# Platonic solid

The Platonic solids
Solid Number of ... Schläfli symbol Image
faces edges vertices
Tetrahedron 4 6 4 {3,3}
Cube 6 12 8 {4,3}
Octahedron 8 12 6 {3,4}
Dodecahedron 12 30 20 {5,3}
Icosahedron 20 30 12 {3,5}

The Platonic solids are a set of 5 polyhedra:

Most notably, they are the only regular convex polyhedra. However they have many notable properties and there are other ways to arrive at this set of five polyhedra.

## As regular polyhedra

The Platonic solids are regular. They are commonly classified as the regular convex polyhedra, there are a number of ways in which they can be considered:

The existence and completeness of the Platonic solids has been known since at least the Ancient Greeks. Some sources attribute their discovery to Pythagoreas, however Theaetetus gives the first known description of all five.

### Proofs of completeness

#### Geometric proof

The following geometric proof sketch demonstrates that there are only 5 finite polyhedra without self intersection. It is trivial to show that a strictly-convex polyhedron must be finite and must not have self-intersection.

Proof —

First we make a couple of observations about regular polyhedra:

1. If a polyhedron is regular then its faces must be regular. Since the symmetry of the polyhedron is flag transitive, it must act transitively on the flags of each face.
2. Every vertex must have the same degree. If two vertices had different degrees, then there could be no symmetry between them, but there must be for the polyhedron to be regular.
3. The angular defect must be the same at every vertex. If two vertices had different angle defects, then there could be no symmetry between them.
4. A polyhedron must have an angular defect greater than 0. If the angle defect is equal to 0 the result is a planar vertex and the resulting polyhedron will not be finite. If it is less than 0 the vertex is hyperbolic, and trying to fit the faces around the vertex results in clefts.
5. The angle defect decreases when you increase either the number of edges per faces or the number of faces per vertex.

From this we can see that each regular polyhedron must have a number of edges per face, and a number of faces per vertex (this is its Schläfli type). Given a Schläfli type there is a unique polyhedron of that type iff the angle defect is greater than 0. To show the completeness we can start with the smallest possible Schläfli type, {3,3} and try every way to increase it. When we find a new regular polyhedron, that is a Schläfli type with angle defect greater than, we will try all ways to increase that as well etc. when we find a Schläfli type with angle defect less than or equal to 0, we stop searching there, since we know there's no way to increase that Schläfli type and get a new regular polyhedron. This creates a type of tree search:

In total we find the 5 regular polyhedra.

This proof is based off a proof by Euclid.[2]

#### Topological proof

An alternative topological proof exists using the Euler characteristic. Both proofs use some common results about convex regular polytopes.

Proof —

First we make a couple of observations about regular polyhedra:

1. If a polyhedron is regular then its faces must be regular. Since the symmetry of the polyhedron is flag transitive, it must act transitively on the flags of each face.
2. Every vertex must have the same degree. If two vertices had different degrees, then there could be no symmetry between them, but there must be for the polyhedron to be regular.

Now we observe a convex polyhedron must be topologically a ball, and thus has Euler characteristic 2. That means ${\displaystyle V-E+F=2}$. So let us have a polyhedron with Schläfli type {p,q}, meaning it has p -gonal faces and q  faces meeting at each vertex.

The diamond property tells us that ${\displaystyle pF=2E=qV}$, thus we can substitute V  and F  in terms of E , p  and q :

${\displaystyle {\dfrac {2E}{q}}-E+{\dfrac {2E}{p}}=2}$

Since E  at least 1, we can divide through to get

${\displaystyle {\dfrac {1}{q}}-{\dfrac {1}{2}}+{\dfrac {1}{p}}={\dfrac {1}{E}}}$

Since E  is positive this can be weakened to the inequality:

${\displaystyle {\dfrac {1}{q}}+{\dfrac {1}{p}}>{\dfrac {1}{2}}}$

From here it can be seen that with the requirement that p  and q  be greater than 2, there are only five solutions.

Since this proof only uses convexity to conclude the polyhedron is spherical, this proof more broadly shows that the platonic solids are the only non-degenerate spherical abstract regular polyhedra.

## Other properties

The Platonic solids also hold properties that are not dependent on them being regular. Many of these properties are weakenings of regularity, that still maintain the Platonic solids as the only solutions.

### As equivelar polyhedra

Abstractly the Platonic solids are the only non-degenerate spherical equivelar polyhedra.[3] This means that any spherical equivelar polyhedron is a distortion of one of the Platonic solids.

As a corollary of this property the Platonic solids are exactly the weakly-regular convex polyhedra.

### As dart-transitive polyhedra

The dart-transitive polyhedra are a broader class of highly symmetric polytopes which includes the regular polytopes. Many of the statements in the section on the Platonic solids as regular polyhedra can be weakened to be statements about dart-transitive polyhedra instead.

• They are the only dart-transitive convex polyhedra.
• They are the only finite dart-transitive polyhedra without self-intersection.
• They are the only topologically spherical non-degenerate dart-transitive abstract polyhedra.[note 1] i.e. they are the only spherical polyhedral regular maps.[4][5]