# 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 $\frac{1}{6}$.

## 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~