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How can I show that for any three points in the plane, four lines can be drawn that separate the three points into distinct enclosed regions?

Can any six points be enclosed in distinct regions formed by five lines?

Clarifications:

Points are distinct, enclosed regions mean bounded regions.

Thank you.

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  • $\begingroup$ any three points in the plane? in that case you could consider that the points are at the same place. $\endgroup$ – Integral Mar 5 '13 at 2:11
  • $\begingroup$ It's easy if the three points lie on a line. Perhaps by a fractional linear transformation we can always reduce to this case? $\endgroup$ – Gerry Myerson Mar 5 '13 at 2:43
  • $\begingroup$ @GerryMyerson If we do that then won't some of the lines be mapped to circles? Or is there a way around that? $\endgroup$ – Chris Brooks Mar 5 '13 at 2:49
  • $\begingroup$ @Joseph, good point. I was thinking circles would be the exceptional case, and that a small perturbation would get us back to lines, but I don't know why I was thinking that. $\endgroup$ – Gerry Myerson Mar 5 '13 at 2:57
  • $\begingroup$ It's tedious, but I can see a relatively straightforward constructive proof by cases. $\endgroup$ – Lepidopterist Mar 5 '13 at 3:25
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Okay, I think this works. By scaling and rotation, we can assume that two of the points are $(0,0)$ and $(0,1)$. Then the other point is $(x,y)$. Now the problem can be solved if the third point is $(1,0)$, with something like

enter image description here

Now if $x\ne 0$, the linear transformation $A=\pmatrix{x&0\\y&1}$ maps the point $(0,1)$ to $(x,y)$ and fixes the other two points, and also maps each green line to some new line, so $A$ applied to each line gives you four lines which enclose the points $(0,1), (0,0)$ and $(x,y)$.

If the third point is collinear with the other two points then it is easy to come up with the four lines that work.

enter image description here

Just make a cone that contains the two top points and another which contains the two bottom points. Then only the middle point will be in the intersection of the cones.

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  • $\begingroup$ Nicely done.${}$ $\endgroup$ – Gerry Myerson Mar 5 '13 at 22:43

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