Now say we wanted to construct an analogous figure in three dimensions - a regular

With a bit of thought, one can see that no more than five regular polyhedrons. A corner of a polyhedron must have at least three faces, the simplest of which is an equilateral triangle. We can form a vertex by putting together three, four, or five equilateral triangles. If we use more than five, the angles total 360 degrees or more and therefore cannot form a corner. So there are three possible ways to construct a regular polyhedra with triangles. Three and only three squares will form a corner, indicating the possibility of one regular solid with square faces. The same reasoning yields one possible polyhedron with three pentagons at each corner. We cannot go beyond the pentagon, because when we try to put three hexagons together at a corner, they equal 360 degrees. So we have a total of five regular polyhedra. A slightly more mathematically rigorous proof is at the bottom of the page.

Are here are the the five platonic solids:

Tetrahedron4 triangular sides Cut-out | |

Hexahedron (cube)6 square sides Cut-out | |

Octahedron8 triangular sides Cut-out | |

Dodecahedron12 pentoganal sides Cut-out | |

Icosahedron20 triangular sides Cut-out |

The slightly-more-rigorous proof:

Euler's formula (for Algebraic Topolgy) states that for a given polyhedron (not neccessarily regular) the number of Faces minus the number of Edges plus the number of Vertices equals two, i.e.

Now we look at a regular polyhedron. It has

We factor this into

Since

1 - (

We simplify this to

(2/

Now for some critical thinking. A

The only remaining values of

k = 3, p = 3 | tetrahedron |

k = 3, p = 4 | octahedron |

k = 3, p = 5 | dodecahedron |

k = 4, p = 3 | cube |

k = 5, p = 3 | isocahedron |

Q.E.D.