There is a problem from M.Gardner book ("Wheels, Life, and other Mathematical Amusements", p. 201):

[Hallard T. Croft] asked if there existed a finite set of points on the plane such that the perpendicular bisector of the line segment joining any two points would always pass through at least two other points of the set.

The known solution to this problem is:

You need 8 points. Take 4 points (A, B, C, D), put them in the corners of a square with side 1 (ABCD). Add 4 more points (E, F, G, H) outside of the square, each corresponds to one side of the square and creates an equilateral triangle with this side (ABE, BCF, CDG, DAH).

I wonder whether there is any other solution to this problem (with at least 2 points in the set)?

  • $\begingroup$ What is Gardner's question exactly ? You set some hypotheses but you didn't ask any question. $\endgroup$ – Gabriel Romon Jul 16 '14 at 9:07
  • $\begingroup$ @G.T.R, Gardner's question? Gardner had a problem to solve, not question. I have a question about this problem. And sorry, I do not have english version of the book to cite it preciselly. $\endgroup$ – klm123 Jul 16 '14 at 9:46
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    $\begingroup$ Well, there is an obvious solution for any number of points divisible by $8$. $\endgroup$ – kaine Jul 16 '14 at 13:24
  • $\begingroup$ @kaine, thanks. Any other solutions? $\endgroup$ – klm123 Jul 16 '14 at 13:45
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    $\begingroup$ Merely I had difficulty generalizing for any number divisible by 8. $\endgroup$ – kaine Jul 16 '14 at 21:22

There are a couple degenerate cases: N=0 and N=1. In those cases, there are zero pairs of points where the perpendicular bisector of the segment between them doesn't pass through at least two other points.

The problem also degenerates if we allow coincident points.

  • $\begingroup$ this is not what i am looking for, of course. $\endgroup$ – klm123 Nov 10 '14 at 17:56

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