Divide an ellipse into three parts

Question

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)?

Answer 1

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).

Answer 2

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 $1$.