My Math Forum Platonic solids - existence proof
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 February 4th, 2011, 11:37 PM #1 Member   Joined: Jun 2009 Posts: 83 Thanks: 1 Platonic solids - existence proof Platonic solids are explained here: http://en.wikipedia.org/wiki/Platonic_solid It can be eassily deduced what the neccessary conditions for the number of vertices, edges and faces are and that only 5 such polyhedrons can exist. But how to prove that such polyhedrons do really exist? I am interested in some "logical" proof. Of course it can be proved geometrically by constructing such polyhedrons but is there another way of proving this without constructing these polyhedrons? Thank you
 February 5th, 2011, 01:15 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Platonic solids - existence proof I look at it as calculating $\sum_{s=3}^\infty\min(0,\lceil\pi/d(s)\rceil-3)$ where $d(s)=\pi-2\pi/s$ is the measure (in radians) of the external angle of an s-sided regular polygon. (You could express it in degrees if you like.) If you wanted to include tilings ("solids" of infinite volume and zero curvature) you'd compute $\sum_{s=3}^\infty\min(0,\lfloor\pi/d(s)\rfloor-2)$ instead which I think would give 7 instead of 5.

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