Find the area of a triangle inscribed in the ellipse

Question

(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.

Answer 1

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 $\color{blue}{\frac{192\sqrt{3}}{49}}$.

Answer 2

$A=(0,3)$ 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 $[ABC]=\color{red}{\frac{192}{49}\sqrt{3}}$.

Answer 3

$A$ is a vertex of the ellipse. If we have $B(-x,y)$ and $C(x,y)$ with $\triangle{ABC}$ equilateral, its area is simply $\sqrt3x^2$. There are several ways to find $x$ such that $B$ and $C$ lie on the ellipse and the triangle is equilateral. One way is to find the intersection of the line $AB$ with the ellipse. The equation of this line is $y=\sqrt3x-3$. Substituting this into the equation of the ellipse and simplifying yields $x(7x+8\sqrt3)=0$, so $x=-\frac{8\sqrt3}7$ and the area of the triangle is ${192\sqrt3\over49}$.