IMO 1978 SL 13

Origin: USA

Problem

Given any point P in the interior of a sphere with ra- dius R, three mutually perpendicular segments PA, PB, PC are drawn terminating on the sphere and having one common vertex in P. Con- sider the rectangular parallelepiped of which PA, PB, PC are coterminal edges. Find the locus of the point Q that is diagonally opposite P in the parallelepiped when P and the sphere are fixed.

Solution

Lemma. If MNPQ is a rectangle and O any point in space, then OM 2 + OP 2 = ON 2 + OQ2. Proof. Let O1 be the projection of O onto MNPQ, and m, n, p, q de- note the distances of O1 from MN, NP, PQ, QM, respectively. Then OM 2 = OO2 1 +q2+m2, ON 2 = OO2 1 +m2+n2, OP 2 = OO2 1 +n2+p2, OQ2 = OO2 1 + p2 + q2, and the lemma follows immediately. Now we return to the problem. Let O be the center of the given sphere S, and X the point opposite P in the face of the parallelepiped through P, A, B. By the lemma, we have OP 2 + OQ2 = OC2 + OX2 and OP 2 + OX2 = OA2 + OB2. Hence 2OP 2 + OQ2 = OA2 + OB2 + OC2 = 3R2, i.e. OQ = \sqrt 3R2 −OP 2 > R. We claim that the locus of Q is the whole sphere (O, \sqrt 3R2 −OP 2). Choose any point Q on this sphere. Since OQ > R > OP, the sphere

with diameter PQ intersects S on a circle. Let C be an arbitrary point on this circle, and X the point opposite C in the rectangle PCQX. By the lemma, OP 2 + OQ2 = OC2 + OX2, hence OX2 = 2R2 −OP 2 > R2. The plane passing through P and perpendicular to PC intersects S in a circle \gamma; both P, X belong to this plane, P being inside and X out- side the circle, so that the circle with diameter PX intersects \gamma at some point B. Finally, we choose A to be the point opposite B in the rectangle PBXA: we deduce that OA2 + OB2 = OP 2 + OX2, and consequently A \inS. By the construction, there is a rectangular parallelepiped through P, A, B, C, X, Q.