ABC is an acute-angled triangle. P is a point inside its circumcircle. The rays AP, BP, CP intersect the circle again at D, E, F. Find P so that DEF is equilateral.
Solution
Let the angle bisector of A meet BC at A'. Let the perpendicular bisector of AA' meet the line BC at X. Take the circle center X through A and A'. Similarly, let the angle bisector of B meet AC at B' and let the perpendicular bisector of BB' meet the line AC at Y. Take the circle center Y through B and B'. The two circles meet at a point P inside the triangle, which is the desired point.
PAB and PED are similar, so DE/AB = PD/PB. Similarly, DF/AC = PD/PC, so DE/DF = (AB/AC)(PC/PB). Thus we need PB/PC = AB/AC. So P must lie on the circle of Apollonius, which is the circle we constructed with center X. Similarly, it must lie on the circle of Apollonius with center Y and hence be one of their points of intersection. It also lies on the third circle and hence we choose the point of intersection inside the triangle.
© John Scholes
jscholes@kalva.demon.co.uk
3 February 2004