Let f(x) = x2 + bx + c. Let C be the curve y = f(x) and let Pi be the point (i, f(i) ) on C. Let Ai be the point of intersection of the tangents at Pi and Pi+1. Find the polynomial of smallest degree passing through A1, A2, ... , A9.
y = f(x) - 1/4.
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 Ai 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 Ai lies on the second degree y = x2 + bx + c - 1/4.
41st Putnam 1980
© John Scholes
3 Nov 1999
Last corrected/updated 21 Nov 02