Square pyramids and stacked spheres.

Let $V_{p} = \dfrac{1}{3}x^2H$ and $V_{s} = \dfrac{4}{3}\pi R^{3}$ .

$\dfrac{x^2}{2} + H_{1}^2 - 2H_{1}R + R^2 = R^2 \implies x^2 = 4H_{1}R - 2H_{1}^2 \implies$

$V_{p} = \dfrac{2}{3}(4H_{1}^2R - 2H_{1}^3) \implies \dfrac{dV_{p}}{dH_{1}} =$ $\dfrac{2}{3}(4H_{1}R - 3H_{1}^2) = 0 \implies H_{1}(4R - 3H_{1}) = 0$

and $x \neq 0 \implies H_{1} = \dfrac{4}{3}R \implies x = \dfrac{4}{3}R = H_{1}$

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The slant height $s = \dfrac{\sqrt{5}}{2}x = \dfrac{4}{3}(\dfrac{\sqrt{5}}{2})R$

and the area of the above triangle is $A = \dfrac{1}{2}x^2 =\dfrac{1}{2}xr_{1} + \dfrac{1}{2}(2sr_{1}) =$

$\dfrac{r_{1}}{2}(x + 2s) \implies r_{1} = \dfrac{x^2}{x + 2s} = \dfrac{x}{\sqrt{5} + 1} =$ $\dfrac{4}{3(\sqrt{5} + 1)}r_{1}$

Let $\phi = \dfrac{\sqrt{5} + 1}{2} \implies r_{1} = \dfrac{2R}{3\phi}$

$\implies R = \dfrac{3\phi}{2}r_{1} \implies H_{1} = \dfrac{4}{3}R = \dfrac{4}{3}(\dfrac{3\phi} {2})r_{1} = 2 \phi R_{1}$

and $H_{2} = H_{1} - 2r_{1} = 2(\phi - 1)r_{1} \implies \dfrac{H_{1}}{H_{2}} = \dfrac{\phi}{\phi - 1} \implies H_{2} = (\dfrac{\phi - 1}{\phi})H_{1}$

Let $j = \dfrac{\phi - 1}{\phi} < 1$ , then $r_{1} = \dfrac{2R}{3\phi} \implies r_{2} = jr_{1} \implies r_{3} = jr_{2} = j^2r_{1}$ and in general

$r_{1} = \dfrac{2R}{3\phi}$ and for each positive integer $n \geq 1 \:\ (r_{n + 1} = j^n r_{1})$

$\therefore$ The Total Volume $V_{T} = \dfrac{4\pi}{3}r_{1}^3 + \dfrac{4\pi}{3}r_{1}^3\displaystyle\sum_{n = 1}^{\infty} (j^3)^n =$

$\dfrac{4\pi}{3}r_{1}^3\displaystyle\sum_{n = 1}^{\infty} (j^3)^{n - 1} =$ $\dfrac{4\pi}{3}r_{1}^3(\dfrac{1}{1 - j^3}) \implies$

$V_{T} = \dfrac{4\pi}{3}(\dfrac{8R^3}{27\phi^3})(\dfrac{\phi^3}{\phi^3 - (\phi - 1)^3}) =$

$(\dfrac{32}{81})(\dfrac{1}{3\phi^2 - 3\phi + 1})\pi R^3 = \dfrac{2^5\pi}{3^5\phi^2 - 3^5\phi + (3^{2})^2}R^3$

$\implies \dfrac{V_{T}}{R^3} = {\dfrac{2^5\pi}{3^5\phi^2 - 3^5\phi + (3^{2})^2}}$ $= \dfrac{a^b\pi}{c^{b}\phi^{a} - c^{b}\phi + (c^{a})^{a}} \implies a + b + c = \boxed{10}$