Let denote the vertices of a cube ABCD (for the bottom face) and A'B'C'D' (for the top face), with A' above A, etc. Let O stand for the center of the cube. What can be said about angle AOC, one of the central angles of the cube?
Angles in a Cube III
Scott E. Brodie
Consider a 2 × 2 × 2 cube centered at the origin of a Cartesian coordinate system. The vertices are located at A(1, 1, 1), B(-1, 1, 1), C(-1, -1, 1), D(1, -1, 1), A'(1, 1, -1), B'(-1, 1, -1), C'(-1, -1, -1), D'(1, -1, -1). A regular tetrahedron can easily be inscribed in the cube by connecting diagonally opposite vertices on the top face with the vertices of the diagonal running in the opposite direction on the bottom face: A(1, 1, 1) to C(-1, -1, 1), and B'(-1, 1, -1) to D'(1, -1, -1).
Now consider a "central angle" of the tetrahedron, say, the angle AOC. By visualizing the tetrahedron in this way, we can readily compute the size of this angle - call it t:
Taking the dot product of the vectors from the origin to the two upper vertices of the tetrahedron, we obtain
√1² + 1² + 1² × √(-1)² + (-1)² + 1² × cos (t) = 1×(-1) + 1× (-1) + 1×1 = -1.
Or
cos (t) = -1/3,
t = arccos (-1/3) ~ 109.47122 degrees.
This angle is an important building block in many crystal structures, including those of diamonds and ice, as well as the basic structure of many organic molecules.
While all central angles in a regular tetrahedron are equal, in a cube, there is another central angle distinct from the above: this is the angle that subtends an edge of a cube, e.g. ∠COC'. We denote it s. While t is obtuse, s is acute. Applying the same technique as for t, it is easily seen that arccos(s) = 1/3. Thus t and s are supplementary: s + t = 180°. The applet helps visualize the situation.
