Prove existence of triangle with vertices in $\mathbb{Z}^2$, each angle of which is $\varepsilon$-close to $\pi/3$.

Question

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 $|\frac{\pi}{3}-\theta_i|<\varepsilon$.

Answer 1

Here's a rather "brute force" approach. Take a big polygon $\mathcal P$ with vertices in $\mathbb R^2$ that has the angles you want, such that each edge of $\mathcal P$ has length $\geq R$. Let $\mathbf{round}(\mathcal P)$ be the polygon whose vertices are the points in $\mathbb Z^2$ nearest to the vertices of $\mathcal P$. Then the angles of $\mathbf{round}(\mathcal P)$ are within $O(1/R)$ of the angles of $\mathcal P$. But you can make $R$ arbitrarily large.

Answer 2

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 ${\pi\over3}$.