TITLE: Prove existence of triangle with vertices in $\mathbb{Z}^2$, each angle of which is $\varepsilon$-close to $\pi/3$.
QUESTION [9 upvotes]: Let $\theta_1, \theta_2$ and $\theta_3$ angles of a triangle. Prove
that for every $\varepsilon>0$, there exists a triangle with vertices
in $\mathbb{Z}^2$ such that $|\frac{\pi}{3}-\theta_i|<\varepsilon$.

REPLY [2 votes]: Take the sector
$$S:=\left\{(x,y)\biggm|x>0, \ {\pi\over6}-{\epsilon\over2}<{\rm Arg}(x,y)<{\pi\over6}+{\epsilon\over2}\right\}\ ,$$
centered at the origin, and obtain $S'$ by reflecting $S$ across the $x$-axis. Since $S$ gets (unboundedly) wider when going out to the right there is a point $(x,y)\in S\cap{\mathbb Z}^2$; furthermore $(x,-y)\in S'\cap{\mathbb Z}^2$. The triangle with vertices $(0,0)$, $(x,y)$, $(x,-y)$ is isosceles, and has all angles $<\epsilon$ from ${\pi\over3}$.