# Archimedes’ Circular Reasoning

Every geometry textbook has formulas for the circumference ($$C = 2 \pi r$$) and area ($$A = \pi r^2$$) of a circle. But where do these come from? How can we prove them?

Well, the first is more a definition than a theorem: the number $$\pi$$ is usually defined as the ratio of a circle’s circumference to its diameter: $$\pi = C/(2r)$$. Armed with this, we can compute the area of a circle. Archimedes’ idea (in 260 BCE) was to approximate this area by looking at regular $$n$$-sided polygons drawn inside and outside the circle, as in the diagram below. Increasing $$n$$ gives better and better approximations to the area.

Look first at the inner polygon. Its perimeter is slightly less than the circle’s circumference, $$C = 2 \pi r$$, and the height of each triangle is slightly less than $$r$$. So when reassembled as shown, the triangles form a rectangle whose area is just under $$C/2\cdot r = \pi r^2$$. Likewise, the outer polygon has area just larger than $$\pi r^2$$. As $$n$$ gets larger, these two bounds get closer and closer to $$\pi r^2$$, which is therefore the circle’s area.

Archimedes used this same idea to approximate the number $$\pi$$. Not only was he working by hand, but the notion of “square root” was not yet understood well enough to compute with. Nevertheless, he was amazingly able to use 96-sided polygons to approximate the circle! His computation included impressive dexterity with fractions: for example, instead of being able to use $$\sqrt{3}$$ directly, he had to use the (very close!) approximation $$\sqrt{3} > 265/153$$. In the end, he obtained the bounds $$3\frac{10}{71} < \pi < 3\frac{1}{7}$$, which are accurate to within 0.0013, or about .04%. (In fact, he proved the slightly stronger but uglier bounds $$3\frac{1137}{8069} < \pi < 3\frac{1335}{9347}$$. See this translation and exposition for more information on Archimedes’ methods.)

These ideas can be pushed further. Focus on a circle with radius 1. The area of the regular $$n$$-sided polygon inscribed in this circle can be used as an approximation for the circle’s area, namely $$\pi$$. This polygon has area $$A_n = n/2 \cdot \sin(360/n)$$ (prove this!). What happens when we double the number of sides? The approximation changes by a factor of $$\frac{A_{2n}}{A_n} = \frac{2\sin(180/n)}{\sin(360/n)} = \frac{1}{\cos(180/n)}.$$ Starting from $$A_4 = 2$$, we can use the above formula to compute $$A_8,A_{16},A_{32},\ldots$$, and in the limit we find that $$\pi = \frac{2}{\cos(180/4)\cdot\cos(180/8)\cdot\cos(180/16)\cdots}.$$ Finally, recalling that $$\cos(180/4) = \cos(45) = \sqrt{\frac{1}{2}}$$ and $$\cos(\theta/2) = \sqrt{\frac{1}{2}(1+\cos\theta)}$$ (whenever $$\cos(\theta/2) \ge 0$$), we can rearrange this into the fun infinite product $$\frac{2}{\pi} = \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}} \cdots$$ (which I found at Mathworld). (It’s ironic that this formula for a circle uses so many square roots!)

# Teach an Old Dog New Trigs

If you studied your Trig tables in high school, you probably remember some facts like \begin{align*} \cos(30) &= \sqrt{3}/2,\\ \cos(45) &= \sqrt{2}/2, \text{ and}\\ \cos(60) &= 1/2. \end{align*} With a few formulas you can get at a few more angles, e.g., $$\cos(15) = \cos(45 – 30) = \cos(45) \cos(30) + \sin(45) \sin(30) = \frac{ \sqrt{6} + \sqrt{2} }{4},$$ and likewise $$\cos(75) = (\sqrt{6} – \sqrt{2})/4$$.

I now offer two more angles that are not as straightforward but nevertheless have surprisingly nice trig values: 18 and 36. Specifically, we have $$\cos(36) = \frac{1 + \sqrt{5}}{4} = \frac{\phi}{2}, \text{ and } \sin(18) = \frac{-1 + \sqrt{5}}{4} = \frac{1}{2 \phi},$$ where $$\phi = (1 + \sqrt{5})/2$$ is the golden ratio. Try to prove these! (Here’s a hint: Draw a 36-72-72 triangle, and bisect one of the large angles to form a 36-36-108 triangle and a smaller 36-72-72 triangle. Both are isosceles! And one of them is similar to the original triangle!)

You may ask for which other angles we can get “nice” answers. Well, this at least lets us get 18 – 15 = 3 degrees, and therefore any multiple of 3 degrees, as long as we don’t mind multiply nested square roots. But if we’re allowing this then there are even more! For example, Gauss showed that \begin{align*}\cos(360/17) &= \frac{1}{16}\bigg( -1 + \sqrt{17} + \sqrt{34-2 \sqrt{17}} \\ & \qquad {} + 2 \sqrt{ 17 + 3 \sqrt{17} – \sqrt{34-2 \sqrt{17}} – 2 \sqrt{34+2 \sqrt{17}} } \bigg),\end{align*} and there are similar (lengthier!) formulas for $$\cos(360/257)$$ and $$\cos(360/65537)$$ using just nested square roots. Why these denominators? You’ll have to ask our friends Fermat and Galois.