Hmm, i think i have an idea, the number of distinct open discs containg atleast k points, for k>2 is bounded by ${n \choose 3}$, since every disc is uniquely defined by the 3 points contained in it and closest to its boundry. every 3 points form a triangle bounded by some disc. suppose two diffrent discs have the same 3 points being closest to the edge, than atleast one disc has a point contained in it, which is not contained inside the other disc, than we can "shrink" the first disc untill the "spare" point is closest to its edge, then it is defined by a diffrent triplet. is there any flaw in my thinking?
Cain
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