2008 AIME II Problems/Problem 13

Problem

A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$ . Then the area of $S$ has the form $a\pi + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ .

Solution

If a point $z = r\text{cis}\,\theta$ is in $R$ , then the point $\frac{1}{z} = \frac{1}{r} \text{cis}\, -\theta$ is in $S$ (where cis denotes $\text{cis}\, \theta = \cos \theta + i \sin \theta$ ). Since $R$ is symmetric about the origin, it suffices to consider the result of the transformation when $-30 \le \theta \le 30$ , and then to multiply by $6$ to account for the entire area.

We note that if the region $S_2 = \left\lbrace\frac{1}{z}|z \in R_2\right\rbrace$ , where $R_2$ is the region outside the circle of radius $1/2$ centered at the origin, then $S_2$ is simply the region inside a circle of radius $2$ centered at the origin. It now suffices to find what happens to the mapping of the region $R_2 - R$ .

The equation of the hexagon side in that region is $x = r \cos \theta = \frac{1}{2}$ , which is transformed to $\frac{1}{r} \cos -\theta = \frac{1}{r} \cos \theta = \frac 12$ . Let $r\cos \theta = a+bi$ where $a,b \in \mathbb{R}$ ; then $r = \sqrt{a^2 + b^2}, \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}$ , so the equation becomes $a^2 - 2a + b^2 = 0 \Longrightarrow (a-1)^2 + b^2 = 1$ . Hence the side is sent to a unit circle centered at $(1,0)$ .

Then $S$ is the union of six unit circles centered at $\cis \frac{k\pi}{6}$ (Error compiling LaTeX. Unknown error_msg), $k = 0,1,2,3,4,5$ , and the region $S_2$ . That is show below.

$[asy] picture p; draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p); [/asy]$

The area of the regular hexagon is just $6 \cdot \frac{(\sqrt{3}^2) \sqrt{3}}{4} = \frac{9}{2}\sqrt{3}$ . The area of each of the $120^{\circ}$ sectors is $6\left(\frac{1}{3}\pi - \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt{3}\right) = 2\pi - \frac{3}{2}\sqrt{3}$ . Their sum is $2\pi + \sqrt{27}$ , and $a+b = \boxed{029}$ .

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2008 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2008 AIME II Problems/Problem 13

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