Circle Circumscribing Pentagon of Dodecahedron and Triangle of Icosahedron in Same Sphere

Proof

 * Euclid-XIV-2.png

Let $AB$ be the diameter of the given sphere.

Let a regular dodecahedron and a regular icosahedron be inscribed within.

Let $CDEFG$ be one pentagonal face of the dodecahedron.

Let $KLH$ be one triangular face of the icosahedron.

It is to be demonstrated that the radii of the circles circumscribing $CDEFG$ and $KLH$ are equal.

Let $DG$ be joined.

From :
 * $DG$ is the side of a cube which has been inscribed within the given sphere.

Let $MN$ be a straight line such that:
 * $AB^2 = 5 \cdot MN^2$

From :
 * the square on the diameter of the given sphere is $5$ times the square on the radius of the circle in which is inscribed the pentagonal base of the pyramidal tip of a regular icosahedron which is inscribed within that sphere.

Therefore $MN$ equals the radius of the circle in which that pentagon is inscribed.

Let $MN$ be cut in extreme and mean ratio at $O$ such that $MO$ is the greater segment.

Then from:

and the converse of:

it follows that:
 * $MO$ is the side of a regular decagon which has been inscribed in the circle whose radius is $MN$.

From :
 * $5 \cdot MN^2 = AB^2 = 3 \cdot DG^2$

Let $DG$ be cut in extreme and mean ratio such that $CG$ is the greater segment.

From the :
 * if two straight lines have been cut in extreme and mean ratio, their segments are in the same ratio.

So:
 * $3 \cdot DG^2 : 3 \cdot CG^2 = 5 \cdot MN^2 : 5 \cdot MO^2$

From :
 * $MO$ is the side of a regular decagon which has been inscribed

and
 * $KL$ is the side of a regular pentagon which has been inscribed

in the circle whose radius is $MN$.

From
 * $MN$ is the side of a regular hexagon which has also been inscribed in that same circle.

Thus from :
 * $5 \cdot MO^2 + 5 \cdot MN^2 = 5 \cdot KL^2$

Therefore:
 * $5 \cdot KL^2 = 3 \cdot CG^2 + 3 \cdot DG^2$

But from :
 * $5 \cdot KL^2$ equals the square on $15$ times the radius of the circle around $KLH$.

From :
 * $3 \cdot DG^2 + 3 \cdot CG^2$ equals the square on $15$ times the radius of the circle around $CDEFG$.

Therefore the two radii are equal.