Matching on sphere to create cycle with chords

Imagine a number of chords of a sphere $S$ which nearly, but not quite, pass through the center of $S$, in such a way that no pair of chords intersect: I would like to connect pairs of chord endpoints by noncrossing paths on the surface of $S$ so that the paths + the chords form a cycle. I can prove (for example, by induction) that this is always possible if I am permitted to use arbitrary paths on the surface. But what I would really like to achieve is connecting the chord endpoints by noncrossing arcs of great circles.

Henceforth consider the chords to pass exactly through the center of $S$, but pretend they do not intersect there. An example is shown left below, where three axes-chords are connected into a cycle. The example right below of coplanar chords shows that what I want to achieve is not always possible. Q. Under what conditions can a set of chords through the center of $S$ be connected into a single cycle by noncrossing arcs of great circles? In particular, can this be achieved if no three chord endpoints lie on a great circle, i.e., are in general position in this sense?

There is a considerable literature on noncrossing geometric matchings in the plane, but I don't see that it applies to my question. Any pointers, ideas, or counterexamples welcomed!

(This arose in an investigation related (nonobviously) to an earlier question, "Untangling entwined rigid chains in 3-space".)

• Is it relevant/necessary that the points are on a sphere and chords are connecting nearly antipodal points? What if there are $2n$ points in "general position" in the plane and they are colored by $n$ colors, 2 points of each color. Can we always find a perfect matching of non-crossing line segments such that a walk that alternates between following a line segment and jumping to the other point of the same color will connect the points to a single cycle? – Johan Wästlund Jun 28 '13 at 14:04
• That's a wonderfully clarifying reformulation of the essence of the discrete geometry question at the heart of this! – Joseph O'Rourke Jun 28 '13 at 14:41
• Answering my own question: When $n\geq 2$, the convex hull is spanned by at least 3 points, so we can find two of them that are adjacent and of different color, say red and green. After pairing them up, we solve the subproblem of the remaining $2n-2$ points under red-green color blindness (treating the other red and green points as having the same color). – Johan Wästlund Jun 28 '13 at 18:03
• One may pose a related question about a finite subset of the projective plane and the arcs of the straight lines--there are just simple points, no notion of any antipodal points. The questions are essentially different but at least they are similar, and the projective version has simpler formulation I'd think (a single projective point is something of a substitute, not exactly, of an antipodal pair). – Włodzimierz Holsztyński Jun 29 '13 at 3:37