Platonic Solids

Platonic Solids

This one is because I'm on a roll with geometry. (See escape velocity)

This post will prove that there are a limited number of regular three-dimensional solids and will determine the nature of these solids.

Euler said that a regular solid obeys the following rule (where V is the number of vertices, E is the number of edges, and F is the number of faces).

This can be expressed using E only by applying a little logic:

Every edge of a regular solid is shared by the sides of 2 adjacent polygons.

Every edge connects two vertices.

Therefore:

(where n is the number of sides each polygon has, and r is the number of polygons that meet at each vertex)

We can divide the new equation by 2E and rearrange to give:

n must be greater or equal to 3 as the simplest polygon is the triangle, and r must be greater than or equal to 3 as at least 3 vertices must meet in a polyhedron.

If both n and r were simultaneously greater than 3, the left hand side of the equation would be less than 2/3 and could not be satisfied if E is a positive integer (which it must be).

If n=3, the equation becomes:

This can give positive integer values of E for when r = 3, 4, and 5.

 If r=3, the equation becomes:

If n=3, we get E=6 again.

In both cases, n and r cannot be greater than or equal to 6. Because n and r must be between 3 and 6, and either n or r must be three for all solids, there is a limited number of valid combinations of n and r. There are only five of these combinations and thus there are only five regular solids.

We will now analyse these results to see what deescribes which solid. To describe the shape we wil use curly brackets to show {n, r}. From earlier we also know that: V=2E/r and F=2E/n, these will be used to determine the nature of these solids.

{3, 3}

When n=3 and r=3, then E=6. Thus V=4 and F=4. Thus we can deduce the following:
The solid consists of four regular triangles, three triangles meet at each of the four vertices forming six edges. This shape is therefore the tetrahedron.

{3, 4}

When n=3 and r=4, then E=12. Thus V=6 and F=8. Thus we can deduce the following:
The solid consists of eight regular triangles, four triangles meet at each of the six vertices forming twelve edges. This shape is therefore the octahedron.

{3, 5}

When n=3 and r=5, then E=20. Thus V=12 and F=20. Thus we can deduce the following:
The solid consists of twenty regular triangles, five triangles meet at each of the twelve vertices forming twenty edges. This shape is therefore the icosahedron.

{4, 3}

When n=4 and r=3, then E=12. Thus V=8 and F=6. Thus we can deduce the following:
The solid consists of four squares, three squares meet at each of the eight vertices forming twelve edges. This shape is therefore the cube.

{5, 3}

When n=5 and r=3, then E=30. Thus V=20 and F=12. Thus we can deduce the following:

The solid consists of twelve regular pentagons, three pentagons meet at each of the twenty vertices forming thirty edges. This shape is therefore the dodecahedron.

Thus, the tetrahedron, octahedron, icosahedron, cube, and the dodecahedron are the only five regular solids.

QED 

These solids were used in Kepler's first attempt at describing at describing a heliocentric model. It worked because the number of known planets at the time was five, and it seemed too much of a coincidence to Kepler. Unfortunately, it was; however, he still went on to discover his three laws of planetary motion.

Aristotle also believed that these solids corresponded to air, earth, fire, water. Apparantly, ordinary people were to be kept ignorant of the dodecahadron - which was not assosiated with an element. However, some philosophers at the time believed in an aether in which the planets moved, and some applied the starry appearance of the dodecahedron to the starry appearance of the night sky.