Difference between revisions of "2016 AMC 12B Problems/Problem 23"

Latest revision as of 11:36, 2 January 2019

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~

@@ 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>
+==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>.
+==Solution 2 (Calculus)==
+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~
+==See Also==
+{{AMC12 box|year=2016|ab=B|num-b=22|num-a=24}}
+{{MAA Notice}}

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

Page

Toolbox

Search