Relative Sizes of Platonic Solids in Same Sphere

Proof
Let:
 * $\mathscr D$ be a regular dodecahedron
 * $\mathscr I$ be a regular icosahedron
 * $\mathscr C$ be a cube

which are inscribed in a given sphere.

Let $e \left({\mathscr C}\right), e \left({\mathscr D}\right), e \left({\mathscr I}\right)$ be the edge (that is, the side) of $\mathscr C, \mathscr D, \mathscr I$ respectively.

Let $s \left({\mathscr C}\right), s \left({\mathscr D}\right), s \left({\mathscr I}\right)$ be the area of the surfaces of $\mathscr C, \mathscr D, \mathscr I$ respectively.

Let $v \left({\mathscr C}\right), v \left({\mathscr D}\right), v \left({\mathscr I}\right)$ be the volumes of $\mathscr C, \mathscr D, \mathscr I$ respectively.

Let $AB$ be cut at $C$ in extreme and mean ratio such that $AC$ is the greater segment.

From :
 * $e \left({\mathscr C}\right)^2 : e \left({\mathscr I}\right)^2 = \left({PQ^2 + PR^2}\right) : \left({PQ^2 + QR^2}\right)$

where:
 * $PQ$ is the radius of the circle which circumscribes the faces of both $\mathscr D$ and $\mathscr I$
 * $R$ is the point at which $PQ$ has been cut in extreme and mean ratio such that $PR$ is the greater segment.

Thus from :
 * $AB : AC = PQ : PR$

Thus:
 * $\left({PQ^2 + PR^2}\right) : \left({PQ^2 + QR^2}\right) = \left({AB^2 + AC^2}\right) : \left({AB^2 + BC^2}\right)$

Hence:
 * $e \left({\mathscr C}\right) : e \left({\mathscr I}\right) = \sqrt {AB^2 + AC^2} : \sqrt {AB^2 + BC^2}$

and so $(1)$ is shown to be true.

From :
 * $s \left({\mathscr D}\right) : s \left({\mathscr I}\right) = e \left({\mathscr C}\right) : e \left({\mathscr D}\right)$

and so $(2)$ is shown to be true.

From :
 * $e \left({\mathscr C}\right) : e \left({\mathscr D}\right) = v \left({\mathscr D}\right) : v \left({\mathscr I}\right)$

But from $(2)$ above:
 * $s \left({\mathscr D}\right) : s \left({\mathscr I}\right) = e \left({\mathscr C}\right) : e \left({\mathscr D}\right)$

Thus from :
 * $v \left({\mathscr D}\right) : v \left({\mathscr I}\right) = s \left({\mathscr D}\right) : s \left({\mathscr I}\right)$

and so $(3)$ is shown to be true.

From $(1)$ above:
 * $e \left({\mathscr C}\right) : e \left({\mathscr I}\right) = \sqrt {AB^2 + AC^2} : \sqrt {AB^2 + BC^2}$

From :
 * $e \left({\mathscr C}\right) : e \left({\mathscr D}\right) = v \left({\mathscr D}\right) : v \left({\mathscr I}\right)$

Thus from :
 * $v \left({\mathscr D}\right) : v \left({\mathscr I}\right) = \sqrt {AB^2 + AC^2} : \sqrt {AB^2 + BC^2}$

and so $(4)$ is shown to be true.

Historical Note
Result $(3)$ appears to have been included in Comparison of the Dodecahedron with the Icosahedron by. This work is now lost.