Every three of $n$ points is the vertices of an isosceles triangle. What is the max of $n$?
Solution 1:
It seems that, if we relax the restriction that the points must not be colinear (a rather artificial restriction IMO), the maximum is 8; with 9 it has been proven impossible. See references here.
The construction for the 8-points set seems to be as follows: 5 points equidistant over a circle or radius $1$ (say, over the $x,y$ plane, with the center at the origin), plus an additional point in the origin, plus two additional points on $(0,0,\pm 1)$