Circle Triangle Hybrid

For the circle:

$\large \displaystyle r_c(\theta) = 1$

Let us consider only the right purple area, since the total purple area will be $3$ times this one. For the triangle, let us define a straight line from the origin with angle $\theta$ with the horizontal. Its equation will be:

$\large \displaystyle y = \tan(\theta) x$

The right side of the triangle has equation:

$\large \displaystyle y = -\sqrt{3} x + 1$

So the point of intersection with the triangle (i.e. a generic point in the right side of the triangle) has coordinates:

$\large \displaystyle x = \frac{1}{\tan(\theta) + \sqrt{3}}, y = \frac{\tan(\theta)}{\tan(\theta) + \sqrt{3}}$

And then we can calculate $r_t(\theta)$ :

$\large \displaystyle r_t^2(\theta) = x^2 + y^2 = \frac{\tan^2(\theta) + 1}{(\tan(\theta) + \sqrt{3})^2}$

$\large \displaystyle r_t^2(\theta) = \frac{\sec^2(\theta)}{(\tan(\theta) + \sqrt{3})^2}$

$\large \displaystyle r_t(\theta) = \frac{\sec(\theta)}{\tan(\theta) + \sqrt{3}}$

So:

$\color{#20A900} \boxed{ \large \displaystyle r_h(\theta) = \frac12 \left(\frac{\sec(\theta)}{\tan(\theta) + \sqrt{3}} + 1 \right ) }$

Let us consider the red marked area in the figure below:

This area will be given by the integral:

$\large \displaystyle \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} \frac12 r_h^2(\theta) d \theta$

So the area $A_p$ of one purple area will be this integral minus the area of the triangle with legs as the two radii of the circle marked in red in the figure, which make an angle of $\frac{2 \pi}{3}$ between them. So, using the area for a triangle knowing two sides and the angle between them:

$\large \displaystyle A_p = \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} \frac12 \left [ \frac12 \left(\frac{\sec(\theta)}{\tan(\theta) + \sqrt{3}} + 1 \right ) \right ] ^2 d \theta - \frac12 \cdot 1 \cdot 1 \cdot \sin \left ( \frac{2\pi}{3} \right)$

$\large \displaystyle A_p = \frac18 \left [\int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{ \sec^2(\theta)}{ (\tan(\theta)+\sqrt{3})^2} d \theta + \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} d\theta + 2 \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{ \sec(\theta)}{ \tan(\theta)+\sqrt{3}} d \theta \right] - \frac{\sqrt{3}}{4}$

The first and second integrals are easy to evaluate. On the third, multiplying above and below by $\cos(\theta)$ :

$\large \displaystyle A_p = \frac18 \left [ \left ( \frac{-1}{(\tan(\theta)+\sqrt{3})} \right) \Bigg |_{-\frac{\pi}{6}}^{\frac{\pi}{2}} + \theta \Bigg |_{-\frac{\pi}{6}}^{\frac{\pi}{2}} + 2 \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{ \sin(\theta)+\sqrt{3}\cos(\theta)} d \theta \right] - \frac{\sqrt{3}}{4}$

Rewriting the remaining intrgral:

$\large \displaystyle A_p = \frac18 \left [ \frac{\sqrt{3}}{2} + \frac{2\pi}{3} + \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{\frac12 \sin(\theta)+ \frac{\sqrt{3}}{2}\cos(\theta)} d \theta \right] - \frac{\sqrt{3}}{4}$

$\large \displaystyle A_p = \frac18 \left [ \frac{\sqrt{3}}{2} + \frac{2\pi}{3} + \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{\cos \left( \theta-\frac{\pi}{6} \right)} d \theta \right] - \frac{\sqrt{3}}{4}$

Make $t = \theta - \frac{\pi}{6}$ , $dt = d \theta$ :

$\large \displaystyle A_p = \frac18 \left [ \frac{\sqrt{3}}{2} + \frac{2\pi}{3} + \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \sec(t) dt \right] - \frac{\sqrt{3}}{4}$

$\large \displaystyle A_p = \frac18 \left [ \frac{\sqrt{3}}{2} + \frac{2\pi}{3} + \ln(\sec(t)+ \tan(t)) \Bigg |_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \right] - \frac{\sqrt{3}}{4}$

$\large \displaystyle A_p = \frac18 \left [ \frac{\sqrt{3}}{2} + \frac{2\pi}{3} + \ln \left ( \frac{2+\sqrt{3}}{2-\sqrt{3}} \right ) \right] - \frac{\sqrt{3}}{4}$

Rationalizing inside the $\ln$ :

$\large \displaystyle A_p = \frac18 \left [ \frac{\sqrt{3}}{2} + \frac{2\pi}{3} + \ln \left ( ( 2+\sqrt{3} )^2 \right ) \right] - \frac{\sqrt{3}}{4}$

$\large \displaystyle A_p = \frac{\pi}{12} + \frac{\ln \left (2+\sqrt{3} \right ) }{4} - \frac{3 \sqrt{3}}{16}$

Since we calculated only one of the $3$ purple areas, $A = 3A_p$ :

$\color{#20A900} \boxed{ \large \displaystyle A = \frac{\pi}{4} + \frac{3 \ln \left (2+\sqrt{3} \right ) }{4} - \frac{9 \sqrt{3}}{16} \approx 0.798838 }$

Thus:

$\color{#3D99F6} \boxed{ \large \displaystyle \lfloor 1000A \rfloor = 798 }$

The triangle can be parameterised in polar coordinates as such:

$r_t(\theta) = \begin{cases} \frac{\sec{\theta}}{\tan{\theta} + \sqrt{3}} & 0 \le \theta \le \frac{\pi}{2} \\ \frac{\sec{\theta}}{\tan{\theta} - \sqrt{3}} & \frac{\pi}{2} \le \theta \le \frac{7\pi}{6} \\ \frac{-0.5}{\sin{\theta}} & \frac{7\pi}{6} \le \theta \le \frac{11\pi}{6}\\ \frac{\sec{\theta}}{\tan{\theta} + \sqrt{3}} & \frac{11\pi}{6} \le \theta \le 2 \pi\\ \end{cases}$

Finally: the required polar curve is:

$r_h = \frac{r_t + 1}{2}$

And the purple area is computed as such:

$A = \int_{0}^{2\pi} \frac{1}{2}r_h^2 \ d\theta - \int_{0}^{2\pi} \frac{1}{2}r_t^2 \ d\theta$ $\implies A \approx 0.7988$

Integrals can be solved using any numerical tool.