2001 AMC 10 Problems/Problem 21
Contents
Problem
A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter and altitude , and the axes of the cylinder and cone coincide. Find the radius of the cylinder.
Solution 1 (video solution)
Solution 2
Let the diameter of the cylinder be . Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, which we solve to find . Our answer is .
Solution 3 (Very similar to solution 2 but explained more)
We are asked to find the radius of the cylinder, or so we can look for similarity. We know that and , thus we have similarity between and by similarity.
Therefore, we can create an equation to find the length of the desired side. We know that:
Plugging in yields:
Cross multiplying and simplifying gives:
Since the problem asks us to find the radius of the cylinder, we are done and the radius of the cylinder is .
~etvat
Solution 4 (graphical)
Assume that a point on a given diameter of the cone is the point on a two-dimensional representation of the cone as shown in Solution 2. The top point of the cone is thus and the line that goes through both points is .
Now we create a second equation. We must choose some point on the line such that , which implies that the cylinder’s diameter, , must be equal to its height, . Solving yields , and the radius is thus .
Solution 5 (Without similar triangles)
Like in Solution 2, we draw a diagram.
It is known that has length and has length , so triangle has area . Also, let be equal to the radius of the cylinder.
Triangles and can be combined into one triangle with base and height . The area of this new triangle is .
Triangle has base and height , so its area is.
Finally, square has area .
Now we can construct an equation to find :
~Dreamer1297
Trivia
This problem appeared in AoPS's Introduction to Geometry as a challenge problem.
See Also
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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