ahg@mentor.cc.purdue.edu (Allen Braunsdorf) (09/11/90)
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 >(although having only a passing acquaintance with topology stemming from >working under a topologist (John Hughes of Foley, van Dam, Feiner, and Hughes) >I have yet to see this proof for myself, so I can't offer pointers to it) I've had to prove it before and it goes like this: A regular polyhedron is one that has some number of (identical) regular polygons for faces. The 5 Platonic solids are the only regular convex polyhedra. Which regular polygons can we use to build a regular polyhedron? To answer this, we must notice that at each vertex of the polyhedron, some number of polygons greater than two share a common vertex. This combination of polygons can't be flat. In fact, the sum of the (identical) angles of the polygons at the common vertex must be less than 360 degrees. This gives us a table of candidates: Sides Angle Number we can join at one vertex ----- ----- ------ -- --- ---- -- --- ------ 3 60 3, 4, or 5 4 90 3 5 108 3 6 120 can't! (3 would be flat, more won't fit) A regular polygon can't have fewer than 3 sides. If you have anything with more sides than a hexagon, things only get worse, so they don't work either. So the only polygons we can use to build out polyhedron are the triangle, square, and pentagon. Combining those polygons in the ways described in the last table gives us: Polygon Number we join at one vertex Polyhedron ------- ------ -- ---- -- --- ------ ---------- Triangle 3 Tetrahedron Triangle 4 Octahedron Triangle 5 Icosahedron Square 3 Cube Pentagon 3 Dodecahedron 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. If you really want more, I can 'splain it better. --- Allen Braunsdorf Purdue University Computing Center ahg@cc.purdue.edu UNIX Systems Programmer
jbm@eos.UUCP (Jeffrey Mulligan) (09/12/90)
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