Difference between revisions of "2017 AMC 10A Problems/Problem 11"

(Solution)
(Solution)
Line 12: Line 12:
  
 
Solving, we find that <math>x = \boxed{\textbf{(D)}\ 20}</math>.
 
Solving, we find that <math>x = \boxed{\textbf{(D)}\ 20}</math>.
 +
 +
==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>
  
 
==Diagram for Solution==
 
==Diagram for Solution==

Revision as of 11:39, 12 February 2019

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

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 $3$ 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 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 $2$ "half spheres", and the radius is $3$, the volume of the $2$ half spheres is $4(3^3)/3 \pi = 36 \pi$. Since we also know that the volume of this whole thing is $216 \pi$, we do $216-36$ to get $180 \pi$ as the area of the cylinder. Thus the height is $180 \pi$ over the base, or $180 \pi/9\pi=20$, $D$

Diagram for Solution

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

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS