# Is there a second solution to points and perpendicular bisectors problem?

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)?

• What is Gardner's question exactly ? You set some hypotheses but you didn't ask any question. Jul 16, 2014 at 9:07
• @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. Jul 16, 2014 at 9:46
• Well, there is an obvious solution for any number of points divisible by $8$. Jul 16, 2014 at 13:24
• I think I may have been wrong about my comment FYI Jul 16, 2014 at 20:48
• Merely I had difficulty generalizing for any number divisible by 8. Jul 16, 2014 at 21:22

## 1 Answer

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.

• this is not what i am looking for, of course. Nov 10, 2014 at 17:56