P is a point inside the triangle ABC. D, E, F are the feet of the perpendiculars from P to the lines BC, CA, AB respectively. Find all P which minimise:
BC/PD + CA/PE + AB/PF.
Solution
We have PD.BC + PE.CA + PF.AB = 2 area of triangle. Now use Cauchy's inequality with x1 = √(PD·BC), x2 = √(PE·CA), x3 = √(PF·AB), and y1 = √(BC/PD), y2 = √(CA/PE), y3 = √(AB/PF). We get that (BC + CA + AB)2 < 2 x area of triangle x (BC/PD + CA/PE + AB/PF) with equality only if xi/yi = const, ie PD = PE = PF. So the unique minimum position for P is the incenter.
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
14 Oct 1998