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.

The point $P(X,Y)$ such that $\triangle{ABX}$ is an equilateral triangle satisfies
$$X+Yi-(1+0i)=(2+2\sqrt 3i-(1+0i))(\cos(\pm 60^\circ)+i\sin(\pm 60^\circ))$$
$$\Rightarrow (X,Y)=(-3/2,3\sqrt 3/2),(9/2,\sqrt 3/2).$$
So, $C(9/2,\sqrt 3/2)$ because $C$ lies in the first quadrant. Hence, the center of $\triangle{ABC}$ is
$$\left(\frac{1+2+(9/2)}{3},\frac{0+2\sqrt 3+(\sqrt 3/2)}{3}\right)=\left(\frac{5}{2},\frac{5}{6}\sqrt 3\right)$$
Thus, 
$$\frac{5}{2}\cdot\frac{5}{6}\sqrt 3=\frac{25}{12}\sqrt 3\Rightarrow p+q+r=25+3+12=40.$$
