Inscribed and Circumscribed Spheres 2

For area of $4n - gon$ :

Let $BC = x$ be a side of the $4n - gon$ , $AC = AB= r^{*}$ , $AD = h^*$ , and $\angle{BAD} = \dfrac{45}{n}$ .

$\dfrac{x}{2} = r^{*} \sin(\dfrac{45}{n}) \implies r^{*} = \dfrac{x}{2 \sin(\dfrac{45}{n})} \implies h^{*} = \dfrac{x}{2} \cot(\dfrac{45}{n}) \implies$

$A_{n - gon} = n \cot(\dfrac{45}{n}) x^2 \implies V_{p} = \dfrac{n}{3} \cot(\dfrac{45}{n}) x^2 H$

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: $AC = H - R, BC = \dfrac{x}{2 \sin(\dfrac{45}{n})}, AB = R \implies$

$R^2 = (H - R)^2 + \dfrac{x^2}{4} \csc^2(\dfrac{45}{n}) \implies x^2 = 4 \sin^2(\dfrac{45}{n}) (2RH - H^3)$

$\implies V_{p}(H) = \dfrac{4n}{3} \sin^2(\dfrac{45}{n}) (2RH^2 - H^3) \implies$

$\dfrac{dV_{p}}{dH} = \dfrac{4n}{3} \sin^2(\dfrac{45}{n}) (H) (4R - 3H) = 0$ , $H \neq 0 \implies H = \dfrac{4R}{3} \implies x^2 = \dfrac{32}{9} \sin^2(\dfrac{45}{n}) R^2 \implies x = \dfrac{4 \sqrt{2} \sin(\dfrac{45}{n}) R} {3}$

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

To find the inscribed sphere: Cut the pyramid and inscribed sphere with a vertical plane passing through the center of the sphere , the cross-section becomes a circle inscribed in an isosceles triangle:

Using the above isosceles triangle we have:

$AC = BC$ is the slant height $s = \dfrac{\sqrt{x^2 \cot^2(\dfrac{45}{n}) + 4h^2}}{2}$ of our square pyramid and $AB = 2 h^{*} = x \cot(\dfrac{45}{n})$ , where from above $h^{*} = \dfrac{x}{2} \cot(\dfrac{45}{n}).$

As an analogy using the regular dodecagon( $n = 3$ ) above, $2 h^{*}$ is the line segment from $10$ to $4$ .

From the isosceles triangle diagram above the Area of the isosceles triangle $A = \dfrac{r(2s + x \cot(\dfrac{45}{n}))}{2}$

$\implies r = \dfrac{2A}{2s + x \cot(\dfrac{45}{n})} \implies r = \dfrac{x \cot(\dfrac{45}{n}) H}{\sqrt{x^2 \cot^2(\dfrac{45}{n}) + 4 H^2} + x \cot(\dfrac{45}{n})}$

Using $H = \dfrac{4R}{3}$ and $x = \dfrac{4 \sqrt{2} \sin(\dfrac{45}{n}) R} {3}$ and after simplifying we obtain:

$r = (\dfrac{\dfrac{4}{3} R}{\sqrt{1 + 2 \sec^2(\dfrac{45}{n})} + 1})$

Converting to radians we have: $r = (\dfrac{\dfrac{4}{3} R}{\sqrt{1 + 2 \sec^2(\dfrac{\pi}{4n}) } + 1})$

$\implies V_{s_2} = \dfrac{64}{27} (\dfrac{1} {\sqrt{1 + 2 \sec^2(\dfrac{\pi}{4n})} + 1})^3 * V_{s} \implies$

$\dfrac{V_{s}}{V_{s_2}} = \dfrac{27}{64} (\sqrt{1 + 2 \sec^2(\dfrac{\pi}{4 n}}) + 1)^{3} =$

$\dfrac{3^3}{2^6} (\sqrt{1 + 2 \sec^2(\dfrac{\pi}{4 n}}) + 1)^{3} =\dfrac{a^3}{b^6} (\sqrt{c + b \sec^2(\dfrac{\pi}{4 n}}) + c)^{3} \implies a + b + c = \boxed{6}$

$$

Notice: For $n = 1$ we obtain: $\dfrac{V_{s}}{V_{s_2}} = \dfrac{27}{64} * (\sqrt{5} + 1)^3 = \dfrac{27}{64} * (\dfrac{\sqrt{5} + 1}{2})^3 * 8 = \dfrac{27}{8} (\phi)^3 = (\dfrac{3 \phi}{2})^3$ , where $\phi$ represents the golden ratio, or $(\phi)^3 = \dfrac{8}{27} \dfrac{V_{s}}{V_{s_{2}}}$