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:
Let O1,O2,O3 be the centers of three spheres, with the sphere centered at O3 having radius 13. Let T1,T2,T3 be the respective points where these spheres are tangent to plane P.
Let point A be the foot of the perpendicular from O3 to the line ℓ. Then O3A is the perpendicular bisector of segment O1O2. Now we need to find the length of segment T3A.
We can see that: O1O2=72,O1O3=O2O3=49
Since O1T1⊥P and O3T3⊥P, these segments are parallel. Hence, points O1,O3,T1,T3 lie in a common plane.
Consider the cross-section O1O3T3T1, which intersects line ℓ at a single point B. Define C as the foot of the perpendicular from O1 to ℓ and D as the foot of the perpendicular from O3 to \overline
Observe the following diagram:
Because O1T1∥O3T3, triangles △O1T1B∼△O3T3B. The similarity ratio is:
O1T1O3T3=3613
Thus:
O1BO3B=3613,O1O3=49,⇒O1B=49+O3B
Solving:
O1B=1336O3B=49+O3B⇒1323O3B=49⇒O3B=23637
Now in △O1DO3, since DO3=2O1O2=36, we apply the Pythagorean theorem:
O1D=492−362=2401−1296=1105
Since ℓ⊥O1C and DO3⊥O1C, we conclude ℓ∥DO3, so triangles △O1CB∼△O1DO3.
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