# All the 2-d polygons that can be used to form 3-d polyhedrons

The requirement is that we start with a 2-d polygon and using just copies of this polygon, cobble them together into a closed 3-d polyhedron (no gaps). I want to find a way to identify all such 2-d polygons.

Here are the ones I'm aware of -

All the platonic solids obviously satisfy this requirement. So, the equilateral triangle which can be used to form the Tetrahedron, Octahedron and Icosahedron. Then there is the square which can be used to form the Cube and finally the regular pentagon which can be used to form the Dodecahedron.

Apart from these, scalene triangles can also be used to form corresponding Tetrahedra and Icosahedra in the same way as equilateral triangles.

Any quadrilateral that has two consecutive sides equal can be used to form a Hexahedron and any pentagon that has two consecutive sides equal to (say) a and the next two consecutive sides equal to (say) b, leaving the third side free can be used to form a Tetartoid.

Some of these pentagons have internal angles greater than 360/3 = 120 degrees.

So, I'm not sure we can rule out some sort of irregular hexagons that can be used in this way as well.

What I have here is a list with no obvious way to proceed in terms of verifying that it is exhaustive. Can someone give me a hint as to how I can make progress?

• Hexagons won't do, unless you want a polyhedron with holes. Nov 20 '18 at 12:18
• Sure, but how can we prove that? Nov 20 '18 at 19:16
• You may want to research "isohedra".
– Blue
Nov 20 '18 at 22:56

Euler says that for convex polyhedra $$V-E+F=2$$. Now if every face is a hexagon, by counting all edges in all faces we get $$6F$$, but in doing so we've counted every edge twice, hence $$E=3F$$. By similar reasoning, if three faces meet at each vertex, we have $$V=2F$$. Hence $$V-E+F=0$$ and not 2, so no luck.