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
