Pi Day!

So it turns out that Pi Day was this past week (3/14), and I forgot all about it as usual. However, entirely by chance, I actually engaged in some appropriately pi-involving activities that day, and thought I'd put it up here.

It started with me and some coworkers at tea-time in the break room, where there was a newspaper-insert advertising a set of nine miniature pizzas for sale. One of my coworkers jokingly said that it was a ploy by the pizza company to sell half the pizza for the same cost.

There was a pause of a few seconds during which I could practically see the wheels turning in his head. I'm sure I looked the same way, as we both mulled over the implications of what he had said. “Actually...”

We both dived for calculation tools, and began scribbling figures and formulae. The result was interesting enough that I thought I'd share it.

First of all, assume the following picture (representing an idealized pizza in a box):

This pizza has an area of \(\pi r^2\). Now, take a look at the following picture, representing nine miniature idealized pizzas in a box of the same size:

The radius of each of these mini-pizzas is \(1/3\) that of the original pizza, or \(r/3\). Thus, the total area of the pizzas is \(9\times\pi(r/3)^2=9\times\pi(r^2/9)=\pi r^2\). Which is exactly the same as the original, single pizza! So the nine mini-pizzas actually give you just as much bang for your buck as the single large one. (In reality I've ignored some details, such as the fact that the nine pizzas have twice the circumference as the single one so you're getting twice as much crust, which means that the actual area covered by condiments will be slightly smaller than the single pizza.).

I noticed something else interesting about this. Let's imagine an arbitrary number of little circles packed into a square in a regular grid, such as that shown in the picture above. We'll let \(n\) stand for the number of circles on a side. Then the area of the pizzas is \(n^2\times\pi\times (r/n)^2=\pi r^2\). The circumference, however, is \(n^2\times2\pi\times(r/n)=2n\pi r\). So if we take the limit as n goes to infinity – if we imagine squashing more and more smaller and smaller circles into the square – the area stays constant, but the circumference goes to infinity. Yet the area of the circles remains \(2\pi r^2\). So if we imagine throwing darts randomly at the whole area, we would always have the same change of hitting the underlying square. (The area of the square is \(4r^2\), the circle area is \(2\pi r^2\), so the ratio of their areas is \(4r^2/\pi r^2=4/\pi\approx1.2732\). Thus the chance of hitting the square rather than one of the circles remains a constant \(1-(1/1.2732)\approx0.2146=21.46\%\).) So even though there could be an infinite number of circles, with an infinitely long cumulative circumference, you'd still have exactly the same chance of hitting one (78.54%) as you would if there were just one circle.

It reminds me a bit of the Koch Snowflake, a geometrical figure with an infinitely long perimeter that encloses a finite area.