Problem:
Three spheres with radii 11,13, and 19 are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at A,B, and C, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that AB2=560. Find AC2.
Solution:
Answer (756):
Let the spheres with radii 11,13 , and 19 have centers P,Q, and R, respectively, and let the three circles have common radius r. Segments AP,BQ, and CR are perpendicular to the plane of △ABC, so AB is the projection of PQ onto that plane. Similarly, AC is the projection of PR onto that plane.
If D is any point on the circle centered at A, then △PAD is a right triangle with AD=r and PD=11, so AP2=121−r2. Similarly, BQ2=169−r2 and CR2=361−r2. The Pythagorean Theorem gives PQ2=AB2+(BQ−AP)2, so from PQ=11+13=24, it follows that
(BQ−AP)2=PQ2−AB2=242−560=16
Thus BQ−AP=4 and BQ2−AP2=(169−r2)−(121−r2)=48. Hence BQ+AP=448=12,BQ=8, AP=4, and r2=105.
