Problem:
In the Cartesian plane let A=(1,0) and B=(2,23). Equilateral triangle ABC is constructed so that C lies in the first quadrant. Let P=(x,y) be the center of △ABC. Then x⋅y can be written as rpq, 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 midpoint of AB is M=(23,3). Then AM=⟨21,3⟩ and MP=31⟨3,−21⟩. It follows that P=M+MP=(25,653). Thus xy=12253. The requested sum is p+q+r=25+3+12=40.
OR
In the complex plane A=1,B=2+23i, and P=x+yi. Note that a 32π counterclockwise rotation centered at P maps B to A. Hence
This is equivalent to the conditions 3x+y3=10 and 3y=x. It follows by simple substitution that (x,y)=(25,653).
OR
Let O=(0,0). Note that ∠AOB+∠APB=60+120=180. Hence AOBP is a cyclic quadrilateral. Let AB=3t. Now by properties of a 30−30−120∘ triangle, t=AP=BP. By Ptolemy's Theorem, t⋅1+t⋅4=t3⋅OP and OP=35. Because AOBP is cyclic, ∠ABP=∠POA=30∘. So xy=OP2cos30⋅sin30=325⋅43=12253.