Polyhedron

[I.N. Galidakis]

I am trying to visualize what looks like a particular polyhedron(?), but I am
having a bout of stupidity and cannot figure something simple about it.

I am describing the shape, section-wise:

Consider the unit circle on the xy-plane. Put vertexes at exp(2*k*Pi/8*i), k \in
{0,1,...,7}

Rotate the unit circle around the x-axis by Pi/4.

Consider the new unit circle on the xy-plane. Put again vertexes on the new unit
circle at exp(2*k*Pi/8*i), k \in {0,1,...,7} (vertex duplicity allowed).

Rotate the new unit circle around the x-axis by Pi/4 (this will make the first
unit circle perpendicular to the xy-plane).

Repeat 7 times, until the first unit circle rotates back to its initial
position.

What is the name of the resultant polyhedron? It has a total of 26 vertexes.

If anyone has any online refs of the resultant polyhedron or can cook up some
Maple code to visualize this polyhedron (preferably with rotational animation),
I'd appreciate it.

Many thanks,
--
I.N. Galidakis

[victor_me...@yahoo.co.uk]

On 6 Oct, 11:53, "I.N. Galidakis" <morph...@olympus.mons> wrote:

> What is the name of the resultant polyhedron? It has a total of 26 vertexes.

I doubt it would have a special name.

> If anyone has any online refs of the resultant polyhedron or can cook up some
> Maple code to visualize this polyhedron (preferably with rotational animation),
> I'd appreciate it.

That's beyond my ability, but consider this description.
It has the north and south poles as vertices. It has eight vertices,
equally spaced, on the equator. These eight vertices define eight
meridians. The remaining 16 vertices lie two on each of these
meridians,
at 45 degrees north and south. Thus the polyhedron has 56 edges and 32
faces. Sixteen of the faces are congruent isosceles triangles, with
one
vertex at a pole. The other sixteen are congruent isosceles trapezia.

Victor Meldrew
"I don't believe it!"

[Robert Israel]

"I.N. Galidakis" <morp...@olympus.mons> writes:

plot3d(1, theta=0..2*Pi,phi=0..Pi,coords=spherical,grid=[9,5]);
P:= %:
plots[display]([seq(plottools[rotate](P, 0, 0, 10*i),
i=0..35)],insequence=true);

<http://www.math.ubc.ca/~israel/problems/poly.gif>

> Many thanks,
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

[I.N. Galidakis]

Many thanks to you and Victor.