Rich wrote:
A^2 + B^2 + C^2 + 2ABC = 1 [...] The linear step is to fix A and B; then C is the root of a quadratic whose two roots sum to -2AB, whence A,B,C' is a solution, C' = -2AB-C.
And yes indeed, this is a simple geometric evolution. Suppose the vertices of your original semicyclic quadrilateral are P,Q,R,S in that order, and the line PS is the diameter of the circle, which Q and R also lie on. Now replace Q by its reflection through the line PS to get Q', also on the circle. The question is whether the length of Q'R is still rational, and indeed it is, and we can calculate it, using Ptolemy's Theorem. In PQRS we had |PQ||RS| + |QR||SP| = |PR||QS|; here |PR| and |QS| are the lengths of the two diagonals, which are not necessraily rational, but the product is. Replacing Q by Q' reshuffles things so that the old diagonals are now outside edges and the new diagonals are Q'R and the diameter PS, but |PQ|=|PQ'| and |SQ|=|SQ'| since we reflected in PS. Ptolemy now says that |PR||QS| + |PQ||RS| = |PS||Q'R|, and comparing this to the preceding, we get |Q'R| = 2 |PQ||RS| / |PS| + |QR|, exactly Rich's linear solution stepping (except that here the sign manifests geometrically in the bow-tie result). --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.