TITLE: Linked circles in R3
QUESTION [15 upvotes]: Two circles in 3-D are linked iff each one passes through the interior of the other.
There are $N$ points in 3-D in general position (no four lie on a plane). Each triple of points defines a unique circle passing through the points.
Does some $N$ guarantee that there must be a linked pair among the $N \choose 3$ circles? If so, what is the smallest such $N$?
(Circles sharing one of the $N$ points do not count. There are ${N \choose 3}{N-3 \choose 3}/2 = {N \choose 3,3,n-6}/2$ eligible pairs of circles.)
I have no idea whatever how to go about finding $N$ (if there is one).
If I had time I would write a Mathematica program to look into this.

REPLY [8 votes]: Is being linked preserved under inversion? It looks so.
If we have six points, make inversion in any of them. We get Infinite point and five normal points. We need to partition them on 3 points $ABC$ and 2 point $DE$ so that line $DE$ passes through the disc of the circle $ABC$. By Radon theorem our five points may be partitioned onto two groups whose convex hulls have a common point. There are two cases: triangle $ABC$ and the segment $DE$ intersect or point $E$ lies inside tetrahedron $ABCD$. Both are ok.