Question

In the Cartesian plane let $A = (1, 0)$ and $B = (2, 2\sqrt{3}).$ 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 $\frac{p\sqrt{q}}{r},$ 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 $p + q + r.$

Answer

Solution:

The midpoint of $\overline{AB}$ is $M = \left(\frac{3}{2}, \sqrt{3}\right),$ and $AB = \sqrt{1 + 12} = \sqrt{13}.$ The third vertex lies at distance $\frac{\sqrt{3}}{2}\sqrt{13}$ from $M$ along a direction perpendicular to $\overrightarrow{AB} = (1, 2\sqrt{3});$ a unit perpendicular is $\frac{1}{\sqrt{13}}(2\sqrt{3}, -1).$ Taking the sign that lands in the first quadrant, $C = M + \frac{\sqrt{3}}{2}\,(2\sqrt{3}, -1) = \left(\frac{9}{2}, \frac{\sqrt{3}}{2}\right)$ (the other choice has negative $x$ -coordinate).

The center of an equilateral triangle is its centroid, the average of the vertices: $P = \left(\frac{1 + 2 + \frac{9}{2}}{3},\ \frac{0 + 2\sqrt{3} + \frac{\sqrt{3}}{2}}{3} \right) = \left(\frac{5}{2}, \frac{5\sqrt{3}}{6}\right).$

Then $x \cdot y = \frac{5}{2} \cdot \frac{5\sqrt{3}}{6} = \frac{25\sqrt{3}}{12},$ so $p + q + r = 25 + 3 + 12 = 40.$

2013 AIME II Problem 4

Solution:

Problem 4 in Other Years