### 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

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

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

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

\(=2 R .\)∎

As we pointed out earlier,

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

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