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

Theorem

In the words of Hypsicles of Alexandria:

As the surface of the dodecahedron is to the surface of the icosahedron, so is the side of the cube to the side of the icosahedron.

(The Elements: Book $\text{XIV}$: Proposition $6$)

Proof

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

From Proposition $3$ of Book $\text{XIV} $: Circle Circumscribing Pentagon of Dodecahedron and Triangle of Icosahedron in Same Sphere:

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 Porism to Proposition $15$ of Book $\text{IV} $: Inscribing Regular Hexagon in Circle:

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

From Proposition $9$ of Book $\text{XIII} $: Sides Appended of Hexagon and Decagon inscribed in same Circle are cut in Extreme and Mean Ratio:

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

From Proposition $1$ of Book $\text{XIV} $: Perpendicular from Center of Circle to Side of Inscribed Pentagon:

$EG = \dfrac {EB + BC} 2$

From Proposition $12$ of Book $\text{XIII} $: Square on Side of Equilateral Triangle inscribed in Circle is Triple Square on Radius of Circle (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 Porism to Proposition $17$ of Book $\text{XIII} $: Construction of Regular Dodecahedron within Given Sphere:

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

So:

\(\ds H : CA\)

\(=\)

\(\ds EG : EF\)

as both are in extreme and mean ratio

\(\text {(1)}: \quad\)

\(\ds \therefore \ \ \)

\(\ds FE \cdot H\)

\(=\)

\(\ds CA \cdot EG\)

Definition of Ratio

We have:

\(\ds H : CD\)

\(=\)

\(\ds FE \cdot H : FE \cdot CD\)

\(\ds \therefore \ \ \)

\(\ds H : CD\)

\(=\)

\(\ds CA \cdot EG : FE \cdot CD\)

substituting from $(1)$

From Proposition $5$ of Book $\text{XIV} $: Ratio of Sizes of Surfaces of Regular Dodecahedron and Regular Icosahedron in Same Sphere:

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

Hence the result.

$\blacksquare$

Historical Note

This proof is Proposition $6$ of Book $\text{XIV}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): The So-Called Book $\text{XIV}$, by Hypsicles