Ratio of Sizes of Surfaces of Cube and Regular Icosahedron in Same Sphere

Proof

 * Euclid-XIV-6.png

Let a regular dodecahedron, a regular icosahedron and a cube be inscribed in a given sphere.

From :
 * the circle which circumscribes the regular pentagon which is the face of the regular dodecahedron is the same size as the circle which circumscribes the equilateral triangle which is the face of the regular icosahedron.

Let $ABC$ be the circle which circumscribes both that regular pentagon and that equilateral triangle.

Let $CD$ be the side of the equilateral triangle.

Let $AC$ be the side of the regular pentagon.

Let $E$ be the center of the circumscribing circle.

Let $EF$ and $EG$ be perpendiculars dropped from $E$ to $CD$ and $AC$ respectively.

Let $EG$ be produced past $G$ to meet the circumference of $ABC$ at $B$.

Let $BC$ be joined.

Let $H$ be equal to the side of the cube.

It is to be demonstrated that the ratio of the surface of the regular dodecahedron to the surface of the regular icosahedron is $H : CD$.

From :
 * $EB$ equals the side of the regular hexagon that can be inscribed in $ABC$.

From :
 * $EB + BC$ is divided at $B$ in extreme and mean ratio where $BE$ is the greater segment.

From :
 * $EG = \dfrac {EB + BC} 2$

From (indirectly):
 * $EF = \dfrac BE 2$

So:
 * $EG : EF = \paren {EB + BC} : BE$

Therefore if $EG$ is divided in extreme and mean ratio, the greater segment is $EF$.

But from :
 * if $H$ is divided in extreme and mean ratio, the greater segment is equal to $CA$.

So:

We have:

From :
 * (surface of dodecahedron) : (surface of icosahedron) = $CA \cdot EG : FE \cdot CD$

Hence the result.