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?
Solution
We can break the octahedron into two square pyramids by cutting it along a plane perpendicular to one of its internal diagonals.
![[asy] import three; real r = 1/2; triple A = (-0.5,1.5,0); size(400); currentprojection=orthographic(1,1/4,1/2); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)^^(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--(0,0,1)^^(0,0,0)--(0,0,1)^^(1,0,0)--(1,0,1)^^(0,1,0)--(0,1,1)^^(1,1,0)--(1,1,1),gray(0.8)); draw((0,r,r)--(r,1,r)--(1,r,r)--(r,0,r)--cycle^^(r,r,0)--(0,r,r)--(r,r,1)--(r,1,r)--(r,r,0)--(1,r,r)--(r,r,1)--(r,0,r)--(r,r,0)); draw((0,r,r)+A--(r,1,r)+A--(1,r,r)+A--(r,0,r)+A--cycle^^(0,r,r)+A--(r,r,1)+A--(1,r,r)+A^^(r,1,r)+A--(r,r,1)+A--(r,0,r)+A); [/asy]](http://latex.artofproblemsolving.com/8/2/8/828d5c82721bf673e5ad6c184d1789de3b1b20d7.png)
The cube has edges of length 1 so all edges of the regular octahedron have length 
. Then the square base of the pyramid has area 
.
We also know that the height of the pyramid is half the height of the cube, so it is 
. The volume of a pyramid with base area 
and height 
is 
so each of the pyramids has volume 
. The whole octahedron is twice this volume, so 
.
See also
| 2006 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
