O is the center of a sphere S. Points A, B, C are inside S, OA is perpendicular to AB and AC, and there are two spheres through A, B, and C which touch S. Show that the sum of their radii equals the radius of S.
Solution
Let D be the circumcenter of ABC. The triangle ABC is in the plane normal to OA. The two spheres both contain through the circumcircle of ABC, so their centers must lie on a line L normal to the plane ABC and hence parallel to OA. Take the plane through OA and the line L. Suppose the centers are P and Q. The sphere center P must touch S at a point X on the line OP. Similarly, the sphere center Q must touch S at a point Y on the line OQ. Since the spheres pass through A, we have PA = PX and hence OP + PA = R, the radius of the sphere S. Similarly OQ + QA = R. Indeed if K is any point on the line L such that OK + KA = R, then the sphere center K will touch S and pass through A. Since it will also have KD perpendicular to DA, it will contain all points on the circumcircle of ABC. But the locus of points such that OK + KA = R is an ellipse with foci O and A. So it meets the line L in just two points, which must therefore be P and Q. Moreover, since the line L is parallel to OA the points must be equidistant from the midpoint of OA (which is the center of the ellipse). Hence OP = AQ and so AQ + AP = R, as required.
© John Scholes
jscholes@kalva.demon.co.uk
21 Aug 2002