# 2006 AMC 10A Problems/Problem 24

## Problem

Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron? $\mathrm{(A) \ } \frac{1}{8}\qquad\mathrm{(B) \ } \frac{1}{6}\qquad\mathrm{(C) \ } \frac{1}{4}\qquad\mathrm{(D) \ } \frac{1}{3}\qquad\mathrm{(E) \ } \frac{1}{2}\qquad$

## Solution

We can break the octahedron into two square pyramids by cutting it along a plane perpendicular to one of its internal diagonals. The cube has edges of length one so the edge of the octahedron is of length $\frac{\sqrt{2}}{2}$. Then the square base of the pyramid has area $\left(\frac{1}{2}\sqrt{2}\right)^2 = \frac{1}{2}$. We also know that the height of the pyramid is half the height of the cube, so it is $\frac{1}{2}$. The volume of a pyramid with base area $B$ and height $h$ is $A=\frac{1}{3}Bh$ so each of the pyramids has volume $\frac{1}{3}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{12}$. The whole octahedron is twice this volume, so $\frac{1}{12} \cdot 2 = \frac{1}{6} \Longrightarrow \mathrm{(B)}$.