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Let $M$ be a lattice polygon on a plane (i.e. its vertices are integer points $(i,j)\in\mathbb Z^2$).

Let us define lattice width in a direction $v=(m,n)\in\mathbb Z^2$ as $w_v(M)=\max\limits_{x,y\in M} v\cdot(x-y)$.

Suppose the $minimal$ lattice width of $M$ equals $d$. It is clear that the area of $M$ should be proportional to $d^2$.

One can prove an inequalities $area\geq d^2/4$ as is written in comments. But it seems far from the best one.

So, the question is what is the best $\alpha$ such that $area(M)\geq \alpha d^2$ for each $M$ with minimal lattice width equals $d$. From my comment below one can extract that $\alpha\leq 3/8$

Added: F. Petrov gave a link to the proof of the fact, but there is no complete proof in the Internet so I rewrote it in the appendix in

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Considering the quadrilateral $Q$ of maximal area inscribed into $M$. THen you may pass to the parallelogram with sides parallel to the diagonals of $Q$ --- its area is at most twice the area of $M$. Hence you may restrict yourself to the lattice parallelograms only. – Ilya Bogdanov Mar 28 '13 at 6:55
Thay gives us $area(M)\geq d^2/4$, it is worse than $g\geq d^2/{2\sqrt{3}}$ – Nikita Kalinin Mar 29 '13 at 12:08
What is the question here? – Vít Tuček Mar 30 '13 at 22:31
@robot: I was looking for an estimation $area(M)\geq cd^2$ – Nikita Kalinin Mar 31 '13 at 7:11

2 Answers 2

up vote 3 down vote accepted

$\alpha=3/8$ is sharp according to, say this article, the authors refer to [L. Fejes-Toth and E. Makai, Jr., On the thinnest non-separable lattice of convex plates, Studia Sci. Math. Hungar. 9 (1974), 191–193.]

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This solution is not true, sorry. I can prove that $area(M)\geq 3d^2/8$. First of all find $v\in \mathbb Z^2$ such that $w_v(M)$ is minimal. Than by an affine transform $a$ put $v$ into vector $(1,1)$.

Now draw lines which give us widths of $a(M)$ in the directions $(1,0),(0,1)$. Now $a(M)$ is inside a horizontal stripe of width at least $d$, inside a vertical stripe of width at least $d$ and inside a diagonal (in the direction $(1,-1)$) stripe of width exactly $d/{\sqrt{2}}$.

Here I use usual widths.

Now look at the picture - we see a 6-gon (or 3,4-gon) $B$ as the intersection of the strips, we know that on each side of $B$ there is a vertex of $a(M)$. By playing with vertices one can find all the extremal cases (so, vertices of $a(M)$ must coincide with vertices of $B$).

Finally, in all the extremal cases the area of $a(M)$ is no less than $3d^2/8$

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You need to be a bit more careful, since the affine transforms do not preserve scalar product. So in fact you need to put the vector orthogonal to $v$ into $(-1,1)$. – Ilya Bogdanov Mar 30 '13 at 8:32
In fact, this bound is tight. Look at the triangle wth vertices $(0,0)$, $(2,1)$ and $(1,2)$. Its area is $3/2$, while its "integral width" is 2. To see that, first notice that all its altitude lengths are greater than 1, hence it is enough to check only the vectors of length less than 2. This check is straightforward. So, what is this bounty for? – Ilya Bogdanov Mar 30 '13 at 8:35
@Ilya: I had started this bounty before I realized that it is a simple problem, and I was very nervous =)) Now I can not cancel it. Concerning the first comment: an affine transformation preserves the lattice width, it is enough for me. – Nikita Kalinin Mar 30 '13 at 12:44
@Ilya: it is not a simple problem, my solution is not true. – Nikita Kalinin Apr 2 '13 at 0:32

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