Andy Latto <andy.latto@pobox.com> wrote:

So there being no discussion on my fallacious proof other than discussion of rot13, I figured it was time to post what's wrong with the proof and the correct solution.

As you saw, I did post my response about an hour before you posted that.

The problem is that even if the set of N points is in general position, the set of N(N-1)/2 perpendicular bisectors is not! In particular, given any 3 of the points, A, B, and C, the three perpendicular bisectors these generate all meet in a point.

Right.

To look at it another way, the perpendicular bisector of AB divides the plane into the "A before B" and "B before A" half-planes, but if you look at the 7 regions generated by 3 lines in the plane that don't intersect in a point, the bounded triangular region is the set of points that are closer to A than B, closer to B than C, and closer to C than A, and that doesn't exist.

I had realized that the bisectors meet at a point, but hadn't thought of what it would mean if they didn't. Your non-transitive ordering is a very intuitive reductio-ad-absurdum proof that perpendicular bisectors of a triangle meet at a point, one that makes it obvious that this rule generalizes to higher dimensions and to non-Euclidian spaces.