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"

m (→‎Problem)
 
(One intermediate revision by one other user not shown)
Line 11: Line 11:
  
 
Let <math>z\rightarrow z-1/2</math>, then we can transform the two inequalities to <math>|x|+|y|+|z-1/2|\le1</math> and <math>|x|+|y|+|z+1/2|\le1</math>. Then it's clear that <math>-1/2\le z \le 1/2</math>, consider <math>0 \le z \le 1/2</math>, <math>|x|+|y|\le 1/2-z</math>, then since the area of <math>|x|+|y|\le k</math> is <math>2k^2</math>, the volume is <math>\int_{0}^{1/2}2k^2 \,dk=\frac{1}{12}</math>. By symmetry, the case when <math>\frac{-1}{2}\le z\le0</math> is the same. Thus the answer is <math>\frac{1}{6}</math>.
 
Let <math>z\rightarrow z-1/2</math>, then we can transform the two inequalities to <math>|x|+|y|+|z-1/2|\le1</math> and <math>|x|+|y|+|z+1/2|\le1</math>. Then it's clear that <math>-1/2\le z \le 1/2</math>, consider <math>0 \le z \le 1/2</math>, <math>|x|+|y|\le 1/2-z</math>, then since the area of <math>|x|+|y|\le k</math> is <math>2k^2</math>, the volume is <math>\int_{0}^{1/2}2k^2 \,dk=\frac{1}{12}</math>. By symmetry, the case when <math>\frac{-1}{2}\le z\le0</math> is the same. Thus the answer is <math>\frac{1}{6}</math>.
 +
 +
==Solution 3==
 +
Do this problem first on the first quadrant. Graph this by using test points and you will see that you will have two tetrahedrons - each of them intersects at the middle, and thus its <math>1/4</math>th of the area of the tetrahedrons (since they are the same area). The area of a tetrahedron is bh/3, which gives us <math>1/6</math>, and then divide that by <math>4</math> to get the bounded area of <math>1/24</math> in the shaded region. Scale this up to the other quadrants now (since they are the same due to the abs value) and you get <math>\textbf{(A) }\tfrac16</math>.
 +
 +
Sol by IronicNinja~
 +
 +
==Video Solution by CanadaMath (Problem 21-25)==
 +
Fast Forward to 20:17 for problem 23
 +
https://www.youtube.com/watch?v=P3jJDLGyF2w&t=1546s
 +
 +
~THEMATHCANADIAN
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=22|num-a=24}}
 
{{AMC12 box|year=2016|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:37, 9 November 2024

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 $\frac{1}{6}$.

Solution 3

Do this problem first on the first quadrant. Graph this by using test points and you will see that you will have two tetrahedrons - each of them intersects at the middle, and thus its $1/4$th of the area of the tetrahedrons (since they are the same area). The area of a tetrahedron is bh/3, which gives us $1/6$, and then divide that by $4$ to get the bounded area of $1/24$ in the shaded region. Scale this up to the other quadrants now (since they are the same due to the abs value) and you get $\textbf{(A) }\tfrac16$.

Sol by IronicNinja~

Video Solution by CanadaMath (Problem 21-25)

Fast Forward to 20:17 for problem 23 https://www.youtube.com/watch?v=P3jJDLGyF2w&t=1546s

~THEMATHCANADIAN

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=232946"