Note: a regular octahedron is a polyhedron with eight faces each of which is an equilateral triangle.
Show that the angle between any two faces of a regular octahedron is \(\arccos \left( -{\frac1 3} \right)\).
Find the ratio of the volume of a regular octahedron to the volume of the cube whose vertices are the centres of the faces of the octahedron.
Solution:
Suppose the vertices are \((\pm1, 0,0), (0,\pm1,0), (0,0,\pm1)\), then clearly this is an octahedron.
We can measure the angle between the faces, by looking at vectors in the same plane and also in two of the faces: \(\langle \frac12, \frac12, - 1\rangle\) and \(\langle \frac12, \frac12, 1\rangle\), then by considering the dot product:
\begin{align*}
&& \cos \theta &= \frac{\frac14+\frac14-1}{\frac14+\frac14+1} \\
&&&= \frac{-2}{6} = -\frac13
\end{align*}
The volume of our octahedron is \(2 \cdot \frac13 \cdot \underbrace{\sqrt{2}^2}_{\text{base}} \cdot \underbrace{1}_{\text{height}} = \frac43\).
The centre of two touching faces are \(\langle \frac13, \frac13, \frac13 \rangle\) and \(\langle \frac13, \frac13, -\frac13 \rangle\) and so the length of the side of the cube is \(\frac23\) and so the volume of the cube is \(\frac8{27}\). Therefore the ratio is \(\frac{2}{9}\)
Show that four vertices of a cube, no two of which are adjacent,
form the vertices of a regular tetrahedron.
Hence, or otherwise, find the volume of a regular
tetrahedron whose edges are of unit length.
Find the volume of a regular octahedron whose edges are of unit length.
Show that the centres of the faces of a cube form the vertices
of a regular octahedron. Show that its volume is half that of the
tetrahedron whose vertices are the vertices of the cube.
\noindent
[{\em A regular tetrahedron (octahedron)
has four (eight) faces, all equilateral triangles.}]