Finding a point using complex geometry 
In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$. Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$. Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$, where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$.

I have seen solutions as follows:
Consider: $A = 1 + 0i$  and $B = 2 + i2\sqrt{3}$
Obviously inside a triangle each angle is $60$ degrees. 
Really we need to consider $B$ since that is the angle of the ray anyway. 
So: $\theta = \arctan(2/2 = 1) = \frac{\pi}{4}$.
The third point must then be $\theta_2 = \frac{\pi}{4} + \frac{\pi}{3} = \frac{7\pi}{12}$
But this cant be since it isnt in the first quadrant.
$\frac{\pi}{4} - \frac{\pi}{3} = -\frac{\pi}{12}$.
but that isnt the first quadrant either.

Let
$$\omega=\frac{1+i\sqrt{3}}{2},\qquad \overline{\omega}=\frac{1-i\sqrt{3}}{2}$$
be primitive sixth roots of unity. Given $A=1,B=2+2i\sqrt{3}$, in order that $ABC$ is an equilateral triangle, $C$ must be the image of $B$ with respect to a $\pm 60^\circ$ rotation with centre in $A$, hence:
$$ C = \frac{1\pm i\sqrt{3}}{2}(B-A)+A $$
so $C=\frac{9}{2}+i\frac{\sqrt{3}}{2}$ or $C=-\frac{3}{2}+i\frac{3\sqrt{3}}{2}$. We have a point in the first quadrant only in the first case, and the center of $ABC$ is given by the centroid:
$$ G = \frac{A+B+C}{3} = \frac{5}{6}\left(3+i\sqrt{3}\right) $$
so:
$$ x\cdot y = \frac{25}{36}\cdot 3\sqrt{3} = \frac{25\sqrt{3}}{12} $$
and $p+q+r=25+3+12=\color{red}{40}$.
