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)?

  • $\begingroup$ The definition of "Platonic solid" requires that the faces are identical regular polygons, but apparently you're looking for some other more general definition that allows irregular polygonal faces. Why is it that you're not counting the triangular bipyramid which doesn't have regular faces? Here's a picture: upload.wikimedia.org/wikipedia/en/thumb/f/f7/… $\endgroup$ Jul 22 '10 at 2:41
  • $\begingroup$ I'm only considering regular polygons...although it would be interesting to consider others. But that is beyond the scope of my question. I'm reluctant to re-define my question, especially after such a good answer has already been given. $\endgroup$
    – Justin L.
    Jul 22 '10 at 5:07
  • $\begingroup$ Oh, I misread your question completely! I thought you were asking, "Why is it that only regular polygons can be faces of the Platonic solids?". $\endgroup$ Jul 25 '10 at 6:47
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    $\begingroup$ @Ben - I re-read my title and saw the source of your confusion, so I edited it. Hope it helps! $\endgroup$
    – Justin L.
    Jul 25 '10 at 7:39
  • $\begingroup$ "…'tessellate' (is that the proper word in this context?)" You can say they tessellate the sphere. (I wouldn't just say tessellate because that implies tessellating the plane, which the pentagon does not do.) $\endgroup$ Oct 11 '15 at 2:02

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.

  • $\begingroup$ Thanks! That makes a lot of sense actually. If it isn't much of a bother for you, do you mind addressing some of the other things I mentioned? Like why triangles are found in more platonic solids than both of the other shapes combined. $\endgroup$
    – Justin L.
    Jul 22 '10 at 1:23
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    $\begingroup$ Because 6 triangles tile a plane. So 5 triangles can be folded to meet at a vertex and make a convex corner. Same with 4 and 3. 2 triangles would just form a 2-sided flat surface, not a polyhedron. $\endgroup$
    – Jason S
    Jul 22 '10 at 1:35
  • $\begingroup$ I might have phrased my question misleadingly; I meant to ask why there are three triangle-based Platonic Solids, and yet only one Square- and one Pentagon-based Platonic Solids? $\endgroup$
    – Justin L.
    Jul 22 '10 at 1:38
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    $\begingroup$ @Justin: More or less as Jason said, it's because the interior angle of an equilateral triangle is small enough that you can fit 3, 4, or 5 of them around a vertex and still have a total measure less than 2&pi;. I don't, however, readily have an answer to the 4D question. $\endgroup$
    – Isaac
    Jul 22 '10 at 1:40
  • $\begingroup$ and @Jason - My apologies! I misread Jason's first reply to mine. It makes perfect sense now; I'm not sure what I was on. $\endgroup$
    – Justin L.
    Jul 22 '10 at 1:45

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