Showing that for every $e > 0$, there exist quasi-equilateral triangles with error $e$ whose vertices have integer coordinates in the plane I'm currently working on the problem stated below, but i'm still kind of a beginner with proofs, so I would appreciate some tips (not a solution) on how to approach this proof.

A triangle is quasi-equilateral with error $e > 0$, if, for every angle $A$, $B$, $C$ in the triangle, we have that $\vert{A-60^{\circ}}\vert < e$, $\vert{B-60^{\circ}}\vert < e$ and $\vert{C-60^{\circ}}\vert < e $. Show that for every $e > 0$, there exist quasi-equilateral triangles with error $e$ that have vertices with integer coordinates in the plane.

I know how to prove that there are no equilateral triangles with integer coordinates, however, this looks much more difficult and I can't find a way to begin.
 A: Idea: set the origin $O$ be the circumcenter of the triangle $ABC$ and use the Pythagorean triple theorem.
Hint 1

Set the vertice $C$ on the axis $\vec{Ox'}$ (the coordinate of $C$ is $(-c,0)$ with $c\in \Bbb N^*)$. The origin $O$ is the circumcenter of the triangle $ABC$. The coordinate of points $A$ and $B$ are $(a,b)$ and $(a,-b)$ with $a,b \in \Bbb N^*$

Hint 2

 The length $OA = OB = \sqrt{a^2+b^2}$ must in $\Bbb N^*$ because $OA = OB = OC$ with $C$ has integer coordinate on the axis $\vec{Ox}$

Hint 3

The solution of Pythagorean triple $a^2+b^2 = c^2$ (link) is, for example, $(a,b,c) = (m^2-n^2,2mn, m^2+n^2)$ with $m,n \in \Bbb N^*, m>n$.

Hint 4

Calculate the $\cos(\widehat{AOx}) = \frac{m^2-n^2}{m^2+n^2}$ which must be in the interval $(v_1,v_2)=\left(\cos(\frac{\pi}{3}+\epsilon), \cos(\frac{\pi}{3}-\epsilon)   \right)$.

Hint 5

Prove there always exists a rational number  $t\in \Bbb Q^+$ such that $$\sqrt{\frac{1+v_1}{1-v_1}} <t < \sqrt{\frac{1+v_2}{1-v_2}}$$ with $(v_1,v_2)$ defined in the hint 4. The couple $(m,n)$ will be defined by $t = \frac{m}{n}$. Hence, you deduce the value of $(a,b,c)$ from $(m,n)$

A: First hint:

It suffices to show this for rational coordinates (then we can just scale up by the largest common denominator).

Second hint:

Since we're rescaling anyway, we might as well take two of our points to be exactly one unit apart: fix them and call them $x$ and $y$.

Third hint:

Consider the set $Z$ of all points $z$ that make an $e$-quasiequilateral triangle.

Fourth hint:

$Z$ is an open subset of the plane.

Fifth hint (if you're struggling with proving the fourth hint):

Take such a point $z$. Then let the angles of the triangle so formed be $A(z), B(z), C(z)$ (for future convenience, I'll make those the angles at $x,y,z$, in that order), and define $f(z) := \max(|A - 60^\circ|,|B - 60^\circ|, |C-60^\circ|)$. Show that $f$ is continuous, and you're done.

Sixth hint (if you're struggling with proving the fifth hint above):

It suffices to show that each of $A,B,C$ are continuous. To do this, take some $z'$ near $z$, and consider the triangle $x,z,z'$.

The end of the proof above:

The angle at $x$ of this triangle is $|A(z) - A(z')|$ by definition, and by the sine rule (with $\theta,\varphi$ the angles at $z,z'$ of this triangle), $|A(z) - A(z')| = |zz'|\frac{\theta}{|xz'|} = |z - z'|\frac{\varphi}{|xz|}$. But neither $\varphi$ nor $\theta$ can be more than $180^\circ$, so this is bounded above by $180^\circ|zz'|\left(\frac{1}{|xz|} + \frac{1}{|xz'|}\right)$. Note that (for $e$ smaller than $30^\circ$ - the final result for larger $e$ is easy):  $z$ cannot lie within a distance of $\frac{1}{2}$ of $x$, as all points within that disk give angles at $y$ of no more than $30^\circ$, and similarly for $z'$. By the triangle rule, we have $|xz| \geq |xz'| - |zz'| \geq \frac{1}{2} - |zz'|$, so for $|zz'| < \frac{1}{4}$, we have $|xz| \geq \frac{1}{4}$, and so $\frac{1}{|xz|} \leq 4$. Thus, $|A(z) - A(z')| \leq 180^\circ|zz'|\left(4 + 2\right) = 1080^\circ|zz'| \to 0$ as $|zz'|\to 0$, so $A$ is continuous.


 The proof for $B$ is the same as for $A$ with $y$ replacing $x$ throughout. For $C$, apply the cosine rule to the triangles $xyz$ and $xyz'$ to obtain $2|xz||yz|C(z) = |xz|^2 + |yz|^2 - 1$ and similarly for $z'$, and either combine these to obtain $$|C(z) - C(z')| =\left|\dfrac{|xz|^2 + |yz|^2 - 1}{2|xz||yz|} - \dfrac{|xz'|^2 + |yz'|^2 - 1}{2|xz'||yz'|}\right|$$ and proceed roughly as above with a moderately ridiculous sequence of approximations, or just notice that the functions $w \mapsto |xw|$ and $w \mapsto |yw|$ are continuous and neither is zero anywhere near either $z$ or $z'$, and hence $C$ is continuous by repeated applications of the product, sum, and quotient rules for continuity.

Fifth hint (if you're struggling with finishing after proving the fourth hint):

Since the rationals are dense in the reals, $\mathbb{Q}^2$ is dense in $\mathbb{R^2}$ (think of squares inside circles), and so there's a rational point in every open subset of $\mathbb{R^2}$, and in particular there's one in the above, so scaling everything by the lowest common multiple of the denominator gives an integer solution.

