Comparison of Sides of Five Platonic Figures

Proof

 * Euclid-XIII-18.png

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

Let $AB$ be bisected at $C$.

Let $AB$ be cut at $D$ where $AD = 2 \cdot DB$.

Let $AEB$ be a semicircle on the diameter $AB$.

Let $CE$ be drawn from $C$ perpendicular to $AB$.

Let $DF$ be drawn from $D$ perpendicular to $AB$.

Let $AF, FB, EB$ be joined.

We have that:
 * $AD = 2 \cdot DB$

Therefore:
 * $AB = 3 \cdot DB$

and so:
 * $AB = \dfrac 3 2 \cdot AD$

We have that $\triangle AFB$ is equiangular with $\triangle AFD$.

So from:

and:

it follows that:
 * $BA : AD = BA^2 : AF^2$

Therefore:
 * $BA^2 = \dfrac 3 2 \cdot AF^2$

But from :
 * the square on the diameter of the given sphere is $1 \frac 1 2$ times the square on the side of the regular tetrahedron inscribed within it.

But $AB$ is the diameter of the given sphere.

Therefore $AF$ is equal to the side of the regular tetrahedron inscribed within.

We have that:
 * $AD = 2 \cdot DB$

Therefore:
 * $AB = 3 \cdot DB$

So from:

and:

it follows that:
 * $AB : BD = AB^2 : BF^2$

Therefore:
 * $AB^2 = 3 \cdot BF^2$

But from :
 * the square on the diameter of the given sphere is $3$ times the square on the side of the cube inscribed within it.

But $AB$ is the diameter of the given sphere.

Therefore $BF$ is equal to the side of the cube inscribed within.

We have that:
 * $AC = CB$

Therefore:
 * $AB = 2 \cdot BC$

But:
 * $AB : BC = AB^2 : BE^2$

But from :
 * the square on the diameter of the given sphere is $2$ times the square on the side of the regular octahedron inscribed within it.

But $AB$ is the diameter of the given sphere.

Therefore $BE$ is equal to the side of the regular octahedron inscribed within.

Let $AG$ be drawn from $A$ perpendicular to $AB$.

Let $AG = AB$.

Let $GC$ be joined.

Let $H$ be the point where $GC$ cuts the semicircle $AEB$.

Let $HK$ be drawn perpendicular to $AB$ from $H$.

We have that:
 * $GA = 2 \cdot AC$

We also have that:
 * $GA : AC = HK : KC$

Also see

 * Five Platonic Solids