37th Putnam 1976 Problem A1 Given two rays OA and OB and a point P between them. Which point X on the ray OA has the property that if XP is extended to meet the ray OB at Y, then XP·PY has the smallest possible value. Solution
: take OX = OY. Let M be the foot of the perpendicular from P to OA, and N be the foot of the perpendicular from P to OB. Obviously X must lie further from O than M - otherwise moving it closer to M would reduce both PX and PY. Similarly Y must lie further from O than N. Let φ be the angle MPX and let θ be the angle MON. Then the angle NPY is θ-φ. Hence PX·PY = PM·PN/(cos φ cos θ-φ). So we minimise PX.PY by maximising (cos φ cos θ-φ). But (cos φ cos θ-φ) = 1/2 (cos(φ+θ-φ) + cos(2φ-θ) ) = 1/2 cos θ + 1/2 cos(2φ-θ). This is obviously maximised by taking φ = θ-φ. But that corresponds to angle PXM = angle PYN and hence OX = OY. Alternatively, PX·PY may remind us of the elementary result that if XX'YY' is cyclic with diagonals XY, X'Y' meeting at P, then PX·PY = PX'·PY'. The problem is to find a suitable circle. With hindsight, it is fairly obvious. Take the circle which touches OA at X and OB at Y. Then any other line through P intersects the circle at X' and Y' before it intersects OA and OB (at X'' and Y'' say). Hence PX'' > PX', PY'' > PY' and PX'·PY' = PX·PY. However, I failed to find the right circle until after I had solved the problem trigonometrically. 37th Putnam 1976 © John Scholes jscholes@kalva.demon.co.uk 23 Jan 2001