11th Putnam 1951 Problem A6 Let C be a parabola. Take points P, Q on C such that (1) PQ is perpendicular to the tangent at P, (2) the area enclosed by the parabola and PQ is as small as possible. What is the position of the chord PQ?
We may take the parabola as y = x 2 and P as (a, a 2 ). The tangent at P has slope 2a, so the normal has slope -1/2a and equation (y - a 2 ) + 1/2a (x - a) = 0. This meets the parabola at x 2 + x/2a - a 2 - 1/2 = 0, so the other root is x = -1/2a - a and Q is (-1/2a - a, (a + 1/2a) 2 ). The area under the curve between P and Q is 1/3 (a 3 + (a + 1/2a) 3 ). The area under the chord PQ is (2a + 1/2a) (a 2 + (a + 1/2a) 2 )/2 = 2a 3 + 3a/2 + 1/2a + 1/16a 3 . Hence the area enclosed is 4/3 a 3 + a + 1/4a + 1/48a 3 . We wish to minimise this. The gradient is 4a 2 + 1 -1/4a 2 - 1/16a 4 = 1/16a 4 (4a 2 + 1) 2 (2a - 1)(2a + 1). Hence there is a minimum at a = 1/2 (and another at a = -1/2). In the minimum position P is (1/2, 1/4) or (-1/2, 1/4), so the minimum positions are the intersection of the parabola with the perpendicular to the axis through the focus (0, 1/4). 11th Putnam 1951 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002