CHAPTER XXI — Critical Commentary & Notes

On Its Twelfth Almost Incomprehensible Effect

Historical and Philological Commentary

  1. Mathematical Verification of the Twelfth Effect (Campanus XIV.9):

    • Let a sphere have diameter $D$. By Euclid XIII.15, the edge of the inscribed cube satisfies $s_{\text{cube}}^2 = \frac{1}{3}D^2$.

    • By Euclid XIII.16.p and Euclid XIII.18, the edge of the inscribed icosahedron satisfies $s_{\text{icosa}}^2 = \frac{5 - \sqrt{5}}{10}D^2$.

    • Meanwhile, in Pacioli's formula with $a+b = 10$, $a = \sqrt{125}-5$, and $b = 15-\sqrt{125}$:

      • $(a+b)^2 + a^2 = 250 - \sqrt{12500} = 250 - 50\sqrt{5} = 50(5 - \sqrt{5})$.

      • By the Fifth Effect (Chapter XIV, Euclid XIII.4), $(a+b)^2 + b^2 = 3a^2 = 450 - \sqrt{112500} = 3 \times 50(3 - \sqrt{5})$!

      • Wait: notice that $\frac{\sqrt{(a+b)^2 + b^2}}{\sqrt{(a+b)^2 + a^2}} = \frac{\sqrt{3a^2}}{\sqrt{(a+b)^2 + a^2}} = \frac{\sqrt{3} \cdot a}{\sqrt{(a+b)^2 + a^2}}$. Since $(a+b)^2 + a^2$ is the square of the hypotenuse of a right triangle with legs $a+b$ (radius of the pentagon’s/icosahedron’s defining circle) and $a$ (the decagon side), by Euclid XIII.10 ($s_5^2 = s_6^2 + s_{10}^2$), $\sqrt{(a+b)^2 + a^2}$ is the side of the pentagon ($s_{\text{icosa}}$) inscribed in the circle of radius $a+b$, while $a$ is the decagon side!

    • In Campanus’s Latin text of Book XIV, Proposition 9 proves this exact relation between the cube edge and icosahedron edge via the segments of a line cut in extreme and mean ratio.

  2. Universal / Bound Roots (*Radici Universali / Radici Legate*): Pacioli introduces the Renaissance algebraic term :term:`radice universale <radice-universale>` or :term:`radice legata <radice-universale>` (written R.V. in the Summa, Distinctio VIII, Tractatus 3, f. 120r), where the radical sign governs an entire binomial or apotome ($\sqrt{250 - \sqrt{12500}}$).

Constantin Winterberg’s Commentary (1889 Edition)

Note 8 (Cap. XXI): Twelfth Effect — Ratio of the Cube Edge to the Icosahedron Edge in the Same Sphere (Sachliche Anmerkungen #8, Cap. XXI, p. 209, l. 13 from bottom)

Denoting by $s$ and $s'$ the edges of the cube and regular icosahedron inscribed in the same sphere, and letting the segment $ab$ be divided in extreme and mean ratio at $c$ ($ac > bc$), Pacioli's claim (following Campanus XIV.9) is:

\[\sqrt{\frac{ab^2 + ac^2}{ab^2 + bc^2}} = \frac{s'}{s} \quad \left(\text{or reciprocally } \sqrt{\frac{ab^2 + bc^2}{ab^2 + ac^2}} = \frac{s}{s'}\right).\]

Proof: Using the extreme and mean ratio relations (a) $ac^2 = ab \cdot bc$ and (b) $bc^2 = (ab - ac)^2 = ab^2 - 2\,ab \cdot ac + ac^2$, we have $ab^2 + bc^2 = 3\,ac^2 = 3\,ab \cdot bc$ (by Note 5) and $ab^2 + ac^2 = ab^2 + ab \cdot bc = ab(ab + bc)$. Hence the quotient under the radical simplifies to:

\[\frac{ab^2 + ac^2}{ab^2 + bc^2} = \frac{ab(ab + bc)}{3\,ab \cdot bc} = \frac{ab + bc}{3\,bc}.\]

Solving $ac^2 = ab \cdot bc = ab(ab - ac)$ gives $bc = \frac{3 - \sqrt{5}}{2}\,ab$, and substituting this into the quotient yields:

\[\sqrt{\frac{ab^2 + ac^2}{ab^2 + bc^2}} = \sqrt{\frac{ab + bc}{3\,bc}} = \sqrt{\frac{5 + \sqrt{5}}{6}} \quad \left(\text{and its reciprocal } \sqrt{\frac{3\,bc}{ab + bc}} = \sqrt{\frac{5 - \sqrt{5}}{6}}\right).\]

The exact same value is obtained when we express $s$ and $s'$ in terms of the circumradius $r$ of the common sphere (Euclid XIII.15, Euclid XIII.16, Euclid XIII.18):

\[s = \frac{2r}{\sqrt{3}}, \qquad s' = \frac{4r}{\sqrt{10 + 2\sqrt{5}}} = r\sqrt{\frac{10 - 2\sqrt{5}}{5}}, \qquad \frac{s}{s'} = \sqrt{\frac{5 + \sqrt{5}}{6}}. \quad \text{Q.E.D.}\]

Anmerkung 8 (Cap. XXI, S. 209): Zwölfter Effect — Verhältniss der Seite des Kubus zur Seite des Ikosaeders

Bezeichnet man mit $s, s'$ resp. die Seiten des Kubus und des regulären Ikosaeders, so wird, wenn die Linie $ab$ in $c$ stetig getheilt ist ($ac > bc$), behauptet:

\[\sqrt{\frac{ab^2 + ac^2}{ab^2 + bc^2}} = \frac{s}{s'}.\]

Beweis: Der Ausdruck unter dem Wurzelzeichen links lässt sich vereinfachen, indem zufolge Voraussetzung: a) $ac^2 = ab \cdot bc$, b) $bc^2 = (ab - ac)^2$. Durch Einsetzen dieser Ausdrücke und Ausführung der Quadratur des letzteren erhält man:

\[\frac{ab^2 + ac^2}{ab^2 + bc^2} = \frac{ab(ab + bc)}{3\,ab \cdot bc} = \frac{ab + bc}{3\,bc}.\]

Aus der Gleichung a), nachdem darin $ac = ab - bc$ gesetzt, erhält man $bc^2 - 3\,ab \cdot bc + ab^2 = 0$, und durch Einsetzen in den vorstehenden Quotienten:

\[\sqrt{\frac{ab^2 + ac^2}{ab^2 + bc^2}} = \sqrt{\frac{ab + bc}{3\,bc}} = \sqrt{\frac{5 + \sqrt{5}}{6}}.\]

Derselbe Werth ergibt sich aber auch, wenn man das Verhältniss $\frac{s}{s'}$ durch den Radius $r$ der umschriebenen Kugel ausgedrückt berechnet, indem bekanntlich ist:

\[s = \frac{2r}{\sqrt{3}}, \qquad s' = \frac{4r}{\sqrt{10 + 2\sqrt{5}}}, \qquad \text{somit} \quad \frac{s}{s'} = \sqrt{\frac{5 + \sqrt{5}}{6}}.\]

Philological and Folio Apparatus

  • 1509 Paganini Witness Folios: 0029, 0030 (f. 7r–f. 7v)

  • Winterberg 1889 Italian Source: winterberg/01_divina_proportione_italian/021_cap_XXI.txt

  • Winterberg 1889 German Source: winterberg/03_divina_proportione_german/021_cap_XXI.txt

  • Marginal Woodcut Diagrams: 1 figure(s) extracted from the 1509 Venice witness.