TITLE: Maximum $C$ such that every shape in $\Bbb R^2$ with area $<C$ can be placed to avoid $\Bbb Z^2$
QUESTION [22 upvotes]: For $C=1$, it has been proved here that every shape in the plane having area less than $1$ can be translated and rotated so that it does not touch any element of $\mathbb Z^2$. (In fact, for $C=1$, only translation is necessary.)
What is the largest $C$ for which this is true?
I think the answer is $\frac{\pi}{2}$. This is an upper bound on the answer as a disk of area $\frac{\pi}{2}$ cannot avoid $\mathbb Z^2$. But I don't know how to prove that the largest $C$ is exactly $\frac{\pi}{2}$.

REPLY [8 votes]: Here is a somewhat simpler proof that the $1\times\sqrt{2}$ rectangle works (see @Peter Woolfitt 's original answer).
If the rectangle is aligned with the grid it is obvious that it hits a grid point.

Assume that the rectangle is tilted with respect to the grid axes by an angle $\alpha$, whereby $0<\alpha\leq{\pi\over 4}$. The longer sides of the rectangle determine an infinite strip of width $1$, shown in red in the above figure. Each vertical grid line intersects the strip in a segment of length $\ell$ with $1<\ell\leq\sqrt{2}$. Therefore each such line has one or two gridpoints within the strip. The rectangle is lying somewhere in the strip. The euclidean distance between two successive red grid points is $1$ or $\sqrt{2}$; therefore their projected distance on the axis of the strip is $\leq\sqrt{2}$. This shows that the rectangle having length $\sqrt{2}$ will hit a grid point in any case.