1st Putnam 1938 Problem B6 What is the shortest distance between the plane Ax + By + Cz + 1 = 0 and the ellipsoid x 2 /a 2 + y 2 /b 2 + z 2 /c 2 = 1. You may find it convenient to use the notation h = (A 2 + B 2 + C 2 ) -1/2 , m = (a 2 A 2 + b 2 B 2 + c 2 C 2 ) 1/2 . What is the algebraic condition for the plane not to intersect the ellipsoid?
The tangent plane to the ellipsoid at (X, Y, Z) is Xx/a 2 + Yy/b 2 + Zz/c 2 = 1. It is parallel to Ax + By + Cz + 1 = 0 iff X/a 2 = kA, Y/b 2 = kB, Z/c 2 = kC for some k. But 1 = X 2 /a 2 + Y 2 /b 2 + Z 2 /c 2 = k 2 (a 2 A 2 + b 2 B 2 + c 2 C 2 ) = k 2 m 2 , so k = ±1/m. There are two values corresponding to two parallel tangent planes (one on either side of the ellipse). The equation of the tangent plane is k(Ax + By + Cz) = 1. The distance of the origin from the plane Ax + By + Cz + 1 = 0 is 1/(A 2 + B 2 + C 2 ) 1/2 = h. The distance of the origin from the tangent plane k(Ax + By + Cz) = 1 is h/|k| = hm. So if m ≥ 1, the plane Ax + By + Cz + 1 = 0 lies between the two tangent planes and hence intersects the ellipse. So in this case the minimum distance is zero. If m < 1, then the distance between the plane Ax + By + Cz + 1 = 0 and the nearer tangent plane is h(1 - m) and that is the required shortest distance. 1st Putnam 1938 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002