Five Platonic Solids

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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.

From Proposition $21$ of Book $\text{XI} $: Solid Angle contained by Plane Angles is Less than Four 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:

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

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.

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 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:

\(\displaystyle n \paren {180^\circ - \dfrac {360^\circ} m}\) \(<\) \(\displaystyle 360^\circ\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle n \paren {1 - \dfrac 2 m}\) \(<\) \(\displaystyle 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle n \paren {m - 2}\) \(<\) \(\displaystyle 2m\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle n \paren {m - 2} - 2 \paren {m - 2}\) \(<\) \(\displaystyle 2m - 2 \paren {m - 2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {m - 2} \paren {n - 2}\) \(<\) \(\displaystyle 4\)


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