Putnam 80/A1 solution

41st Putnam 1980

Problem A1

Let f(x) = x² + 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 A₁, A₂, ... , A₉.

Solution

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 - (i² + ib + c) = (2i + b)(x - i), or y - (2i + b)x = c - i². Solving, we find that A_i is (i + 1/2, i² + 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 = x² + bx + c - 1/4.

41st Putnam 1980