Problem:
Two spheres with radii 36 and one sphere with radius 13 are each externally tangent to the other two spheres and to two different planes P and Q. The intersection of planes P and Q is the line ℓ. The distance from line ℓ to the point where the sphere with radius 13 is tangent to plane P is nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Answer (335):
Let A be the center of the sphere with radius 13, B be the center of one of the spheres with radius 36, and C be the point where the two spheres with radius 36 are tangent. The centers of the three spheres are equidistant to P and Q, so the plane containing these three points bisects the dihedral angle created by P and Q and contains point C and line ℓ. Let points D and E be the points on ℓ such that AD and BE are perpendicular to ℓ. Let F and G be the points where P is tangent to the spheres centered at A and B, respectively.
Because AB=13+36=49 and BC=36, the distance AC is 492−362=(49+36)(49−36)=85⋅13. Segments AD and BE lie in the same plane and are both perpendicular to ℓ, so AD∥BE. Segments DF and EG lie in plane P and are both perpendicular to ℓ, so DF∥EG. It follows that △ADF∼△BEG. Thus
AFAD=BGBE=BGAD+AC so 13AD=36AD+85⋅13.
It follows that AD=231385⋅13. Then the required distance DF is