2014 AMC 10B Problems/Problem 23
A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?
First, we draw the vertical cross-section passing through the middle of the frustum. Let the top base have a diameter of , and the bottom base have a diameter of . (If you don't get why the blue hypotenuse is , reference this link: https://www.ck12.org/geometry/Tangent-Lines/lesson/Tangent-Lines-GEOM/)
Then using the Pythagorean theorem we have: , which is equivalent to: . Subtracting from both sides,
Solving for , we end up with Next, we can find the volume of the frustum (truncated cone) and of the sphere. Since we know , we can solve for using we get: Using , we get so we have: Dividing by , we get which is equivalent to by the Quadratic Formula, , so
Similar to above, draw a smaller cone top with the base of the smaller circle with radius and height . The smaller right triangle is similar to the blue highlighted one in Solution . Then where is the radius of the sphere. Then .
From the Pythagorean theorem on the blue triangle in Solution , we get similarly that .
From the volume requirements, we get that which yields .
The small right triangle on top is similar to the big right triangle of the entire big cone. So .
Substituting yields .
Substituting and yields which yields .
Solving in yields so .
Solution 3: Another Way of Simplifying the Equation in Solution 2
Similar to the previous solutions, we let , , and be the radii of the larger base, smaller base, and the sphere, respectively. By the Pythagorean Theorem, we have . Let , be the heights of the large and small cones, respectively, then we have and Thus we have the equation Substituting the expressions for , , in order into the equation yields The numerator can be factored to yield Solving the quadratic to obtain the ratio
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