Problem:
In the Cartesian plane let A=(1,0) and B=(2,23β). Equilateral triangle ABC is constructed so that C lies in the first quadrant. Let P=(x,y) be the center of β³ABC. Then xβ y can be written as rpqββ, 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.
This is equivalent to the conditions 3x+y3β=10 and 3βy=x. It follows by simple substitution that (x,y)=(25β,653ββ).
OR
Let O=(0,0). Note that β AOB+β APB=60+120=180. Hence AOBP is a cyclic quadrilateral. Let AB=3βt. Now by properties of a 30β30β120β triangle, t=AP=BP. By Ptolemy's Theorem, tβ 1+tβ 4=t3ββ OP and OP=3β5β. Because AOBP is cyclic, β ABP=β POA=30β. So xy=OP2cos30β sin30=325ββ 43ββ=12253ββ.