32nd Eötvös 1928 Problem 3 Given two points A, B and a line L in the plane, find the point P on the line for which max(AP, BP) is as short as possible.
Let B be the point further from the line (or one of them). Let B' be the foot of the perpendicular from B to the line. If BB' > AB', then we must obviously take P = B'. If not, then let X be the point where the perpendicular bisector of AB meets the line. We take P to be X. For X must lie between A' and B'. If we take P to the right of X then max(AP,BP) = AP and AP > AX. If we take P to the left of X, then max(AP, BP) = BP and BP > BX. 32nd Eötvös 1928 © John Scholes jscholes@kalva.demon.co.uk 1 Nov 2003 Last corrected/updated 1 Nov 03