P is a point inside a sphere. Three mutually perpendicular rays from P intersect the sphere at points U, V and W. Q denotes the vertex diagonally opposite P in the parallelepiped determined by PU, PV, PW. Find the locus of Q for all possible sets of such rays from P.

**Solution**

Suppose ABCD is a rectangle and X any point inside, then XA^{2} + XC^{2} = XB^{2} + XD^{2}. This is most easily proved using coordinates. Take the origin O as the center of the rectangle and take OA to be the vector __a__, and OB to be __b__. Since it is a rectangle, |__a__| = |__b__|. Then OC is -__a__ and OD is -__b__. Let OX be __c__. Then XA^{2} + XC^{2} = (__a__ - __c__)^{2} + (__a__ + __c__)^{2} = 2__a__^{2} + 2__c__^{2} = 2__b__^{2} + 2__c__^{2} = XB^{2} + XD^{2}.

Let us fix U. Then the plane k perpendicular to PU through P cuts the sphere in a circle center C. V and W must lie on this circle. Take R so that PVRW is a rectangle. By the result just proved CR^{2} = 2CV^{2} - CP^{2}. OC is also perpendicular to the plane k. Extend it to X, so that CX = PU. Then extend XU to Y so that YR is perpendicular to k. Now OY^{2} = OX^{2} + XY^{2} = OX^{2} + CR^{2} = OX^{2} + 2CV^{2} - CP^{2} = OU^{2} - UX^{2} + 2CV^{2} - CP^{2} = OU^{2} - CP^{2} + 2(OV^{2} - OC^{2}) - CP^{2} = 3OU^{2} - 2OP^{2}. Thus the locus of Y is a sphere.

Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

© John Scholes

jscholes@kalva.demon.co.uk

12 Oct 1998