Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
I'm reminded of a result that any (topological) embedding of the complete graph on $6$ vertices has a linked pair of triangles. en.wikipedia.org/wiki/Linkless_embedding –  Douglas Zare Mar 25 at 4:38
I'm guessing that 6 points suffice. Perhaps if they are almost at the vertices of an octahedron? –  Sam Nead Mar 25 at 5:27
Another random thought - it may be enough to work in $S^3$. In that case, you can reduce the number of parameters using the Mobius group. –  Sam Nead Mar 25 at 5:28
There is an arbitrarily numerous (even infinite) set of points in general position on a sphere. Then each circle lies on the sphere, hence no two circles are linked. Maybe you want to assume that no five points lie on a sphere...? –  Wlodek Kuperberg Mar 28 at 18:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.