Difference between revisions of "2014 AMC 10B Problems/Problem 23"
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Solving for s, we end up with | Solving for s, we end up with | ||
<cmath>s=\sqrt{r}</cmath>. | <cmath>s=\sqrt{r}</cmath>. | ||
− | Next, we can find the | + | Next, we can find the volume of the frustum and of the sphere. Since we know <math>V_{frustum}=2V_{sphere}</math>, we can solve for <math>s</math> |
using <math>V_{frustum}=\frac{\pi*h}{3}(R^2+r^2+Rr)</math> | using <math>V_{frustum}=\frac{\pi*h}{3}(R^2+r^2+Rr)</math> | ||
we get: | we get: |
Revision as of 21:33, 21 February 2014
Problem
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?
(Diagram edited from copeland's diagram)
Solution
First, we draw the vertical cross-section passing through the middle of the frustum. Let the top base have a diameter of 2, and the bottom base have a diameter of 2r. (diagram by copeland?)
Then using the Pythagorean theorem we have: , which is equivalent to: . Subtracting from both sides,
Solving for s, we end up with . Next, we can find the volume of the frustum 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 , so
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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