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2006 AMC 12B Problems/Problem 24

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Problem

Let S be the set of all point (x,y) in the coordinate plane such that 0 \le x \le \frac{\pi}{2} and 0 \le y \le \frac{\pi}{2}. What is the area of the subset of S for which

\sin^2x-\sin x \sin y + \sin^2y \le \frac34?

\mathrm{(A)}\ \dfrac{\pi^2}{9}\qquad\mathrm{(B)}\ \dfrac{\pi^2}{8}\qquad\mathrm{(C)}\ \dfrac{\pi^2}{6}\qquad\mathrm{(D)}\ \df...

Solution

We start out by solving the equality first. \begin{align*}\sin^2x - \sin x \sin y + \sin^2y &= \frac34 \\\sin x &= \frac{\sin y \pm \sqrt{\sin^2 y - 4 ( \sin^2y ... We end up with three lines that matter: x = y + \frac\pi3, x = y - \frac\pi3, and x = \pi - (y + \frac\pi3) = \frac{2\pi}{3} - y. We plot these lines below.

size(5cm);D((0,0)--(3,0)--(3,3)--(0,3)--cycle);D((1,-0.1)--(1,0.1));D((2,-0.1)--(2,0.1));D((-0.1,1)--(0.1,1));D((-0.1,2)--(0.... Note that by testing the point (\pi/6,\pi/6), we can see that we want the area of the pentagon. We can calculate that by calculating the area of the sqaure and then subtracting the area of the 3 triangles.

\begin{align*}A &= \left(\frac{\pi}{2}\right)^2 - 2 \cdot \frac12 \cdot \left(\frac{\pi}{6}\right)^2 - \frac12 \cdot \lef...

See also

2006 AMC 12B (ProblemsResources)
Preceded by
Problem 23
Followed by
Problem 25
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
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