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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 \(f(P) = 0 .\)