Two distinct regular tetrahedra – entirely contained within an unit cube - have all their vertices among the vertices of the same unit cube.
Determine the volume of the region formed by the intersection of the tetrahedra.
The intersetion of the tetrahedra is a regular octahedron with each of its six vertices at the center of one of the faces of the cube.
The octahedraon can be considered as consisting of two square-based pyramids with equilateral slant faces. As the opposite pairs of vertices are on opposite faces of the cube, their separations are 1 unit each, and the base-to-base pyramids each have height 1/2.
The edges of the pyramid, both slant and along the square base are going from the center of a square face of the original cube to the center of an adjacent face, and have the length sqrt(2)/2. Thus the square base of each pyramid had area 2/4 = 1/2.
The volume of each pyramid is then (1/2)*(1/2)/3 = 1/12.
Since the octahedron in question consists of two of these pyramids, it has volume 1/6.
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Posted by Charlie
on 2014-08-27 19:39:03 |
solution
