ABC is an equilateral triangle. P is a variable interior point such that ∠APC = 120o. The ray CP meets AB at M, and the ray AP meets BC at N. What is the locus of the circumcenter of the triangle MBN as P varies?
Solution
Answer: the segment of the perpendicular bisector of BG (where G is the center of the triangle) which forms a rectangle with AC.
∠MPN = ∠APC = 120o and ∠MBN = 60o, so MBNP is cyclic, in other words, P lies on the circumcircle of BMN.
P also lies on the circle AGC, so ∠CPG = ∠CAG (if P is on the same side of AG as A) = 30o = ∠MBG. So PMBG is cyclic. In other words, G also lies on the circumcircle of BMN. If P lies on the other side, the same conclusion follows from considering ∠APG.
Since B and G lie on the circumcircle, the center O must lie on the perpendicular bisector of BG. But it is clear that the extreme positions of O occur when P is at A and B and that these are the feet of the perpendiculars from A and B to the perpendicular bisector.
© John Scholes
jscholes@kalva.demon.co.uk
19 Oct 2002