3rd Putnam 1940

Problem A2

y = f(x) is continuous with continuous derivative. The arc PQ is concave to the chord PQ. X is a point on the arc PQ for which PX + XQ is a maximum. Prove that XP and XQ are equally inclined to the tangent at X.




Take s to be the arc-length PX and z to be PX + XQ. Suppose the tangent is WXY, so that angle WXP = θ and angle YXQ = φ. Then to first order if we vary X, the change δz = δs (cos θ - cos φ), which is can be made positive (by choosing the sign of δs appropriately), thus contradicting the maximality of X, unless θ = φ.

The official solution (Gleason et al, see foot of question paper, points out that almost the entire hypothesis is unnecessary. Suppose S is an entirely arbitary set of points in the plane, and that A and B are any two points in the plane (not necessarily in S). If X is a point of S such that PX + XQ ≥ PY + YQ for all points Y of S, then we can easily show that S lies in the half-plane bounded by external angle bisector of PXQ. If X lies on PQ, then S must be a subset of the segment PQ and the result is trivial. So assume it does not. Now reflect Q in the bisector to get Q', with PXQ' a straight line. Then if Y is any point in the same half-plane as Q', we have YQ' < YQ and hence PY + YQ > PY + YQ' ≥ PQ' = PX + XQ' = PX + XQ, so Y is not in S.

This argument depends upon X achieving a global maximum. The original wording of the question, "a point ... for which ... is a maximum" (which on this point I quoted exactly), is somewhat ambiguous. Does it mean a local or a global maximum? If it means a local, then we have to take S to be a small arc for which the maximum is global. Then that arc lies on one side of the line. Since it also has a point in common (X) and is differentiable, the line must be the tangent to the arc at X.

This solution is somewhat harder and less obvious (unless you have seen it before), so I prefer the simpler solution.



3rd Putnam 1940

© John Scholes
15 Sep 1999