The following interesting problem was asked at [Aops](https://artofproblemsolving.com/community/u738252h2922192p26122750) and I wonder if it was based on some research paper:



> 
> Let $K$ be a convex body in $\mathbb R^2$, such that the diameter of $K$ is less than $\sqrt2$.  
> 
> Prove that there is a lattice-point-free translation of $K$.
> 
> 
> 


Is the 'convex condition' necessary?  

Can this problem be generalized to $\mathbb R^n$?



Lattice-point-free body diameter
For higher dimension, this does not hold, see the answer by Sergei Ivanov [here](https://mathoverflow.net/q/42774/4312).


For dimension 2, in the closed set $K$ without lattice-point free translate (in other words, such that $K$ has a point in every translate $\mathbb{Z}^2+c$ of the lattice) we may even find two points for which both coordinates differ at least by 1. This may be proved as follows. Fix real $a\in [0,1]$ and consider the union of vertical lines $V\_a:=\{(x, y) \colon x-a\in \mathbb{Z}\}$. In this set $V\_a$, the ordinates take all values modulo 1, thus there exist two points $p\_a, q\_a\in V\_a$ with ordinates differing at least by 1. If they have different abscissas, that's what we need. So, these $p\_a, q\_a$ have equal abscissa $v\_a$, congruent to $a$ modulo 1. Therefore there exist two values $a, b$ for which $|v\_a-v\_b|$ is more than 1 or arbitrarily close to 1. Between points $p\_a, q\_a, p\_b, q\_b$ let $p\_a$ be the lowest, then $p\_a$ and the highest of $q\_a, q\_b$ is (almost) necessary pair of points.


