Tuesday, 18 February 2014

Using Euler's Gem to find Platonic Solids

I was recently watching a video by Norm Wildberger of New South Wales University on Algebraic Topology and he did a nice explanation with algebra of why there can only be the five platonic solids we know using Euler's beautiful V+F-E=2 for closed polyhedra.  It was simple enough (as was much of this "beginners" course) for a bright high school student.
I was struck by the fact that the only explanations I had ever seen, and given, was the geometric argument that went something like, "Well, six hexagons will lie flat, so they can't form a solid, and n-gons bigger than that won't fit at all."  It always included a little more hand-waving than I was comfortable with, but it was what I knew.

Since many teachers and students might not choose to stumble onto a course called Algebraic Topology, I wanted to replicate his explanation. 

He begins by defining the Regular polyhedra with the common statements then proceeds to his proof:
 We let F, E, and V represent the number of Faces, Edges and Vertices as usual.

Let n = Number of edges (and vertices) on each regular polygonal face
Let m = Number of edges (or faces) that meet at each vertex
 and both of these must be greater than or equal to 3.

The number of edges will be \( E = \frac{n*F}{2} \) since every edge is counted twice when counting all the faces.
The number of vertices will be \(V= \frac{n*F}{m} \) since each vertex is counted m times when counting all faces.

Replacing V and E in Eulers formula with these expressions on F we have

V +  F - E = 2    becomes  \( \frac{n*F}{m} + F - \frac{n*F}{2} =2 \)

Now if we solve for F we get \(F=\frac{4m}{2m + 2n - mn} \)

If we require that the number of faces is a positive integer, then the denominator, 2m+ 2n-mn must be greater than zero.  From this we can extract \(2n + 2m > mn \)  and from this \(2n  >  mn - 2m \) and then to \( 2n > m (n-2) \) 

Finally we can divide by n-2 to get \(\frac{2n}{n-2} > m \)  but calling on the fact that both m & n are greater than or equal to 3, we can add \(\frac{2n}{n-2} > m \ge 3 \)  and applying the transitive law of inequalities, we can get \(\frac{2n}{n-2} >  3 \).

This becomes\( 2n > 3n-6 \) and the trivial algebra brings us to  \(n <  6 \)  so there can be no Platonic solids with faces with more than five sides.

Returning for a moment's glance at  \( 2n + 2m > mn \) we see that it is symmetrical in m & n, and it must also be true that \(m <  6\) .  m and n are now limited to the possible sets n={3,4,5} and m={3,4,5}  (I find this use of appeal to symmetry to tie m and n together as one of the better reasons to show this to students.)

If we return now to  \(F=\frac{4m}{2m + 2n - mn} \) and try different values for n and m from the set nXm  we will see that the only possible solutions are the five we have.
Letting n=3 ( regular triangle faces  ) we get  \( \frac{4m}{2m+6-3m} = \frac{4m}{6-m}\)   and testing the values m= 3,4,5 we get 12/3 = 4, 16/2 = 8, and 20/1 = 20 faces, the tetrahedron, octahedron, and icosahedron are all that are possible with triangular faces.

If we use n=4 we get  \( \frac{4m}{2m+8-4m} = \frac{4m}{8-2m}\) since m can only be in the range 3,4,5, only 3 will produce a positive integer in the denominator, so we must have 12/2 = 6 faces, or a cube as the only possibility.

Returning to check n=5 we have \( \frac{4m}{2m+10-5m} = \frac{4m}{10-3m}\ ) only m=3 produces a positive integer in the denominator, and so there must be 12 pentagonal faces.

I think this would be a wonderful addition to a presentation of the Platonic solids because it utilizes a beautiful theorem, but in a way that provides support for the beautiful way that geometry and algebra compliment each other (and perhaps the mathematical spelling "complement" is even more meaningful here).  It is important to point out to students that the algebra tells us that these are all the possible  Platonic solids.  We must still construct them. 


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