Not Three Points, But Four Points!

The first thing to note is that, by symmetry, $A$ and $D$ are diametrically opposite points. Since points $A,B,C,D$ all lie on the unit circle, we can write their coordinates as $A(\cos q,\sin q),\;\;B\left(\cos \left(q-\frac{2\pi}{3}\right),\sin \left(q-\frac{2\pi}{3}\right)\right)\;\;C\left(\cos \left(q+\frac{2\pi}{3}\right),\sin \left(q+\frac{2\pi}{3}\right)\right)\;\;D(-\cos q,-\sin q)$

where $q=\theta+\frac{2\pi}{3}$ .

All the points are also on the parabola; starting with $A$ and $D$ , $a_2 \cos^2 q + a_1 \cos q + a_0 = \sin q$ and $a_2 \cos^2 q - a_1 \cos q + a_0 = -\sin q$

Subtracting these gives $a_1=\tan q$

Adding gives $a_0=-a_2 \cos^2 q$

Plugging in these results and the coordinates of $B$ , $a_2 \cos^2 \left(q-\frac{2\pi}{3}\right) + \tan q \cos \left(q-\frac{2\pi}{3}\right) - a_2 \cos^2 q = \sin \left(q-\frac{2\pi}{3}\right)$

This solves to give $a_2=\sec \left(\frac{\pi}{6}-2q\right) \sec q$

Similarly, from point $C$ we find $a_2=-\sec \left(\frac{\pi}{6}+2q\right) \sec q$

Equating these, $\begin{aligned}\sec \left(\frac{\pi}{6}-2q\right) \sec q &=-\sec \left(\frac{\pi}{6}+2q\right) \sec q \\ \sec \left(\frac{\pi}{6}-2q\right) &=-\sec \left(\frac{\pi}{6}+2q\right) \\ \cos \left(\frac{\pi}{6}-2q\right) +\cos \left(\frac{\pi}{6}+2q\right) &=0 \\ 2\cos \left(\frac{\pi}{6}\right) \cos 2q &=0 \\ \cos 2q&=0 \end{aligned}$

so $q=\frac{(2k+1)\pi}{4}$ for some integer $k$ . The different roots give reflections of the given diagram; from the diagram and given conditions we want $q=\frac{3\pi}{4}$ so that $\theta=\frac{\pi}{12}$ and the answer is $\boxed{12}$ .