Find the surface comprising the curves which satisfy dx/(yz) = dy/(zx) = dz/(xy) and which meet the circle x = 0, y2 + z2 = 1.
We have x dx = y dy = z dz. Integrating, y2 = x2 + h, z2 = x2 + k (*). We are told that some point of the circle x = 0, y2 + z2 = 1 belongs to the curve, so h and k must be non-negative with sum 1. Hence the curve (*) lies in the surface 2 x2 + 1 = y2 + z2, 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
5 Mar 2002