Let T be a tangent plane to the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1. What is the smallest possible volume for the tetrahedral volume bounded by T and the planes x = 0, y = 0, z = 0?
Answer: √3 abc/2.
The normal at (x0, y0, z0) is (xx0/a2, yy0/b2, zz0/c2). So the tangent plane is x x0/a2 + y y0/b2 + z z0/c2 = 1. This cuts the three axes at x = a2/x0, y = b2/y0, z = c2/z0. We can regard two of these lengths as defining the base of the tetrahedron, and the third as forming its height. Hence its volume is a2b2c2/(6x0y0z0).
We wish to maximize x0y0z0. That is equivalent to maximising x02/a2 y02/b2 z02/c2. But we know that the sum of these three numbers is 1, so their maximum product is 1/27 (achieved when they are all equal - the arithmetic/geometric mean result). Hence x0y0z0 has maximum value abc/(3√3).
6th Putnam 1946
© John Scholes
14 Dec 1999