2016 AMC 12B Problems/Problem 23

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 12 Problems and Solutions

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