41st Putnam 1980 Problem A1 Let f(x) = x 2 + bx + c. Let C be the curve y = f(x) and let P i be the point (i, f(i) ) on C. Let A i be the point of intersection of the tangents at P i and P i+1 . Find the polynomial of smallest degree passing through A 1 , A 2 , ... , A 9 .
y = f(x) - 1/4. Easy The answer suggests there ought to be a one-line solution, but I cannot see it. The equation of the tangent at x = i is y - (i 2 + ib + c) = (2i + b)(x - i), or y - (2i + b)x = c - i 2 . Solving, we find that A i is (i + 1/2, i 2 + i + bi + b/2 + c). Clearly these do not lie on a straight line, so the degree is at least 2. But we see at once that any A i lies on the second degree y = x 2 + bx + c - 1/4. 41st Putnam 1980 © John Scholes jscholes@kalva.demon.co.uk 3 Nov 1999 Last corrected/updated 21 Nov 02