Divide an ellipse into three parts I'm looking at this problem:

Let N be the number of ordered pairs (x,*y) of integers such that $x^2+xy+y^2\le 2012$. Prove that N is not divisible by 3.

from https://www.imomath.com/index.php?options=585 Problem 3
The answer shows that all integer points matching this inequality can be divided into 3 parts as {(x,y),(-x-y,x),(y,-x-y)} except (0,0).
The conclusion seems also correct for all real points inside the ellipse $x^2+xy+y^2=A$. So the ellipse is divided into 3 parts except the center point.
My question is: what are the areas of these 3 parts? Is it related to measure theory (or something like Banach–Tarski paradox)?
 A: Let $\omega = e^{\frac{2\pi i}{3}}$ be a cubic root of unity.
For each $(x,y) \in \mathbb{Z}^2$, associate with it a complex number $z = x - y\omega$. We have
$$|x - y\omega|^2 = x^2 + xy + y^2$$
The collection of these $z$ form a triangular lattice in $\mathbb{C}$
(known as Eisenstein integers).
Notice
$$\begin{array}{rll}
 (-x-y,x) & \to (-x-y) - x\omega &= \omega^2 z\\
  (y,-x-y) &\to y - (-x-y)\omega &= \omega z
\end{array}$$
The relation among the three pairs $(x,y), (-x-y,x), (y,-x-y)$
isn't anything exotic. They are simply images of each other under
$120^\circ$ rotations of the Eisenstein integers (embedded in the complex plane).
A: Here's a picture of your ellipse: when point $A=(x,y)$ lies in the first quadrant (blue region), then point $B=(-x-y,x)$ lies in the red region and point $C=(y,-x-y)$ lies in the green region.
Switching the sign of coordinates for $A$ (third quadrant) causes the same to happen for $B$ and $C$. And areas are indeed the same:
$$
\text{blue area}=\text{red area}=\text{green area}=
\frac{2012 \pi }{3 \sqrt{3}}.
$$
That the area is the same can be proved just by noticing that the transformation sending $A$ to $B$ (or to $C$) is a linear mapping with determinant equal to $1$.

