Re: Polyhedra with congruent faces

On May 2, 1:29 pm, Thomas Mautsch <maut...@xxxxxxx> wrote:

[snip]


I posted a (somehow) similar problem a year ago under the title
"The Impossible Hexahedron" <news:443af7f3@xxxxxxxxxxxxx> -
to find a (convex) polyhedron combinatorially equivalent to the cube
such that all six faces have side lengths a,b,c,d (in changing orders):

b
+-----------------+
|\ /|
| \ / |
| \a c/ |
| \ / |
| \ d / |
| +-----+ |
| | | |
c| b| |a |d
| | | |
| +-----+ |
| / c \ |
| / \ |
| /d b\ |
| / \ |
|/ a \|
+-----------------+

The only examples I received
(from Dave Rusin in <news:e1kia4$eln$1@xxxxxxxxxxxxxxxxx> )
were weird-looking ones
with non-convex faces and intersecting edges.


Try this one:

http://img514.imageshack.us/img514/3264/impossiblehexahedrondy8.gif

This was calculated numerically, so I don't have exact analytic
expressions for the lengths and angles. I've only shown numerically
that it comes right to about 14 d.p. (unless I goofed of course, which
has been known!).


.

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