Here's a better search cutoff than the area bound, though I still don't see how to make the search finite. Theorem: Any convex lattice N-gon that contains K collinear lattice points (x,y), (x+a,y+b), (x+2a,y+2b), ..., (x+(K-1)a,y+(K-1)b) has area at least (K-1)(ceiling(N/2)-1)/2. The proof relies on two lemmas that probably have better proofs than these: Lemma 1: If a and b are relatively prime integers, not both zero, then the lines through lattice points in the direction of the vector (a,b) are at a distance of 1/sqrt(a^2+b^2) from each other. Proof: Take a square WXYZ with no lattice points on its boundary and with WX a translate of the vector (a,b). Every line parallel to WX passing through WXYZ that touches any lattice point must touch exactly one lattice point in WXYZ. There are a^2+b^2 lattice points in WXYZ, so that many lines, with a total separation of sqrt(a^2+b^2), so the separation between any two is 1/sqrt(a^2+b^2) QED. Lemma 2: Suppose a convex plane figure meets parallel lines R,S,T such that the intersection with line R is a line segment of length X, and lines S and T are at a distance of Y from each other. Then the area of the figure is at least XY/2. Proof: Let the intersection with line R be segment AB, and let C,D be points in the figure on S,T respectively. Then by convexity the figure contains triangles ABC and ABD. If R lies between S and T then the triangles are disjoint and have total area XY/2. If R is outside S and T then the larger triangle has total area greater than XY/2 QED. Proof of theorem: Suppose the convex lattice N-gon contains K collinear lattice points on a line parallel to vector (a,b), with a,b relatively prime, not both zero. Consider the lattice lines parallel to vector (a,b) that meet vertices of the N-gon. Since no more than two vertices lie on any line, there are at least ceiling(N/2) such lines. By Lemma 1, two of the lines must be at least (ceiling(N/2)-1)/sqrt(a^2+b^2) apart. Furthermore, the line containing the K collinear lattice points meets the N-gon in a segment at least (K-1)sqrt(a^2+b^2) long. By Lemma 2, the N-gon has an area of at least (K-1)(ceiling(N/2)-1)/2 QED. In searching for an 11-gon of area less than 21.5, we can then rule out any line of 9 collinear lattice points. Furthermore, once we have two vertices X,Y of the polygon, lattice points ruled out by applying the theorem to X will shadow other points from Y, which will be ruled out by convexity, and those points may cast shadows from X, and so on. Furthermore, any P for which the triangle PXY contains a prohibited point will then be prohibited. If we can surround the partial polygon with a ring of points that are angularly close, this will reduce the problem to finitely many cases of choosing vertices from finite sets. Unfortunately, I suspect this not is enough to make the whole problem finite. Using Dylan's observation, it looks like we still have infinitely many points to choose from for the third vertex (though I haven't worked it out in detail). Still, we may be closing in on a final answer to the problem. Dan