4th Putnam 1941

Problem A5

The line L is parallel to the plane y = z and meets the parabola y2 = 2x, z = 0 and the parabola 3x = z2, y = 0. Prove that if L moves freely subject to these constraints then it generates the surface x = (y - z)(y/2 - z/3).



We may take a point on the first parabola to be (2s2, 2s, 0) and a point on the second to be (3t2, 0, 3t). The line through these two points has the general point k (2s2, 2s, 0) + (1 - k) (3t2, 0, 3t). We require that this does not meet the plane y = z (we consider the exceptional case where the line lies entirely in the plane y = z later). The general point has y = z if 2ks = 3(1 - k)t or (2s + 3t)k = 3t. Hence we must have 2s + 3t = 0. In this case we find that the general point on the line is (3/2 kt2 + 3t2, -3kt, 3t - 3kt). Hence it has (y - z) = - 3t, (y/2 - z/3) = -kt/2 - t, and (y - z)(y/2 - z/3) = 3/2 kt2 + 3t2 = x, so the line certainly lies entirely in the surface x = (y - z)(y/2 - z/3). Conversely, given any point (x, y, z) on that surface (with y not equal to z), we may take t = (z - y)/3, k = y/(y - z) and then the point lies on the line given by t, k. Hence the lines generate the surface.

Finally, we need to consider the special case y = z. If L lies in the plane y = z, then it must meet the two parabolas at the origin. But any line in the plane y = z which passes through the origin meets the conditions. Such a line does not, however, lie in the surface x = (y - z)(y/2 - z/3), because for points on the surface if y = z, then necessarily x = 0. Thus the question is in error on this point.



4th Putnam 1941

© John Scholes
5 Mar 2002