2016 AMC 12B Problems/Problem 23

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 $\frac{1}{6}$ .

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AMC 12 Problems and Solutions

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2016 AMC 12B Problems/Problem 23

Contents

Problem

Solution 1 (Non Calculus)

Solution 2 (Calculus)

See Also