1977 AMC 12 Problem 27

Attempt Problem 27 of the 1977 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1977 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

27.

There are two spherical balls of different sizes lying in two corners of a rectangular room, each touching two walls and the floor. If there is a point on each ball which is 55 inches from each wall which that ball touches and 1010 inches from the floor, then the sum of the diameters of the balls is
2020 inches
3030 inches
4040 inches
6060 inches
not determined by the given information
Answer: C
Concepts:spherecoordinate geometryquadratic
Small hint:
Place the corner at the origin with the walls and floor as coordinate planes
Big hint:
A sphere of radius rr tangent to all three planes has center (r,r,r)(r,r,r)
Solution:
For radius r,r, the center is (r,r,r)(r,r,r) and the given point is (5,5,10).(5,5,10). Thus 2(5−r)2+(10−r)2=r2, 2(5-r)^2+(10-r)^2=r^2, which simplifies to r2−20r+75=0,r^2-20r+75=0, or (r−5)(r−15)=0.(r-5)(r-15)=0. The two radii are 55 and 15,15, so the sum of the diameters is 2(5+15)=402(5+15)=40 inches.
Therefore, the correct answer is C.
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