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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.

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    $\begingroup$ 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.) $\endgroup$ – Christopher Carl Heckman Aug 18 '15 at 21:30
  • $\begingroup$ @CarlHeckman Yes, I see it now. Please make it an answer, so I can vote and close it. $\endgroup$ – HolgerFiedler Aug 18 '15 at 21:36
  • $\begingroup$ This is the tetrakis cuboctahedron. $\endgroup$ – Akiva Weinberger Aug 20 '15 at 15:58
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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.

When you add the 4 smaller triangles, you are adding 12 new vertices. Each of these vertices is adjacent to two of the original vertices, and 4 new vertices, so they have degree 6.

Not all the vertices have the same degree, so it's not a platonic solid. $~\square$

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In this case, each four congruent (regular) triangles are on each of 8 original faces of the octahedron. Since each four congruent triangles do not form any new face (surface) of the solid hence the solid with 32 congruent triangles is originally the octahedron it is not a new platonic solid.

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  • $\begingroup$ There is some misunderstanding. The four new triangles are made originally on the globe and divide the spherical surface of the original triangle. The new four flat triangles are not located on the surface of the bigger one surface. Anyway Carl Heckman shows in his comment, that this solid isn't a Platonic solid. $\endgroup$ – HolgerFiedler Aug 19 '15 at 5:55
  • $\begingroup$ @HolgerFiedler If they're all equilateral, then every four faces are coplanar. In your construction, they don't end up being equilateral. $\endgroup$ – Akiva Weinberger Aug 20 '15 at 16:10

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