## Proof of the area of a circle using elementary methods

One of the must fundamental discoveries of ancient natural philosophy was a formula for the area of a circle.

Theorem: Given any circle specified by its center and radius $r$, the area of the circle $A = \pi r^2$ for a specific real number $\pi$.

Yet, how many of us can give a convincing arguement to a skeptic that this formula is true? You may have learned how to use calculus to prove this formula, but calculus itself is a very fancy tool that confounds rather than clarifies. So, let's not use calculus. Instead, here is my favorite form of argument, that requires little more than elementary-school mathematics and a tad of algebra. It's the kind of argument that you might come up with on your own, though Carl Gauss was perhaps the first to do it this particular way using Rene Descarte's coordinate system.

Proof:

First, we should agree that if we are given the center and the radius of a circle, the circle is determined uniquely. A circle, after all, is the set of points equidistant from a given center, and the radius determines this distance. Second, since Euclidean spaces are translationally invariant, two circles with the same radius but different centers are "congruent", and have the same areas. So the area of a circle must be uniquely determined by it's radius. This is not true for other geometric objects like parallelograms and triangles, but it is for circles.

So, there is a function $A(r)$ which tells us the area of a circle, depending on it's radius r. We would like to determine this function.

As a simple constraint, we can construct upper and lower bounds on the area by inscribing a square and circumscribing a larger square. We know, for instance, that a square with side-length r can be completely contained in our circle, and a square with side-length $2r$ can completely contain our circle. Thus, $r^2 < A(r) < 4 r^2$ If we try a little harder, we find that the largest inscribed square has an edge with length $r \sqrt{2}$ , so $2 r^2 < A(r) < 4 r^2$

This SUGGESTS that $A(r) = K r^2$ for some K, with $K \approx 3$, but there are many other possibilities. For instance, $A(r) = \frac{2 r^2 + 3.5 r^3}{1+r}$ At this point there are many ways to proceed. You might try to show that this particular guess is wrong by numerically estimating the area of a large circle, but then I'd just come up with a new function that matches your estimate, but differs from $\pi r^2$ almost everywhere else -- you would never all the possibilities one-by-one. Other appealing methods make use of mathematical ideas that were derived after this famous result was known, and thus threaten to entangle us in a web of circular reasoning rather than resolving our question from first principles. Some proofs make use of the formula for the circumference of a circle, but that also assumes we already understand the concept of $\pi$. We could proceed with the inscription of a regular polygon, but finding the area of regular polygons involves the use of trigonometric functions. We could also make use of calculus, but that's a further-sophistication that would involve a whole new set of ideas. We should be able to understand this in very simple terms like the Greek natural philosophers.

So let us build directly on the idea of fitting squares into a circle and proceed by the method of exhaustions using two questions.

1) How many squares with edge length $r/n$ can we fit inside a circle of radius r without overlapping (n is an integer)?

If it takes $p(n)$ squares, $A(r) > p(n) (\frac{r}{n})^2$

2) How few squares with edge length r/n does it take to cover a circle of radius r completely (n is an integer)?

If it takes q(n) squares, $A(r) < q(n) (\frac{r}{n})^2$

From these two formulas, $\frac{p(n)}{n^2} < \frac{A(r)}{r^2} < \frac{q(n)}{n^2}$ Now we squeeze. If $\lim_{n \rightarrow \infty} [p(n)-q(n)]/n^2 = 0$, then $A(r) = K r^2 \quad \text{where}\quad K = p(\infty)/\infty^2 = q(\infty)/\infty^2.$ Each step of the way, $q(n)-p(n) < n^2$, but also, we can expect $q(n) < p(n)+2n$, since the number of squares contacting the circle is never more than twice the number on a side. We can actually determine the exact difference, ($q(n)-p(n) = 2 n -1$) using a geometric argument. Start with a vertical line of squares along the far edge. Move each square towards the center until it overlaps the circle for the first time. Do the same along the top. By convexity (I will leave it to you to think about why it is useful to know that a circle is convex), this will completely cover our circle's edge, using exactly 2n-1 squares.

Then as $n \rightarrow \infty$, the error represented by the difference between our upper and lower bounds shrinks like $\frac{q(n)-p(n)}{n^2} = \frac{2n-1}{n^2} < \frac{2}{n}$ So $p(n)/n^2$ converges to $q(n)/n^2$, and there is a unique real number $\pi$ such that as $n \rightarrow \infty$, $\frac{p(n)}{n^2} < \pi < \frac{q(n)}{n^2} < \frac{p(n)}{n^2} + \frac{2}{n}.$ We must conclude $A(r) = \pi r^2$. We not only have a formula, but also a way to calculate $\pi$. $\Box$

With formulas for the upper and lower bounds, we can resort to some calculation at this point. Let $n = 2^m$ for $m = 1,2,3,\ldots$ This is convenient because it lets us use our last set of squares and subdivide only those that overlap the circle. We can do all this using some algebra in Cartesian coordinates, rather than tedious geometric construction, which is a small and worth-while cheat.

$m$ $n$ Inscribed Circumscribing Lower bound Upper bound
0 1 0 1 0.000000 4.000000
1 2 1 4 1.000000 4.000000
2 4 8 15 2.000000 3.750000
3 8 41 56 2.562500 3.500000
4 16 183 214 2.859375 3.343750
5 32 770 833 3.007812 3.253906
6 64 3149 3276 3.075195 3.199219
7 128 12730 12985 3.107910 3.170166
8 256 51209 51720 3.125549 3.156738
9 512 205356 206379 3.133484 3.149094
10 1024 822500 824547 3.137589 3.145397
11 2048 3292134 3296229 3.139624 3.143529
12 4096 13172634 13180825 3.140601 3.142554
13 8192 52698912 52715295 3.141100 3.142076
14 16384 210812207 210844974 3.141347 3.141835
15 32768 843281848 843347383 3.141470 3.141714
16 65536 3373193506 3373324577 3.141531 3.141653
17 131072 13492906143 13493168286 3.141562 3.141623
18 262144 53971888157 53972412444 3.141577 3.141608
But when you use it, you'll find that it's not a very fast algorithm. With nearly 54 billion points counted, we still only have 4 digits pinned down. 2/n, after all, approaches 0 very slowly. Better ways are stories for other days, but we have been able to show that $K = p(\infty)/\infty^2 = \pi = 3.14159265358979323846264338327...$ Once we understand that $A(r) = \pi r^2$, we can easily use the method of exhaustion to establish the circumference. If we unfold the circle into thin slices like Archimedes, we see that it's area must be $r C / 2$, were C is the circumference. So $r C / 2 = \pi r^2$ $C = 2 \pi r$ We now have a new tool to bootstrap our mathematics.

This is a very constructive approach to the problem. Historically, many mathematical studies focused on the question of how to construct a square with the same area as a given circle using a compass and straight-edge in a few steps. Eventually, squaring-the-circle was proven to be impossible.