In the Poincaré model, the hyperbolic plane H is viewed as an open unit disk D in the (ordinary) plane, with the geodesics being any arc of a circle that's perpendicular to the circular boundary bd(D). For any geodesic polygon in H, its area is A = [sum of its exterior angles] - 2 pi. One can also have ideal (geodesic) polygons, all of whose vertices lie on "the circle at infinity", i.e., bd(D). Since each exterior angle of such a polygon must be pi, its area is a function only of the number of sides N: A = (N-2) pi. Also, all the sides of an ideal polygon are of infinite length. QUESTION: When are two such ideal hyperbolic N-gons isometric (i.e., congruent in H) ? Note that the space of all such N-gons is N-dimensional (depending on the vertices on bd(D), whereas the group of isometries of H is the same as the group of conformal (or anti-conformal) automorphisms of D -- a 3-dimensional group. Hence there are N-3 dimensions of congruence classes of such ideal N-gons.
From this it's easy to see that all ideal triangles in H are congruent. Given the 4 vertices on bd(D) of each of two ideal squares in H, what's a concise algorithm to determine whether the two squares are congruent?
--Dan