Five Platonic Solids/Proof 1
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
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.
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.
$\blacksquare$
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XIII}$. Propositions