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Given a set A composed by six non collinear distinct points on the plane, let us consider only the partitions whose elements are pairs of points in A. Then, we call the set of such partitions by P(A). So, fix a partition in P(A). Then, take each pair in this partition and consider the 3 segments joining the two points in A belonging to each pair. For some partitions those 3 segments intersect in one single point. Let S(P(A)) be the set of such partitions. Then can you provide an example of 6 not collinear points on the plane such that there are two partitions P, P^’ in S(P(A))=\nonempty, whose associated intersection points are different.

I am looking for an example where six non collinear distinct points on the plane are located in such a way that the above configuration holds. I do not know if this configuration exists and I guess it doesn't.

Thanks.

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    $\begingroup$ The tag "discrete-geometry" would be more appropriate. $\endgroup$ – Jan Kyncl May 15 '17 at 20:24
  • $\begingroup$ Does "non-collinear" just mean that the six aren't all on the same line? or does it mean that no three are on the same line? $\endgroup$ – Gerry Myerson May 15 '17 at 22:41
  • $\begingroup$ @GerryMyerson Hello Gerry, you are right. I mean that the six aren't all on the same line. However, an example where any 3 points of them are not collinear would be better. Such a more general example should show more the generality of this lack of uniqueness. However I am reading your example and I am checking if it is true. $\endgroup$ – Francesco Ciardiello May 15 '17 at 23:19
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For every set $A$ of six points in general position (no three on a line) there is at most one such partition, which can be seen as follows.

First, if three segments determined by $A$ cross at the same point, then each of the segments is a halving segment, which means that the line determined by the segment splits the remaining four points evenly in two halfplanes.

Second, the convex hull of $A$ has at least three extremal points (vertices of the polygon). Each extremal point is incident to exactly one halving segment, so this determines at least two pairs in the partition. Since $A$ only has six points, the whole partition is determined uniquely.

If $A$ is not in general position, a similar argument shows that for each extremal point $x$ of $A$, the line extending the segment with endpoint $x$ is determined uniquely. There are at least two extremal points for which the lines are different, and this determines the intersection point of the three segments.

There are examples where two different partitions are possible but with the same intersection point. For example $A=\{(-2,0),(-1,0),(1,0),(2,0),(0,1),(0,-1)\}$.

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Let $Q=(-6,0)$, $R=(6,0)$, $S=(0,2)$, $T=(-3,3)$, $U=(3,3)$, $V=(0,6)$. The partition $QU,RT,SV$ gives the point $S$, the partition $QT,RU,SV$ gives the point $V$.

EDIT: I just noticed that the question specifies line segments, not lines. If that's what's wanted, the this example fails, as $QT$ and $RU$ don't intersect, unless extended.

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  • $\begingroup$ True. That's very true. The problem has a convexity nature more than a linear one. I start to think that such a configuration simply does not exist. I think that the proof provided by @JanKyncl. It seems that Jan uses a Caratheodory convex argument to prove this. Then I suppose that there is some chance to have such desired configuration when |A|/2 > 3 and |A| is even. In the case made explicit in my question. If |A|=<3 then the desired configuration seems not to exist, i.e. |A|=6, then 3=3. $\endgroup$ – Francesco Ciardiello May 16 '17 at 11:13

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