CHAPTER XXX — Critical Commentary & Notes

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

Historical & Mathematical Commentary

  • Paganini 1509 Marginal Diagram: Folio 0035 (f. 10r) displays diagram 0035.g1.png, showing the unfolded squares of the cube with their bisected medians and the pentagonal face \(bvxcu\) pitched like a hipped roof (coperto / colmo) over the cube edge \(bc\).

  • Why Section 2 Proved Euclid XIII.4, Euclid XIII.6, and Euclid XIII.7: Every step of Euclid’s construction of the dodecahedron in Euclid XIII.17 calls upon a specific proposition from earlier in Book XIII:

    • Euclid XIII.4 (\(L^{2} + b^{2} = 3a^{2}\)) proves both that the roof edges equal the ridge \(2a\) and that the twelve roof vertices lie at the same distance \(\sqrt{3}\cdot(\frac{1}{2}s_{\text{cube}})\) from the center of the cube as the eight cube corners!

    • Euclid XIII.7 proves that the pentagon \(bvxcu\) is equiangular because three of its angles are subtended by equal chords (\(bc\) and the two diagonals meeting at \(u\)).

    • Euclid XIII.6 proves that the edge \(a_{12} = \frac{\sqrt{5}-1}{2} s_{\text{cube}}\) is an Apotome (Euclid X.73).

  • Verification of Pacioli’s Numerical Example: For \(D = 6\), \(s_{\text{cube}} = \sqrt{12}\). The greater segment of \(\sqrt{12}\) cut in extreme and mean ratio is \(\frac{\sqrt{5}-1}{2}\sqrt{12} = \sqrt{3}(\sqrt{5}-1) = \sqrt{15} - \sqrt{3}\).

Constantin Winterberg’s Commentary (1889 Edition)

Note 13 (Cap. XXX): Verification of the Pentagon Diagonal in the Dodecahedron Construction (Euclid XIII.17) (Sachliche Anmerkungen #13, Cap. XXX, p. 224, l. 4 from bottom)

Retaining the lettering of the figure in Chapter XXX (where $ke$ is half the cube edge, divided in extreme and mean ratio at $l$ with $kl > le$, and $an = hn = 2\,kl$ is the cube edge): by the three-dimensional Pythagorean projection,

\[ln^2 = an^2 - ke^2 - le^2 = 4\,kl^2 - ke^2 - le^2.\]

Since $kl^2 = ke \cdot le$, we can rewrite this as:

\[ln^2 = 2\,kl^2 + 2\,ke \cdot le - ke^2 - le^2 = 2\,kl^2 - (ke - le)^2 = 2\,kl^2 - kl^2 = kl^2,\]

whence $ln = kl$, proving that the five vertices lie at equal radii and form an equilateral and equiangular pentagon (Euclid XIII.17).

Anmerkung 13 (Cap. XXX, S. 224): Nachweis der Diagonalengleichung beim Dodekaeder (Euklid XIII.17)

Dies ergibt sich unter Beibehaltung der Figur des Textes folgendermassen: Es ist der Construction zufolge (da $an = 2\,kl$ als Kante des Würfels gleich $hn$ u. s. w. ist):

\[ln^2 = an^2 - ke^2 - le^2 = 4\,kl^2 - ke^2 - le^2.\]

Da $kl^2 = ke \cdot le$ ist, kann man diese Gleichung auch schreiben:

\[ln^2 = 2\,kl^2 - (ke^2 - 2\,ke \cdot le + le^2) = 2\,kl^2 - (ke - le)^2 = kl^2. \quad \text{w. z. b. w.}\]

Note 14 (Cap. XXX): Semicircle Proof of the Fifth Effect Used in the Dodecahedron Sphere Inscription (Sachliche Anmerkungen #14, Cap. XXX, p. 225, l. 5 from top)

This relation ($el^2 + ke^2 = 3\,kl^2$, already proved algebraically in Note 5) is seen most simply geometrically if we describe a semicircle on $ek$ as diameter, erect at $l$ a perpendicular meeting the semicircle at $m$, and draw the chords $em$ and $km$:

  • By hypothesis, $kl^2 = ke \cdot el$, and by the right triangle $emk$ in the semicircle, $em^2 = ke \cdot el$, so $em = kl$.

  • Moreover, $el^2 = em^2 - ml^2$ and $ke^2 = em^2 + mk^2$. Adding both gives:

    \[el^2 + ke^2 = 2\,em^2 + (mk^2 - ml^2) = 2\,em^2 + kl^2 = 3\,em^2 = 3\,kl^2. \quad \text{Q.E.D.}\]

Anmerkung 14 (Cap. XXX, S. 225): Geometrischer Beweis der Relation $el^2 + ke^2 = 3\,kl^2$

Dies ergibt sich am einfachsten, wenn man sich über $ek$ als Durchmesser einen Halbkreis denkt und im Punkte $l$ eine Senkrechte bis zum Durchschnitt $m$ mit ihm, sowie die Sehnen $em, km$ zieht. Dann ist nach Voraussetzung $kl^2 = ke \cdot el$, und ferner nach bekannten Sätzen:

\[el^2 = em^2 - ml^2, \qquad ke^2 = em^2 + mk^2,\]

also addirt:

\[el^2 + ke^2 = 2\,em^2 + mk^2 - ml^2 = 2\,em^2 + lk^2.\]

Da ferner $em^2 = el \cdot ek = lk^2$, so folgt: $el^2 + ke^2 = 3\,em^2 = 3\,kl^2$. w. z. b. w.

  • Winterberg Footnotes 1–4 (Cap. XXX, pp. 222–224): (1) Mathematical abbreviations are introduced occasionally here and in the following chapters for clearer overview. (2) Namely, because of the similarity of the two triangles $ksh$ and $qrh$ following from the proportion, since the sum of the two adjacent angles plus the dihedral angle enclosed by the two cube faces equals $180^\circ$. (3) For brevity, the expression ‘continuous division’ (stetige Theilung) is used for Pacioli's ‘proportion having a mean and two extremes’. (4) Compare Chapter LXXI. (Winterberg Fussnoten 1–4 (Cap. XXX, S. 222–224): 1) Es sei erlaubt, hier und im Folgenden einige mathematische Abkürzungen der bessern Uebersicht wegen gelegentlich anzubringen. 2) scil. wegen der aus obiger Proportion folgenden Aehnlichkeit der beiden Dreiecke ksh und qrh, da die Summe der beiden anstossenden Winkel plus dem durch die beiden Kubusflächen eingeschlossenen zusammen gleich 180° sind. 3) Der Kürze wegen ist hier und im Folgenden dieser Ausdruck für den gleichbedeutenden: „nach der Proportion, die einen mittlern und zwei äussere Abschnitte hat“ gesetzt worden. 4) cfr. Cap. LXXI.)

Philological and Folio Apparatus

  • 1509 Paganini Witness Folios: 0036, 0037, 0038 (f. 10v–f. 11v)

  • Winterberg 1889 Italian Source: winterberg/01_divina_proportione_italian/030_cap_XXX.txt

  • Winterberg 1889 German Source: winterberg/03_divina_proportione_german/030_cap_XXX.txt

  • Marginal Woodcut Diagrams: 0 figure(s) extracted from the 1509 Venice witness.