Inscribed and Circumscribed Spheres 3

For area of $n - gon$ :

Let $PQ = x$ be a side of the $n - gon$ , $PB = QB= m$ , $AB = h^*$ , and $\angle{QBA} = \dfrac{\pi}{n}$ .

$\dfrac{x}{2} = m \sin(\dfrac{\pi}{n}) \implies m = \dfrac{x}{2 \sin(\dfrac{\pi}{n})} \implies h^* = \dfrac{x}{2} \cot(\dfrac{\pi}{n}) = \dfrac{rj}{2} \cot(\dfrac{\pi}{n})$

In right $\triangle{ABC}$ $OE \perp BC, OE = OD = OA = r, OC = r + j, AB = h^{*} = \dfrac{rj}{2} \cot(\dfrac{\pi}{n}) \implies EB = \dfrac{rj}{2} \cot(\dfrac{\pi}{n})$ and let $CE = y$ .

$\triangle{ABC} \sim \triangle{COE} \implies \dfrac{j \cot(\dfrac{\pi}{n})}{2} = \dfrac{rj \cot(\dfrac{\pi}{n}) + 2y}{2(r + j)} \implies rj\cot(\dfrac{\pi}{n}) + j^2\cot(\dfrac{\pi}{n}) = rj\cot(\dfrac{\pi}{n}) + 2y \implies$ $y = \dfrac{j^2\cot(\dfrac{\pi}{n})}{2}$ .

In right $\triangle{COE}$ the pythagorean theorem $\implies \dfrac{j^4\cot^2(\dfrac{\pi}{n})}{4} + r^2 = r^2 + 2rj + j^2 \implies j^4\cot^2(\dfrac{\pi}{n}) = 8rj + 4j^2 \implies r = (\dfrac{j^2\cot^2(\dfrac{\pi}{n}) - 4}{8}) j \implies$

$h = AC = 2r + j = \dfrac{j^3\cot^2(\dfrac{\pi}{n})}{4}$ and $h^{*} = AB = \dfrac{(j^2cot(\dfrac{\pi}{n}) - 4)\cot(\dfrac{\pi}{n})}{16}$

In $\triangle{ABC}$ the pythagorean theorem $\implies s = BC = \dfrac{j^2\cot(\dfrac{\pi}{n})(j^2\cot^2(\dfrac{\pi}{n}) + 4)}{16} = \dfrac{5}{4}j^2\cot(\dfrac{\pi}{n}) = \dfrac{20j^2\cot(\dfrac{\pi}{n})}{16} \implies$

$\dfrac{j^2\cot(\dfrac{\pi}{n})}{16} (j^2\cot(\dfrac{\pi}{n}) - 16) = 0 \implies j = 4\tan(\dfrac{\pi}{n})$ for $j > 0 \implies r = 2\tan(\dfrac{\pi}{n}) (4 - \tan^2(\dfrac{\pi}{n}))$ .

For the circumscribed sphere:

Let $O$ be center of the circumscribed sphere and $R$ be the radius.

In right $\triangle{AOH}, AH = m, OA = h - R,$ and $OH = R \implies$

$(h - R)^2 + m^2 = R^2 \implies h^2 - 2hR + m^2 = 0 \implies R = \dfrac{m^2 + h^2}{2h}$

From above $m = \dfrac{rj}{2 \sin(\dfrac{\pi}{n})} = 4\sec(\dfrac{\pi}{n})\tan(\dfrac{\pi}{n}) (4 - \tan^2(\dfrac{\pi}{n}))$ and $h = 16\tan(\dfrac{\pi}{n}) \implies$ $R = \dfrac{(\tan(\dfrac{\pi}{n}))[(4 - \tan^2(\dfrac{\pi}{n}))^2 \sec^2(\dfrac{\pi}{n}) + 16]}{2}$

$\dfrac{R(n)}{r(n)} = \dfrac{(4 - \tan^2(\dfrac{\pi}{n}))^2 \sec^2(\dfrac{\pi}{n}) + 16}{4(4 - \tan^2(\dfrac{\pi}{n}))}$ and $\lim_{n \rightarrow \infty} \dfrac{R(n)}{r(n)} = 2 \implies$ $\lim_{n \rightarrow \infty} \dfrac{V_{2}(n)}{V_{1}(n)} = 2^3 = \boxed{8}$ .

$$