It depends, then, on whether you consider the condition for the count of faces at a vertex to be the number of faces containing that vertex, or the number of faces overlapping with an arbitrarily-small neighborhood of the vertex. But for the ordinary finite Platonic solids, either definition works, and thus either can validly be considered as an extension of the concept.
Oh, and I missed one in my list of potential “Platonic solids” in post #7: You could also have a single vertex with any number of rays arranged around it at equal angles.
Also, I realized there’s another very natural way to define a Platonic solid, on which a sphere would count as an infinitary instance. We can think of a Platonic solid as a solid with a certain symmetry property: given any two vertices, there’s a symmetry of the solid taking one to the other, and similarly for edges or faces. Indeed, given any two “flags” (a vertex contained in an edge contained in a face), there’s a unique symmetry taking the one to the other.
If we consider a sphere to be like a surface with many infinitesimally small circular faces corresponding to each point, these in turn with infinitesimally small edges corresponding to some direction around that point, and these edges in turn with vertices corresponding to one of two directions along them, then we have perfect symmetry: any two “faces” are related by rotations of the sphere, and indeed, any “flag” in this context has a unique rotational or reflection symmetry of the sphere taking it to any other particular “flag”.
This is actually the way Platonic solids are often generalized in math to higher dimensions: in terms of regular polytopes, “regularity” meaning every flag is taken to every other flag by some symmetry of the polytope. If we remove the ordinary constraint that polytopes be understood as finitary objects, and allow a sphere to be considered a polytope as above, then it is certainly regular in just this way.
Allowing ourself to consider a sphere to be an infinitary polytope at all, in this way, still carries such niggles as I noted above: the vertices in this interpretation arise only from one edge, and the edges themselves only from one face; nothing actually straightforwardly touches anything else.
But I suppose this is just the same niggle as in considering a circle to be an infinitary polygon, with an infinitesimal edge corresponding to each direction, say, and two nominal vertices to the two directions along these edges, but no vertex shared by two edges. Nothing actually straightforwardly touches anything else; there’s infinitely much stuff between.
These difficulties aren’t anything to do with the regularity that’s the most distinct part of Platonic solids; just with considering spheres as having polytopeyness at all in the first place.
The easy way to define the Platonic Solids is “Equal faces, equal edges and equal vertices, with the faces being regular polygons”
Even easier: Dragon Dice
Fun at home project: get 12 pieces of 2x4 and cut the ends to a point, 45º and at 45º … make sure they’re all the same length … nail together … viola … a skeletal cube … perhaps the math gurus can cipher out the angles for the other solids …
The point is for a Platonic Solid, each piece of wood is identical in every way … d20 is 30 pieces, each one exactly like the other … just the nails won’t hold as well …
Sure if you were making the definition you could do that, and you can say its got infinite sides.
But you could say that platonic solids have countable , real number, sides, rather than this “not a number” of infinity. One trouble with infinitity is that the interior angle is 0 and the exterior angle is 180 … OR if they are not, ok, so how far wrong am I ?
Another way we can call the sphere a Platonic solid: A convex solid can also be defined as a tessellation of a sphere, and a Platonic solid can then be defined as a tessellation of a sphere where all faces, edges, and vertices are respectively identical. Well, a sphere is definitely a tessellation of a sphere, and all one of its faces are identical, as are all zero of its vertices and edges.
By this definition, also qualifying would be the pair of hemispheres (with two faces, one edge, and zero vertices), and an infinite family of wedge-slicings (each with n faces, n edges, and two vertices, for any n >= 3).
Perhaps we should consider the sphere (with 2 polar points, 1 meridian edge, and 1 all-encompassing self-bordering face) to be the n = 1 case of the hosohedron.
“Platonic solids” as a term has the connotations of the traditional 5, but anytime I have reason to think about the mathematical concept (regular spherical polytopes/finite (2, m, n) triangle groups), I do consider hosohedra and dihedra along with the rest [though normally in a way which excludes the n = 1 case]. I also normally think of even the classical Platonic solids not in terms of flat pointy objects, but in terms of spherical tilings (and these, in turn, in terms of suitably structured groups, as before).
