What property of certain regular polygons allows them to be faces of the Platonic Solids? It appears to me that only Triangles, Squares, and Pentagons are able to "tessellate" (is that the proper word in this context?) to become regular 3D convex polytopes.
What property of those regular polygons themselves allow them to faces of regular convex polyhedron?  Is it something in their angles?  Their number of sides?
Also, why are there more Triangle-based Platonic Solids (three) than Square- and Pentagon- based ones? (one each)
Similarly, is this the same property that allows certain Platonic Solids to be used as "faces" of regular polychoron (4D polytopes)?
 A: The regular polygons that form the Platonic solids are those for which the measure of the interior angles, say α for convenience, is such that $3\alpha<2\pi$ (360°) so that three (or more) of the polygons can be assembled around a vertex of the solid.
Regular (equilateral) triangles have interior angles of measure $\frac{\pi}{3}$ (60°), so they can be assembled 3, 4, or 5 at a vertex ($3\cdot\frac{\pi}{3}<2\pi$, $4\cdot\frac{\pi}{3}<2\pi$, $5\cdot\frac{\pi}{3}<2\pi$), but not 6 ($6\cdot\frac{\pi}{3}=2\pi$--they tesselate the plane).
Regular quadrilaterals (squares) have interior angles of measure $\frac{\pi}{2}$ (90°), so they can be assembled 3 at a vertex ($3\cdot\frac{\pi}{2}<2\pi$), but not 4 ($4\cdot\frac{\pi}{2}=2\pi$--they tesselate the plane).
Regular pentagons have interior angles of measure $\frac{3\pi}{5}$ (108°), so they can be assembled 3 at a vertex ($3\cdot\frac{3\pi}{5}<2\pi$), but not 4 ($4\cdot\frac{3\pi}{5}>2\pi$).
Regular hexagons have interior angles of measure $\frac{2\pi}{3}$ (120°), so they cannot be assembled 3 at a vertex ($3\cdot\frac{2\pi}{3}=2\pi$--they tesselate the plane).
Any other regular polygon will have larger interior angles, so cannot be assembled into a regular solid.
