CHAPTER XXVII — Critical Commentary & Notes

How the Second Regular Body, Called the Exacedron or Cube, Is Formed and Placed in the Sphere.

Historical & Mathematical Commentary

• Campanus vs. Heiberg Ordering: In Campanus’s Latin Elements, the Cube is Proposition 14 of Book XIII and the Octahedron is Proposition 15 (whereas in Heiberg’s Greek text the Octahedron is Euclid XIII.14 and the Cube is Euclid XIII.15). Pacioli follows Campanus’s order here and explicitly explains the reason for this transposition at the end of Cap. XXXI.

• Relation to the Tetrahedron: Using the same semicircle \(adb\) cut at \(c\) with \(ac = 2\,cb\), the longer chord \(da\) is the edge of the inscribed tetrahedron (\(da^{2} = \frac{2}{3}ab^{2}\)), the shorter chord \(db\) is the edge of the inscribed cube (\(db^{2} = \frac{1}{3}ab^{2}\)), and therefore the edge of the tetrahedron is the face diagonal of the cube (\(da^{2} = 2\,db^{2}\)).

Constantin Winterberg’s Commentary (1889 Edition)

Note 11 (Cap. XXVII): Division of the Sphere Diameter by the Inscribed Cube (Euclid XIII.15) (Sachliche Anmerkungen #11, Cap. XXVII, p. 217, l. 17 from top)

The square of the diagonal of a square is double the square of its side ($ad^2 = 2\,bd^2$). Passing a plane through one of the three main space diagonals $ab$ of the cube and an edge $bd$, this plane intersects the circumscribed sphere in a great circle whose diameter is $ab$, while the two chords drawn from its endpoints to $d$ are the cube edge $bd$ and the face diagonal $ad$. By the right-triangle projection theorem (Euclid VI.8):

\[bd^2 = bc \cdot ab \quad \text{and} \quad ad^2 = ac \cdot ab.\]

Since $ad^2 = 2\,bd^2$, dividing the two equations gives:

\[\frac{bc}{ac} = \frac{bd^2}{ad^2} = \frac{1}{2}, \qquad \text{and } ab^2 = 3\,bd^2. \quad \text{Q.E.D.}\]

Anmerkung 11 (Cap. XXVII, S. 217): Theilung des Kugeldurchmessers beim Kubus (Euklid XIII.15)

Bekanntlich ist das Quadrat der Diagonale eines Quadrats doppelt so gross als das Quadrat der Seite. Denkt man sich nun durch eine der drei Hauptaxen $ab$ des Kubus und eine Kante $bd$ eine Ebene gelegt, so schneidet sie die umschriebene Kugel in einem Kreise, dessen Durchmesser jene Axe $ab$ ist, während die zwei von ihren Endpunkten ausgehenden Sehnen die Seite $bd$ und Flächendiagonale $ad$ des Quadrats sind. Daher hat man nach bekannten Sätzen:

\[bd^2 = bc \cdot ab \quad \text{und} \quad ad^2 = ac \cdot ab.\]

Somit, da $ad^2 = 2\,bd^2$ ist, erhält man durch Division vorstehender Gleichungen:

\[\frac{bc}{ac} = \frac{1}{2}. \quad \text{w. z. b. w.}\]

Philological and Folio Apparatus

• 1509 Paganini Witness Folios: 0032, 0033 (f. 8v–f. 9r)

• Winterberg 1889 Italian Source: winterberg/01_divina_proportione_italian/027_cap_XXVII.txt

• Winterberg 1889 German Source: winterberg/03_divina_proportione_german/027_cap_XXVII.txt

• Marginal Woodcut Diagrams: 3 figure(s) extracted from the 1509 Venice witness.