Finding the formula for the circumference of a circle with sequences Today I had an idea on how to find the formula for the circumference of a circle: $C = 2\pi r$, where $r$ is the radius of a circle. The idea is that we start with an equilateral triangle ($3$ sides) with a side length of $a$. The perimeter is $3a$. Notice that the distance of the "center" of the triangle is the same from all sides. Then we look at a square, with a side length of less than $a$, say $b$, and its center is also the same distance from all the sides. Its perimeter is $b$. We continue in this fashion, and form a sequence for the perimeters of those shapes: $a_n = n s_n$, where $s_n$ is the side length of the nth shape. Now we calculate $a_n$'s limit, and we get $2 \pi r$, where $r$ is the limit of the sequence of distances of the center of the shape from the sides: $r_n$. The only problem is that I don't know how small to make the side each time... Will this idea work? If so, how small should I make the side each time?  
Thanks!
 A: Your idea is right and, as legend has it (I think!), a very similar thing is how Archimedes derived the area formulas for the circle back in the day, dabbling in the first calculus. I believe he also uses calculus-type arguments to derive the surface area of a sphere, by projecting from a cylinder.
Consider ($n\ge3$) a regular $n$-gon $A_n$ centred on the centre of a circle of radius $r$, with all $n$ vertices on the circumference. Also consider a regular $n$-gon $B_n$, centred at the centre of the same circle, with vertices arranged so that the midpoint of every side touches the circumference of the circle.
Assume it makes sense to talk about a perimeter length $\ell(C)$ for the circle $C$. As in Lucho's answer, the length $\ell(A_n)$ is equal to $2nr\cdot\sin(\pi/n)$. You can derive this as follows:

Since $A_n$ is regular, the perimeter is $n$ multiplied by the length of one side. Each side sits inside an isosceles triangle with legs of length $r$ and angle $2\pi/n$ opposite the desired side, which is of length $h$. From the cosine law and the double angle formula: $$h^2=r^2+r^2-2r^2\cos(2\pi/n)=2r^2(2\sin^2(\pi/n))=(2r\sin(\pi/n))^2$$So: $$h=2r\cdot\sin(\pi/n)$$Then: $$\ell(A_n)=2nr\cdot\sin(\pi/n)$$

Let's make a calculation for $\ell(B_n)$.

Since $B_n$ is regular we need only consider the length of one side, of length $h$. Now the central angle is still $2\pi/n$, but the length of the adjacent legs of the isosceles triangle is less obvious. I can find $h/2$ by dropping a perpendicular from the centre of the triangle to the midpoint of the desired side: this induces a right triangle with base $r$, height $h/2$ and angle $\pi/n$. We get: $$\frac{1}{2}h=r\cdot\tan(\pi/n)$$Concluding: $$\ell(B_n)=2nr\cdot\tan(\pi/n)$$

Very importantly: $$\ell(A_n)<\ell(C)<\ell(B_n)$$For all $n$. This is geometrically clear, though a fully rigorous proof of this is hard (I think the proof that a straight line is a shortest 'path' requires higher mathematics). We get: $$2\pi r\cdot\frac{n}{\pi}\sin(\pi/n)<\ell(C)<2\pi r\cdot\frac{n}{\pi}\tan(\pi/n)$$For all $n$. Letting $n\to\infty$: $$2\pi r\cdot1\le\ell(C)\le2\pi r\cdot1,\,\ell(C)=2\pi r$$
You can (more interestingly) do the same for the area. It is important to have a lower and upper bound, as approximating a circle (or any smooth surface, really) with small subdivisions can sometimes go wrong...
A: Yes, it works. Consider a circumference of radius $R$ and inscribe an n regular polygon concentric to the circle.
You can show that the perimeter of the regular polygon will be
$$P=2nr\sin{\left(\dfrac{\pi}{n}\right)}=(2\pi r) \cdot \dfrac{n}{\pi}\sin{(\dfrac{\pi}{n})} \to 2\pi r $$ as $n \to \infty$.
