Given a plane k, a point P in the plane and a point Q not in the plane, find all points R in k such that the ratio (QP + PR)/QR is a maximum.
Solution
Consider the points R on a circle center P. Let X be the foot of the perpendicular from Q to k. Assume P is distinct from X, then we minimise QR (and hence maximise (QP + PR)/QR) for points R on the circle by taking R on the line PX. Moreover, R must lie on the same side of P as X. Hence if we allow R to vary over k, the points maximising (QP + PR)/QR must lie on the ray PX. Take S on the line PX on the opposite side of P from X so that PS = PQ. Then for points R on the ray PX we have (QP + PR)/QR = SR/QR = sin RQS/sin QSR. But sin QSR is fixed for points on the ray, so we maximise the ratio by taking ∠RQS = 90o. Thus there is a single point maximising the ratio.
If P = X, then we still require ∠RQS = 90o, but R is no longer restricted to a line, so it can be anywhere on a circle center P.
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
12 Oct 1998