2013 AIME II Problems/Problem 4

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, $M = (\dfrac{3}{2},\sqrt{3})$ , $\overline{AB}$ can be parameterized by $(\dfrac{2\sqrt{3}}{\sqrt{13}}, \dfrac{-1}{\sqrt{13}}) t + (\dfrac{3}{2},\sqrt{3})$ . 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 lenght of $\overline{MP}$ to be $\dfrac{\sqrt{13}}{2\sqrt{3}}$ . Therefore, $P =( \dfrac{2\sqrt{3}}{\sqrt{13}},\dfrac{-1}{\sqrt{13}})(\dfrac{\sqrt{13}}{2\sqrt{3}}) + (\dfrac{3}{2},\sqrt{3}) = (\dfrac{5}{2}, \dfrac{5}{2\sqrt{3}})$ . Therefore $xy = \dfrac{25\sqrt{3}}{12}$ yielding an answer of $p + q + r = 25 + 3 + 12 = 040$ .

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2013 AIME II Problems/Problem 4