CHAPTER XXXVIII

How the Exacedron or Cube Is Formed in the Octocedron.

The fourth proposition of the fifteenth book (Campanus XV.4) demonstrates the converse: how to inscribe the cube (6 square bases) inside the octocedron (8 triangular bases).

Inscribing the Cube in the Octahedron (Campanus XV.4)

In a given regular body of eight equilateral triangular bases (octocedron), to inscribe the cube (exacedron) so that the eight solid angles of the cube stand at the centers of the eight triangular faces of the octocedron.

Construction and Proof ([IV.4], [VI.2], [I.47], [XIII.12])

  1. In each of the eight equilateral triangular faces of the given octocedron (four in the upper pyramid and four in the lower pyramid sharing the equatorial square \(abcd\)), find the center of that equilateral triangle ([IV.4], [IV.5]) by drawing the perpendiculars (medians/angle-bisectors) from two corners to the midpoints of the opposite sides; their intersection is the center of the triangle (see the marginal diagram 0039.g1.png on Folio 0039).

  2. Let \(e, f, g, h\) be the four centers of the four upper triangular faces of the octocedron, and let \(k, l, m, n\) be the four centers of the four lower triangular faces.

  3. Join the four upper centers in order (\(ef, fg, gh, he\)), join the four lower centers in order (\(kl, lm, mn, nk\)), and join each upper center to the corresponding lower center across the four equatorial edges (\(ek, fl, gm, hn\)).

  4. In any equilateral triangle, the center divides each median (altitude) from a vertex to the midpoint of the opposite edge in the ratio \(2 : 1\), so its distance from the midpoint of the edge is one-third of the altitude of the triangle.

  5. By the second and fourth of the sixth, the four upper centers \(e, f, g, h\) lie in a horizontal plane parallel to the equatorial square \(abcd\) of the octocedron, and the square \(efgh\) is similar to the square Formed by the midpoints of the four equatorial edges at \(\frac{2}{3}\) scale. Since the midpoint square of the equatorial square of side \(s_{\text{octa}}\) has side \(\frac{\sqrt{2}}{2}s_{\text{octa}}\), each edge of \(efgh\) (and likewise \(klmn\) and the four vertical edges \(ek, fl, gm, hn\)) equals two-thirds of the half-diagonal, or:

    \[s_{\text{cube}} = \frac{\sqrt{2}}{3}\,s_{\text{octa}}, \qquad s_{\text{octa}}^{2} : s_{\text{cube}}^{2} = 9 : 2\]

Therefore the eight centers \(e, f, g, h, k, l, m, n\) of the eight triangular faces of the octocedron are the vertices of an inscribed cube, and the square on the edge of the octocedron has to the square on the edge of its inscribed cube the ratio \(9 : 2\) (quadrupla sesquialtera).