# 2001 AMC 10 Problems/Problem 21

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## 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 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.