How Many Sides Does A Circle Have?

One? Zero? A bunch?

To answer this question, we're going to use some trigonometry, some precalculus and some calculus. Don't worry if you don't know it yet. I'll explain what you need to know.

First, the trig. The angle around the center of a polygon is 360 degrees (i.e. a circle). In our math, we'll be using radians (not degrees). All we need to know is that 360 degree = 2π. In a right triangle, the longest side (opposite the right angle) is known as the hypotenuse. The sine of an angle (sin θ) is the opposite side divided by the hypotenuse. The cosine of an angle (cos θ) is the adjacent side divided by the hypotenuse. The only other thing we need to know is:

2(sin(θ))(cos(θ)) = sin(2θ). Don't worry about why, just know that it is (we may cover this another time).

Next, the precalculus. The limit of a function is a value. If I say "The limit of y as x approaches infinity," I want to know what y looks like as x gets really big.

Here's an example. y = 5x. What is the limit of y as x approaches 2? In some cases, we can plug in. If we do that here, we get y = 10. So the limit of y as x approaches 2 is 10.

Now, y = 1/x. What's the limit of y as x approaches zero? As x approaches infinity?

If something is divided by zero, we say it approaches infinity. If something is divided by infinity, we say it approaches zero.

Now for the calculus. The two words I'll use are L'Hopital and derivative. If I get a limit that becomes infinity/infinity or 0/0, I can use L'Hopital's rule to calculate the limit. The works by using derivatives. To follow my work, you don't need to know what a derivative is. In fact, if you'd like, just substitute the word "ice cream" for derivative. After all, who doesn't love ice cream?

That covers all you need to know. Let's get started.

First, recall from last time that the area of a regular convex polygon is given by (1/2)(a)(p) and that p is equal to (n)(s), where a is the apothem, p is perimeter, n is the number of sides and s is the side length.

When thinking about circles, we don't usually come across the term "apothem," so we'll want a way to write our equations without it. Unfortunately, we also don't hear side lengths in reference to circles, we need to get rid of that as well. To do this, we'll use the other side of our little triangles running from the center to a corner. We'll call this longest side (the hypotenuse) r, for reasons that will become obvious later. Here, the hypotenuse is r, the opposite side is s/2 and the adjacent side is a. Using the definitions for sine (opposite/hypotenuse) and cosine (adjacent/hypotenuse) for our angle θ, we get the following:

Now that we have expressions for s and a, we need a way to write θ in terms of other things we're using. We know the central angle is 2π radians and that there are 12 little triangles that are all the same. This means the top angle of a little triangle is 2π/12 (or π/6).

For an n-gon, the central angle is still 2π, but we now have 2n little triangles (from yesterday's post), with means our angle is going to be π/n for each one.

Using the the equations for s and a found earlier, we can rewrite the equation for area.

Remember the other trig equation from earlier? We can now use it to make sine and cosine into one function, making the math that comes later much easier.

Now we substitute in our π/n.

We take r to be a constant, so our only variable is n. This means area is a function of n and can be written as A(n) (Read: A of n). Now, we know that if there are no sides, there can be no enclosed area, so that case isn't very interesting. But what if there are infinitely many sides? To find out, let's take the limit of A(n) as n goes to infinity.

We ended up with infinity times zero, which is no directly solvable using limits. Since r and 2 aren't dependent upon n, I can bring them outside to cleanup my function before I mess with it.

Remember when I mentioned L'Hopital? I said it would let me work with infinity/infinity or 0/0. Well, now I have infinity times 0. Fortunately, there's a trick I can use to make it 0/0!

Think of it like this. Consider 2. Now, consider 1/2. Finally, consider 1/(1/2). The last is the same as 2, but it is written differently. In our case, we have n (which is becoming infinity). If I rewrite it as 1/n, I've got 0 (on the top), but my expression is different. To fix that, I have to bring 1/n into the denominator. It's like earlier. I started with 2. I flipped it, but it was different. To make it the same, I brought it to the bottom of the fraction. My old expression and my new expression are the same thing, but they're written in different languages. Now, I have 0/0.

L'Hopital's rule works like this, but only if I'm dealing with 0/0 or infinity/infinity. It says that the limit of a fraction of two functions equals the limit of a fraction of two function's derivatives (or ice creams). If our expression is f(n)/g(n) (Read: f of n over g of n), then the derivatives (ice creams) are f'(n) and g'(n) (Read: f prime of n and g prime of n). I've calculated the derivatives (ice creams) below. Don't get caught up in how they are what they are (for today), just keep up with which is which.

Since f(n)/g(n) gave me 0/0, let's see my new expression using f'(n) and g'(n).

There's a -1/n^2 term in both the top and bottom, so it simplifies away.

Now I plug my new expression back into my limit. Don't forget the r^2/2 outside of the limit.

Since 2π is a constant, I can bring it outside the limit and simplify.

Since cosine is a function, I can move my limit inside the parentheses. So now we're taking the limit of 2π/n as n approaches infinity. What did we say about this earlier?

By plugging in and dividing by infinity, we create a fraction that becomes 0. We now take the cosine of 0, which has a value of 1. Since we had πr^2 outside the limit, we multiply this by 1 to get the expression πr^2.

So we can say that the limit of the area of a regular convex polygon as the number of sides approaches infinity is equal to πr^2 (Read: pi r squared), which is formula for the area of a circle. This implies that a circle is like a polygon with infinitely many sides!

This is one of my favorite problems to work out and show people. It's one of those things that doesn't really have a practical purposes, but it's kinda cool to see.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I've been doing this problem for a couple of years now, but I'm not sure when I started. At the earliest, it was during my senior year of high school (after I learned L'Hopital's rule from Mrs. J. Haynes in Calculus). At the latest, it was this past December.

The images used here were created by me using Noteshelf for iPad.