19th BMO 1983 Problem 1 Given points A and B and a line l , find the point P which minimises PA 2 + PB 2 + PN 2 , where N is the foot of the perpendicular from P to l . State without proof a generalisation to three points.
Let M be the midpoint of AB. We have the familiar result that PA 2 + PB 2 = 2PM 2 + 2MA 2 . Hence P must minimise 2PM 2 + PN 2 and hence lie on MC, the perpendicular from M to the line. Let G be the point on the segment MC such that GC = 2MG. Suppose P is a distance x from G, taking positive distances towards C. Then PC 2 + 2PM 2 = (GC-x) 2 + 2(GM+x) 2 = GC 2 + 2GM 2 + 2x(2GM-GC) + 2x 2 = GC 2 + 2GM 2 + 2x 2 , which is minimised by taking x = 0. So we should take P at G. For 3 points and a line, take D to be the centroid of the three points and D' to be the foot of the perpendicular from D to the line. Then P should be taken at the centroid of the three points and D'. 19th BMO 1983 © John Scholes jscholes@kalva.demon.co.uk 10 February 2004 Last corrected/updated 10 Feb 04