Deceptively Complex

It is easy to verify the coordinates $Q(2+\sqrt{3}, 1)$ , $R(2+\sqrt{3}, 1+\sqrt{3})$ and $S(3+\sqrt{3}, 3+\sqrt{3})$ .

Next, we find $|OS|$ . The readers are invited to verify that $|OS|=\sqrt{(2){(3+\sqrt{3})}^{2}}=\sqrt{18}+\sqrt{6}$ .

Since triangle $OPQ$ is isosceles, the perpendicular bisector through $P$ bisects the $\angle OPQ$ as well. Since $O$ and $P$ lie on the circle, $|CO|=|CP|$ and so $\angle CPO = \angle COP = 75 ^ \circ$ .

Therefore, for point $C$ to touch the real axis, the whole polygon rotates by $105 ^ \circ$ about point $O$ .

Note that the ray $\overline{OS}$ makes an angle of $45 ^ \circ$ with the positive real axis. Upon a rotation of $105 ^ \circ$ about $O$ , the ray $\overline{OS}$ now makes a degree of $30 ^ \circ$ with the negative real axis.

It remains for us to evaluate the new position of $S$ , but this is simple. It is just

$(-(\sqrt{6}+\sqrt{18})\cos 30 ^ \circ, (\sqrt{6}+\sqrt{18})\sin 30 ^ \circ)$ , which simplifies to

$(-\frac{3}{2}(\sqrt{6}+\sqrt{2}), \frac{3\sqrt{2}+\sqrt{6}}{2} i)$ ,

from which the answer $3+2+6+2+3+2+6+2=\boxed{26}$ follows.