### 3rd Putnam 1940

Problem B4

Let S be the surface ax2 + by2 + cz2 = 1 (a, b, c all non-zero), and let K be the sphere x2 + y2 + z2 = 1/a + 1/b + 1/c (known as the director sphere). Prove that if a point P lies on 3 mutually perpendicular planes, each of which is tangent to S, then P lies on K.

Comment. The original question also asked for a proof that every point of K had this property, which is (a) false unless we allow planes which are asymptotes (or tangents at infinity), and (b) unreasonably hard - at least, I cannot see a neat proof.

Solution

Moderately hard.

Let f(x, y, z) = ax2 + by2 + cz2. Then the normal vector is the vector grad f, so the tangent plane at (u, v, w) is a.u.x + b.v.y + c.w.z = 1.

Note that this does not pass through the origin. The general plane not through the origin has equation p.(x - λp) = 0 or p.x = λ, where x is the vector (x, y, z) representing a general point on the plane, p = (p, q, r) is a unit vector normal to the plane, and λ > 0 is the distance of the plane from the origin. If this is a tangent plane at some point of the quadric, then a(p/λa)2 + b(q/λb)2 + c(r/λc)2 = 1, or p2/a + q2/b + r2/c = λ2.

So suppose P is the point of intersection of three perpendicular tangent planes pix + qiy + riz = λi, i = 1, 2, 3, where pi are orthonormal vectors. The squared distance of P from the origin is λ12 + λ22 + λ32 = (p12 + p22 + p32)/a + (q12 + q22 + q32)/b + (r12 + r22 + r32)/c = 1/a + 1/b + 1/c, since pi are orthonormal. [This is the key trick: if the rows of a matrix are orthonormal vectors, then the matrix is orthogonal and hence its columns are also orthonormal vectors.].

[For a proof of the italicized second part, see the book referred to below.]

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.

© John Scholes
jscholes@kalva.demon.co.uk
15 Sep 1999