AMC12 2016 B

Answer (A): In the first octant, the first inequality reduces to $x+y+z\le 1$, and the inequality defines the region under a plane that intersects the coordinate axes at $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. By symmetry, the first inequality defines the region inside a regular octahedron centered at the origin and having internal diagonals of length $2$. The upper half of this octahedron is a pyramid with altitude $1$ and a square base of side length $\sqrt{2}$, so the volume of the octahedron is $2\cdot \frac{1}{3}\cdot (\sqrt{2})^2\cdot 1=\frac{4}{3}$. The second inequality defines the region obtained by translating the first region up $1$ unit. The intersection of the two regions is bounded by another regular octahedron with internal diagonals of length $1$. Because the linear dimensions of the third octahedron are half those of the first, its volume is $\frac{1}{8}$ that of the first, or $\frac{1}{6}$.