Circle Circumscribing Pentagon of Dodecahedron and Triangle of Icosahedron in Same Sphere
Theorem
In the words of Hypsicles of Alexandria:
- The same circle circumscribes both the pentagon of the dodecahedron and the triangle of the icosahedron inscribed in the same sphere.
(The Elements: Book $\text{XIV}$: Proposition $2$)
Lemma
In the words of Hypsicles of Alexandria:
- If an equilateral and equiangular pentagon be inscribed in a circle, the sum of the squares on the straight line subtending two sides and on the side of the pentagon is five times the square on the radius.
(The Elements: Book $\text{XIV}$: Proposition $2$ : Lemma)
Proof
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 Proposition $17$ of Book $\text{XIII} $: Construction of Regular Dodecahedron within Given Sphere:
Let $MN$ be a straight line such that:
- $AB^2 = 5 \cdot MN^2$
- 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 Proposition $15$ of Book $\text{XIII} $: Construction of Cube within Given Sphere:
- $5 \cdot MN^2 = AB^2 = 3 \cdot DG^2$
Let $DG$ be cut in extreme and mean ratio such that $CG$ is the greater segment.
- 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 Proposition $16$ of Book $\text{XIII} $: Construction of Regular Icosahedron within Given Sphere:
- $MO$ 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 Porism to Proposition $15$ of Book $\text{IV} $: Inscribing Regular Hexagon in Circle
- $MN$ is the side of a regular hexagon which has also been inscribed in that same circle.
- $5 \cdot MO^2 + 5 \cdot MN^2 = 5 \cdot KL^2$
Therefore:
- $5 \cdot KL^2 = 3 \cdot CG^2 + 3 \cdot DG^2$
- $3 \cdot DG^2 + 3 \cdot CG^2$ equals the square on $15$ times the radius of the circle around $CDEFG$.
Therefore the two radii are equal.
$\blacksquare$
Historical Note
This proof is Proposition $2$ of Book $\text{XIV}$ of Euclid's The Elements.
This was proved by Aristaeus the Elder.
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