a 3d geometry problem with tetrahedrons and cube

Question

Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?

So at first thought I tried to drew the figure by hand as this problem is supposed to be done in a non- graphing utility environment -- but it didn't really work out so I want to ask is there anyway to do this without a diagram? I know the volume of tetrahedrons but I just don't get where they intersect. Thanks in advance.

Answer 1

Here's an alternative approach. Label the vertices of the cube as follows. $O:(0,0,0)$, $A:(1,0,0)$, $B:(0,1,0)$, $C:(1,1,0)$, $D:(0,0,1)$, $E:(1,0,1)$, $F:(0,1,1)$, $G:(1,1,1)$. One of the tetrahedra has vertices $ABDG$. This is obtained from the cube by removing four tetrahedra congruent to $OABD$. Tetrahedron $OABD$ has volume $1/6$, so tetrahedron $ABDG$ has volume $1-4/6=1/3$.

The region we seek is octahedral and obtained from $ABDG$ by removing four tetrahedra similar to it, and with half the side-lengths. For instance one of these little tetrahedra has vertices $A$, and the mid-points of $AB$, $AD$ and $AG$. Each little tetrahedron has volume $1/8$ of the volume of $ABDG$, so the volume of the sought octahdedron is half that of $ABDG$, that is $1/6$.

Answer 2

By hand sketching you should be able to convince yourself that the sides of a tetrahedron are all diagonals of faces of the cube. One side lies in each face of the cube. The two tetrahedrons use all the twelve face diagonals between them. The drawing below shows three slices through the cube.

The first is one face and the two diagonal lines are the edges of the two tetrahedra. The second is $1/6$ way up the cube. The two rectangles are the cross sections of the tetrahedra in that plane and the diagonal square in the middle is the intersection. The third is the midplane of the cube. The diagonal square in the middle is the common cross section of the tetrahedra. The side of this square is $\frac {\sqrt 2}2$ and its area is $\frac 12$. The short side of the rectangle, and therefore the side of the intersection square increases linearly from the bottom to the midplane, so the side of the intersection is $\sqrt 2 x$ when $0 \le x \le \frac 12$ and $\sqrt 2(1-x)$ when $\frac 12 \le x \le 1$. We can integrate the area over $[0,\frac 12]$ and double it to get the volume. $$V=2\int_0^{\frac 12}(\sqrt 2 x)^2dx=4\int_0^{\frac 12}x^2dx=\frac 16$$

Answer 3

The intersection is a regular octahedron with its vertices at the face centers of the cube. Its volume then is ${1\over6}$.