2006 AMC 12B Problem 24

Question

Let

S

be the set of all points

(x, y)

in the coordinate plane such that

0 \le x \le \dfrac{\pi}{2}

and

0 \le y \le \dfrac{\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{3}{4}?

\dfrac{\pi^2}{9}

\dfrac{\pi^2}{8}

\dfrac{\pi^2}{6}

\dfrac{3\pi^2}{16}

\dfrac{2\pi^2}{9}

Treat the expression as a quadratic in

\sin x

and find where it equals

\dfrac34

The boundary reduces to

\sin x = \sin\!\left(y \pm \dfrac{\pi}{3}\right),

giving straight lines in

S

Fixing

y,

solve

\sin^2 x - \sin x \sin y + \sin^2 y = \dfrac34

as a quadratic in

\sin x:

\begin{aligned} &\sin x = \frac{1}{2}\sin y \\ &\quad {}\pm \frac{\sqrt3}{2}\cos y \\ &\quad = \sin\!\left(y \pm \frac{\pi}{3}\right). \end{aligned}

Within

S,

\sin x = \sin\!\left(y - \tfrac{\pi}{3}\right)

gives the line

x = y - \tfrac{\pi}{3},

while

\sin x = \sin\!\left(y + \tfrac{\pi}{3}\right)

gives

x = y + \tfrac{\pi}{3}

for

y \le \tfrac{\pi}{6}

and

x = -y + \tfrac{2\pi}{3}

for

y \ge \tfrac{\pi}{6}.

These lines split

S

into regions; testing the corners shows the inequality holds only in the middle band. Its area is

\begin{aligned} &\left(\frac{\pi}{2}\right)^2 - \frac{1}{2}\left(\frac{\pi}{3}\right)^2 \\ &\quad {}- 2 \cdot \frac{1}{2}\left(\frac{\pi}{6}\right)^2 \\ &\quad = \frac{\pi^2}{6}. \end{aligned}

Thus, the correct answer is C.

Problem 24 in other years