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Difference between revisions of "2001 AMC 10 Problems/Problem 21"

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Let's say you don't know similar triangles but still want to solve this problem. You can combine triangles <math>\triangle DBG</math> and triangle <math>\triangle ECF</math> into one triangle with base <math>10-x</math> and height <math>x</math>. The area of this new triangle is
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Triangles <math>\triangle DBG</math> and <math>\triangle ECF</math> can be combined into one triangle with base <math>10-2x</math> and height <math>2x</math>. The area of this new triangle is <math>\frac{2x(10-2x)}{2} = 2x(5-x)</math>.
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<math>\frac{2x(10-2x)}{2} = 2x(5-x)</math>.
 
  
 
Triangle <math>\triangle ADE</math> has base <math>2x</math> and height <math>12-2x</math>, so its area is<math>\frac{2x(12-2x)}{2} = 2x(6-x)</math>.
 
Triangle <math>\triangle ADE</math> has base <math>2x</math> and height <math>12-2x</math>, so its area is<math>\frac{2x(12-2x)}{2} = 2x(6-x)</math>.
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Now we can construct an equation to find <math>x</math>:
 
Now we can construct an equation to find <math>x</math>:
  
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<cmath>2x(5-x) + 2x(6-x) + (4x^2) = 60</cmath>
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<cmath>2x(5-x) + 2x(6-x) + 4x^2 = 60</cmath>
 
<cmath>\Rightarrow 2x(5-x+6-x+2x) = 60</cmath>
 
<cmath>\Rightarrow 2x(5-x+6-x+2x) = 60</cmath>
 
<cmath>\Rightarrow 2x(11) = 60</cmath>
 
<cmath>\Rightarrow 2x(11) = 60</cmath>

Latest revision as of 22:14, 10 February 2024

Problem

A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

$\textbf{(A)}\ \frac{8}3\qquad\textbf{(B)}\ \frac{30}{11}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{25}{8}\qquad\textbf{(E)}\ \frac{7}{2}$

Solution 1 (video solution)

https://youtu.be/HUM035eNKvU

Solution 2

[asy] draw((5,0)--(-5,0)--(0,12)--cycle); unitsize(.75cm); draw((-30/11,0)--(-30/11,60/11)); draw((-30/11,60/11)--(30/11,60/11)); draw((30/11,60/11)--(30/11,0)); draw((0,0)--(0,12)); label("$2r$",(0,30/11),E); label("$12-2r$",(0,80/11),E); label("$2r$",(0,60/11),S); label("$10$",(0,0),S); label("$A$",(0,12),N); label("$B$",(-5,0),SW); label("$C$",(5,0),SE); label("$D$",(-30/11,60/11),W); label("$E$",(30/11,60/11),E); [/asy]




Let the diameter of the cylinder be $2r$. Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, $\frac{12-2r}{12}=\frac{2r}{10}$ which we solve to find $r=\frac{30}{11}$. Our answer is $\boxed{\textbf{(B)}\ \frac{30}{11}}$.

Solution 3 (Very similar to solution 2 but explained more)

We are asked to find the radius of the cylinder, or $r$ so we can look for similarity. We know that $\angle BEF = \angle BDA$ and $\angle FBE = \angle ABD$, thus we have similarity between $\triangle BFE$ and $\triangle BAD$ by $AA$ similarity.

Therefore, we can create an equation to find the length of the desired side. We know that:

$\frac{BE}{BD}=\frac{FE}{AD}.$


Plugging in yields:


$\frac{12-2r}{12}=\frac{r}{5}.$

Cross multiplying and simplifying gives:


$5(12-2r)=12r$

$\Downarrow$


$r=\frac{30}{11}.$

Since the problem asks us to find the radius of the cylinder, we are done and the radius of the cylinder is $\boxed{\textbf{(B)}\ \frac{30}{11}}$.

~etvat

Solution 4 (graphical)

Assume that a point on a given diameter of the cone is the point $(0,0)$ on a two-dimensional representation of the cone as shown in Solution 2. The top point of the cone is thus $(5,12)$ and the line that goes through both points is $y=\frac{12}{5}x$.

Now we create a second equation. We must choose some point $(x,y)$ on the line $y=\frac{12}{5}x$ such that $y=10-2x$, which implies that the cylinder’s diameter, $10-2x$, must be equal to its height, $y$. Solving yields $x=\frac{25}{11}$, and the radius is thus $\frac{10-2x}{2}=\frac{\frac{60}{11}}{2}=\boxed{\textbf{(B)}\ \frac{30}{11}}$.

Solution 5 (Without similar triangles)

Like in Solution 2, we draw a diagram.

[asy] draw((5,0)--(-5,0)--(0,12)--cycle); unitsize(.75cm); draw((-30/11,0)--(-30/11,60/11)); draw((-30/11,60/11)--(30/11,60/11)); draw((30/11,60/11)--(30/11,0)); draw((0,0)--(0,12)); label("$2x$",(0,30/11),E); label("$2x$",(0,60/11),S); label("$H$",(0,0),S); label("$A$",(0,12),N); label("$B$",(-5,0),SW); label("$C$",(5,0),SE); label("$D$",(-30/11,60/11),W); label("$E$",(30/11,60/11),E); label("$F$",(30/11,0),S); label("$G$",(-30/11,0),S); [/asy]

It is known that $\overline{AH}$ has length $12$ and $\overline{BC}$ has length $10$, so triangle $\triangle ABC$ has area $60$. Also, let $x$ be equal to the radius of the cylinder.


Triangles $\triangle DBG$ and $\triangle ECF$ can be combined into one triangle with base $10-2x$ and height $2x$. The area of this new triangle is $\frac{2x(10-2x)}{2} = 2x(5-x)$.

Triangle $\triangle ADE$ has base $2x$ and height $12-2x$, so its area is$\frac{2x(12-2x)}{2} = 2x(6-x)$.

Finally, square $DEFG$ has area $4x^2$.


Now we can construct an equation to find $x$:

\[2x(5-x) + 2x(6-x) + 4x^2 = 60\] \[\Rightarrow 2x(5-x+6-x+2x) = 60\] \[\Rightarrow 2x(11) = 60\] \[\Rightarrow 22x = 60\] \[\Rightarrow x = \frac{60}{22} = \boxed{(B) \frac{30}{11}}\]

~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
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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