Points on a cube

Question

Professor Halfbrain has spent his entire weekend by placing colored dots on the surface of a huge wooden cube. His objective was to find large groups of dots that form the vertices of a regular polygon (= a convex polygon whose angles all are the same and whose edges all have the same length).

The professor managed to prove the following two deep theorems.

Professor Halfbrain's first theorem: There exist three points on the surface of a cube that form the vertices of a regular polygon and that do not all lie on the same face.

Professor Halfbrain's second theorem: There is no group of 37 points on the surface of a cube that form the vertices of a regular polygon and that do not all lie on the same face.

This puzzle asks you to improve the two theorems of professor Halfbrain and to make them even deeper. Find an integer $x$, so that "three points" in the first theorem may be replaced by "$x$ points", and so that "37 points" in the second theorem may be replaced by "$x+1$ points" (again yielding true statements, of course).

Answer 1

You can get

a 12-gon

This can be done by:

You can slice a cube to get a regular hexagon

Then, inscribe a regular 12-gon in the hexagon. This can be done by extending every other edge of the 12-gon until they meet in a hexagon.

This is optimal because:

Consider the plane that the polygon lies in. The intersection of the cube and the slice forms a polygon that contains at most 6 edges, one per face in the cube. The regular polygon must lie in this polygon, and must have at most to vertices on each edge, since otherwise an edge contains three consecutive vertices, which is impossible as there are in a line. So, the maximum number of vertices of the polygon is 6*2=12.