Solution of the problem #24 in AMC 10 A 2011

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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 tetrahedral?
(A) 1/12    (B) sqrt(2)/12    (C) sqrt(3)/12    (D) 1/6    (E) sqrt(2)/6

Solution:

Let’s assign the following numbers to the vertices of the unit cube:
numbers 1, 2, 3, 4 to the four vertices of the bottom face of the cube (clockwise) and numbers 5, 6, 7, 8 to the vertices of the top face of the cube.

Let's position the two tetrahedra that satisfy the conditions of this problem so that the first tetrahedron has vertices 4, 5, 7, 2 and the second  tetrahedron has the remaining four vertices 1, 3, 6, 8. The pyramid with vertices 4, 5, 7, 8 shares one face with the first tetrahedron (the triangle with vertices 4,5,7).

Obviously, the other three faces of the first tetrahedron have their adjacent pyramids, congruent to the pyramid with vertices 4, 5, 7, 8. If we remove all four adjacent pyramids from the cube, the only part left will be the first tetrahedron.

It makes calculation of the volume of each of the two tetrahedra simple: it is the volume of the cube minus four times volume of the pyramid with vertices 4, 5, 7, 8.
The volume of this pyramid is the area of its face with vertices 4, 5, 8 (S = 1*1* 1/2) multiplied by the length of its altitude (the edge between vertices 7 and 8 whose length is 1) and divided by 3. It is: S*1*1/3 = 1*1*1/2*1/3 = 1/6. The volume of the cube equals 1*1*1 = 1. So, the volume of the tetrahedron is: 1 – 4*1/6 = 1/3.

Let’s observe that the face of the first tetrahedron with vertices 4, 5, 7 cuts the three edges of the other tetrahedron in the middle of each edge between the vertices: (1,8), (6,8), (3,8). It is easy to see if we realize that the edges of the second tetrahedron are the other diagonals of the square faces of the cube! For example, two diagonals (1,8) and (4,5) of the square (1,5,8,4) are the edges of two different tetrahedra!

We can see from this observation that each tetrahedron cuts off four top pyramids from the other tetrahedron, and each top pyramid is similar to the entire tetrahedron with the ratio of their edges = 1/2. Therefore, the ratio of the volumes of each top pyramid to the entire tetrahedron is (1/2)^3 = 1/8.
So, each top pyramid of a tetrahedron cut by the other tetrahedron has volume equal to 1/8 of the volume of tetrahedron. Now, we need to subtract the volumes of four top pyramids from the volume of any one tetrahedron to calculate the volume of intersection of two tetrahedra. It is: 1/3 - 4*(1/3)*(1/8) = 1/6   (D)