MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|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. – Douglas Zare Mar 25 '14 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 '14 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 '14 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 '14 at 18:15
Permit me to cite an earlier, related question: "Random rings linked into one component?" This asked if a random collection of unit-radius circles are linked into one component (under certain conditions), and the answer is, essentially: Yes. – Joseph O'Rourke Aug 26 '15 at 23:09

$\def\ov{\overline}$As Wlodek Kuperberg mentioned in the comments, it is reasonable to assume that no five points lie on a sphere. Under this condition, I claim that 6 points suffice.

1. We may choose a sphere passing through four given points and separatinf the other two.

To prove this, choose a face $ABC$ of the convex hull of given points, start with a very large sphere passing through $A$, $B$, $C$ and containing no other points inside it. Blow it down and then up again till it becomes a large sphere on the other side containing the other three points. It will take these points one by one, so when it passes through the second point we catch the required moment.

2. Now let $A$, $B$, $C$, $D$ be our points on the sphere $\Omega$, $E$ be the point outside $\Omega$, and $F$ be the point inside $\Omega$. Let the line $EF$ meet the sphere at $X$ and $Y$ (with the order $E$, $X$, $F$, $Y$ on the line). Then there exists a point $R$ on the segment $XF$ such that $\ov{RE}\cdot \ov{RF}=\ov{RX}\cdot \ov{RY}$; notice that $R$ lies inside $\Omega$.

This point is magical in the following sense: If $AR$ meets $\Omega$ at $A'$, then the points $A$, $E$, $F$, and $A'$ are concyclic (since $\ov{RA}\cdot \ov{RA'}=\ov{RX}\cdot \ov{RY}=\ov{RE}\cdot \ov{RF}$). Thus, If the plane $BCD$ separates $A$ and $A'$, then the circle $\omega=(AEF)$ is linked with $(BCD)$ (indeed, the arc $AFA'$ of $\omega$ crosses the disk bounded by $(BCD)$ exactly once, while the other arc $AEA'$ does not). In this case we say that the point $A$ works.

3. So it remains to show that one of the four points $A$, $B$, $C$, $D$ works. If $R$ lies inside the tetrahedron $ABCD$, then every its vertex works. Otherwise, $R$ is (non-strictly) separated from this tetrahedron by a plane containing one of its faces (say, $BCD$). In this case, the (infinite) cone with apex $A$ and base $(BCD)$ contains $R$ (since it contains the intersection of the interior of $\Omega$ with the corresponding half-space of $BCD$), so $A$ works.

share|cite|improve this answer

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.

share|cite|improve this answer
Cool. This doesn't contradict Wlodek Kuperberg's counterexample of all point lying on a sphere because that translates of points $A,B,C,D,E$ lying on the same plane, leading to convex hulls $ABC$ and $DE$ correspond to intersecting rather than linked circles. – Michael Aug 26 '15 at 22:25
That's beautiful. Yes, being linked is preserved under inversion; it's a property of circles on $S^3$, which are preserved by the group of Möbius transformations on $S^3$ (including inversion). I suspect that Ilya's proof is a disguised version of this one. – Dylan Thurston Aug 26 '15 at 22:28

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.