Ratio of a regular polygon's radius to its perimeter in terms of number of sides

Question

how would I derive an equation in which y represents the ratio of a regular polygon's diameter to its radius, and x represents the number of sides of the regular polygon? I'm thinking that as x goes to infinity, y should approach 2pi. Please tell me if I'm misunderstanding something basic. Thanks!

Answer 1

Let $L $ be the side length.

then by the well-known formula in a triangle : $$L^2=R^2+R^2-2R.R.\cos (\frac {2\pi}{x}) $$

thus $$L=R\sqrt {2}\sqrt {1-\cos (\frac {2\pi}{x})} $$ $$=2R\sin (\frac {\pi}{x}) $$

and $$\boxed {y=\frac {xL}{R}=2x\sin (\frac {\pi}{x}) }$$

where $xL $ is the perimeter and $R $ the radius. $$\lim_{x\to+\infty}y=\lim_{x\to+\infty}2\pi \frac {\sin (\frac {\pi}{x})}{\frac{\pi}{x}}=2\pi $$

Answer 2

Construct a right triangle from the center to a vertex to the midpoint of a side. The angle at the center is then $\frac \pi x$. If the radius is $r$, half the side is $r\sin \frac \pi x$ so the perimeter is $2xr \sin \frac \pi x$ Then $y=\frac 1r(2xr \sin \frac \pi x)=2x \sin \frac \pi x \to 2\pi$ because $\sin z \approx z$ for $z \ll 1$