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Max. # of convex/concave hexagons possible using 6 points
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In the plane you get to choose six points.
By choosing these six points in one of an infinite number of appropriate configurations,
what is the maximum number of combined convex and concave (but not selfintersecting)
hexagons that can be formed, if each of the six points is to be one of the vertices for every
hexagon so formed?
For reference sake, the points could be labeled A, B, C, D, E, and, F.
I am not asking for any coordinates. However, if you want to volunteer a set of them to
illustrate your work/solution, then that would be fine.

Here is an example with four points: A, B, C, and, D.
Place them in the plane. You choose where.
What is the maximum number of combined convex and concave (but not selfintersecting)
quadrilaterals that can be formed, if each of the four points is to be one of the vertices for
every quadrilateral so formed?
If you do not choose all of the points to be distinct and/or you place at least three of them
on the same lime, you won't get any quadrilaterals formed.
If you place the four points so that they are the vertices of a convex quadrilateral, then
you will get one total (convex) quadrilateral.
If you place three of the points as vertices of a triangle and the fourth point as an interior
point of that triangle, then three total (concave) quadrilaterals can be formed.
"Three" is the answer, unless there is some other general configuration for four points that
has been overlooked with a higher total number of quadrilaterals.
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