Date: Feb 28, 2013 5:43 AM
Author: David Petry
Subject: Re: Deformable platonic "solids"

On Wednesday, February 27, 2013 11:21:32 AM UTC-8, Frederick Williams wrote:


> Suppose the platonic solids aren't solid at all but are made of rigid
> line segments with completely flexible hinges at the vertices. The cube
> can be flattened into a... um... non cube. The tetrahedron, octahedron
> and icosahedron cannot be deformed at all. But what about the
> dodecahedron, can it be deformed?


The best way to think about it is in terms of "degrees of freedom". Any point in 3-space has three degrees of freedom, so 'n' points have '3n' degrees of freedom. Any line segment joining two points reduces the total number of degrees of freedom of the system of points and line segments by one. So start by fixing the positions of two points joined by a line segments (and hence zero degrees of freedom), multiply the number of remaining points by 3, and subtract the number of line segments. If the result is equal to 1 or less, the system is rigid. Otherwise it is not.