In that case, is there any reason the points need to be polar (or, in general, antipodal)? Couldn’t you just as well pick any two arbitrary points on the sphere and call them vertices, and draw an edge along either of the geodesics between those two points?
The simplest answer is that a sphere is not a polyhedron. More interesting question: what is a polyhedron? Part of the answer is that all faces are polygons. But there is more (not to mention, what’s a polygon?) There is a wonderful book by Imre Lakatos called Proofs and Refutations that discusses this at great length, in connection with Euler’s formula V - E + F = 2. But you always need finitely many vertices, edges, and faces.
If you want to expand the definition, go ahead, but understand that you are changing the definition–and there should be a payoff in doing so, which there probably is not in this case.
I suppose you could; I just took it that way as limiting case of the other hosohedra (for which it is important to use polar opposite points, or else the symmetry is upset).
I normally don’t think of Platonic solids in terms of metric properties anyway, just topological ones (but any topologically regular polytope can straightforwardly be realized by a metrically regular one in a space of the right curvature).
I don’t quite grasp why this wasn’t the end of the discussion. The smaller the circle, the closer the situation is to the plane, and in the plane the smallest proportion of “gap” to cover when using equally sized circles is when you have three circles touching. You can use infinitely small circles if you want, but there will always be almost 10 % gap.
Though I guess that means you’d get a smaller and smaller proportion of gaps if you try to tessellate a sphere using very tiny equilateral triangles … Did the real mathematicians switch away from using tiny circles without me noticing?
What do circles have to do with tessellation? Unless you mean the Apollonian circles mentioned in passing above?
The thing is, unlike the two-dimensional case where there exist regular polygons with an arbitrarily large number of sides, the only tessellations of a sphere by equilateral triangles are obtained from the tetrahedron, octahedron, and icosahedron, as in Euclid. It is not a limit of Platonic solids in that sense.
The post I quoted, and a lot of the following posts, were focusing on covering the sphere with tiny circles. I used tessellation because it seemed obvious what that meant in the context of the discussion.
And in the context of my description of the gaps I assumed it was obvious I was thinking of taking a sphere of radius r, covering it with equilateral triangles of side length 1/x as efficiently as possible, and guessing that the optimal configuration will, on average, increase the percentage of the sphere covered as x increases.
OK, it is not too hard to cover a fixed percentage of a sphere with small equilateral triangles, but could you please explain or give a hint how to do it so that the percentage approaches 100% as the equilateral triangles get smaller?
Okay, this took a while, as I wasn’t sure of the correctness of my initial approaches and wasn’t sure what “we” thought of as the requirements. Do all the triangles have to touch the inscribed sphere? Do they have to touch in the center of the triangle?
I think I’ve found an approach that fulfills both those criteria.
From any polygon n, where n is an odd number you can create a band of equilateral triangles. Take each edge, make a triangle at ninety degrees to the plane of the polygon. All the top points then make up another polygon in a parallel plane, turned half the smallest rotational symmetry angle.
Now it seems correct to me, but I could be wrong, that if you then add a band at 90 degrees to the first one, and at 45 and 135, and at 22.5, 67.5, 112.5 and 157.5 and so on, removing whatever much is necessary to avoid overlap when crossing the original equator and all previous bands, and stopping when the gaps between bands are smaller than the height of the triangles, you get a system where the proportion of cover to gaps decreases on average as n increases.
I think I understand what you’re saying, but some parts of your phrasing are puzzling. It sounds like you’re saying to make such bands along meridian lines, taking each band as close to the poles as it can get without overlapping other bands. But what do you mean by “From any polygon n, where n is an odd number…”?
I’m also not convinced that this method (if that is indeed what you’re proposing) would approach total coverage in the limit of small triangles. If your triangle size is chosen such that there are no gaps between bands at the equator, then as soon as you get off of the equator, half of your bands will be excluded, and you’ll have wedge-shaped gaps extending all the way up (and down) to ±60º. And at the same point where those gaps close, you need to end even more bands, and so have another range of latitudes with gaps, and so on. I think that this actually works out to 50% coverage overall.
