By that I mean I had a weird brain fart. A weird and wrong brain fart.
I think you’re right. :smack:
“Tessellating” a sphere with same sized equilateral triangles is just not going to work. Look at the section of the Wikipedia article where it talks about angle deficit.
A vertex is either going to have 5 or less triangles meeting there and the surface angle is fixed and will never approach 0. Or it has 6 triangles meeting and the surface is perfectly flat. Not locally flat, but flat for the entire object.
The later could be used to argue something weird about the plane being a Platonic solid. (Did I just type that?)
You could give up on having the vertices of the triangles not meet in general and hope something nice happens in the limit, but then the edges won’t line up right, etc.
You can never jam together more than 5 at a vertex, but the “surface angle” (actually I am thinking of spherical triangles) will approach zero as the triangles get smaller. Equilateral triangles of different sizes on a sphere have different angles…
Limit of what, though? An explicit construction would be one way to solve the problem, assuming it is possible.
Well, yes, the plane is a Platonic solid, on all the most natural generalizations beyond the traditional concept. If there’s to be any generalizing at all, you’ll soon want to include the plane. Basically all the mathematics of Platonic solids (e.g., the theory of Coxeter/triangle groups) applies to regular planar tilings as well; indeed, the theory of Platonic solids might as well be the theory of planar tilings, just for different kinds of “plane”.
In more detail, to each pair (n, m) of integers > 1, there is a freely generated regular tiling of n-gons, m of which meet at each vertex; all that changes is the curvature of the space it lives in. If 1/n + 1/m > 1/2, the angles of the faces around a vertex sum to less than 360 degrees, and it lives in elliptic space (i.e., as a tessellation of a sphere, wrapping around on itself finitely; i.e., corresponds to a familiar sort of solid); if 1/n + 1/m = 1/2, the angles of the faces around a vertex sum to precisely 360 degrees, and it lives in Euclidean space (i.e., as a tessellation of the infinite Euclidean plane, and can also be quotiented into a finite torus at various scales); and if if 1/n + 1/m < 1/2, the angles of the faces around a vertex sum to more than 360 degrees, and it lives in hyperbolic space (i.e., as a tessellation of the infinite hyperbolic plane, and can also be quotiented into various exotic finite structures).
Yep, you could even have 7 or more equilateral triangles meeting at each vertex; it just takes you into a hyperbolic plane (much like having 5 or less takes you out of the familiar Euclidean plane and into an elliptic plane; i.e., sphere). See Order-7 triangular tiling - Wikipedia.
(Well, strictly, the sphere double covers the elliptic plane and is not identical to it, but putting that niggle to the side…)
The phrase “spherical triangles” takes you completely out of the topic of Platonic solids. Those involve flat regular polygons only.
The phrase “of different sizes” takes you completely out of the topic of Platonic solids. Those involve identical polygons only.
More precisely, the plane is three different Platonic solids, one each with triangular, square, and hexagonal faces.
I wouldn’t say completely out of the topic; any classical Platonic solid can be inscribed in a sphere and is inextricably related to a regular tessellation of the sphere (just join the vertices by spherical geodesics rather than in Euclidean space). You should be amenable to this point of view if you want the plane to be a Platonic solid as well, and also regular tessellations of the hyperbolic plane. There is also something to be said for everything happening on the sphere rather than in some ambient space.
True, the packing problem is a bit of a sidetrack, but not because of polygons of different sizes. Consider a sphere of radius r, and fix a number x. Now try to pack the sphere with many identical equilateral triangles of area 4πr[sup]2[/sup]/x.
How many can you fit?
Good point!