.. the hyperbolic plane ... ha[s] ideal (geodesic) polygons, all of whose vertices lie on "the circle at infinity"... When are two such ideal hyperbolic N-gons isometric (i.e., congruent...)? --Dan
Two ideal hyperbolic quadrilaterals are congruent iff they have equal angles between their diagonals. - Scott
I think Scott's observation is it. Ideal triangles are always congruent. If N>3: Given an N-sided ideal hyperbolic polygon, take any four consecutive vertices A B C D, and consider the diagonals AC and BD. Call the point of intersection E, and find the angle BEC. Associate that angle with the polygon-side BC. There are as many such angles as sides: N of them. Clearly, if two such N-gons are congruent, the associated angles are the same. It is less obvious that if two such N-gons are incongruent, some of the associated angles are different. But if it's true for 4-gons, one should be able to construct an inductive proof on the number of sides. Contrapositively, if the sequences of associated angles are the same [different], the polygons are congruent [incongruent]. ----- A vertex of an ideal N-gon can be represented as its direction (in some co-ordinate system) from the origin. Let A, B, C, D be the directions of four consecutive vertices. Then the angle X (associated with BC) is cos [(D-B)/2] cos [(C-A)/2] - cos [(D-C+B-A)/2] cos X = ----------------------------------------------- sin [(D-B)/2] sin [(C-A)/2] (Can someone double-check?) ----- When are the sequences of associated angles the same? Here are two such sequences (not necessarily in radians :-) 2 6 4 8 2 9 3 8 4 6 2 3 9 2 They might took different, but they're really the same. (In the second, start with the first '2', and go backwards.) Is there an O(n) algorithm for cyclically comparing two sequences of numbers? -- Don Reble djr@nk.ca