Path: utzoo!utgpu!watserv1!watmath!att!occrsh!uokmax!apple!agate!eos!jbm
From: jbm@eos.UUCP (Jeffrey Mulligan)
Newsgroups: comp.graphics
Subject: Re: Why are there only 5 regular convex polyhedra?
Message-ID: <7243@eos.UUCP>
Date: 12 Sep 90 02:13:56 GMT
References: <1990Sep9.095906.26612@rice.edu> <49433@brunix.UUCP> <13811@mentor.cc.purdue.edu>
Organization: NASA Ames Research Center, California
Lines: 64

ahg@mentor.cc.purdue.edu (Allen Braunsdorf) writes:

>In article <49433@brunix.UUCP> dbc@cs.brown.edu (Brook Conner) writes:
>>Dwayne,
>>
>>The Platonic solid are the only regular solids because they are :)
>>Seriously, I'm sure this is a result of some result in topology  somewhere

[ perfectly good enumerative proof omitted ]

>A "real" proof would bog down in the details more, of course, but
>that's how you prove it.  If you use F+V=E+2 you can show it to people
>with less hand-waving and figure-drawing.

I wasn't going to get into this, but the topological proof
is just *so* elegant!

Let us define the following symbols:

	f	number of edges per face (>=3, assumed equal for all faces)
	v	number of edges per vertex (>=3, assumed equal for all vertices)

(The number of vertices per edge, and the number of faces per edge
are both obviously equal to 2).

Just to be sure there's no confusion, let's define the quantities
in Euler's equation (which Allen B. kindly gave us):

	F	total number of faces
	V	total number of vertices
	E	total number of edges

(Euler's equation has a more general form, something like
F+V-E-1= genus , where genus = 1 for a simply connected object,
genus = 2 for a torus or a teacup, genus = 3 for a soup tureen
with two handles, etc.  And a reference that I just consulted tells
me that Schlaefli generalized the equation to structures of arbitrarily
high dimension exactly 100 years after Euler - in 1852!)

The constraints are:

1.	F+V=E+2		(Euler's equation)
2.	V = F*f/v	(number of vertices =
				number of faces * vertices per face
				/ number of times each vertex is counted
3.	E = F*f/2	( number of edges =
				number of faces * edges per face
				/ number of times each edge is counted )

So now all that has to be done is find all positive integer solutions
of these equations!  This can be done simply by enumerating all
pairs of f and v with f,v >= 3, and seeing if positive integer
solutions for V,E and F exist.  The above constraints can be expressed
in a form which is completely symmetric in F and V (and f and v),
so that a solution generated by f=x and v=y implies the existence
of a dual solution with f=y and v=x.

Enumeration of solutions left as an exercise to the reader.

-- 

	Jeff Mulligan (jbm@eos.arc.nasa.gov)
	NASA/Ames Research Ctr., Mail Stop 262-2, Moffet Field CA, 94035
	(415) 604-3745