Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Topic: Deformable platonic "solids"
Replies: 23   Last Post: Mar 12, 2013 8:11 PM

 Messages: [ Previous | Next ]
 David Petry Posts: 1,104 Registered: 12/8/04
Re: Deformable platonic "solids"
Posted: Feb 28, 2013 1:44 PM

On Thursday, February 28, 2013 4:48:04 AM UTC-8, Richard Tobin wrote:
> david petry <david_lawrence_petry@yahoo.com> wrote:

> >Any line segment joining two points reduces the
> >total number of degrees of freedom of the system of points and line
> >segments by one.

> Not in general. Consider a deformable solid with a square face.
> Joining two opposite corners will make that square rigid. Joining
> the other two will have no further effect, while adding a line
> somewhere else in the solid may.

Yes, of course, if the degrees of freedom are already minimal, they can't be reduced further. But the answer I gave does show us how to answer the original question.

"THEOREM" If a platonic solid has V vertices and E edges, then it will be rigid in the sense of Frederick Williams if and only if 3V - E = 6.

Examples

Tetrahedron: V = 4, E = 6, 3V-E = 6 (rigid)
Octahedron: V = 6, E = 12, 3V-E = 6 (rigid)
Cube: V = 8, E = 12, 3V-E = 12 (not rigid)
Dodecahedron: V = 20, E = 30 3V-E = 30 (not rigid)
Icosahedron: V = 12, E = 30, 3V-E = 6 (rigid)