imc

WLOG, let the centroid of $△ABC$ be $I=(-1,-1)$. The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, $A=(1,1)$, so $AI=BI=CI=2\sqrt{2}$, so since $△AIB$ is isosceles and $\angle AIB=120^{\circ }$, then by the Law of Cosines, $AB=2\sqrt{6}$. Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to $\frac{s}{\sqrt{3}}$. Therefore, the area of the triangle is $\frac{(2\sqrt{6}{)}^2\sqrt{3}}{4}=6\sqrt{3}$, so the square of the area of the triangle is $\boxed{108}$. Note: We could've also noticed that the centroid divides the median into segments of ratio $2:1.$