Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
m (→‎Solution 5 (Without similar triangles))
 
(8 intermediate revisions by 3 users not shown)
Line 41: Line 41:
  
 
==Solution 3 (Very similar to solution 2 but explained more)==
 
==Solution 3 (Very similar to solution 2 but explained more)==
−
 
−
<math>\text{We can begin by drawing a diagram with the given information}</math>:
 
−
 
−
[[File:2001amc10solution.jpg]]
 
  
 
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.  
 
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.  
Line 71: Line 67:
  
 
~etvat
 
~etvat
 +
 +
==Solution 4 (graphical)==
 +
Assume that a point on a given diameter of the cone is the point <math>(0,0)</math> on a two-dimensional representation of the cone as shown in Solution 2. The top point of the cone is thus <math>(5,12)</math> and the line that goes through both points is <math>y=\frac{12}{5}x</math>.
 +
 +
Now we create a second equation. We must choose some point <math>(x,y)</math> on the line <math>y=\frac{12}{5}x</math> such that <math>y=10-2x</math>, which implies that the cylinder’s diameter, <math>10-2x</math>, must be equal to its height, <math>y</math>. Solving yields <math>x=\frac{25}{11}</math>, and the radius is thus <math>\frac{10-2x}{2}=\frac{\frac{60}{11}}{2}=\boxed{\textbf{(B)}\ \frac{30}{11}}</math>.
 +
 +
==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 <math>\overline{AH}</math> has length <math>12</math> and <math>\overline{BC}</math> has length <math>10</math>, so triangle <math>\triangle ABC</math> has area <math>60</math>. Also, let <math>x</math> be equal to the radius of the cylinder.
 +
 +
 +
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>.
 +
 +
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>.
 +
 +
Finally, square <math>DEFG</math> has area <math>4x^2</math>.
 +
 +
 +
Now we can construct an equation to find <math>x</math>:
 +
 +
<cmath>2x(5-x) + 2x(6-x) + 4x^2 = 60</cmath>
 +
<cmath>\Rightarrow 2x(5-x+6-x+2x) = 60</cmath>
 +
<cmath>\Rightarrow 2x(11) = 60</cmath>
 +
<cmath>\Rightarrow 22x = 60</cmath>
 +
<cmath>\Rightarrow x = \frac{60}{22} = \boxed{(B) \frac{30}{11}}</cmath>
 +
 +
~Dreamer1297
  
 
==Trivia==
 
==Trivia==

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_21&oldid=215213"