12th Putnam 1952 Problem B2 Find the surface comprising the curves which satisfy dx/(yz) = dy/(zx) = dz/(xy) and which meet the circle x = 0, y 2 + z 2 = 1.
We have x dx = y dy = z dz. Integrating, y 2 = x 2 + h, z 2 = x 2 + k (*). We are told that some point of the circle x = 0, y 2 + z 2 = 1 belongs to the curve, so h and k must be non-negative with sum 1. Hence the curve (*) lies in the surface 2 x 2 + 1 = y 2 + z 2 , which is a one-sheet hyperboloid with the x-axis as an axis of symmetry. However, not all points of this surface lie on a curve (*). Clearly we require |y| >= |x| and |z| ≥ |x|. Equally, it is clear that this is a sufficient condition for we can then find h, k to satisfy (*). 12th Putnam 1952 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002