Rotate, if rotated is repetition, flip 2, rotate ?

Very good, MCarreira. But what is the natural linear order of this set of permutations?If I'm a little Rogers Veraarts bug who wants to visit all permutations with the least amount of walking, then I might choose: ABC, ACB, CAB, CBA, BCA, BAC. Is that better? If the bug is forced to walk the entire triangle (e.g., ABC requires the walk A->B->C->A) then things are a little different. Or one might consider the last vertex visited as automatically being the first vertex of the next permutation.

Yes, I usually sort them too. But in this case, I choose rotation for the first three (ABD,BCA,CAB), a reflection of the triangle (from ABC to ACB), and then rotations for the last three (ACB, CBA, BAC). That seemed more elegant in this geometric context because the permutations of the triangle could be defined by geometric operators of rotation and reflection.Here's how I got 8 instead of 4. Ignoring the permutation aspect (assuming ABC==BCA==CAB==ACB==CBA==BAC), I get 8 different spherical triangles because each AB, AC and BC edge can be independently chosen as either the shortest great-circle distance or the longest great-circle distance. Admittedly, this creates some strange spherical triangles.Come to think, some of the 8 figures constructed in this way are probably not valid because if you have more than one of the long great-circle edges, they will intersect on the backside of the sphere. So we are now down to 4 triangles. And yet every one of the 4 has a complementary spherical triangle. For example triangle ABC with angles a, b, and c has a spherical complement triangle with angles 2π-a, 2π-b, and 2π-c. So we're back to 8, again.

like this?

Yes, the long complements of the short segments do intersect at antipodal locations to the original city locations.

Only 1 what? The long complement of segment AB passes through antipode of A which is located at the intersection of the long complement of segment AB and the long complement of segment AC. The long complement of segment AB also passes through antipode of B which is located at the intersection of the long complement of segment AB and the long complement of segment BC.

And yet it satisfies the definition of a spherical triangle. Admittedly, it's not a nice little convex spherical triangle but it is "a figure formed on the surface of a sphere by three great circular arcs intersecting pairwise in three vertices"

I guess one way to look at these abherent spherical triangles is to consider them from a topological viewpoint. If we start with a nice "normal" spherical triangle and then smoothly shift the cities, we can create these other types of triangles. The coloured triangular area can cover the sphere in ways that don't seem triangular to a someone used to planar triangles but which do follow more technical definitions of a spherical triangle.