Fancy Triangle

Suppose that the centroid of such an equilateral triangle has coordinates $(x,y)$ , so that the three vertices of the equilateral triangle have coordinates $P_j \;\; \big(x + r\cos\big(\theta + \tfrac{2\pi j}{3}\big),y + r\sin\big(\theta + \tfrac{2\pi j}{3}\big)\big) \hspace{2cm} j = 0,1,2$ for some $r > 0$ and $0 \le \theta < 2\pi$ . Then $x,y,r,\theta$ must satisfy the equations $F_0(x,y,r,\theta) \; = \; F_1(x,y,r,\theta) \; = \; F_2(x,y,r,\theta) \; = \; 1$ where $F_j(x,y,r,\theta) \; = \; \frac{\big(x + r\cos\big(\theta+\frac{2\pi j}{3}\big)\big)^2}{a^2} + \frac{\big(y + r\sin\big(\theta+\frac{2\pi j}{3}\big)\big)^2}{b^2}$ The equation $F_1(x,y,r,\theta) = F_2(x,y,r,\theta)$ leads to the equation $- 2b^2x\sin\theta + 2a^2y\cos\theta \; = \; (a^2-b^2)r\sin\theta\cos\theta$ while the equation $2F_0(x,y,r,\theta) = F_1(x,y,r,\theta) + F_2(x,y,r,\theta)$ leads to the equation $4b^2x\cos\theta + 4a^2y\sin\theta \; = \; (a^2-b^2)r\cos2\theta$ and hence we deduce that $x \; = \; \frac{a^2-b^2}{4b^2}r\cos3\theta \hspace{2cm} y \; = \; \frac{a^2-b^2}{4a^2}r\sin3\theta$ The equation $F_0(x,y,r,\theta) = 1$ can now be solved to obtain $r$ as a function of $\theta$ , and the end result of all this is that the coordinates of the centroid lie on the ellipse $\frac{x^2}{A^2} + \frac{y^2}{B^2} \; = \; 1$ where $A \; = \; \frac{a(a^2-b^2)}{a^2+3b^2} \hspace{2cm} B \; = \; \frac{b(a^2-b^2)}{3a^2+b^2}$ and so the area enclosed by this ellipse is $\pi AB \; = \; \boxed{\frac{\pi ab(a^2-b^2)^2}{(3a^2+b^2)(a^2+3b^2)}}$