AIME 2013 II, Problem 4

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 $p+q+r$.

Solution: The distance from point $A$ to point $B$ is $\sqrt{13}$. The vector that starts at point A and ends at point B is given by $B - A = (1, 2\sqrt{3})$. Since the center of an equilateral triangle, $P$, is also the intersection of the perpendicular bisectors of the sides of the triangle, we need first find the equation for the perpendicular bisector to $\overline{AB}$. The line perpendicular to $\overline{AB}$ through the midpoint, $M =  \left(\dfrac{3}{2},\sqrt{3}\right)$, $\overline{AB}$ can be parameterized by $\left(\dfrac{2\sqrt{3}}{\sqrt{13}}, \dfrac{-1}{\sqrt{13}}\right) t + \left(\dfrac{3}{2},\sqrt{3}\right)$. At this point, it is useful to note that $\Delta BMP$ is a 30-60-90 triangle with $\overline{MB}$ measuring $\dfrac{\sqrt{13}}{2}$. This yields the length of $\overline{MP}$ to be $\dfrac{\sqrt{13}}{2\sqrt{3}}$. Therefore, $P =\left(\dfrac{2\sqrt{3}}{\sqrt{13}},\dfrac{-1}{\sqrt{13}}\right)\left(\dfrac{\sqrt{13}}{2\sqrt{3}}\right) + \left(\dfrac{3}{2},\sqrt{3}\right) = \left(\dfrac{5}{2}, \dfrac{5}{2\sqrt{3}}\right)$. Therefore $xy = \dfrac{25\sqrt{3}}{12}$ yielding an answer of $p + q + r  = 25 + 2 + 12 = \fbox{40}$.

Answer: 40