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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
h
0
=
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
C
0
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
C
0
is a distance
AC
0
=
a/
√
3. If we draw
a line from
C
0
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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 Fall '08
 SPRUNGER
 Physics

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