Oh, my ... how bad those numbers are. For $n$ points in the plane (three by three non-collinear), any complete triangulation of their convex hull - a polygon having $3\leq k \leq n$ vertices - is made by precisely $t = 2n-k-2$ triangles. The least $t=n-2$ occurs when $k=n$. But then, if the points are contained in a circle of diameter $1$, the area of their convex hull is less than $\pi/4$, thus there will exist a triangle of the triangulation having area less than $\pi/4(n-2)$. For $n=59$ already we have $\pi/4(n-2) < 1/72$, and so we are done, with only $59$ points. More advanced methods, like those from here http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1030132, may be used to lower the number of points to $39$. This is known not to be sharp either, but we don't have general methods to further decrease this bound. A good idea this C3 problem hasn't been chosen for the competition (rather C2 has been).