# The Japanese Theorem for Nonconvex Polygons - Extreme Values for the Radial Sum Function

Author(s):
David Richeson

### Extreme Values for the Radial Sum Function

Now that we have a formula for the total inradius function we can look for extreme values of the function; in other words we can determine which polygons yield the largest and smallest total inradius.

Theorem. Let $\mathcal{P}_n^c = \mathcal{P}_{R,n}^c$ be the space of convex cyclic $n$-gons with circumradius $R$, and let $f: \mathcal{P}_n^c \rightarrow {\mathbb R}$ be the function given by $f(P) = r_P .$

1. The unique absolute maximum of $f$ is the regular $n$-gon $P_n .$

2. The set of absolute minima of $f$ is the 1-skeleton of $\mathcal{P}_n^c .$

3. The function $f$ has no relative, non-absolute extrema.

Proof. From the definition of $\mathcal{P}_n^c$ and the representation of $f$ given on the previous page, it suffices to investigate the extrema of the function

$f(P) = f(\theta_1, \ldots, \theta_n) = R \left( 2 - n + \sum_{k=1}^n \cos\left(\frac{\theta_k}{2} \right) \right)$

subject to the constraints $g ( \theta_1, \ldots, \theta_n) = 2 \pi$ and $\theta_k \geq 0 .$ By the method of Lagrange multipliers, extreme values occur when $\nabla f = \lambda \nabla g$ for some constant $\lambda$ or when $(\theta_1, \ldots, \theta_n)$ is on the boundary of the simplex $\mathcal{P}_n^c .$

Observe that $\nabla f = ( -R \sin (\theta_1 / 2)/2, \ldots, -R \sin(\theta_n / 2)/2)$ and $\nabla g = (1, \ldots, 1) .$ Thus, interior extrema can only occur when $\sin(\theta_1 /2) = \cdots = \sin(\theta_n / 2)$ and $\theta_1 + \cdots + \theta_n = 2 \pi .$ Clearly $\theta_1 = \cdots = \theta_n =2\pi/n$ is one such point (this is the regular $n$-gon $P_n$). In this case $f(P_n) = R(2-n(1 - \cos(\pi / n))) .$ We claim that there are no other extreme values in the interior of the simplex and that this is the absolute maximum.

Suppose that $(\theta_1, \ldots, \theta_n)$   is another extreme value. Then there must be $k$ and $j$ such that $\theta_k < \theta_j .$ Because $\theta_k, \theta_j \in [0, 2 \pi] ,$  and if $\sin(\theta_k / 2) = \sin(\theta_j / 2) ,$ it must be the case that $\pi / 2 - \theta_k/2 = \theta_j / 2 - \pi / 2 .$ However, this implies that $\theta_k + \theta_j = 2 \pi .$ Thus $(\theta_1, \ldots, \theta_n)$ is not in the interior of the simplex. In fact in this case, the polygon is a vertex or is on an edge of the simplex and $f(P) = 0 .$

David Richeson, "The Japanese Theorem for Nonconvex Polygons - Extreme Values for the Radial Sum Function," Convergence (December 2013)