## Angles in a Cube III

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?

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Discussion ### 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. ### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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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.

It is apparent that triangle ACC' is right, with legs 2 and 22 and hypotenuse 23. (To see that apply the Pythagorean theorem to a couple of right triangles.) The median to the hypotenuse is just a half of the latter, 3, drawing our attention once more to the fact that AC' is a straight line. • 