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AMC 12 Mock Exam 3 Problem 12

Problem 12 of 25

A right circular cone has base radius 66 and height 88. A sphere is inscribed in the cone: it is tangent to the base and tangent to the lateral surface along a circle. A second, smaller sphere, whose center also lies on the axis of the cone, lies above the first sphere; it is tangent to the first sphere and tangent to the lateral surface along a circle. What is the radius of the smaller sphere?

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Solution

Take the cross-section through the axis: an isosceles triangle with base 1212, height 88, and slant sides 1010. The first sphere appears as the incircle, with radius areasemiperimeter=4816=3\frac{\text{area}}{\text{semiperimeter}}=\frac{48}{16}=3; its center is on the axis at height 33, i.e., at distance 8−3=58-3=5 from the apex VV.

The half-angle θ\theta at the apex satisfies sin⁡θ=610=35\sin\theta=\frac{6}{10}=\frac35. A sphere tangent to the lateral surface with center on the axis at distance dd from VV has radius dsin⁡θ=35dd\sin\theta=\frac35d (check: 35⋅5=3\frac35\cdot5=3). For the smaller sphere at distance d<5d<5, tangency to the first sphere means 5−d=3+35d5-d=3+\frac35d, so d=54d=\frac54 and the radius is 35⋅54=34\frac35\cdot\frac54=\boxed{\tfrac34}.

(Alternatively, the homothety centered at VV taking the small sphere to the large one has ratio 1+sin⁡θ1−sin⁡θ=4\frac{1+\sin\theta}{1-\sin\theta}=4, so the radius is 34\frac34.)

Tagged: spheres in a cone (cross-section, similar configurations)