# 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.

## 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.

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.

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.

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.

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.

$\blacksquare$