## Five Regular Solids, Part Three

Parts One and Two of this series of blogs established the following facts about solids that have F flat faces, E straight edges and V vertices.

1.

$F + V - E = 2$

The second fact concerns those solids that are regular. By regular we mean that each face has the same number, m, of edges and each vertex has the same number, n, of edges meeting there. We showed that

2. In a regular solid

$E= \frac {mF} {2}$

and

$E = \frac {nV} {2}$

Each regular solid is defined by m and n, as we shall show. And this demonstration will lead to the conclusion that there are only five such solids. The argument is short, but algebraic.

First, fact 2 means that

$mF = nV$

so

$F = \frac {n} {m} V$

and

$V = \frac {m} {n} F$

Now we have enough to write fact 1 in terms of F only.

$F + V - E = 2$

$F + \frac {m} {n} F - \frac {m}{2} F =2$

$F\left ( 1+\frac{m}{n} - \frac {m} {2} \right ) = 2$

$F \left( \frac {2m + 2n - nm} {2n} \right ) = 2$

$F = \frac {4n} {2m + 2n - mn}$

It follows from this that in order to be regular, a solid must have

$2m + 2n - mn > 0$

If we take into account that both m and n must be at least 3, we have very few pairs that satisfy this condition. We start with m = 3 and n = 3. If the condition is satisfied, we compute

$F = \frac {4n}{2m + 2n - nm}$

$V = \frac {4m}{2m + 2n - nm}$

$E = \frac {2mn}{2m + 2n - nm}$

and move on to the next value of n.

Here’s the list.

____________________

### m=3, n=3

$2m + 2n - nm = 3$.

The condition is satisfied. F = 4, V = 4, E = 6. We have the tetrahedron.

____________________

### m = 3, n = 4

$2m + 2n - nm = 2$

F = 8, V = 6, E = 12. The octahedron.

____________________

### m = 3, n = 5

$2m + 2n - nm = 1$

F = 20, V = 12, E = 30. The icosahedron.

____________________

### m = 3, n = 6

$2m + 2n - nm = 0$

The condition is not met. There is no solution for F, V or E. In fact, for greater n, 2m + 2n -mn is negative. We choose the next value of m and begin again with n =3.

____________________

### m = 4, n = 3

$2m + 2n - nm = 2$

F = 6, V = 8, E = 12. The cube.

____________________

### m = 4, n = 4

$2m + 2n - nm = 0$

Another stopping point. For values of n greater than or equal to 4, no solutions exist.

____________________

### m = 5, n = 3

$2m + 2n - nm = 1$

F = 12, V = 20, E = 30. The dodecahedron.

____________________

### And that’s it.

And that’s it. Greater values of m and n fail the test. We have exhausted the supply of regular solids. There are five, and no more.