Nested Spheres and Pyramids

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 and perpendicular to the base of the pyramid, 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: $r = (\dfrac{\dfrac{4}{3} R}{\sqrt{1 + 2 \sec^2(\dfrac{\pi}{4n}) } + 1})$

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Now let $\alpha_{n} = \dfrac{1}{\sqrt{1 + 2 \sec^2(\dfrac{\pi}{4n}) } + 1}$ and $w_{n} = \sin(\dfrac{\pi}{4n})$ .

Let $r1 = R, r_{2} = r = \dfrac{4}{3} \alpha_{n} R , x_{1} = x = \dfrac{4\sqrt{2}}{3} w_{n} R, H_{1} = H = \dfrac{4R}{3} \implies$

$x_{2} = \dfrac{4\sqrt{2}}{3} w_{n} r_{2} = \dfrac{4\sqrt{2}}{3} w_{n} (\dfrac{4}{3} \alpha_{n}) R$ and $H_{2} = \dfrac{4}{3} r^2 = \dfrac{4}{3} (\dfrac{4}{3} \alpha_{n}) R$

$r_{3} = \dfrac{4}{3} \alpha_{n} r_{2} = (\dfrac{4}{3} \alpha_{n})^2 R \implies$

$x_{3} = \dfrac{4\sqrt{2}}{3} w_{n} (\dfrac{4}{3} \alpha_{n})^2 R$

$H_{3} = \dfrac{4}{3} (\dfrac{4}{3} \alpha_{n})^2 R$

In General:

$r_{m} = (\dfrac{4}{3} \alpha_{n})^{m - 1} R , x_{m} = \dfrac{4 \sqrt{2}}{3} w_{n} (\dfrac{4}{3} \alpha_{n})^{m - 1} R , h_{m} = \dfrac{4}{3} (\dfrac{4}{3} \alpha_{n})^{m - 1} R$

$\implies V_{p}(m) = \dfrac{32 n}{27\pi} \cot(\dfrac{\pi}{4n}) w_{n}^2 (\dfrac{64}{27} \alpha_{n}^3)^{m - 1} V_{s}(1)$ and $V_{s}(m) = (\dfrac{64}{27} \alpha_{n}^3)^{m - 1} V_{s}(1)$

$\implies V^{*}_{s}(n) = \sum_{m = 1}^{\infty} V^{*}_{s}(m) = \dfrac{27}{27 - 64 \alpha_{n}^3} V_{s}(1)$

and

$V^{*}_{p}(n) = \sum_{m = 1}^{\infty} V_{p}(m) = \dfrac{32 n}{27\pi} \cot(\dfrac{\pi}{4n}) w_{n}^2 (\dfrac{27}{27 - 64 \alpha_{n}^3}) V_{s}(1)$

Using $w_{n} = \sin(\dfrac{\pi}{4n}) \implies V^{*}_{p}(n) = \dfrac{16 n}{27\pi} \sin(\dfrac{\pi}{2n}) (\dfrac{27}{27 - 64 \alpha_{n}^3}) V_{s}(1)$

Let $w^{*}_{n} = \sqrt{1 + 2 sec^2(\dfrac{\pi}{4n}}) + 1 \implies \alpha_{n} = \dfrac{1}{w^{*}_{n}} \implies$

$V^{*}_{s}(n) * V^{*}_{p}(n) = \dfrac{16 n}{27\pi} \sin(\dfrac{\pi}{2 n}) (\dfrac{27 (w^{*}_{n})^3}{27 (w^{*}_{n})^3 - 64})^2 (V_{s}(1))^2$

Let $j(n) = \dfrac{16 n}{27\pi} sin(\dfrac{\pi}{2})$

To obtain $\lim_{n \rightarrow \infty} j(n)$ :

Using the inequality: $\cos(x) < \dfrac{\sin(x)}{x} < 1$ we obtain:

$\cos(\dfrac{\pi}{2 n}) < \dfrac{2 n}{\pi} \sin(\dfrac{\pi}{2 n}) < 1 \implies$ $\dfrac{8}{27} \cos(\dfrac{\pi}{2 n}) < j(n) < \dfrac{8}{27} \implies \lim_{n \rightarrow \infty} j(n) = \dfrac{8}{27}.$

and

$\lim_{n \rightarrow \infty} (\dfrac{27 (w^{*}_{n})^3}{27 (w^{*}_{n})^3 - 64})^2 = \dfrac{27^2 (52 + 30\sqrt{3})}{(103 + 81\sqrt{3})^2}$

Letting $V_{s}(1) = 1$ and $k(n) = (\dfrac{27 (w^{*}_{n})^3}{27 (w^{*}_{n})^3 - 64})^2 \implies$

$\lim_{n \rightarrow \infty} V^{*}_{s}(n) * V^{*}_{p}(n) = \lim_{n \rightarrow \infty} j(n) * \lim_{n \rightarrow \infty} k(n) = \dfrac{216 (52 + 30\sqrt{3})}{(103 + 81\sqrt{3})^2} = \boxed{0.3793639}$ to seven decimal places.

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Using right circular cones:

Note: From above(changing degrees to radians) $\dfrac{x}{2} = r^{*} \sin(\dfrac{\pi}{4n}) \implies x = 2 r^{*} \sin(\dfrac{\pi}{4n})$

$\therefore V_{p}(n) =\dfrac{n}{3} \cot(\dfrac{\pi}{4 n}) x^2 H = \dfrac{2 n}{3} \sin(\dfrac{\pi}{2n}) {r^{*}}^2 H$ .

Let $m(n) = \dfrac{2n}{3} \sin(\dfrac{\pi}{2n})$ .

Using the inequality $\cos(x) < \dfrac{\sin(x)}{x} < 1 \implies \cos(\dfrac{\pi}{2n})< \dfrac{2n}{\pi} * \sin(\dfrac{\pi}{2n}) < 1 \implies \dfrac{\pi}{3} * \cos(\dfrac{pi}{2n})< m(n) < \dfrac{\pi}{3}$

$\implies \lim_{n \rightarrow \infty} V_{p}(n) = \boxed{\dfrac{1}{3} \pi {r^{*}}^2 H = V_{cone}}$ $LaTeX$