Difference between revisions of "2001 AMC 10 Problems/Problem 21"
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<math> \textbf{(A)}\ \frac{8}3\qquad\textbf{(B)}\ \frac{30}{11}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{25}{8}\qquad\textbf{(E)}\ \frac{7}{2} </math> | <math> \textbf{(A)}\ \frac{8}3\qquad\textbf{(B)}\ \frac{30}{11}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{25}{8}\qquad\textbf{(E)}\ \frac{7}{2} </math> | ||
− | ==Solution== | + | ==Solution 1 (video solution)== |
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+ | https://youtu.be/HUM035eNKvU | ||
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+ | ==Solution 2== | ||
<asy> | <asy> | ||
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Let the diameter of the cylinder be <math> 2r </math>. Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, <math> \frac{12-2r}{12}=\frac{2r}{10} </math> which we solve to find <math> r=\frac{30}{11} </math>. Our answer is <math> \boxed{\textbf{(B)}\ \frac{30}{11}} </math>. | Let the diameter of the cylinder be <math> 2r </math>. Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, <math> \frac{12-2r}{12}=\frac{2r}{10} </math> which we solve to find <math> r=\frac{30}{11} </math>. Our answer is <math> \boxed{\textbf{(B)}\ \frac{30}{11}} </math>. | ||
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+ | ==Solution 3 (Very similar to solution 2 but explained more)== | ||
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+ | We are asked to find the radius of the cylinder, or <math>r</math> so we can look for similarity. We know that <math>\angle BEF = \angle BDA</math> and <math>\angle FBE = \angle ABD</math>, thus we have similarity between <math>\triangle BFE</math> and <math>\triangle BAD</math> by <math>AA</math> similarity. | ||
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+ | Therefore, we can create an equation to find the length of the desired side. We know that: | ||
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+ | <math>\frac{BE}{BD}=\frac{FE}{AD}.</math> | ||
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+ | Plugging in yields: | ||
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+ | <math>\frac{12-2r}{12}=\frac{r}{5}.</math> | ||
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+ | Cross multiplying and simplifying gives: | ||
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+ | <math>5(12-2r)=12r</math> | ||
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+ | <math>\Downarrow</math> | ||
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+ | <math> r=\frac{30}{11}.</math> | ||
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+ | Since the problem asks us to find the radius of the cylinder, we are done and the radius of the cylinder is <math>\boxed{\textbf{(B)}\ \frac{30}{11}}</math>. | ||
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+ | ~etvat | ||
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+ | ==Trivia== | ||
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+ | This problem appeared in AoPS's Introduction to Geometry as a challenge problem. | ||
==See Also== | ==See Also== | ||
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{{AMC10 box|year=2001|num-b=20|num-a=22}} | {{AMC10 box|year=2001|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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+ | [[Category: Introductory Geometry Problems]] |
Revision as of 12:24, 12 October 2021
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
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 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.