AMC12 2020 B

Answer (B): First consider the graph of the equation $\sin^2(\pi x)+\sin^2(\pi y)=1$ in the unit square of the $xy$-coordinate plane. The equation implies that $\sin^2(\pi y)=\cos^2(\pi x)=\sin^2\!\left(\pi\left(\frac12-x\right)\right),$ so for $0\le x\le\frac12$ and $0\le y\le\frac12$, the equation is equivalent to $y=\frac12-x$. Because $\sin(\pi x)=\sin(\pi(1-x))$ and $\sin(\pi y)=\sin(\pi(1-y))$, the graph is symmetric about the lines $x=\frac12$ and $y=\frac12$. Thus within the unit square, the graph is a square with vertices $(0,\frac12)$, $(\frac12,0)$, $(1,\frac12)$, and $(\frac12,1)$. The solutions of the given inequality correspond to the points inside this square. For a fixed value of $a$, the set of all possible ordered pairs $(x,y)$ corresponds to an $a\times1$ rectangle, which has area $a$. If $a\le\frac12$, then the solutions of the inequality lie inside a triangle with altitude $a$ and base $2a$, which has area $a^2$, so $P(a)=\frac{a^2}{a}=a\le\frac12.$ If $a>\frac12$, then the solutions of the inequality lie inside the union of a triangle with area $\frac14$ and a trapezoid with altitude $a-\frac12$ and base lengths $1$ and $2(1-a)$. The total area of this region is $\frac14+\left(a-\frac12\right)\left(\frac{1+2(1-a)}{2}\right)=-a^2+2a-\frac12,$ so $P(a)=-a+2-\frac{1}{2a}.$ The value of $P(a)$ is maximized when the value of $a+\frac{1}{2a}$ is minimized. By the Arithmetic Mean–Geometric Mean Inequality, $a+\frac{1}{2a}\ge 2\sqrt{a\cdot\frac{1}{2a}}=\sqrt2,$ and the minimum value occurs when $a=\frac{1}{2a}$. This means $a=\frac12\sqrt2$ (which is indeed greater than $\frac12$), and the corresponding value of $P(a)$ is $2-\sqrt2$ (which is greater than the maximum in the case $a\le\frac12$).