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Equilateral triangle at (1,0) and (2,2rt3)

Source: AIME II 2013, Problem 4

April 4, 2013
analytic geometrytrigonometrygeometrygeometric transformationrotationAMCAIME

Problem Statement

In the Cartesian plane let A=(1,0)A = (1,0) and B=(2,23)B = \left( 2, 2\sqrt{3} \right). Equilateral triangle ABCABC is constructed so that CC lies in the first quadrant. Let P=(x,y)P=(x,y) be the center of ABC\triangle ABC. Then xyx \cdot y can be written as pqr\tfrac{p\sqrt{q}}{r}, where pp and rr are relatively prime positive integers and qq is an integer that is not divisible by the square of any prime. Find p+q+rp+q+r.