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.

The points of the ellipse are parameterised by $(2 \cos \theta, 3 \sin \theta)$ & by symmetry it is clear that the points $B$ and $C$ will have coordinates  $(2 \cos \theta, 3 \sin \theta)$ & $(-2 \cos \theta, 3 \sin \theta)$. So the requirement of equilateralness gives
\begin{eqnarray*}
(4 \cos \theta)^2=  (4 \cos \theta)^2+ (3 (1-\sin \theta))^2 \\
4 \cos^2 \theta=  3 (1-\sin \theta))^2  \\
4 (1+\sin \theta) =  3 (1-\sin \theta) \\
\end{eqnarray*}
The factor $(1-\sin \theta)$ can be discarded as it corresponds to the the points being concurrent. So we have $\sin \theta =- \frac{1}{7}$ and $\cos \theta =- \frac{4\sqrt{3}}{7}$. The length of the sides of the triangle are  $\frac{16\sqrt{3}}{7}$ and the area of the triangle is $\color{blue}{\frac{192\sqrt{3}}{49}}$.
