CHAPTER XXXIX

How the Tetracedron Is Formed in the Octocedron.

The fifth proposition of the fifteenth book (Campanus XV.5) shows how to inscribe the tetracedron (4 triangular bases) inside the octocedron (8 triangular bases) by means of the cube already inscribed in the octocedron.

Inscribing the Tetrahedron in the Octahedron (Campanus XV.5)

In a given regular body of eight triangular bases (octocedron), to inscribe the regular body of four triangular bases (tetracedron) so that the four vertices of the tetracedron stand at four alternating centers of the eight faces of the octocedron.

Construction and Proof (via Campanus XV.4 and XV.2)

  1. First, in the given octocedron inscribe the cube \(efghklmn\) by the method of Chapter XXXVIII (Campanus XV.4), so that the eight vertices of the cube stand at the eight centers of the triangular faces of the octocedron (see the marginal diagram 0039.g2.png on Folio 0039).

  2. Second, in this inscribed cube \(efghklmn\), inscribe the tetracedron by the method of Chapter XXXVI (Campanus XV.2), drawing the six face diagonals across the six square faces of the cube to join four non-adjacent vertices of the cube (for instance \(e, g, l, n\)).

  3. Because the eight vertices of the cube lie at the centers of the eight faces of the octocedron, the four vertices \(e, g, l, n\) of this tetracedron likewise stand at four alternating face-centers of the octocedron!

  4. Moreover, since by Chapter XXXVIII the square on the edge of the octocedron has to the square on the edge of its inscribed cube the ratio \(9 : 2\), and by Chapter XXXVI the square on the edge of the tetracedron inscribed in that cube is double (\(2 : 1\)) the square on the edge of the cube, by ex aequali ([V.22]) the square on the edge of the octocedron has to the square on the edge of its inscribed tetracedron the ratio \(9 : 4\):

    \[s_{\text{octa}}^{2} : s_{\text{tetra}}^{2} = 9 : 4 \implies s_{\text{tetra}} = \frac{2}{3}\,s_{\text{octa}}\]

Thus the edge of the tetracedron inscribed in the octocedron is commensurable in length with the edge of the octocedron, being exactly two-thirds (\(\frac{2}{3}\)) of it!