More fun with points

[jay_shark]

Determine all integers n>3 for which there exist n points A1, A2...An in the plane , no three are collinear and real numbers r1, r2, ...rn such that for any distinct i,j,k , the area of the triangle AiAjAk =ri+rj+rk

[gumpzilla]

I'll be very surprised if the answer to this is anything other than n=4, but I haven't a proof yet. EDIT: With 4 points it's clearly possible (just pick your four points to form a square). A couple of questions pop to mind: Is it possible to have 4 points with this property without them being in a square? I suspect it isn't but again, haven't a proof. Proving such a lemma would prove the entire conjecture that it is only possible for 4 points. EDIT AGAIN: Square can be relaxed to at least at least rectangle, though I don't think to parallelogram, since the diagonals have to be equal in order for all four possible triangles to have the same area (Heron's formula)