10th Putnam 1950

Problem B5

Do either (1) or (2):

(1)   Show that if ∑(an + 2an+1) converges, then so does ∑ an.

(2)   Let S be the surface 2xy = z2. The surface S and the variable plane P enclose a cone with volume πa3/3, where a is a positive real constant. Find the equation of the envelope of P. What is the envelope in the case of a general cone?



(1) Suppose an + 2an+1 converges to 3k. We show that anconverges to k.

Given any ε > 0, take N so that an + 2an+1 is within ε of 3k for all n ≥ N. Take a positive integer M such that aN is within (2M + 1)ε of k.

Then aN+1 is within ( (2M + 1)ε + ε)/2 = (2M-1 + 1)ε of (3k - k)/2 = k. By a trivial induction aN+M is within 2ε of k. Then aN+M+1 is within (2ε + ε)/2, and hence within 2ε, of k. So by a trivial induction, an is within 2ε of k for all n > N + M.

(2) Answer: the 2-sheet hyperboloid u2 = v2 + z2 + a2.

The nature of the surface S becomes much clearer if we change coordinates to u, v, z with z unchanged and the x, y axes rotated through π/4 to u = (x + y)/√2, v = (-x + y)/√2 (or inversely, x = (u - v)/√2, y = (u + v)/√2 ). Now S becomes u2 - v2 = z2 or v2 + z2 = u2, which is evidently a right circular cone with vertex the origin and axis the u-axis.

Evidently the plane u = a forms a cone volume 1/3 πa3 with the surface (the base is a circle radius a and the height is a). If the plane moves round to be almost tangent to the sides of the cone it should still cut off a cone with the same volume, so we might suspect that the envelope is a two sheet hyperboloid asymptotic to the cone and passing through the points u = ±a, v = z = 0. This has equation u2 = v2 + z2 + a2. This is a surface of revolution formed by rotating the hyberbola u2 = v2 + a2 about the u-axis, so it is sufficient to look at tangent planes that are parallel to (ie do not intersect) the z-axis. We need to show that these cut off cones volume 1/3 πa3.

The tangent to the hyperbola at u = a cosh t, v = a sinh t is (v - a sinh t) = coth t (u - a cosh t) or v sinh t - u cosh t + a = 0 (*). It cuts v = u at u = v = a/(cosh t - sinh t) and it cuts v = -u at u = -v = a/(cosh t + sinh t). So the corresponding plane cuts the hyperboloid in an ellipse with major axis length 2A, where 2A is the distance between these two points of intersection, which is 2a cosh 2t. At the centre of the ellipse (which we find as midway between the two points of intersection) we have u = a cosh t, v = a sinh t, so the extremities of minor axis have z2 = u2 - v2 = a2. Thus the minor axis has length 2a. Hence the area of the ellipse is πa2√(2 cosh 2t). The distance of the origin from the plane is its distance from the line (*) which is a/√(2 cosh 2t). Hence the volume of the cone is 1/3 area of base x height = 1/3 πa3, as we had guessed.

Finally, if we have a general cone (which does not have circular cross-section or right-angle), we can transform it into a right circular cone by an affine transformation. Such transformations preserve the ratios of distances and hence the ratios of volumes. They also preserve tangency, so the envelope in this more general case will still be a hyperboloid asymptotic to the cone.



10th Putnam 1950

© John Scholes
5 Mar 2002