jmc

Call the length of a side of the cube x. Thus, the volume of the cube is $x^3$. We can then find that a side of this regular octahedron is the square root of $(\frac{x}{2}{)}^2$+$(\frac{x}{2}{)}^2$ which is equivalent to $\frac{x\sqrt{2}}{2}$. Using our general formula for the volume of a regular octahedron of side length a, which is $\frac{a^3\sqrt{2}}{3}$, we get that the volume of this octahedron is... $(\frac{x\sqrt{2}}{2}{)}^3\to \frac{x^3\sqrt{2}}{4}\to \frac{x^3\sqrt{2}}{4}\ast \frac{\sqrt{2}}{3}\to \frac{2x^3}{12}=\frac{x^3}{6}$ Comparing the ratio of the volume of the octahedron to the cube is… $\frac{\frac{x^3}{6}}{x^3}\to \boxed{\frac{1}{6}}$