# Five Platonic Solids/Proof 2

## Theorem

There exist exactly five platonic solids:

- $\paren 1: \quad$ the regular tetrahedron
- $\paren 2: \quad$ the cube
- $\paren 3: \quad$ the regular octahedron
- $\paren 4: \quad$ the regular dodecahedron
- $\paren 5: \quad$ the regular icosahedron.

In the words of Euclid:

*I say next that no other figure, besides the said five figures, can be constructed which is contained by equilateral and equiangular figures equal to one another.*

(*The Elements*: Book $\text{XIII}$: Proposition $18$ : Endnote)

## Proof

Consider a convex regular polyhedron $P$.

Let $m$ be the number of sides of each of the regular polygons that form the faces of $P$.

Let $n$ be the number of those polygons which meet at each vertex of $P$.

From Internal Angles of Regular Polygon, the internal angles of each face of $P$ measure $180^\circ - \dfrac {360^\circ} m$.

The sum of the internal angles must be less than $360^\circ$.

So:

\(\displaystyle n \paren {180^\circ - \dfrac {360^\circ} m}\) | \(<\) | \(\displaystyle 360^\circ\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle n \paren {1 - 2 m}\) | \(<\) | \(\displaystyle 2\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {m - 2} \paren {n - 2}\) | \(<\) | \(\displaystyle 4\) |

But $m$ and $n$ are both at greater than $2$.

So:

- if $m = 3$, $n$ can only be $3$, $4$ or $5$
- if $m = 4$, $n$ can only be $3$
- if $m = 5$, $n$ can only be $3$

and $m$ cannot be greater than $3$.

There are $5$ possibilities in all.

Therefore all platonic solids have been accounted for.

$\blacksquare$

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)