Nested Spheres and Square Pyramids

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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 $V_{s}(1) = \dfrac{4}{3} \pi R^3$ and the volume of the square pyramid $V_{p}(1) = \dfrac{1}{3} x^2 h$ .

For right triangle $COH$ we have:

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

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

$h = \dfrac{4R}{3}$ maximizes $V_{p(1)}(h)$ since $\dfrac{d^2V_{p}(1)}{dh^2}|_{(h = \dfrac{4R}{3})} =\dfrac{-8R}{3} < 0$
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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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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 $x = \dfrac{4R}{3} = h \implies r = \dfrac{4R}{3(\sqrt{5} + 1)} = \dfrac{2R}{3\phi}$

Now let $r(1) = R, r(2) = r = \dfrac{2R}{3\phi}$ and $x(1) = \dfrac{4}{3} R = h(1)$ .

$x(2) = \dfrac{4}{3} r(2) = \dfrac{4}{3} (\dfrac{2}{3\phi}) R = h(2)$

$r(3) = \dfrac{2}{3\phi} r(2) = (\dfrac{2}{3\phi})^2 R$

$x(3) = \dfrac{4}{3} r^3 = \dfrac{4}{3} (\dfrac{2}{3\phi})^2 R = h(3)$

$r(4) = \dfrac{2}{3\phi} r(3) = (\dfrac{2}{3\phi})^3 R$

$x(4) = h(4) = \dfrac{4}{3} (\dfrac{2}{3\phi})^3 R$

For each positive integer $n$ :

$r(n) = (\dfrac{2}{3\phi})^{n - 1} R$ and $x(n) = \dfrac{4}{3}(\dfrac{2}{3\phi})^{n - 1} R = h(n) \implies$

$V_{s}(n) = (\dfrac{8}{27 \phi^3})^{n - 1} V_{s}(1)$ and $V_{p}(n) = \dfrac{16}{27\pi} (\dfrac{8}{27 \phi^3})^{n - 1} V_{s}(1)$

$\implies$

$V_{s} =V_{s}(1) \sum_{n = 0}^{\infty} (\dfrac{8}{27\phi^3})^{n} = (\dfrac{27 \phi^3}{27 \phi^3 - 8}) V_{s}(1)$ and $V_{p} = \dfrac{16}{27\pi} (\dfrac{27 \phi^3}{27 \phi^3 - 8}) V_{s}(1)$ ,

$\implies V_{s} * V{p} = \dfrac{16}{27\pi} (\dfrac{27 \phi^3}{27 \phi^3 - 8})^2 (V_{s}(1))^2 = \dfrac{a}{b\pi} (\dfrac{b \phi^3}{b \phi^3 - c})^2 (V_{s}(1))^2 \implies$

$a + b + c = \boxed{51}$

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Can someone please inform me what program to use to draw diagrams like the two above and whether it is available for free online.