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At the end of the Elements Book XIII Euclid gives a demonstration that there are only five regular solids, i.e. 'no other figure, besides the said five figures, can be constructed by equilateral and equiangular figures equal to one another' e.g. here. The demonstration given of this is essentially that any vertex comprised of a number of equal polygons can only be formed by 3, 4 or 5 triangles or 3 squares or 3 pentagons, the reason being that any more of any of these leads to a total angle at the vertex of 360° or more.

It's been pointed out that Euclid also needs the condition of convexity. This was probably assumed by Euclid, not least because much of the previous discussion in Book XIII was about inscribing solids in spheres. The essence of my question is that if we don't assume this then the argument appears potentially to fail, i.e. we can easily have more than 5 triangles meeting at a vertex if it can be nonconvex, e.g. as at one of the inner vertices of a solid similar to the small stellated dodecahedron

Stell dodec

(from Wikipedia) but composed of 60 equilateral triangular faces rather than 60 isosceles triangle (not thinking of it as twelve intersecting pentagrams either).

The standard way to then show that there are only five Platonic solids is using Euler's formula $V-E+F=2$ applied to a polyhedron comprising $n$ $m$-gons, say, and showing that the only cases that work with the formula are those corresponding to the Platonic solids.

I note that without convexity then there are some other possible solids, e.g. the icosahedron with one vertex pushed in such as

dented icosa

(taken from Richeson's book 'Euler's Gem'). In a sense this is just a varied icosahedron, however, and loses the regularity that we probably want from any Platonic solid so I don't think anyone would call it a 'sixth' Platonic solid.

As above with the small stellated dodecahedron, I'm not thinking of cases where faces 'intersect' such as the Kepler-Poinsot polyhedra, thinking of these as better described by a set of smaller polygons, and also note that in such solids we don't have the same number of smaller polygons meeting at each vertex so they aren't candidates for a sixth Platonic solid. I'm also not thinking of cases where there are holes in the solid, i.e. genus not equal to 2, or where a cube might be stuck into the middle of a face etc.

To phrase one definite question then: is there a quick, knock down argument as to why a set of nonconvex vertices formed from more than 6 equilateral triangles, or more than 4 squares or pentagons, as described above, could not form a regular solid, meaning a solid with the same number of the same regular polygons meeting at each vertex?

Of course, it doesn't 'feel' as though this would be possible, and we can justify this with Euler's formula, but Euler's formula was not known to Euclid / requires algebraic manipulation not known at that time - I'm wondering if there is a justification of convexity rather than just assuming it.

Any ideas very welcome (or requests for clarification of what I mean, of course).

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  • $\begingroup$ On the general vagaries found in definitions of "regular polyhedron" (and "polyhedron" and even "polygon"), see works by Branko Grünbaum. In particular, "Are your polyhedra the same as my polyhedra?" (PDF link via washington.edu). $\endgroup$
    – Blue
    Feb 3 '21 at 20:03
  • $\begingroup$ Many thanks - that illustrates well how there is no common definition of regular polyhedron as you say. $\endgroup$
    – John1970
    Feb 4 '21 at 18:39
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If you are demanding that faces do not intersect, then the idea of angular deficit offers your proof.

Count the polygonal angles around each vertex of some convex polyhedron and see how much they fall short of 360 deg. For example the cube has 3 x 90 deg corners so is 90 deg short.

Then count the vertices and multiply that by the corner's deficit. For the cube this is 8 x 90 deg = 720 deg or 2 full turns. This is the polyhedron's angular deficit.

It is the same value for all convex polyhedra. In fact it is the same for all polyhedra which have a simple sphere-like surface, such as your dimpled icosahedron (we say they are topological spheres), regardless of any valleys, dimples, etc. and is directly related to the Euler sum V - E + F = 2.

Any skew regular vertex has an angular sum of 360 deg or more, i.e. an angular excess of 0 or more. You cannot obtain an angular deficit using only skew regular vertices. Hence you cannot obtain the necessary Euler value for your solid.

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    $\begingroup$ You say "A non-convex vertex has an angular sum greater than 360 deg": this is false. Consider the icosahedron with one vertex pushed in, as described in the original post. $\endgroup$ Feb 15 '21 at 16:16
  • $\begingroup$ Thanks. I have corrected that bit. $\endgroup$ Feb 15 '21 at 17:55
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    $\begingroup$ Why should a non-convex regular vertex have a large angle sum? Again, consider the center of the "dimple" in the icosahedron with one vertex pushed in. $\endgroup$ Feb 15 '21 at 18:41
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    $\begingroup$ @RavenclawPrefect You aren't the ghost of Branko Grünbaum, are you? He used to be adept at picking up on these woopsies. Have I got it right this time? $\endgroup$ Feb 15 '21 at 20:04
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    $\begingroup$ I think your current answer rules out things like the Kepler-Poinsot solids, but doesn't show that the dimpled icosahedron fails? (Of course, it does fail, because some of the vertices have inconsistent dihedral angles, but you need some way of showing that these inconsistencies arise.) $\endgroup$ Feb 15 '21 at 20:11
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As per Guy's post, Descartes's observation that the total angular deficit of solids which are homeomorphic to a sphere is 720 deg justifies it not being possible to form such a solid where more than five equilateral triangles meet at each vertex (or more than three squares or pentagons) even though such vertices are possible in polyhedra. To justify the total angular deficit sum being 720 deg needs Euler's formula, or similar reasoning, and so would not have been know to Euclid. An assumption of convexity therefore seems to be needed for Euclid's argument for only five Platonic solids at the end of Book XIII of the Elements to work.

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  • $\begingroup$ Many thanks - I agree with all you say. I voted up Guy's post as giving the answer for this reason. Thank you for adding your confirming of this too. $\endgroup$
    – John1970
    Jun 6 '21 at 14:35

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