Distance from lattice point to line What is the smallest $d> 0$ such that for any line in the plane not parallel to the $x$- or $y$-axis, the distance of some lattice point to the line does not exceed $d$?
If the line has slope $1$ and passes through point $(1/2,0)$, it has distance at least $\frac{1}{2\sqrt{2}}$ from any lattice point. This might also be the minimum $d$. There's of course an explicit formula for the distance from a line to any point, but here we have infinitely many points, which can generate infinitely many distances, for example if the slope is irrational.
 A: The minimum $d$ is $\frac{1}{2\sqrt2}$, in the case you showed. The way to prove it is relatively easy. Choose four neighboring lattice points $(x_0,y_0)$, $(x_0+1,y_0)$, $(x_0,y_0+1)$, $(x_0+1,y_0+1)$ such as the line $y=ax+b$ goes in between. The distance from such a lattice point to the line can be written in terms of Pythagoras' theorem. We know the $x$ and $y$ components in the triangles formed by the lattice points, and the line. Along $x$ you have either $\{x\}$ or $\{1-x\}$ and along $y$ you have either $\{ax+b\}$ or $1-\{ax+b\}$ . Here the $\{\}$ notation means the fractional part. 
We can write the areas of one of these triangles as either $0.5\{x\}\{ax+b\}$ or $0.5d\sqrt{\{x\}^2+\{y\}^2}$. From here $$d=\frac{\{x\}\{ax+b\}}{\sqrt{\{x\}^2+\{y\}^2}}$$
and similar combinations, with $\{x\}$ replaced with $1-\{x\}$ and/or $\{ax+b\}$ replaced with $1-\{ax+b\}$
Since we want to find the maximum distance for any line, it means that the $x$ and $y$ component are independent. The maximum value is reached when $\{x\}=1/2$ and $\{ax+b\}=1/2$. The minimum distance is then $$d=\frac{\frac{1}{2}\frac12}{\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}}=\frac{1}{2\sqrt2}$$
A: I'm giving an additional solution that ultimately uses the same kind of idea, but is to me a little clearer on all the min/maxing of distances.
So we have a line $y=ax+b$ with $a \neq 0$. If the slope $a$ is negative, we can replace it with $y=-ax+b$, which corresponds to mirroring on the $y$-axis, which is a bijection between lattice points, so if we can prove it for positive $a$, it follows for all $a$.
Similiarly, if $a > 1$, we can replace it with $y=\frac1a x-\frac{b}a$, which corresponds to mirroring on the $y=x$ axis, which is also a bijection between lattice points.
So, for symmetry reasons, we need to consider $0 < a \le 1 $ only. Since $a>0$, our line is intersecting any parallel to the $x$-axis.

The line $y=ax+b$ is red, the lattice of integer points in the plane is partially shown. The distance of 2 such lattice points to the line is shown in blue, the angle under which the line hits the parallel to the $x$-axis is $\alpha$. Since $0 < a = \tan(\alpha) \le 1$ we have $0 < \alpha \le 45°$.
The intersection point of the red line with the parallel to the $x$-axis is somewhere between the 2 lattice points, the $x$ distances are $l_1$ and $l_2$ with $l_1+l_2=1$ (this corresponds to $\{x\}$ and $\{1-x\}$ in Andrei's solution). 
For the distance $d_1$ of the left lattice point to the line we get $d_1=l_1\sin(\alpha)$ and similiarly $d_2=l_2\sin(\alpha)$.
Now we know that at least one of $l_1,l_2$ is $\le \frac12$ and from $0 < \alpha \le 45°$ it follows that $\sin(\alpha) \le \frac1{\sqrt{2}}$, so 
$$\min(d_1,d_2) \le \frac12 \frac1{\sqrt{2}} = \frac1{2\sqrt{2}}$$
which shows that $\frac1{2\sqrt{2}}$ is indeeded the highest possible value for the minimal distance.
