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August 6[edit]

Great circles[edit]

If a number of great circles are drawn onto the surface of a sphere then the surface is partioned into a number of shapes, which may be spherical triangles or other spherical polygons. My question is whether, given any arbitrary number of great circles, configurations in which all partitions are spherical triangles are abundant or rare (or very rare). I can see that there is an infinite family of equator plus arbitrary number of circles through the poles. I know there are a few other individual cases. What I'm not sure about is whether there are other infinite families, or many more (even infinitely more) individual cases, or whether the cases very quickly run out as is the case, say, with regular polyhedra. Anybody have any thoughts on this? 109.156.50.162 (talk) 00:57, 6 August 2015 (UTC)[reply]

With ~~2 or~~ 3 great circles (not all passing through the same 2 poles), I believe nothing is possible except spherical triangles. With 4 or more, it seems to require quite a rare event, where three or more great circles share two common "poles". StuRat (talk) 01:13, 6 August 2015 (UTC)[reply]

I think with 2 great circles you get 4 digons or lunes, not triangles. --RDBury (talk) 06:50, 6 August 2015 (UTC)[reply]

Agreed. Comment updated. StuRat (talk) 20:54, 7 August 2015 (UTC)[reply]

If you look at Spherical polyhedron there are a few, such as those with vertex figure 4.2q.2p, that have triangular faces and edges that lie on great circles. There are only three of them but you could perhaps generate more by adding more great circles that pass through vertices. I don’t know if any other named polyhedra are suitable, or even form a series. There are maybe more at commons:Category:Spherical_polyhedra which could be used.--JohnBlackburne^words_deeds 01:31, 6 August 2015 (UTC)[reply]

It's not clear that the OP is talking about regular configurations, but for n≥3 you essentially get polyhedra and since the regular polyhedra peter out these would as well if you only count regular ones. In general position you can assume that no more than two of the circles meet at any point and that determines the number of vertices, edges and faces for n≥2 circles as V = n(n-1), E = 2n(n-1), F = 2+n(n-1) and the average number of sides per face is 4-8/(n²-n+2). This is <4 so there must be at least one triangle for n≥3. If n>3 then the average is >3 so there must be at least one non-triangle. In other words a configuration with all triangles requires that at least three of the circles meet at the same point, which is unlikely for random circles. --RDBury (talk) 06:50, 6 August 2015 (UTC)[reply]

I am not talking (only) about regular configurations. I am also not talking about random configurations, but ones that are specially devised or selected to have the desired property. 109.156.50.162 (talk) 11:40, 6 August 2015 (UTC)[reply]

Other than the case of n meridians all cut by a single great circle (a degenerate configuration), are there any examples where the configuration is not regular? I cannot think of any, and it seems very likely that any non-degenerate configuration of great circles that forms only triangles is necessarily regular. Sławomir
Biały 14:06, 11 August 2015 (UTC)[reply]