The Japanese Theorem for Nonconvex Polygons - Regular Polygons

Author(s):
David Richeson

Regular Polygons

In our previous theorem we saw that for a fixed $$n$$ and fixed $$R$$ the maximum value of $$r_P$$ is attained by the regular $$n$$-gon, $$P_n$$. Thus we may ask, how large can $$r_P$$ get for all $$n$$?

Theorem. If $$P_n$$ is a regular $$n$$-gon inscribed in a circle of radius $$R$$, then $$(r_P)_{n=3}^{\infty}$$  is an increasing sequence and $$\lim_{n \rightarrow \infty} r_{P_n} = 2R .$$

We illustrate this theorem using the applet below. The $$n$$-gons are inscribed in a circle of radius 1. Use the slider to increase the number of sides of the polygon and observe the value $$r_{P_n}$$  approaching 2.




Proof. By our previous theorem, the radial sum of the regular polygon $$r_{P_n}$$ is

$r_{P_n} = R \left(2 - n + \sum_{k=1}^n \cos \left( \frac{\theta_k}{2} \right) \right) = R \left( 2 - n + n \cos \left( \frac{\pi}{n} \right)\right) .$

It is straightforward to show that this is an increasing function of $$n$$ when $$n > 2$$. Moveover, by a change of variables and an application of l'Hospital's rule we have

$\lim_{n \rightarrow \infty} r_{P_n} = \lim_{n \rightarrow \infty} R \left(2 + n \left( \cos \left( \frac{\pi}{n} \right) - 1 \right) \right)$

$= R \left( 2 + \lim_{m \rightarrow 0^+} \frac{\cos(m \pi) - 1}{m} \right)$

$= R \left( 2 + \lim_{m \rightarrow 0^+} - \pi \sin(m \pi) \right)$

$$=2 R .$$∎

$r_P = 2R - \sum_{k=1}^n (R - d_k) .$

For regular polygons this implies that $$r_{P_n} = 2R - n(R - D_n) ,$$ where $$D_n$$ is the distance from the center of the circle to a side of the regular $$n$$-gon $$P_n$$. This theorem says that $$D_n$$ approaches $$R$$ fast enough that the term $$n (R - D_n)$$ goes to zero.

Although the proof and the applet show that the total inradius tends toward the diameter, it may not be apparent why this is so. In Figure 9 we see that this result is more clear if we choose a different triangulation.

Figure 9