y)} \right)> 0$$

$$2\sin{\left( \frac{ \pi x + \frac{\pi}{2} - \pi y }{2} \right)} \cos{\left( \frac{ \pi x - \frac{\pi}{2} + \pi y }{2} \right)} 2 \sin{\left( \frac{ \pi x - \frac{\pi}{2} + \pi y }{2} \right)} \cos{\left( \frac{ \pi x + \frac{\pi}{2} - \pi y }{2} \right)} > 0$$

$$2\sin{\left( \frac{ \pi x + \frac{\pi}{2} - \pi y }{2} \right)} \cos{\left( \frac{ \pi x + \frac{\pi}{2} - \pi y }{2} \right)} 2 \sin{\left( \frac{ \pi x - \frac{\pi}{2} + \pi y }{2} \right)} \cos{\left( \frac{ \pi x - \frac{\pi}{2} + \pi y }{2} \right)} > 0$$

$$\sin{\left(\pi x + \frac{\pi}{2} - \pi y\right)} \sin{\left( \pi x - \frac{\pi}{2} + \pi y \right)} > 0$$

If $\sin{\left(\pi x + \frac{\pi}{2} - \pi y\right)} > 0$ and $\sin{\left( \pi x - \frac{\pi}{2} + \pi y \right)} > 0$

$$0 < \pi x + \frac{\pi}{2} - \pi y < \pi \quad \text{and} \quad 0 < \pi x - \frac{\pi}{2} + \pi y < \pi$$

$$0 < x + \frac{1}{2} - y < 1 \quad \text{and} \quad 0 < x - \frac{1}{2} + y < 1$$

$$x - \frac{1}{2} < y < x + \frac{1}{2} \quad \text{and} \quad -x + \frac{1}{2} < y < -x + \frac{3}{2}$$

If $\sin{\left(\pi x + \frac{\pi}{2} - \pi y\right)} < 0$ and $\sin{\left( \pi x - \frac{\pi}{2} + \pi y \right)} < 0$

$$\pi < \pi x + \frac{\pi}{2} - \pi y < 2\pi \quad \text{and} \quad \pi < \pi x - \frac{\pi}{2} + \pi y < 2\pi$$

$$1 < x + \frac{1}{2} - y < 2 \quad \text{and} \quad 1 < x - \frac{1}{2} + y < 2$$

$$x - \frac{3}{2} < y < x - \frac{1}{2} \quad \text{and} \quad -x + \frac{3}{2} < y < -x + \frac{5}{2}$$

Notice that $\sin{\left(\pi x + \frac{\pi}{2} - \pi y\right)} < 0$ and $\sin{\left( \pi x - \frac{\pi}{2} + \pi y \right)} < 0$ cannot be true, because that means $x$ is in the interval $[1, 2]$. Thus, $y$ is in the the boundaries $x - \frac{1}{2} < y < x + \frac{1}{2}$, and $-x + \frac{1}{2} < y < -x + \frac{3}{2}$.

Finishing like Solution 1, 3, or 4 gives $P(a) =\boxed{\textbf{(B)}\ 2 - \sqrt{2}}$

## Solution 3 (Calculus finish)

We find the same region as in the first solution or the second solution, and again notice we must have $a \geq \frac{1}{2}$.

We now express $P$ as a function of $b=(1-a)$. The triangle on the right of the line $x=b$ is an isosceles right triangle with altitude $b$, so it has area $b^2$. The total area of the region to the left of $x=b$ has area $1-b$. So $$P(b) = \frac{1/2 - b^2}{1-b}$$ Differentiating using the quotient rule, we find $P$ has local extrema at $$P'(b) = \frac{(1-h)(-2h)-(-1/2 - h^2)(-1)}{(1-h)^2} = 0$$ Setting the numerator equal to $0$ and solving the quadratic, we find $P$ has extrema at $b = 1 \pm \frac{\sqrt{2}}{2}$. Only $b=1-\frac{\sqrt 2}{2}$ is within the desired region. Plugging in, we get $P(1-\frac{\sqrt 2}{2}) = \boxed{\textbf{(B)}\ 2 - \sqrt{2}}$ as our solution. We also need to check $P(b=0) = 1/2$. But $1/2 < 2 - \sqrt{2}$, and if this isn't immediately obvious, $1/2$ isn't an answer choice anyways.

~jd9 (AMGM scary)

## Solution 4 (Calculus Finish)

We find the same region as in the first solution or the second solution, and again notice $a \geq \frac{1}{2}$.

The numerator of $P(a)$ is the area of the triangle with vertices $(0, \frac12), (\frac12, 0), (\frac12, 1)$ plus the area of the rectangle with vertices $(\frac12,0), (a,0), (a,1), (\frac12,1)$ subtract the area of the isosceles right triangle between $x=a, y=0,x - \frac{1}{2}$ and the isosceles right triangle between $x=a, y=1, -x + \frac{3}{2}$. The denominator of $P(a)$ is the area of rectangle with vertices $(0,0), (a,0), (a,1), (0,1)$.

$$P(a) = \frac{ \frac14 + a - \frac12 -(a - \frac12)^2 }{a} = \frac{a - \frac14 -a^2 + a - \frac14 }{a} = \frac{-a^2 + 2a - \frac12 }{a} = -a + 2 - \frac{1}{2a}$$

By the power rule, $P'(a) = -1 - \frac12(-1)\frac{1}{a^2} = -1 + \frac{1}{2a^2}$. Solving $P'(a) = 0$, we find that $a = \frac{ \sqrt{2} }{2}$

$\because P'(a) = 0$ when $a = \frac{ \sqrt{2} }{2}$, $P'(a)$ is positive when $a < \frac{ \sqrt{2} }{2}$, and $P'(a)$ is negative when $a > \frac{ \sqrt{2} }{2}$

$\therefore P(a)$ has a local maximum when $a = \frac{ \sqrt{2} }{2}$.

$$P(\frac{ \sqrt{2} }{2}) = - \frac{\sqrt{2} }{2} + 2 - \frac{1}{2 \cdot \frac{\sqrt{2} }{2}} =\boxed{\textbf{(B)}\ 2 - \sqrt{2}}$$

## See Also

 2020 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast 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.