# 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," Loci (December 2013)