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

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

@@ Line 7: / Line 7: @@
 ==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 <math>3</math> 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.
@@ Line 14: / Line 14: @@
 ==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==
 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

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