Relative Sizes of Platonic Solids in Same Sphere
Summary
In the words of Hypsicles of Alexandria:
- If $AB$ be any straight line divided at $C$ in extreme and mean ratio, $AC$ being the greater segment, and if we have a cube, a dodecahedron and an icosahedron inscribed in one and the same sphere, then:
- $(1) \quad \left({\text{side of cube} }\right) : \left({\text{side of icosahedron} }\right) = \sqrt {AB^2 + AC^2} : \sqrt {AB^2 + BC^2}$;
- $(2) \quad \left({\text{surface of dodecahedron} }\right) : \left({\text{surface of icosahedron} }\right)$
- $ = \left({\text{side of cube} }\right) : \left({\text{side of icosahedron} }\right)$;
- $(3) \quad \left({\text{content of dodecahedron} }\right) : \left({\text{content of icosahedron} }\right)$
- $ = \left({\text{surface of dodecahedron} }\right) : \left({\text{surface of icosahedron} }\right)$;
- and $(4) \quad \left({\text{content of dodecahedron} }\right) : \left({\text{content of icosahedron} }\right)$
- $ = \sqrt {AB^2 + AC^2} : \sqrt {AB^2 + BC^2}$
(The Elements: Book $\text{XIV}$: Proposition $8$ : Summary)
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 $\map e {\mathscr C}, \map e {\mathscr D}, \map e {\mathscr I}$ be the edge (that is, the side) of $\mathscr C, \mathscr D, \mathscr I$ respectively.
Let $\map s {\mathscr C}, \map s {\mathscr D}, \map s {\mathscr I}$ be the area of the surfaces of $\mathscr C, \mathscr D, \mathscr I$ respectively.
Let $\map v {\mathscr C}, \map v {\mathscr D}, \map v {\mathscr I}$ 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.
- $\map e {\mathscr C}^2 : \map e {\mathscr I}^2 = \paren {PQ^2 + PR^2} : \paren {PQ^2 + QR^2}$
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.
- $AB : AC = PQ : PR$
Thus:
- $\paren {PQ^2 + PR^2} : \paren {PQ^2 + QR^2} = \paren {AB^2 + AC^2} : \paren {AB^2 + BC^2}$
Hence:
- $\map e {\mathscr C} : \map e {\mathscr I} = \sqrt {AB^2 + AC^2} : \sqrt {AB^2 + BC^2}$
and so $(1)$ is shown to be true.
$\Box$
- $\map s {\mathscr D} : \map s {\mathscr I} = \map e {\mathscr C} : \map e {\mathscr D}$
and so $(2)$ is shown to be true.
$\Box$
- $\map e {\mathscr C} : \map e {\mathscr D} = \map v {\mathscr D} : \map v {\mathscr I}$
But from $(2)$ above:
- $\map s {\mathscr D} : \map s {\mathscr I} = \map e {\mathscr C} : \map e {\mathscr D}$
Thus from Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:
- $\map v {\mathscr D} : \map v {\mathscr I} = \map s {\mathscr D} : \map s {\mathscr I}$
and so $(3)$ is shown to be true.
$\Box$
From $(1)$ above:
- $\map e {\mathscr C} : \map e {\mathscr I} = \sqrt {AB^2 + AC^2} : \sqrt {AB^2 + BC^2}$
- $\map e {\mathscr C} : \map e {\mathscr D} = \map v {\mathscr D} : \map v {\mathscr I}$
Thus from Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:
- $\map v {\mathscr D} : \map v {\mathscr I} = \sqrt {AB^2 + AC^2} : \sqrt {AB^2 + BC^2}$
and so $(4)$ is shown to be true.
$\blacksquare$
Historical Note
This proof is Proposition $8$ of Book $\text{XIV}$ of Euclid's The Elements.
Result $(3)$ appears to have been included in Comparison of the Dodecahedron with the Icosahedron by Apollonius of Perga. This work is now lost.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous): The So-Called Book $\text{XIV}$, by Hypsicles
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {A}.6$: Apollonius (ca. $\text {262}$ – $\text {190}$ B.C.)