Inscribed and Circumscribed Spheres

For inscribed sphere:

Using the above isosceles triangle we have:

$CG = BC = \dfrac{5j^2}{4}$ , $GB = x = rj$ , $AC = h = 2r + j$ and $r = OA = OE = OF = OD$ .

From the diagram the Area of the isosceles triangle is $A = \dfrac{(2r + j)rj}{2} = \dfrac{1}{2}r^2 j + \dfrac{5}{4}r j^2 \implies 4r^2j + 2rj^2 = 2r^2j + 5rj^2$

$\implies 3rj^2 - 2r^2j = 0 \implies rj(3j - 2r) = 0 \implies r = \dfrac{3j}{2}$ for $r,j > 0 \implies AB = \dfrac{x}{2} = \dfrac{3j^2}{4},$ $AC = h = 4j$ and $BC = s = \dfrac{5j^2}{4}$ .

$\therefore$ For right $\triangle{ABC}$ we have: $\dfrac{25j^4}{16} = \dfrac{9j^4}{16} + 16j^2 \implies j^2(j^2 - 16) = 0 \implies j = 4$ for $j > 0$ $\implies x = 24$ , $h = 16$ and $r = 6$ .

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For the circumscribed sphere:

Let $O$ be center of the circumscribed sphere and $R$ be the radius.

In right $\triangle{AOH}, AH = \dfrac{x}{\sqrt{2}}, OA = h - R,$ and $OH = R \implies$

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

From above: $x = 24$ and $h = 16$ $\implies R = 17 \implies \dfrac{V_{1}}{V_{2}} = (\dfrac{R}{r})^3 = (\dfrac{17}{6})^3 = (\dfrac{a}{b})^3 \implies a + b = \boxed{23}$