### IMO 1978

Problem A2

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 XA2 + XC2 = XB2 + XD2. 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 XA2 + XC2 = (a - c)2 + (a + c)2 = 2a2 + 2c2 = 2b2 + 2c2 = XB2 + XD2.

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 CR2 = 2CV2 - CP2. 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 OY2 = OX2 + XY2 = OX2 + CR2 = OX2 + 2CV2 - CP2 = OU2 - UX2 + 2CV2 - CP2 = OU2 - CP2 + 2(OV2 - OC2) - CP2 = 3OU2 - 2OP2. 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.