At last I've got a bound that makes minimizing the area of convex lattice polygons a finite problem. There are two problems. First, the bound isn't terribly good. Second, the proof has a little hand- waving in it. I invite any improvements. Let us choose a convex lattice N-gon of minimum area. As we know, all images of the polygon under SL(2,Z) have the same area. Let us chose the representative polygon P with minimum diameter D. I will show that the area of P is at least D min( (ceiling(N/2)-1)/2, sqrt(1/10)D, 1) . For N>6, this implies that the area is at least D. (partial) Proof: Take vertices G,G' that realize the diameter of P. Using isometries of the plane, take G=(0,0), G'=(a,b), where b >= a >= 0 (i.e., in the second octant of the plane). If a=0, the theorem from 19 December shows that the area is at least D(ceiling(N/2)-1)/2. Otherwise we have b >= a > 0. Consider the transform F: (x,y) |-> (x,y-x) in SL(2,Z). |F((a,b))|^2 = 2aa - 2ab + bb = D^2 + a(a-2b) <= D^2 + a(a-2a) < D^2 . By our choice of P, F(P) must have diameter at least D, so there must be vertices H=(r,s) and H'=(r+c, s+d) for which |F((c,d))| >= d . This turns out to mean that point (c,d) lies outside the ellipse { (x,y) : (T+1)(Tx-y)^2 + (1/(T+1))(x+Ty)^2 = (T+2)D^2 } where T=(sqrt(5)+1)/2 is the golden ratio. In addition, since a^2+b^2 is the diameter of the polygon, we have |(c,d)| <= D, so (c,d) lies inside the circle c^2+d^2 = D^2. These requirements confine (c,d) to two crescent-shaped areas. In particular, HH' has slope at most 1/2. We may assume the slope is finite (d is not zero) by the first paragraph of this proof. Now I come to the point of using Lemma 2 from 19 December: Suppose a convex plane figure meets parallel lines R,S,S' such that the intersection with line R is a line segment of length X, and lines S and S' are at a distance of Y from each other. Then the area of the figure is at least XY/2. Line R will be GG', and lines S, S' will be the lines parallel to R through H, H' respectively. It remains to measure the distance between S and S'. Here I wave my hands to show that the distance is minimized by two cases: Ta > b, where the distance between S and S' is at least sqrt(2/5) D, approached when (c,d) is approximately (sqrt(4/5)D, sqrt(1/5)D). Ta < b, where the distance between S and S' is at least 2, approached when (c,d) is approximately (-1, D-1). I'm fairly sure I could prove this on a good day, but I'm somewhat bogged down in it now. Anyway, this completes the (partial) proof. This shows that if we examine all convex lattice N-gons of diameter at most D, and the minimum area is at most D, then we have found the minimum area N-gon. The reason I think this is not a very good bound is that the bound is based on a polygon that lies within the strip { (x,y) : -1 <= x <= 1 }, which we know has no convex N-gons for N > 6. I think the real bound is proportional to N D, meaning that we should have to search only through N-gons of diameter at most Constant.Area/N . Dan