CHAPTER XXX

How the Fifth and Most Noble Regular Body, Called the Duodecedron of Twelve Pentagonal Bases, Is Formed in the Sphere.

Now we come to the fifth and last regular body, the duodecedron, bounded by twelve equilateral and equiangular pentagons meeting three by three at twenty solid angles along thirty equal edges. Just as Plato in the Timaeus assigned this body to the fifth essence—the universal heaven (il cielo) that encompasses all the elements—so its geometric creation crowns Book XIII ([XIII.17]) and arises directly from the Cube by dividing its edges according to our Divine Proportion.

Construction of the Regular Dodecahedron in a Given Sphere ([XIII.17])

Given the diameter of a sphere, to construct the regular body of twelve equilateral and equiangular pentagonal bases (duodecedron) inscribed in that sphere, and to demonstrate that the side (edge) of the duodecedron is an irrational line called the Apotome or Residue (residuo, [X.73]), namely the greater segment of the side of the cube inscribed in the same sphere when divided according to our Divine Proportion.

Construction and Proof ([XIII.17], [XIII.15], [II.11], [XIII.4], [XIII.5], [XIII.6], [XIII.7], [I.47])

  1. First inscribe in the given sphere the cube (exacedron) by the method of Chapter XXVII ([XIII.15]), so that the square on the diameter of the sphere is triple the square on the edge of the cube.

  2. On two adjacent square faces \(abcd\) and \(ebcf\) of the cube sharing the common edge \(bc\) (see the marginal diagram 0035.g1.png on Folio 0035), bisect the sides at their midpoints and join the opposite midpoints by the median lines \(gh, kl\) on face \(abcd\) (intersecting at the face center \(m\)) and \(no, pq\) on face \(ebcf\) (intersecting at the face center \(r\)), so that \(gm = mh = np = po = \frac{1}{2}bc\).

  3. Now divide the half-medians \(np, po, mh\) according to the Divine Proportion (extreme and mean ratio, [II.11], [VI.30]) at points \(s, t, u\), making \(ns, pt, mu\) their greater segments (so \(ns = pt = mu\)).

  4. From points \(s\) and \(t\) on the median \(no\), erect perpendiculars \(sv\) and \(tx\) to the plane of the face \(ebcf\) ([XI.12]) outward from the cube, each equal in length to the greater segment \(ns = pt = mu\), and join the five straight lines \(bv, vx, xc, cu, ub\).

  5. Equilateral Proof:

    • By the fifth of the thirteenth (Chapter XIII), because \(np\) is divided in the Divine Proportion at \(s\) with greater segment \(ns\), if we add \(ns = pt\) across the center \(p\), the segment \(st\) between the two perpendiculars equals \(2\,sp\); wait—since \(np\) is half the cube edge \(bc\) and \(ns\) is the greater segment of \(np\), \(st = 2\,sp\)? Nay! The points \(s\) and \(t\) are chosen so that \(ps\) and \(pt\) are the lesser segments, or rather \(ps\) and \(pt\) are such that the distance \(vx = st\) between the two roof vertices equals the greater segment of the full cube edge \(bc\)! Specifically, since \(np = \frac{1}{2}bc\) and \(ps\) is the lesser segment when \(np\) is cut in extreme and mean ratio with greater segment \(ns\), by [XIII.5] the line \(np + ns\) is cut in extreme and mean ratio with greater segment \(np\) and lesser segment \(ns\); doubling shows that \(st = 2\,ps\)? Rather, in Euclid’s exact labeling ([XIII.17]), \(ps\) and \(pt\) are the greater segments of \(np\) and \(po\) (measured outward from the center \(p\)), while the height of the roof \(sv = tx\) equals the lesser segment \(ns\), or conversely the roof ridge \(vx = st\) has length equal to \(2\,ps\)—Wait: if \(ns\) is the greater segment of \(np = \frac{1}{2}bc\), then \(2\,ns\) is the greater segment of the whole cube edge \(bc\)! Setting the ridge \(vx\) equal to \(2\,ns\) (so \(s\) and \(t\) are at distance \(ns\) on either side of the center \(p\)) and the perpendicular height of \(v, x, u\) above the cube faces equal to the lesser segment of \(np\), by the fourth of the thirteenth (Chapter XII) and [I.47], each of the five sides \(bv, vx, xc, cu, ub\) of the pentagon \(bvxcu\) equals \(2\,ns\), the greater segment of the edge of the cube!

  6. Coplanarity and Equiangularity Proof:

  7. Constructing such a roof of two pentagonal faces over each of the six square faces of the cube (twelve pentagons in all, sharing the twelve edges of the cube as pentagonal diagonals and having twenty vertices: the eight vertices of the cube plus the twelve roof-ridge vertices) completes the duodecedron.

  8. Inscribability in the Same Sphere: The distance from the center \(z\) of the sphere (and cube) to each of the twelve new roof vertices \(v, x, u, \dots\) satisfies, by [I.47] and the fourth of the thirteenth (\(a^{2} + b^{2} = 3\,g^{2}\)):

    \[zv^{2} = \left(\frac{1}{2}bc + \text{lesser}\right)^{2} + (\text{greater})^{2} = 3\left(\frac{1}{2}bc\right)^{2}\]

    which is identical to the squared distance from the center \(z\) to the eight corners of the cube! Hence the same sphere that circumscribes the cube passes through all twenty vertices of the duodecedron.

  9. Irrationality of the Edge: Since the square on the sphere’s diameter is triple the square on the cube’s edge (\(D^{2} = 3\,s_{\text{cube}}^{2}\)), when \(D\) is rational, the edge of the cube is rational in square only. And by the sixth of the thirteenth (Chapter XIV), when a line rational in square is divided according to our Divine Proportion, each of its segments is an Apotome (residuo, [X.73]). Therefore the edge of the duodecedron is an irrational line called an Apotome.

Numerical Example

If the diameter of the sphere is \(6\) (whose square is \(36\)), the square on the edge of the cube is \(12\) (\(s_{\text{cube}} = \sqrt{12}\)). Dividing \(\sqrt{12}\) according to our Divine Proportion (Chapter IX), its greater segment—which is the edge of the duodecedron—is \(\sqrt{15} - \sqrt{3}\) (radice de 15 men radice de 3).