Pegs and Holes, Part II

I was going to go and prove this whole thing formally (in a way that would make Dr. Viswanath Ramakrishna -- my multivariable calculus professor at UTD whose IQ was probably in the quadruple digits -- proud), but I'm lazy. So I'll do the abbreviated version:

First, you know that the areas of the square and the circle are equal, so you know x² = π * r² are equal (where x is the length of any side of the square and r is the radius of the circle). Now you have a way to relate x with r.

Second, assume that the circle and square have the same center.

So to prove that a square peg can't fit in a round hole, you have to show that the distance from the center of the square to one of its corners is always greater than the radius of the circle. (You can do this because of our first assumption, from which you can relate x with r.) This makes sense because the corner is that stubborn part of the square that wouldn't fit into the round hole.

Likewise, to show that the round peg won't fit in the square hole, you need to show that the distance from the center of the square to the midpoint of one of its sides (so that it forms a right angle) is always less than the radius of the circle. Again, this makes sense because that's the stubborn part of the circle that wouldn't fit into the square hole.

So there you have it. Eat it, Co-Worker #1. :) Add some equations, inequalities, the Pythagorean theorem, and some fancy diagrams to the above paragraphs and you have the type of thing you would need on every (and I mean every) homework assignment I had in those college mathematics courses. It got old, let me tell you, but this type of thing is fun to think about on the rare occasion, especially for an over-analytical mind like mine.

(NOTE: for another, perhaps more interesting glimpse into the world of mathematics, see my earlier post on Pascal's triangle.)