37. The description in the problem statement implies that an atom is at the centerpoint C of the regular tetrahedron, since its four neighbors are at the four vertices. The side length for the tetrahedron is given as a = 388 pm. Since each face is an equilateral triangle, the “altitude” of each of those triangles (which is not to be confused with the altitude of the tetrahedron itself) is h0 = 1 2 a √ 3 (this is generally referred to as the “slant height” in the solid geometry literature). At a certain location along the line segment representing “slant height” of each face is the center C0 of the face. Imagine this line segment starting at atom A and ending at the midpoint of one of the sides. Knowing that this line segment bisects the 60 ◦ angle of the equilateral face, then it is easy to see that C0 is a distance AC0 = a/ √ 3. If we draw a line from C0 all the way to farthest point on the tetrahedron (this will land on an atom we label B ), then this new line is the altitude
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