MathsIsFun.com

# Platonic Solids - Why Five?

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 that for any convex polyhedron (which includes the platonic solids) that 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

 To see why this works, imagine taking the cube and adding an edge (say from corner to corner of one face). You will have an extra edge, plus an extra face: 7 + 8 - 13 = 2 Likewise if you included another vertex (say half way along a line) you would get an extra edge, too. 6 + 9 - 13 = 2. "No matter what you do, you always end up with 2" (But for when it doesn't read Euler's Formula)

## Faces Meet

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

 The faces can be triangles (3 sides), squares (4 sides), etc. Let us call this "s", the number of sides each face has.

 Also, at each corner, how many faces meet? For a cube 3 faces meet at each corner. For an octahedron 4 faces meet at each corner. Let us call this "m" (how many faces meet at a corner).

(Those two are actually enough to show what type of solid it is)

## 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. But for every corner, there is an edge! So it will also equal 2E.

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!

s m 1/s+1/m > 0.5 ?
3 3 0.666...
3 4 0.583...
4 3 0.583...
4 4 0.5
5 3 0.533...
3 5 0.533...
5 4 0.45
4 5 0.45
5 5 0.4
etc... ... ...

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

Example: s=5, m=5

"1/s + 1/m - 1/2 = 1/E" becomes "1/5 + 1/5 - 1/2 = 1/E" or "-0.1 = 1/E",

which makes E (number of edges) = -10, And you can't have a negative number of edges!

## Real?

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

s m what it means solid
3 3 triangles meeting 3-at-a-corner tetrahedron
3 4 triangles meeting 4-at-a-corner octahedron
4 3 squares meeting 3-at-a-corner cube
5 3 pentagons meeting 3-at-a-corner dodecahedron
3 5 triangles meeting 5-at-a-corner icosahedron

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:

Example: The Octahedron's Schläfli symbol is {3,4}, and the Icosahedron's is {3,5}, can you work out the rest?