Return to Question

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 $\alpha\leq 3/8$

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 http://arxiv.org/abs/1306.4688

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 $d^2$.

The best estimation I found is $d\leq \sqrt{2(g+1)\sqrt{3}}$ where $g$ is the number of lattice points inside $M$. (Some Metric Inequalities for Lattice Polygons, Stanley Rabinowitz, 1989).

Does anybody know a better estimation? (after 24 years of progress =))

${\bf ADDED:}$ The inequality above is only for $usual$ width, so it does not work here.

One can prove an inequality $area\geq d^2/4$ 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 $d$.

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 $\alpha\leq 3/8$

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 $d^2$.

The best estimation I found is $d\leq \sqrt{2(g+1)\sqrt{3}}$ where $g$ is the number of lattice points inside $M$. (Some Metric Inequalities for Lattice Polygons, Stanley Rabinowitz, 1989).

Does anybody know a better estimation? (after 24 years of progress =))

${\bf ADDED:}$ The inequality above is only for $usual$ width, so it does not work here.

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

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 $d^2$.

The best estimation I found is $d\leq \sqrt{2(g+1)\sqrt{3}}$ where $g$ is the number of lattice points inside $M$. (Some Metric Inequalities for Lattice Polygons, Stanley Rabinowitz, 1989).

Does anybody know a better estimation? (after 24 years of progress =))

${\bf ADDED:}$ The inequality above is only for $usual$ width, so it does not work here.

One can prove an inequality $area\geq d^2/4$ 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 $d$.