Brother Brocard's Brilliant Biological Bond

Time for a general approach using trilinear coordinates. Let us set up the (by now) standard angle $\theta = \angle BAC$ , so that $A,B,C$ have coordinates $(-1,0)$ , $(1,0)$ and $(\cos2\theta,\sin2\theta)$ respectively, and so the triangle $ABC$ has sides $a=2\sin\theta$ , $b=2\cos\theta$ , $c=2$ . Consider a point $P$ which has exact trilinear coordinates $u:v:w$ , so that $u,v,w$ are the perpendicular distances from $P$ to the sides $BC,AC,AB$ respectively. We note that $\tfrac12(au + bv + cw) = \Delta$ , the area of $ABC$ , and hence $u\sin\theta + v\cos\theta + w \; = \; \sin2\theta$ Since $CDPE$ is a rectangle, we deduce that $\overrightarrow{OP} \; = \; \overrightarrow{OC} - u\binom{\cos\theta}{\sin\theta} + v\binom{\sin\theta}{-\cos\theta}$ and hence $P$ has coordinates $\big(X(\theta),Y(\theta)\big) \; = \; \big(\cos2\theta -u\cos\theta +v\sin\theta,w\big)$ The two Brocard points $\Omega$ and $\Omega'$ have relative trilinear coordinates $\frac{c}{b}\,:\, \frac{a}{c}\,:\, \frac{b}{a} \hspace{1cm} \frac{b}{c}\,:\, \frac{c}{a}\,:\, \frac{a}{b}$ respectively, and so $\Omega$ has coordinates $\big(X(\theta),Y(\theta)\big) \; = \; \left(\frac{8(\cos^4\theta + \cos^2\theta - 1)}{9 - \cos4\theta}\,,\,\frac{16\sin\theta\cos^3\theta}{9 - \cos4\theta}\right)$ The loci of $\Omega$ and $\Omega'$ are shown in blue and purple respectively here. The locus of $\Omega'$ is the reflection of the locus of $\Omega$ in the $y$ -axis. Thus the desired area is equal to $4$ times the area of the interior of the locus of $\Omega$ that lies in the first quadrant. Thus we want to calculate $A \;=\; 4\int_{\theta_0}^0 X'(\theta)Y(\theta)\,d\theta \; = \; 2048 \int_0^{\theta_0} \frac{\cos^4\theta (11 + 4\cos2\theta + \cos4\theta) \sin^2\theta}{(9 - \cos4 \theta)^3}\,d\theta$ where $\theta_0 = \cos^{-1}\sqrt{\tfrac12(\sqrt{5}-1)}$ is the value of the parameter $\theta$ where the locus of $\Omega$ meets the positive $y$ -axis. A $t = \tan\theta$ substitution, followed by partial fractions, will handle this integral, and we obtain $A \; =\; \frac15\left[5 \sqrt{2 (\sqrt5-1)} - 10 \tan^{-1}\sqrt{\sqrt5 -2} + 2 \sqrt5 \tan^{-1}\sqrt{5 (2 + \sqrt5)}\right]$ and hence $\lfloor 10^6 A \rfloor \; =\; \boxed{1881336}$ .