Brazilian Math Olympiad

First consider $n-1$ points in the circumference of a circle and its center. Since any two points in the circumference and the center determine an isosceles triangle, the total number of isosceles triangles in this set of points is at least $(n-1)(n-2)/2>e{n}^2$ for some small $e$.

In another hand, let's consider any set of $n$ points in the plane, no three collinear. There are $n(n-1)/2$ choices for two distinct points among the $n$. For each, say $A$ and $B$, there are at most two points $M$ among the given ones such that $AMB$ is isosceles at $M$, because such points $M$ belong to the perpendicular bisector of $AB$ and no three of the points can belong to this line. Thus the total number of isosceles triangles is at most $n(n-1)<n^2$. Therefore $f(n)<n^2$.