2006 AMC 10A Problems/Problem 24

From AoPSWiki

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 $\frac{1}{12} \cdot 2 = \frac{1}{6} \Longrightarrow \mathrm{(B)}$ .

2006 AMC 10A (Problems)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Personal tools

Navigation

Search

Toolbox

2006 AMC 10A Problems/Problem 24

From AoPSWiki

Problem

Solution

See also