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)?