1st Putnam 1938

Problem A1

A solid in Euclidean 3-space extends from z = -h/2 to z = +h/2 and the area of the section z = k is a polynomial in k of degree at most 3. Show that the volume of the solid is h(B + 4M + T)/6, where B is the area of the bottom (z = -h/2), M is the area of the middle section (z = 0), and T is the area of the top (z = h/2). Derive the formulae for the volumes of a cone and a sphere.

Solution

Easy

Let the polynomial be az3 + bz2 + cz + d. Then the volume is ∫-h/2h/2 (az3 + bz2 + cz + d) dz = bh3/12 + dh. But B + T = bh2/2 + 2d, M = d, so h(B + 4M + T)/6 = bh3/12 + dh, which proves the formula. For a sphere radius R, we have h = 2R, B + T = 0 and M = πR2, so the formula gives 4/3 πR3, as usual. For a cone height h, base area A, we have B = A, T = 0, M = A/4, so the volume is hA/3, as usual.

The original text of the problems and the official solutions are available in: A M Gleason, R E Greenwood & L M Kelly, The William Lowell Putnam Mathematical Competition, Problems and Solutions, 1938-1964, MAA 1980.

Out of print, but available in some university libraries.