Let $S$ be the set of all points $(x,y)$ in the coordinate plane such that $0\leq x\leq \frac\pi 2$ and $0\leq y\leq \frac\pi 2$. What is the area of the subset of $S$ for which $\sin^2 x - \sin x\sin y + \sin^2 y\le \frac 34$? $\mathrm{(A)}\ \frac {\pi^2}9 \qquad$ $ \mathrm{(B)}\ \frac {\pi^2}8 \qquad$ $ \mathrm{(C)}\ \frac {\pi^2}6 \qquad$ $ \mathrm{(D)}\ \frac {3\pi^2}{16} \qquad$ $ \mathrm{(E)}\ \frac {2\pi^2}9$
Solution
Put $P=x+y$ and $M=x-y$. Then $\sin^2x+\sin^2y=1-\cos P\cos M$ and $\sin x\sin y=\frac12(\cos M-\cos P)$.
So $\sin^2x-\sin x\sin y+\sin^2y-\frac34=\frac14+\frac12\cos P-\cos P\cos M-\frac12\cos M=\left(\cos P+\frac12\right)\left(\frac12-\cos M\right)$.
The condition is $\left(\cos P+\frac12\right)\left(\frac12-\cos M\right)\le0$. Now $\cos P+\frac12\ge0\iff P\le\frac{2\pi}3$ (as $0\le P\le\pi$), and $\frac12-\cos M\ge0\iff|M|\ge\frac\pi3$ (as $|M|\le\frac\pi2$).
So the region is (a) $x+y\le\frac{2\pi}3$ and $|x-y|\le\frac\pi3$, or (b) $x+y\ge\frac{2\pi}3$ and $|x-y|\ge\frac\pi3$.
Use units of $\frac\pi6$, so $S$ becomes the square $[0,3]^2$ (and areas get multiplied by $\frac{\pi^2}{36}$ at the end); (a) is $x+y\le4$, $|x-y|\le2$ and (b) is $x+y\ge4$, $|x-y|\ge2$.
Region (b): $x-y\ge2$ forces $(x,y)$ into the triangle with vertices $(2,0),(3,0),(3,1)$, where $x+y\le4$; so (b) (and its mirror) has area $0$.
Region (a): from the square of area $9$ remove the triangle $x+y>4$ (vertices $(1,3),(3,1),(3,3)$, area $2$) and the two triangles $|x-y|>2$ (area $\frac12$ each). These do not overlap, so the area is $9-2-1=6$.
The required area is $6\cdot\frac{\pi^2}{36}=\frac{\pi^2}6$, answer (C).
Let $S$ be the square $0\le x\le\frac\pi2$, $0\le y\le\frac\pi2$ in the coordinate plane. What is the area of the region of $S$ where $\cos^2x+\cos^2y+\sqrt2\cos x\cos y\le\frac12$?
