IMO 1979 SL 25

Origin: USA

Problem

Given a point P in a given plane \pi and also a given point Q not in \pi, show how to determine a point R in \pi such that QP+PR QR is a maximum.

Solution

Let us first look for such a point R on a line l in \pi going through P. Let \angleQPR = 2\theta. Consider a point Q′ on l such that Q′P = QP. Then we have QP + PR QR = RQ′ QR = sin Q′QR sin QQ′R . Since QQ′P is fixed, the maximal value of the expression occurs when \angleQQ′R = 90◦. In this case (QP + PR)/QR = 1/sin \theta. Looking at all possible lines l, we see that \theta is minimized when l equals the projection of PQ onto \pi. Hence, the point R is the intersection of the projection of PQ onto \pi and the plane through Q perpendicular to PQ.