8th USAMO 1979 Problem 4 P lies between the rays OA and OB. Find Q on OA and R on OB collinear with P so that 1/PQ + 1/PR is as large as possible.
Let the line through P parallel to OA meet OB at C. Note that C is fixed. Let the line through C parallel to QR meet OP at D. D varies as Q varies. Triangles ODC, OPR are similar, so CD/PR = OD/OP. Also triangles OPQ and PDC are similar, so CD/PQ = DP/OP. Adding, CD/PR + CD/PQ = 1. Hence we maximise 1/PQ + 1/PR by making CD as small as possible. That is achieved by making angle CDP = 90 o and hence QR perpendicular to OP. 8th USAMO 1979 © John Scholes jscholes@kalva.demon.co.uk 20 Aug 2002