Hoci Foci Loci

Let the circle have radius $2$ . Simple angle-chasing and trigonometry tell us that $\begin{aligned} r_1(\theta) & =\; OA \; = \; \frac{2\cos2\theta}{\cos\theta} \\ r_2(\theta) & = \; OB \; = \; \frac{\sin4\theta}{\sin3\theta} \end{aligned}$ for $|\theta| \le \tfrac14\pi$ . Thus the blue area is $\tfrac12\int_{-\frac14\pi}^{\frac14\pi} r_1(\theta)^2\,d\theta \; = \; 8 - 2\pi$ while the red area is $\tfrac12\int_{-\frac14\pi}^{\frac14\pi}r_2(\theta)^2\,d\theta \; =\; \tfrac43 - \tfrac49\sqrt{3}\tanh^{-1}\tfrac{1}{\sqrt{3}}$ which makes the ratio of areas $R = 0.4813775219...$ and hence $\lfloor 10^6 R \rfloor = \boxed{481377}$ .

Let the circle be a unit circle with an equation of $x^2 + y^2 = 1$ , and label the diagram as follows:

Let the $x$ -coordinate of $C$ be $k$ . Since it is on the circle $x^2 + y^2 = 1$ , it has a $y$ -coordinate of $\sqrt{1 - k^2}$ , so its coordinates are $C(k, \sqrt{1 - k^2})$ .

Then $CA$ has an equation of $y=\cfrac{\sqrt{1-k^{2}}}{k+1}(x+1)$ , $CO$ has an equation of $y=\cfrac{\sqrt{1-k^{2}}}{k}x$ , and $CB$ has an equation of $y=\cfrac{\sqrt{1-k^{2}}}{k-1}(x-1)$ .

Since $BD$ is the angle bisector of $\angle CBA$ , by the half angle formula $\tan \angle CBA = \cfrac{\sqrt{1-k^{2}}}{k-1} = \cfrac{2 \tan \angle DBA}{1 - \tan^2 \angle DBA}$ , which rearranges to $\tan \angle DBA = \cfrac{\sqrt{1-k^{2}}-\sqrt{2}\sqrt{1+k}}{k+1}$ .

Then $DB$ has an equation of $y = \cfrac{\sqrt{1-k^{2}}-\sqrt{2}\sqrt{1+k}}{k+1}(x - 1)$ .

Combining $y = \cfrac{\sqrt{1-k^{2}}-\sqrt{2}\sqrt{1+k}}{k+1}(x - 1)$ and $y=\cfrac{\sqrt{1-k^{2}}}{k+1}(x+1)$ and eliminating $k$ gives $y=\pm \sqrt{\cfrac{-x^{3}+x^{2}+x-1}{x-3}}$ , the equation of the blue curve, which contains an area of $\displaystyle 2 \int_{-1}^{1} \sqrt{\cfrac{-x^{3}+x^{2}+x-1}{x-3}} dx = 8 - 2\pi$ .

Combining $y = \cfrac{\sqrt{1-k^{2}}-\sqrt{2}\sqrt{1+k}}{k+1}(x - 1)$ and $y=\cfrac{\sqrt{1-k^{2}}}{k}x$ and eliminating $k$ gives $y=\pm \sqrt{(x-1)(x-3)-2\sqrt{(x-1)^{2}(x^{2}-2x+2)}}$ , the equation of the red curve, which contains an area of $\displaystyle 2 \int_{-\frac{1}{3}}^{1} \sqrt{(x-1)(x-3)-2\sqrt{(x-1)^{2}(x^{2}-2x+2)}} dx \approx 0.8264360025$ .

The ratio of the red area to the blue area is then $R \approx \cfrac{0.8264360025}{8 - 2\pi} \approx 0.4813775$ , so $\lfloor 10^6 R \rfloor = \boxed{481377}$ .