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 "2017 AMC 10A Problems/Problem 11"

m (→‎Problem)
m (→‎Solution 2)
 
(17 intermediate revisions by 11 users not shown)
Line 5: Line 5:
 
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24</math>
 
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24</math>
  
−
==Solution==
+
==Solution 1==
  
−
In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within <math>3</math> units of a point would be a sphere with radius <math>3</math>.  However, we need to find the region containing all points within 3 units of a segment.  It can be seen that our region is a cylinder with two hemispheres on either end.  We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal <math>216 \pi</math>):
+
In order to solve this problem, we must first visualize what the region looks like. We know that, in a three dimensional space, the region consisting of all points within <math>3</math> units of a point would be a sphere with radius <math>3</math>.  However, we need to find the region containing all points within <math>3</math> units of a segment.  It can be seen that our region is a cylinder with two hemispheres/endcaps on either end.  We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal <math>216 \pi</math>):
  
 
<math>\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi</math>, where <math>x</math> is equal to the length of our line segment.
 
<math>\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi</math>, where <math>x</math> is equal to the length of our line segment.
Line 14: Line 14:
  
 
==Solution 2==
 
==Solution 2==
−
Because this is just a cylinder and <math>2</math> "half spheres", and the radius is <math>3</math>, the volume of the <math>2</math> half spheres is <math>4(3^3)/3 \pi = 36 \pi</math>. Since we also know that the volume of this whole thing is <math>216 \pi</math>, we do <math>216-36</math> to get <math>180 \pi</math> as the area of the cylinder. Thus the height is <math>180 \pi</math> over the base, or <math>180 \pi/9\pi=20</math>, <math>D</math>
+
Because this is just a cylinder and <math>2</math> hemispheres ("half spheres"), and the radius is <math>3</math>, the volume of the <math>2</math> hemispheres is <math>\frac{4(3^3)\pi}{3} = 36 \pi</math>. Since we also know that the volume of this whole thing is <math>216 \pi</math>, we do <math>216 \pi-36 \pi</math> to get <math>180 \pi</math> as the volume of the cylinder. Thus the height is <math>180 \pi</math> divided by the area of the base, or <math>\frac{180 \pi}{9\pi}=20</math>, so our answer is <math>\boxed{\textbf{(D)}\ 20}.</math>
 +
 
 +
~Minor edit by virjoy2001 and slamgirls
  
 
==Diagram for Solution==
 
==Diagram for Solution==
 
http://i.imgur.com/cwNt293.png
 
http://i.imgur.com/cwNt293.png
 +
 +
==Video Solution==
 +
https://youtu.be/s4vnGlwwHHw
  
 
==See Also==
 
==See Also==

Latest revision as of 13:24, 13 April 2024

Problem

The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$. What is the length $\textit{AB}$?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$

Solution 1

In order to solve this problem, we must first visualize what the region looks like. We know that, in a three dimensional space, the region consisting of all points within $3$ units of a point would be a sphere with radius $3$. However, we need to find the region containing all points within $3$ units of a segment. It can be seen that our region is a cylinder with two hemispheres/endcaps on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal $216 \pi$):

$\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi$, where $x$ is equal to the length of our line segment.

Solving, we find that $x = \boxed{\textbf{(D)}\ 20}$.

Solution 2

Because this is just a cylinder and $2$ hemispheres ("half spheres"), and the radius is $3$, the volume of the $2$ hemispheres is $\frac{4(3^3)\pi}{3} = 36 \pi$. Since we also know that the volume of this whole thing is $216 \pi$, we do $216 \pi-36 \pi$ to get $180 \pi$ as the volume of the cylinder. Thus the height is $180 \pi$ divided by the area of the base, or $\frac{180 \pi}{9\pi}=20$, so our answer is $\boxed{\textbf{(D)}\ 20}.$

~Minor edit by virjoy2001 and slamgirls

Diagram for Solution

http://i.imgur.com/cwNt293.png

Video Solution

https://youtu.be/s4vnGlwwHHw

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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=2017_AMC_10A_Problems/Problem_11&oldid=218158"