Platonic Solids - Why Five?

Simplest Reason: Angles at a Vertex

The simplest reason there are only 5 Platonic Solids is this:

	At each vertex at least 3 faces meet (maybe more).
	When you add up the internal angles that meet at a vertex, it must be less than 360 degrees (at 360° the shape would flatten out).

We also know that a Platonic Solid's faces are all identical regular polygons:

	A regular triangle has internal angles of 60°, so we can have: 3 triangles (3×60°=180°) 4 triangles (4×60°=240°) or 5 triangles (5×60°=300°)
	A square has internal angles of 90°, so there is only: 3 squares (3×90°=270°)
	A regular pentagon has internal angles of 108°, so there is only: 3 pentagons (3×108°=324°)

And this is the result:

Another Reason (using Toplogy)

In a nutshell, it is impossible to have more than 5,
because any other possibility would violate simple rules about
the number of edges, corners and faces you can have together.

Euler's Formula

Do you know about Euler's Formula? It says: for any convex polyhedron (which includes the Platonic Solids) the Number of Faces plus the Number of Vertices (corner points) minus the Number of Edges always equals 2

It is written: F + V - E = 2

Try it on the cube: A cube has 6 Faces, 8 Vertices, and 12 Edges, so: 6 + 8 - 12 = 2

 

Faces Meet

Next, think about a typical platonic solid. What kind of faces does it have, and how many meet at a corner (vertex)?

Exploding Solids!

Now, imagine we pulled a solid apart, cutting each face free.

You would have all these little flat shapes. And there would be twice as many edges (because you cut along the edge).

Example: the cut-up-cube would now be six little squares. Each square would have 4 edges, for 24 edges in total (versus 12 edges when joined up to make a cube).

So, how many edges? Twice as many as the original number of edges "E", or simply 2E

But this will also be the same as counting all the edges of the little shapes. There will be s (number of sides per face) times F (number of faces).

This can be written as sF = 2E

 

Likewise, when we cut it up, what was one corner will now be several corners. In the case of a cube there will be three times as many corners.

• The new number of corners is: how many faces that meet at a corner (m) times how many vertices of the original solid (V), which is mV
• The new number of edges is: twice as many as the original solid, which is 2E

And because we now have a collection of polygons there is the same number of corners as edges (a square has 4 corners and 4 edges, a pentagon has 5 corners and 5 edges, etc.)

This can be written as mV = 2E

Bring Equations Together

That is all the equations we need, let us use them together:

sF = 2E, hence F = 2E/s
mV = 2E, hence V = 2E/m

Now let us put those into "F+V-E=2":

F + V - E = 2
2E/s + 2E/m - E = 2

Next, some rearranging ... divide the lot by "2E":

1/s + 1/m - 1/2 = 1/E

Now, "E", the number of edges, cannot be less than zero, so "1/E" cannot be less than 0:

1/s + 1/m - 1/2 > 0

Or, more simply:

1/s + 1/m > 1/2

So, all that remains is to try different values of:

• "s" (number of sides each face has, cannot be less than 3), and
• "m" (number of faces that meet at a corner, cannot be less than 3),

and we are done!

The Possibilities!

The possible answers are:

Result: There are only 5 that work! All the rest are just not possible in the real world.

Real?

And the last step is to see if those solids are real:

So, only 5, and they all exist.

Job Done.

Schläfli !

And just to keep you well educated ... the "s" and "m" values put together inside curly braces {} make what is called the "Schläfli symbol" for polyhedra: