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 "2016 AMC 12B Problems/Problem 23"

(→‎Solution)
m (→‎Problem)
Line 1: Line 1:
−
=Problem=
+
==Problem==
  
 
What is the volume of the region in three-dimensional space defined by the inequalities <math>|x|+|y|+|z|\le1</math> and <math>|x|+|y|+|z-1|\le1</math>
 
What is the volume of the region in three-dimensional space defined by the inequalities <math>|x|+|y|+|z|\le1</math> and <math>|x|+|y|+|z-1|\le1</math>
  
 
<math>\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ 1</math>
 
<math>\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ 1</math>
 +
 
=Solution 1 (Non Calculus)=
 
=Solution 1 (Non Calculus)=
 
The first inequality refers to the interior of a regular octahedron with top and bottom vertices <math>(0,0,1),\ (0,0,-1)</math>. Its volume is <math>8\cdot\tfrac16=\tfrac43</math>. The second inequality describes an identical shape, shifted <math>+1</math> upwards along the <math>Z</math> axis. The intersection will be a similar octahedron, linearly scaled down by half. Thus the volume of the intersection is one-eighth of the volume of the first octahedron, giving an answer of <math>\textbf{(A) }\tfrac16</math>.
 
The first inequality refers to the interior of a regular octahedron with top and bottom vertices <math>(0,0,1),\ (0,0,-1)</math>. Its volume is <math>8\cdot\tfrac16=\tfrac43</math>. The second inequality describes an identical shape, shifted <math>+1</math> upwards along the <math>Z</math> axis. The intersection will be a similar octahedron, linearly scaled down by half. Thus the volume of the intersection is one-eighth of the volume of the first octahedron, giving an answer of <math>\textbf{(A) }\tfrac16</math>.

Revision as of 20:16, 19 June 2017

Problem

What is the volume of the region in three-dimensional space defined by the inequalities $|x|+|y|+|z|\le1$ and $|x|+|y|+|z-1|\le1$

$\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ 1$

Solution 1 (Non Calculus)

The first inequality refers to the interior of a regular octahedron with top and bottom vertices $(0,0,1),\ (0,0,-1)$. Its volume is $8\cdot\tfrac16=\tfrac43$. The second inequality describes an identical shape, shifted $+1$ upwards along the $Z$ axis. The intersection will be a similar octahedron, linearly scaled down by half. Thus the volume of the intersection is one-eighth of the volume of the first octahedron, giving an answer of $\textbf{(A) }\tfrac16$.

Solution 2 (Calculus)

Let $z\rightarrow z-1/2$, then we can transform the two inequalities to $|x|+|y|+|z-1/2|\le1$ and $|x|+|y|+|z+1/2|\le1$. Then it's clear that $-1/2\le z \le 1/2$, consider $0 \le z \le 1/2$, $|x|+|y|\le 1/2-z$, then since the area of $|x|+|y|\le k$ is $2k^2$, the volume is $\int_{0}^{1/2}2k^2 \,dk=\frac{1}{12}$. By symmetry, the case when $\frac{-1}{2}\le z\le0$ is the same. Thus the answer is $\frac{1}{6}$.

See Also

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 12 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=2016_AMC_12B_Problems/Problem_23&oldid=86113"