Thursday, September 21, 2006

BOUTS: Best of UseTheSource

Long, long ago, OK, 1999, I registered the domain name usethesource.com (as in Use the source, Luke!) and used it to start a web site which would these days be called a blog. The site was powered by Slashdot's code Slashcode and featured a mix of my commentary on the news and original articles. You can still read the old site at archive.org. The site even got me an appearance with Leo Laporte on The Screen Savers.

Most of the articles published are irrelevant today. The commentary is often on start ups that have fizzled out long ago, and I shut down the site in 2004. But some of the articles are worth repeating. So, from time to time, I'll be republishing original pieces from UseTheSource as BOUTS entries in this blog.

To get things rolling here's an article I wrote back in 2002 about calculating the area of an annulus based on the length of a tangent.

Originally published June 12, 2002

Take a look at the following shape. It's an annulus: two concentric circles, something like a simple washer or donut. Imagine that you know only one fact about this shape, the length of a tangent of the inner circle where it touches the edges of the outer circle. Call that length x. Can you calculate the area of the yellow shaded part? This problem was presented on the NPR radio show CarTalk a few weeks back and after I solved it I realized that there were a couple of interesting ways of calculating the area. Both require knowledge of the formula for the area of a circle: πr2, where r is the radius of the circle. One requires remembering Pythagoras' Theorem, the other a little logical reasoning.

Solution by logical reasoning

Insight: there must be many such concentric circles where it's possible that the tangent has length x.

In fact if I start with the small circle in the middle it must always be possible to choose the size of the outer circle so that the tangent is x. So what if I make then inner circle have size zero. Then all I need is an outer circle with diameter x. Since we know there's only one solution (surely the person posing this question knew that there was only one solution), then we can just calculate the area of the outer circle when the inner circle has zero radius.

The outer circle in that case has diameter x or radius x/2 and so the area is π(x/2)2 or πx2/4.

Solution by Pythagoras

To calculate the area of the annulus we need to calculate the area of the big circle and subtract the area of the small circle. If we name the radius of the big circle r and the radius of the small circle s then we need to calculate πr2 - πs2 or π(r2 - s2). Hmm. That r2 - s2 bit looks a lot like something we might get from Pythagoras' Theorem (the square on the hypotenuse is equal to the sum of the squares on the other two sides). For Pythagoras we need a right angle triangle. Low and behold we have one. Since we have a tangent we know it's at right angles to a radius of the inner circle. The complete triangle has sides r, s and x/2. So run through Pythagoras on this triangle and we get r2 = (x/2)2 + s2. Subtract s2 from both sides and you've got r2 - s2 = (x/2)2. Now we know how to calculate r2 - s2, it's (x/2)2 and so the area of the annulus is π(x/2)2 or πx2/4.