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Platonic Solids

Problem

It is well known that the five Platonic solids are the regular tetrahedron (four equilateral triangle faces), cube (six square faces), regular octahedron (eight equilateral triangle faces), regular dodecahedron (twelve regular pentagon faces), and the regular icosahedron (twenty equilateral triangle faces).

Prove that no more than five regular (convex) polyhedra exist.


Solution

We begin by noting that for a polyhedron to be regular each face must be a regular polygon, also we note that at least three faces must meet at a common vertex. It shall also be assumed that we already know that a regular tetrahedron is formed from four equilateral triangles, similarly a cube is formed from six squares, a regular octahedron is formed from eight equilateral triangles, a regular dodecehedron is formed from twelve regular pentagons, and a regular icosahedron is formed from twenty equilateral triangles.

It can be seen that no more than five equilateral triangles can be placed around a central vertex before we form a (flat) planar shape.


Consider the diagram on the left: as we pull together the two radii on the horizontal diagonal we form an equilateral triangle base pyramid. That is, we create the first of the Platonic solids: the regular tetrahedron.


Four equilateral triangles "pull together" to form a square base pyramid and two of these combine to create a regular octahedron.


(Note that we have not proved that a regular octahedron is actually being formed, rather that a solid consisting of eight equilateral triangles is made.)

By using five equilateral triangles we form a pentagon base pyramid and four of these combine to produce a regular icosahedron.


It can be seen that six equilateral triangles will tesselate perfectly around the central vertex to a regular hexagon, and therefore cannot be a candidate for creating a regular polyhedron.

Next we consider placing squares around a vertex. It should be clear that it is only possible to fit three squares together (four would tessalate perfectly). Two of these would combine to form a cube.


In a similar way it should be clear that only three pentagons can be placed around a central vertex (three interior angles of 108o add to 324o); four of these "crowns" fit togther to form a regular dodecahedron.


As the interior angle of a regular hexagon is 120o, three of these will tesselate perfectly around a central vertex, and clearly for any polygon containing more edges we will not be able to fit at least three around a central vertex at all.

Hence there can be no more than five regular (convex) polyhedra.

Prove that eight equilateral triangles must necessarily form a regular octahedron.
What about the other Platonic solids?

Problem ID: 246 (28 Oct 2005)     Difficulty: 3 Star

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