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.

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About aharmlessdrudge

Way back during the late Bronze age -- actually it was the 1950s -- all of us in high school had to take a vocational test to determine our interests and, supposedly, our future careers. I cannot remember the outcome, but I do recall one question that gave me pause. "If you were to win a Nobel prize, would it be in literature or in physics?" I hesitated over the question: although I enjoyed mathematics and science more than English class, I did have a couple of unfinished (and very bad) novels hidden away at home. I cannot remember what I chose back then, but the dilemma followed me to university, where I switched from mathematics to English and -- after a five-year stint in journalism -- back to mathematics. I recently retired as a professor of statistics. Retirement. What a good chance to revive my literary ambitions. I have finished a novel -- more about that in good time -- and a rubble of drafts of articles about mathematics and statistics is taking up space on my hard disk.
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