Inscribed and Circumscribed Spheres

For $(1):$

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Sorry, but the above picture was the only picture that I could find.

Using the above picture:

Let $x$ be the side of the square base and $AH$ be the height $h$ of the pyramid. Let $OA$ and $OC$ be radius $R$ of the sphere, so $OH$ is $h - R$ and half the diagonal $CH$ of the square is $\dfrac{x}{\sqrt{2}}$ . The volume of the sphere $V_{s} = \dfrac{4}{3} \pi R^3$ and the volume of the square pyramid $V_{p} = \dfrac{1}{3} x^2 h$ .

For right triangle $COH$ we have:

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$R^2 = \dfrac{x^2}{2} + (h - R)^2 = \dfrac{x^2}{2} + h^2 - 2hR + R^2 \implies x^2 + 2h^2 - 4hR = 0 \implies x^2 = 4hR - 2h^2 \implies$ $$

$V_{p}(h) = \dfrac{1}{3} (4h^2R - 2h^3) \implies \dfrac{dV_{p}}{dh} = \dfrac{2h}{3}(4R - 3h) = 0$ and $h \neq 0 \implies h = \dfrac{4R}{3}.$ $$

$h = \dfrac{4R}{3}$ maximizes $V_{p}(h)$ since $\dfrac{d^2V_{p}}{dh^2}|_{(h = \dfrac{4R}{3})} =\dfrac{-8R}{3} < 0$
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$h = \dfrac{4R}{3} \implies x^2 = \dfrac{16 R^2}{9} \implies x = \dfrac{4R}{3} = h.$

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For $(2):$

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 two opposing sides of the base, the cross-section becomes a circle inscribed in an isosceles triangle:

Again, this is the only picture I could find.

Using the above isosceles triangle we have:

$AC = BC$ is the slant height $s = \dfrac{\sqrt{x^2+ 4h^2}}{2}$ of our square pyramid and $AB = x$ .

From the diagram the Area of the isosceles triangle $A = sr + \dfrac{1}{2} xr = \dfrac{r(2s + x)}{2}$

$\implies r = \dfrac{2A}{2s + x } = \dfrac{xh}{2s + x} = \dfrac{xh}{\sqrt{x^2 + 4h^2} + x}$

From $(1)$ we had $x = \dfrac{4R}{3} = h \implies r = \dfrac{4R}{3(\sqrt{5} + 1)} \implies$ $\dfrac{V_{s}}{V_{s_{2}}} = \dfrac{R^3}{r^3} = \dfrac{27}{64} * (\sqrt{5} + 1)^3 = \dfrac{a}{b} * (\sqrt{c} + d)^3 \implies$

$\boxed{a + b + c + d = 97}$