Find the area of a triangle inscribed in the ellipse (Unicamp Mathematics Olympiad) Let $\zeta$ denote the Carteasian region given by $\zeta = \{(x,y): \frac{x^2}{4} + \frac{y^2}{9} = 1\}$. Let $A = (0,3), B = (x,y), C = (z,w)$ be points such that $A,B,C \in \zeta%$ and $\Delta ABC$ is equilateral. What is the area of $\Delta ABC$?
My approach: Since $B, C$ are homogeneous, I tried to do enough algebra so everything "cuts out". I used Gauss' formula for the area of a triangle given tree points, the fact that ${AB}^2 = {BC}^2 = {AC}^2$ and used the explicit equation of $\zeta$, but it ended up being too much algebra and I couldn't find an explicit result.

$A=(0,3)$ is a vertex of the given ellipse. If $ABC$ is equilateral, by symmetry we have that $B$ and $C$ share the same $y$-coordinate, hence they are points of the form
$$ B=\left(-2\sqrt{1-\frac{y^2}{9}},y\right),\qquad C=\left(2\sqrt{1-\frac{y^2}{9}},y\right) $$
with $y\in(-3,3)$.  In order that $ABC$ really is equilateral, we must have
$$ 3-y = \sqrt{3}\cdot 2\sqrt{1-\frac{y^2}{9}} $$
hence $y=-\frac{3}{7}$ and $BC=\frac{16}{7}\sqrt{3}$. It follows that $[ABC]=\color{red}{\frac{192}{49}\sqrt{3}}$.

