dan hoey:
By considering the point at infinity, I understand this to mean that if we have points 01234567 in cocircular quartets 0123, 4567, 0145, 2367, and 0246, then quartet 1357 must also be cocircular. I didn't
in fact this is easy to see directly. the three angles at each vertex, suitably interpreted, add to 2 pi . use this relation at vertices 0 , 3 , 5 and 6 , along with the fact that opposite angles of a cyclic quadrilateral add to pi to see that if five of those quadrilaterals are cyclic, then so is the sixth. alternatively, use the cross ratio, which keeps track of lengths as well as angles. if k(a, b, c, d) = (a - b)(c - d) / ((c - b)(a - d)) is the usual cross ratio, then for eight distinct complex numbers, we have k(a,b,c,d) k(a,e,f,b) k(a,d,h,e) k(g,f,e,h) k(g,c,b,f) k(g,h,d,c) = 1 (the cross ratio of four distinct complex numbers is real iff the four points are concyclic.) the cross ratio is nice in that it automatically handles the "suitable interpretation".
Are all such patterns known? Say, given a mapping of a hypergraph to the plane, where vertices are mapped to points, and the images of the vertices of each hyperedge are cocircular, can we tell what other subsets of vertices are forced to map to cocircular points?
interesting question! i might have speculated that any such theorem (that a particular pattern holds) could be deduced by looking at the angles. but i don't see that conway's configuration (see below) can be. maybe i just didn't see it. the question of determining all such patterns seems ambitious to me. perhaps first we should try to find a collection of patterns from which all others can be built up. i'll give the first of these: if two of the quadruples ABCD , BCDE , CDEA , DEAB , EABC are concyclic, then so are the remaining three. (one needs to assume that the five points are distinct.) these extra conditions to avoid degenerate cases are annoying, and perhaps their complexity is not bounded over all such configurations. the condition above, that the five points be distinct, is not unreasonable. but conway's configuration has a different kind of restriction. john conway: ) For example, if all but one of the quadrilaterals ) AaBb,AaCc,AaDd,BbCc,BbDd,CcDd are cyclic, then so is ) that last one. there must be an extra condition required here. the condition that five of the six quadrilaterals are cyclic, and their circumcircles are all distinct, is sufficient. perhaps there's a more elegant restriction, but that seems like the most natural. (one could say that at least four of the five circumcircles are distinct, but that would imply that all five are.) mike