Is there a proof for Archimedes' pictorial proof for the approximation of pi? Is there a proof for Archimedes' pictorial proof for the approximation of pi? Maybe a link that someone can provide please?  The picture I am referring to is found on page 3 of this link: http://scipp.ucsc.edu/~haber/ph116A/pi-2010.pdf
 A: The picture referred to by the OP is a proof only if you have already established the first hundred decimal digits of $\pi$.  It's really not a proof at all, and it certainly isn't Archimedes' proof.  If you really want to see how Archimedes proved the upper and lower bounds
$$3{10\over71}\lt\pi\lt3{1\over7}$$
one place to get the gory, longwinded details is here.
A: The perimeter of a regular $n$-gon circumscribed in a circle of of radius $1$ is given by $n\sqrt{2-2\cos(2\pi/n)}$ (you can get this by breaking the $n$-gon into $n$ isosceles triangles and calculating the length of the base). Taking the limit of this as $n$ goes to infinity we have $$\begin{align}
\lim_{n\to \infty}n\sqrt{2-2\cos(2\pi/n)} &= 
\lim_{n\to\infty}\frac{\sqrt{2-2\cos(2\pi/n)}}{1/n} \\ 
&=\lim_{n\to\infty}\frac{\sqrt{2}\sqrt{1-1\cos(2\pi/n)}}{1/n} \\ 
&=\lim_{n\to\infty}\frac{\sqrt{2}\sqrt{1-(1-2\sin^2(\pi /n))}}{1/n} \\ 
&=\lim_{n\to\infty}\frac{2\sin(\pi/n)}{1/n} \\ 
&=\lim_{x\to \infty} \frac{-\frac
{2\pi\cos(\pi/x)}
{x^2}}
{-\frac{1}{x^2}} \\ &=2\pi\end{align}$$
With the penultimate line following from L'Hopital's rule.
A: It is instructive to demonstrate the spirit of Archimedes idea, so anachronistic liberties can be tolerated. So let us examine the unit circle from first principles (calculus not required).
Since our unit circle can be inscribed in a square, the perimeter of any regular pentagon inscribed in this unit circle is bounded above. So we can take the supremum of the perimeters of the polygons as our definition of circumference.
To calculate it, consider this picture:

We can start with the diameter as initial chord with length $a = 2$. The sagitta now is the same as the radius, so is equal to $1$. If we bisect the chord we can view this as inscribing a 4-gon (square), with each side of length $\sqrt 2$ inside the circle. Now just rinse and repeat, updating your sagitta. We only need to take square roots to get to $\pi$. Just like Archimedes, all we need is the Pythagorean Theorem and some simple Triangle/Circle Theorems.
Google spreadsheets gave us more precision than Archimedes:

Looking at the output, it appears that accuracy might be compromised when we multiply very small numbers by very big numbers. Also, the method did not impress us as a quick way of calculating the number $\pi$. So maybe we can discover new ways of calculating it. Calculus anyone?
