Mike Reid writes: << just by chance, i was working on something related to this the other day. the formula for the circumradius of a convex cyclic quadrilateral, with side lengths a , b , c and d is: /------------------------------------------------------------------ / (ab + cd) (ac + bd) (ad + bc) R = \ / ------------------------------------------------------------------ \/ (-a + b + c + d) (a - b + c + d) (a + b - c + d) (a + b + c - d) the formula for the circumradius of a triangle can be obtained by putting d = 0 into this formula.
This is a nice formula! Not having thought about this before, what I find particularly surprising is that the quadrilateral circumradius formula is independent of the cyclic order of the edge-lengths. Is this easy to see a priori ? Is it also true for convex polygons of > 4 sides ? Wait -- consider any 3 consecutive (ccw) vertices A,B,C of a convex n-gon. If the triangle ABC is cut from the n-gon and re-glued so that they now appear in ccw order C,B,A, then it's easy to see that doesn't change the circumcircle (by its symmetry about the perpendicular bisector of AC). Since all permutations of the cyclic ordeer of the edges are generated by ones of this sort, that shows the circumcircle and circumradius are functions purely of the multiset of edge lengths. I think. (If true, I never realized this independence before.) --Dan