What is the ratio of volumes of a regular tetrahedron and the smallest cube that can encompass it?
The smallest cube around a tetrahedron is equivalent to the largest tetrahedron in a cube.
The tetrahedron with the largest volume will have verticies coinciding with the verticies of the cube. The four verticies of the cube are all sqrt 2 units apart. A tetrahedron with an edge of sqrt 2 has a volume of (sqrt 2)^3*(sqrt 2)/12=1/3. The cube has volume 1^3=1. The ratio of tetrahedron to cube is (1/3)/1=1/3.
solution
