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In other words, nested regular Platonic solids.

What the radius ratio of the innermost insphere divided by the outermost circumsphere?

The numerator of the fraction has the form, when simplified, of $5+2\sqrt5$ . The desired answer is the denominator of that fraction.

The radius of a circumsphere is the distance from the center of the solid to one of its vertices.

The radius of an insphere is the distance from the center of the solid to the center of one of its faces.

Nesting occurs when the insphere radius of a containing solid is equal to the circumsphere radius of the contained solid.

The answer is 135.

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$\begin{array}{l} \text{Cube}\to \frac{1}{\sqrt{3}} \\ \text{Dodecahedron}\to \frac{\sqrt{110 \sqrt{5}+250}}{5 \left(\sqrt{3}+\sqrt{15}\right)} \\ \text{Icosahedron}\to \frac{3 \sqrt{3}+\sqrt{15}}{3 \sqrt{2 \sqrt{5}+10}} \\ \text{Octahedron}\to \frac{1}{\sqrt{3}} \\ \text{Tetrahedron}\to \frac{1}{3} \\ \end{array}$

In the same sense that $\frac12$ of $\frac23$ is $\frac13$ , the combination of the shrinkage factors is a product. The order of the factors is unimportant as multiplication of real numbers is communative.

$\frac{\sqrt{\frac{110 \sqrt{5}+250}{2 \sqrt{5}+10}} \left(3 \sqrt{3}+\sqrt{15}\right)}{135 \left(\sqrt{3}+\sqrt{15}\right)}$ is $\frac{1}{135} \left(2 \sqrt{5}+5\right)$ .

The shrinkage is to about 7%.