Size of Surface of Regular Icosahedron

Theorem

In the words of Hypsicles of Alexandria:

If $ABC$ be an equilateral triangle in a circle, $D$ the centre, and $DE$ perpendicular to $BC$,
$30 BC . DE =$ (surface of icosahedron).

(The Elements: Book $\text{XIV}$: Proposition $4$)

Proof

Let $ABC$ be the equilateral triangle which is the face of a regular icosahedron.

Let the circle $ABC$ be circumscribed around the equilateral triangle $ABC$.

Let $D$ be the center of the circle $ABC$.

Let $DE$ be the perpendicular dropped from $D$ to $BC$.

Let $BD$ and $CD$ be joined.

Then:

\(\ds DE \cdot BC\)

\(=\)

\(\ds 2 \cdot \triangle DBC\)

Area of Triangle in Terms of Side and Altitude

\(\ds \therefore \ \ \)

\(\ds 3 \cdot DE \cdot BC\)

\(=\)

\(\ds 6 \cdot \triangle DBC\)

\(\ds \)

\(=\)

\(\ds 2 \cdot \triangle ABC\)

\(\ds \therefore \ \ \)

\(\ds 30 \cdot DE \cdot BC\)

\(=\)

\(\ds 20 \cdot \triangle ABC\)

The result follows from the definition of a regular icosahedron as having $20$ such faces.

$\blacksquare$

Historical Note

This proof is Proposition $4$ 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