# Is a solid from 32 congruent regular triangles a Platonic solid?

Imagine a ball (a globe) divided by three circles, rectangular to each over. If one smooth the surface between the circles one get 8 triangles and this is the platonic solid octahedron.

Going back to the globe it seems possible to divide each of the eight curved surfaces into four congruent triangles, again using circles for this. If one smooth the surfaces again, we get 8 * 4 triangles and the solid satisfies the definition of Platonic solids.

• Not all the vertices have the same "degree", though. Some have 4, and some have 6. This is thus not a Platonic solid. (There are in fact only five Platonic solids; this has been proven.) – Christopher Carl Heckman Aug 18 '15 at 21:30
• @CarlHeckman Yes, I see it now. Please make it an answer, so I can vote and close it. – HolgerFiedler Aug 18 '15 at 21:36
• This is the tetrakis cuboctahedron. – Akiva Weinberger Aug 20 '15 at 15:58

More precisely: If $v$ is one of the vertices you get when dividing the sphere into 8 equal parts, $v$ will be adjacent to exactly 4 other vertices (its "degree" is 4). Dividing the triangular regions into 4 smaller triangular regions will not change the degrees of these vertices.
Not all the vertices have the same degree, so it's not a platonic solid. $~\square$