Area of a circle

Yes! Today is the $\pi$ day! So I think I should write something about $\pi$ a little bit.

Obviously, the area of a circle is one of the most popular topics. So, sure, I'm writing it here. Well, why would I write it when it's already popular? Because somehow basic proofs are misleading in some sense.

The famous basic proof is in this form:

1. Divide the circle into $n$ sectors.
2. Rearrange the sectors in an alternating form.
3. Each sector resembles a triangle, i.e., it can be approximated by a triangle.
4. The height of each triangle is $r$ and the width is $\frac{2\pi r}{n}$, so the area is $\frac{1}{2} \frac{2\pi r}{n} r = \frac{\pi r^2}{n}$.
5. Since there are $n$ triangles, the area is $\frac{\pi r^2}{n} n = \pi r^2$. Therefore, the area is $\pi r^2$.

I want to picture this as clear as possible, so here's a figure (the red triangle has height tends to $r$ and width tends to $\frac{2\pi r}{n}$):

The reason I don't like this proof is because of the word resembles, or can be approximated, because to what extent can someone say something resembles another thing? To what degree can you say a shape is approximated by another? This is not so clear, so I'll present another proof, because I like this one more.

Note. Both proofs, in modern sense, are not fully rigorous. In order to claim a proof to be rigorous, one must rigorously define the area of a shape first, which is actually a more complicated task in the modern mathematics.

The proof I like

Surely, I'm not the author, but I've already forgotten who wrote it (If I'm not mistaken, I believe it was Archimedes who proved using this, and the method is called the method of exhaustion), so let me present it here.

Consider a circle. Draw a regular triangle inside the circle such that each vertex touches the circle. Draw another regular triangle outside the circle such that each side touches the circle. Let $p_3$ denote the area of the inner triangle. Let $q_3$ denote the area of the outer triangle. Repeat this for a square, a pentagon, a hexagon, etc., so that we have two sequences $(p_3, p_4, \dots)$ and $(q_3, q_4, \dots)$.

The inner shape is in yellow, and the outer shape is in green. Now let us calculate $p_k$ and $q_k$ for any integer $k \geq 3$.

Consider an inner $k$-gon. It's made up of $k$ isosceles triangles, suppose each triangle has width $w_k$ and height $h_k$, then each triangle has area $\frac{1}{2} w_k h_k$, and the total area of the $k$-gon is $\frac{1}{2} w_k h_k k$.

In the same fashion, suppose the outer $k$-gon is divided to $k$ triangles where each has width $W_k$ and height $H_k$, then the total area is $\frac{1}{2} W_k H_k k$.

This gives us the relation

$$\frac{1}{2}h_k w_k k \leq A \leq \frac{1}{2} H_k W_k k$$ for all integer $k \geq 3$, where $A$ is the area of the circle.

Here, we skip the rigorous step and conclude that both sides (i.e. $\frac{1}{2}h_k w_k k$ and $\frac{1}{2} H_k W_k k$) tend to the same value as $k$ tends to infinity, so $A$ must be that value. This can be convinced using the picture above: as $k$ tends to infinity, both the inner polygon and outer polygon become closer and closer.

Now, consider the triangles from the standard proof: its width is $\frac{2\pi}{k}$ and its height is $r$. Observe that $w_k \leq \frac{2 \pi r}{k} \leq W_k$ and $h_k \leq r \leq H_k$ (in particular, $H_k = r$ always). The second inequality is obvious, but the first one might not be so obvious. How do I convince myself? Take a look at the perimeter of the inner polygon, which is $k w_k$. It is surely less than the perimeter of the circle (which, by definition, is $2 \pi r$). And the perimeter of the circle is surely less than the perimeter of the outer polygon, which is $k W_k$. This convinces us that $k w_k \leq 2 \pi r \leq k W_k$, so $w_k \leq \frac{2 \pi r}{k} \leq W_k$.

Since

$$0 \leq w_k \leq \frac{2 \pi r}{k} \leq W_k \\ 0 \leq h_k \leq r \leq H_k$$

We deduce that $0 \leq w_k h_k \leq \frac{2 \pi r^2}{k} \leq W_k H_k$. This proves that $\frac{1}{2} w_k h_k k \leq \pi r^2 \leq \frac{1}{2} W_k H_k k$.

Wait! We still cannot deduce that $A = \pi r^2$. What we deduce now is that both $A$ and $\pi r^2$ lies between the area of the inner polygon and the outer polygon. And since both polygons tends to the same area, then $A$ must, in fact, be $\pi r^2$. (Because $A$ can't be smaller than $\pi r^2$ and $A$ can't be bigger than $\pi r^2$.) This completes the convince. $\square$

Filling in the gaps

I'm not sure if this is possible or not, but I think one can fill in the gaps of this proof by a basic framework of analysis. This includes:

• Theorem. A pair of sequences $(a_n)_n$ and $(b_n)_n$ is said to be adjacent if $(a_n)_n$ is nonincreasing, $(b_n)_n$ is nondecreasing, and $a_n - b_n$ tends to $0$ as $n$ tends to infinity. And if two sequences are adjacent, then they converge, and converge to the same limit.
• Theorem. (Squeeze theorem) Suppose $(a_n)_n, (b_n)_n$ and $(c_n)_n$ are sequences of real numbers such that for all $n \in \mathbb{N}$, $a_n \leq b_n \leq c_n$. If $\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = \ell$, then $\lim_{n \to \infty} b_n = \ell$ also.

And yes, to actually fill in all the gaps, one need a stronger statement/assumption. For example, that the ratio between the circumference and the diameter of any circle is constant, and that there is a way to define area (at least for "simple" shapes, to make sense of the area of a circle). I think Euclid's five postulates is also needed? (not sure)