Find a+b in Regular Hexagon Complex Plane Problem

[ehrenfest]

Homework Statement

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 = $\{ 1/z |x \in R}$. Then the area of S has the form a $\pi +\sqrt b$, where a and b are positive integers. Find a+b.

Homework Equations

The Attempt at a Solution

This should only require high school math although there is probably a solution using the fact that 1/z is a Mobius transformation or something else in complex analysis.

The hexagon is contained in the closed of radius 1/sqrt(3) center at the origin, which means that S is contained inside of the closed disk of radius \sqrt3 centered at the origin. So basically we need to figure out what to subtract off of 3 \pi. Anyone know how to do that?

Please just give a hint.

 

[Tedjn]

What you are trying to do is a good idea, but there doesn't seem to be an easy way to find the cutoffs, as you've noticed.

I caved and peeked at what others did. To solve it without calculus, you need to know what an inversion is. That said, there is a fairly simple solution with calculus that a lot of high school students are probably more familiar with than with inversions.

 

[Architeuthis Dux]

The key to solving this problem is to use the fact that the opposite pairs of sides of the regular hexagon are one unit apart. This means that the hexagon can be inscribed in a circle of radius one, with the center of the circle also at the origin.

Since S is the region outside the hexagon, we can think of it as the region outside the circle of radius one centered at the origin. This region can be divided into six equal sectors, each with a central angle of 60 degrees.

To find the area of S, we can first find the area of one sector and then multiply it by six. The area of a sector with central angle 60 degrees is given by (1/6) * pi * r^2, where r is the radius of the circle. In this case, r=1, so the area of one sector is (1/6) * pi.

Since there are six sectors, the total area of S is (6/6) * pi = pi. Therefore, a=1 and b=0, and a+b=1.

So the final answer is 1.