# 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!)

## 6 thoughts on “Archimedes’ Circular Reasoning”

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2. Kevin says:

It seems to me that 360 and 180 are reversed in middle part of the third-to-last equation. A_n =(n/2)*sin(360/n), so the larger central angle (360/n) should go with the polygon with smaller number of sides (A_n). If we have 2sin(180/n)/sin(360/n) instead, then sin(360/n)=2sin(180/n)cos(180/n) can cancel with the numerator and give the desired result.

1. Zachary Abel says:

Thanks, Kevin! You are absolutely right. This has now been fixed.

3. Brian says:

I love it!

Gems from mathematics shine like no other (except maybe physics gems).

Question: Can I get an RSS feed?

1. Zachary Abel says:

Thanks!

The RSS feed is located here, and I have included a link in the top menu bar. Thank you for pointing out this omission.