Existence of a real which allows a observer to see a square in a circle In $\mathbb{R}^2$ was disposed on each point coordinates in $\mathbb{Z}$, except $(0,0)$, a small open square of side  $2r$ where $r\in]0,\frac{1}{2}[$. 
Prove that there is a real $R>0$ such that for any real $\theta$, an observer located at $(0,0)$ and looking in the direction $(\cos\theta,\sin\theta)$ manages to see a square located in a circle with centre coordinates $(0,0)$ and radius $R$ (Euclidean)
Thanks in advance
 A: Assume wlog that $\theta$ is in the first quadrant.
Start with $A=(1,0)$, $B=(0,1)$, so the ray given by $\theta$ is in $\angle AOB$. Repeat the following procedure:


*

*Complete to a parallelogram $OACB$. 

*Otherwise, the ray is either in $\angle AOC$ or in $\angle COB$ (if the ray hits $C$ precisely, it doesn't matter which alternative you choose). In the first case let $B:=C$, in the second let $A:=C$. Then go back to step 1.


Note that the parallelograms always have area $1$ and the triangles $AOC$, $COB$ have area $\frac12$ throughout this nonterminating procedure.
Therefore, as soon as $|OC|>\frac1r$ (which it does because $|OC|\to \infty$ during the procedure), the heights of these triangle become $<\frac r2$ so that the squares around $A$ and $B$ block all viewing directions in $\angle AOB$ (each covers at least the distance to $OC$; this works even if we replace the squares with disks of radius $r$). We conclude that a lattice point as asked lies within $R=\frac 1r$. 

Another way to look at this problem is to find a "good enough" rational approximation $\tan \theta\approx \frac yx$ with integers $x,y$. The procedure described above does so essentially by starting with $\frac01<\tan\theta<\infty=\frac10$ and repeatedly splitting the interval $\frac ab<\tan\theta<\frac cd$ at the Farey sum $\frac{a+c}{b+d}$.

Yet another variation of the same theme.
Assume for some $\theta$ that the ray does not hit any of the squares within $R=\frac1r$. Let $X$ be the (non-lattice) point of intrsection of the given ray with the circle of radius $R$.
If the line segment $OX$ contains a lattice point (apart from $O$) we are done, so assume it does not.
Of the finitely many lattice points in that disk pick $A$ such that $OAX$ is positively oriented and $\angle AOX$ is minimal, pick $B$ such that $OXB$ is positively oriented and $\angle XOB$ is minimal. Then both these angles are positive.
By choice of $A,B$, there is no lattice point inside the area bounded by $OA$, $OB$ and the circle of radius $R$. By the Pieck formula, $\Delta OAB$ has area $\frac12$.
Let $C=A+B$ be the point that makes $OACB$ a parallelogram.
Then $OC$ is strictly between $OA$ and $OB$, hence the lattice point $C$ is not inside the $R$ disk. We conslude $|OC|\ge R$. Since the area of $OACB$  is $1$,
the triangles $OCB$ and $COA$ have height $\le \frac r2$ (with respect to $OC$ as base line). Therefor the $r$ disks arouond $A,B$ each at least touch $OC$ and hence one of them intersects $OX$.
