## Wednesday, February 18, 2009

### Trig or Treat

Dear Dr. Math,
This is a question I thought of while pondering the air intake of a wood stove. The air intake is a series of holes, covered or uncovered by a sliding metal plate with equal sized holes.
Imagine 2 circles with equal radius, R. Slide one circle over the other. Express, in terms of R, how far one circle has to occlude the other such that half of the area is covered.

Bob H., Ashland, OR

Dear Bob,

Here's a picture of the problem, if I understand it correctly:

For legal reasons, before we get to the solution, I feel I should warn all you readers out there: what follows may involve some high-school level trigonometry, which I understand many of you have intentionally purged from your brains to make room for Grey's Anatomy plots. Part of the reason I like this question so much is that it shows that these concepts may very well have some relevance (outside of the very important pursuit of measuring the heights of buildings using a sextant) despite your high school math teacher's best attempts to convince you otherwise. Those readers who are subject to trigonometry-induced seizures should turn back now.

OK, with that out of the way, let's blow up part of the picture and label some of the relevant objects. The goal is to get a handle on this shady part of town:

First, there's the radius, R, which we've assumed is the same for the two circles. Let's call the angle formed by the center and two points of intersection $\theta$. Note: this has nothing to do with thetans (or does it?). Splitting this angle down the middle forms two right triangles with an angle of $\theta / 2$. According to the rules of trigonometry, the height of each triangle is $R \sin(\theta / 2)$ and the width is $R \cos(\theta / 2)$, as I've labeled here:

Now, the strategy I'd like to employ to compute the area of that funny little almond-shaped region, which I'll call C, is to think of it as consisting of two pieces, each of which is the difference between a pie-slice of the circle, A, and a triangle, B. In pictures:

The reason this helps is that circles and triangles are shapes whose areas we know how to compute. "Funny little almond shapes," not so much.

The area of the circular slice is in proportion to the whole area of the circle as the angle $\theta$ is to the whole angle of a circle, 360° (a quarter of a circle takes up 90°, for example). So in terms of R and $\theta$, that's $\frac{A}{\pi R^2} = \frac{\theta}{360}$ and so $A = \frac{\theta}{360} * \pi R^2$.
Now, the area of the whole triangle is $\frac{1}{2} (\text{base})*(\text{height})$, which in this case is $\frac{1}{2} R \cos(\theta/2) * 2 * R \sin(\theta/2)$. So, $B = R^2 \cos(\theta/2) \sin(\theta/2)$. By a sneaky trick I learned in trigonometry class, I can rewrite this as $B = \frac{1}{2}R^2\sin(\theta)$.

If we throw all these things into the hopper, we get that the area of the almond-shaped piece, C, is $2*(A-B)$, and so $C = 2*\left(\frac{\theta}{360}*\pi R^2 - \frac{1}{2} R^2 \sin(\theta) \right)$. I can feel some of you starting to panic out there, but just take a deep breath and try to relax. Put on some Enya or something--maybe that song she wrote about trigonometry.

What were we doing? Oh yeah, right; now we have a formula for computing the area of the overlapping part of the two circles, which only depends on the angle $\theta$. The question was, When is this area equal to half the area of the circle?, so we need to solve for $\theta$. Half the area of the circle is $\frac{1}{2}\pi R^2$; therefore, the equation we need to solve is:

$\frac{1}{2}\pi R^2 = 2*\left(\frac{\theta}{360}*\pi R^2 - \frac{1}{2} R^2 \sin(\theta)\right)$

which, after we divide through by $R^2$ and clean up a bit leaves us with:

$\frac{\pi}{2} = \frac{\pi}{180} *\theta - \sin(\theta)$.

It's interesting to pause here and note that R completely vanished from the equation. This means whatever configuration we come up with as an answer must have the same angle, independent of the radius.

OK. So, how do we solve this equation?

Actually... we don't.

The problem is that we have a $\theta$ and a $\sin(\theta)$, and the thing about those two is that they're like Sydney and Cristina on Grey's Anatomy; they just don't mix well. Unfortunately, there's no way to get any further with this equation using the rules of algebra. So, here's where we cheat and approximate the solution with a calculator (in my case, a TI-89). Maybe someday I'll tell you all about what goes on inside a calculator when it does these approximations, or about how we could use calculus to solve the problem if the shape were something else. But anyway, for now, the answer is $\theta \approx 132.4^{\circ}$.

Lastly, we should translate this answer into a more meaningful form, for example by figuring out what the distance is between the two centers of the circles. Using trigonometry one last time, we can write this distance as $2*R\cos(\theta/2)$, which for the magic $\theta$ above is $0.808*R$. The final picture, then, is:

Put that in your wood stove and smoke it.

-DrM

#### 1 comment:

sowhat said...

Hi Dr math，

what if the two circles are not euqal to each other? Suppose two overlapped circles with radius r and R, respectively, if the overlaping area is exactly 0.5*pi*r^2, what should be the distance between the two circles?