Pyramid Craze

For area of $n - gon$ :

Let $BC = x$ be a side of the $n - gon$ , $AC = AB= r$ , $AD = h^*$ , and $\angle{BAD} = \dfrac{180}{n}$ .

$\dfrac{x}{2} = r \sin(\dfrac{180}{n}) \implies r = \dfrac{x}{2 \sin(\dfrac{180}{n})} \implies h^* = \dfrac{x}{2} \cot(\dfrac{180}{n}) \implies A_{\triangle{ABC}} = \dfrac{1}{4} \cot(\dfrac{180}{n}) x^2 \implies$

$A_{n - gon} = \dfrac{n}{4} \cot(\dfrac{180}{n}) x^2 \implies V_{p} = \dfrac{n}{12} \cot(\dfrac{180}{n}) x^2 H$

The volume of the sphere $V_{s} = \dfrac{4}{3} \pi R^3$

Let $H$ be the height of the given pyramid.

In the right triangle above: $AC = H - R, BC = \dfrac{x}{2 \sin(\dfrac{180}{n})}, AB = R \implies$

$R^2 = (H - R)^2 + \dfrac{x^2}{4} \csc^2(\dfrac{180}{n}) \implies x^2 = 4H(2R - H) \sin^2(\dfrac{180}{n})$

$\implies V_{p}(H) = \dfrac{n}{6} \sin(\dfrac{360}{n}) (2RH^2 - H^3) \implies$

$\dfrac{dV_{p}}{dH} = \dfrac{n}{6} \sin(\dfrac{360}{n}) (H) (4R - 3H) = 0$ , $H \neq 0 \implies H = \dfrac{4R}{3} \implies x^2 = \dfrac{32}{9} \sin^2(\dfrac{180}{n}) R^2 \implies x = \dfrac{4 \sqrt{2}}{3} \sin(\dfrac{180}{n}) R$

$H = \dfrac{4R}{3}$ maximizes $V_{p}(H)$ since $\dfrac{d^2V_{p}}{dH^2}|_{(H = \dfrac{4R}{3})} =\dfrac{-2n}{3} \sin(\dfrac{360}{n}) R < 0$

$H = \dfrac{4R}{3}$ and $x^2 = \dfrac{32}{9} \sin^2(\dfrac{180}{n}) R^2 \implies$ $V_{p} = \dfrac{4n}{27\pi} \sin(\dfrac{360}{n}) V_{s} \implies$ $\dfrac{V_{p}}{V_{s}} = \dfrac{2^2 n}{3^3\pi} \sin(\dfrac{360}{n}) = \dfrac{a^2 n}{b^3\pi} \sin(\dfrac{c}{n}) \implies a + b + c = \boxed{365}.$