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Let $A$ be a set composed by an even number $n$ of distinct points in $\mathbf{R}^{k}$, such that any 3 points in $A$ are non-collinear in $\mathbf{R}^{k}$. Let us consider the set $P_{2}(A)$ of partitions of $A$, whose elements are pairs of points in $A$.

Fix a partition $\Omega\in P_{2}(A)$. Take $(i,j)\in \Omega $ and consider the segment joining $(i,j)$. Then we generate $\frac{n}{2}$ segments for each $\Omega$. Let $P^{*}_{2}(A)\subset P_{2}(A)$ be a subset of partitions such that the associated $\frac{n}{2}$ segments have an intersection point in $\mathbf{R}^{k}$.

Is the subset $P^{*}_{2}(A)$ reduced to at most a unique partition? If yes, any idea about a simple proof?

P.S. If $k=2$ the result seems to be quite intuitive. The result seems to be true for any cardinality of $A$ and for higher dimensions $k>2$. This question is related to a similar post of mine, even it is different.

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  • $\begingroup$ Anyway if you take in the complex plane, 0 and the 3 cubic roots of 1, no $2+2$ partition yields an intersection... $\endgroup$
    – YCor
    Commented May 28, 2017 at 15:29
  • $\begingroup$ Well, your property is invariant under projections, so if what you want is true in $\mathbb R^2$, it is true in any $\mathbb R^k$. On the other hand, any bad planar configuration can be embedded into any higher dimension. So it is a pure plane geometry problem. $\endgroup$
    – fedja
    Commented May 28, 2017 at 15:50
  • $\begingroup$ @fedja not exactly, if the projections of 2 segments intersect, it does not mean that the 2 original segments intersect. Anyway your comment applies if the question is to show that there is at most one 2-partition for which the segments pairwise intersect, since indeed it boils down to dimension 2 after choosing a generic projection. $\endgroup$
    – YCor
    Commented May 28, 2017 at 15:54
  • $\begingroup$ @Ycor Of course. I just was a bit sloppy. On the other hand, I would be way more comfortable with replacing "unique" by "at most one" than with replacing "intersecting" with "pairwise intersecting" when editing this post ;-) $\endgroup$
    – fedja
    Commented May 28, 2017 at 15:58
  • $\begingroup$ @Ycor There is no existence if you remove "pairwise" (and "pairwise" does not help any: two generic lines in $\mathbb R^3$ don't intersect either) . Uniqueness, on the other hand, gets only easier. Anyway, let's try to solve the problem before nitpicking at each other any more :-). $\endgroup$
    – fedja
    Commented May 28, 2017 at 16:07

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A full answer was given in the comments. It is (almost) exactly the same argument as in the linked thread, so the only purpose of typing this "answer" is to remove the question from the "Unanswered" list. Retyping it here for the convenience of the reader:

1) It is enough to do the planar case, since we can project any bad higher-dimensional configuration by a generic planar projection.

2) The uniqueness (on the plane) holds for exactly the same reason as before, even with pairwise intersections: pick a point $A$ on the boundary of the convex hull of the given points. Its matching point $B$ should satisfy the property that the line $(AB)$ splits the set into two parts of the same cardinality. Such $B$ is unique, so one segment is known. Proceed by induction.

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