# Five Platonic Solids

## 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 1

A solid angle cannot be constructed from only two planes.

Therefore at least three faces need to come together to form a vertex.

Let $P$ be a platonic solid.

Let the polygon which forms each face of $P$ be a equilateral triangles.

We have that:

- each vertex of a regular tetrahedron is composed of $3$ equilateral triangles
- each vertex of a regular octahedron is composed of $4$ equilateral triangles
- each vertex of a regular icosahedron is composed of $5$ equilateral triangles.

$6$ equilateral triangles, placed together at a vertex, form $4$ right angles.

- a solid angle is contained by plane angles which total less than $4$ right angles.

Thus it is not possible to form $P$ such that its vertices are formed by $6$ equilateral triangles.

For the same reason, it is not possible to form $P$ such that its vertices are formed by more than $6$ equilateral triangles.

Hence there are only $3$ possible platonic solids whose faces are equilateral triangles.

We have that each vertex of a cube is composed of $3$ squares.

$4$ squares, placed together at a vertex, form $4$ right angles.

Thus, again from Proposition $21$ of Book $\text{XI} $: Solid Angle contained by Plane Angles is Less than Four Right Angles:

For the same reason, it is not possible to form $P$ such that its vertices are formed by more than $4$ squares.

Hence there is only $1$ possible platonic solid whose faces are squares.

We have that each vertex of a regular dodecahedron is composed of $3$ regular pentagons.

From Lemma to Proposition $18$ of Book $\text{XIII} $: Comparison of Sides of Five Platonic Figures:

- the vertices of a regular pentagon equal $1 \dfrac 1 5$ right angles.

$4$ regular pentagons, placed together at a vertex, form $4 \dfrac 4 5$ right angles.

Thus, again from Proposition $21$ of Book $\text{XI} $: Solid Angle contained by Plane Angles is Less than Four Right Angles:

- it is not possible to form $P$ such that its vertices are formed by $4$ regular pentagons.

For the same reason, it is not possible to form $P$ such that its vertices are formed by more than $4$ regular pentagons.

Hence there is only $1$ possible platonic solid whose faces are regular pentagons.

$3$ regular hexagons, placed together at a vertex, form $4$ right angles.

Thus, again from Proposition $21$ of Book $\text{XI} $: Solid Angle contained by Plane Angles is Less than Four Right Angles:

- it is not possible to form $P$ such that its vertices are formed by $3$ or more regular hexagons.

Regular polygons with more than $6$ sides have vertices which are greater than those of a regular hexagon.

Therefore $3$ such regular polygons, placed together at a vertex, form more than $4$ right angles.

- it is not possible to form $P$ such that its vertices are formed by $3$ or more regular polygons with more than $6$ sides.

Hence the $5$ possible platonic solids have been enumerated and described.

$\blacksquare$

## Proof 2

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:

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

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

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

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

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

But $m$ and $n$ are both 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.

$\blacksquare$

## Historical Note

Euclid's proof that there exist exactly Five Platonic Solids appears to have originated with Theaetetus of Athens.

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $5$ - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.) - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $5$ - 2008: Ian Stewart:
*Taming the Infinite*... (previous) ... (next): Chapter $2$: The Logic of Shape: The golden mean