UNIT V — HISTORICAL, PHILOLOGICAL & DEDUCTIVE NOTES

Commentary on (13, 14, 15), Pre-Malfatti Circle Packing, Truncated Polyhedra & the Two Colophons

1. Why Every Renaissance Surveyor Learned (13, 14, 15)

From Heron's Metrica (I.8) through Al-Khwārizmī, Leonardo Pisano's Practica Geometriae (1220), and Part II, fol. 9v of Pacioli’s Summa, the scalene triangle $(13, 14, 15)$ served for fifteen centuries as the canonical benchmark of Heron’s formula because it is the smallest non-right Heronian triangle formed by gluing two primitive/scaled Pythagorean triangles—$(5, 12, 13)$ and $(9, 12, 15)$—along their shared cathetus $AD = 12$. Every single intermediate quantity in Pacioli’s computation ($s = 21$, excesses $6, 7, 8$, area $84$, altitude $12$, segments $5$ and $9$, inradius $r = 4$, and circumradius $R = 65/8$) is rational, allowing the surveyor to verify Heron’s fourth-degree radical $\sqrt{s(s-a)(s-b)(s-c)}$ directly against Euclid [II.13]!

2. Pacioli’s 1494 Three-Circle Problem (fol. 57r) vs. Malfatti (1803)

On Part II, fol. 57r (scan n0580), Pacioli asks for the three largest equal mutually tangent circles inside the isosceles $(10, 10, 12)$ triangle (altitude $8$, area $48$, inradius $3$). Setting the unknown radius (or scaled segment) as 1.co. and applying the Pythagorean theorem to the triangle of circle centers reduces the configuration to the Canon 4 quadratic $5\frac{3}{4}\text{.ce.} + 16\text{.co.} = 48$ ($23r^2 + 64r = 192 \implies r = \frac{8\sqrt{85} - 32}{23} \approx 1.8155$). Historically, this problem in Piero della Francesca’s and Pacioli’s abbaco corpus is the direct Renaissance ancestor of Jakob Bernoulli’s (1744) and Gian Francesco Malfatti’s (1803) famous three-circle marble-cutting problem.

3. Derivation of $\sqrt{6\frac{6}{11}}$ on fol. 69v & the Two November 1494 Dates

Why does the truncated tetrahedron inscribed in a sphere of diameter $D = 6$ ($R = 3$) have exact edge $\sqrt{6\frac{6}{11}}$ on fol. 69v? Trisecting each edge of a regular tetrahedron of side $3e$ removes four corner tetrahedra of side $e$, leaving 4 equilateral triangles and 4 regular hexagons of common edge $e$. In Cartesian coordinates where the parent tetrahedron has vertices $(\pm \frac{3\sqrt{2}}{4}e, \pm \frac{3\sqrt{2}}{4}e, \dots)$, the 12 vertices of the truncated tetrahedron are all permutations with even sign changes of $\left(\frac{3\sqrt{2}}{4}e, \frac{\sqrt{2}}{4}e, \frac{\sqrt{2}}{4}e\right)$, whose squared circumradius from the origin is:

$$R^2 = \left(\frac{9}{8} + \frac{1}{8} + \frac{1}{8}\right) e^2 = \frac{11}{8} e^2 \implies e^2 = \frac{8}{11} R^2 = \frac{2}{11} D^2$$

Setting $D = 6$ gives $e^2 = \frac{72}{11} = 6\frac{6}{11}$, matching Pacioli’s printed value to the letter!

Finally, note the fascinating bibliographical chronology preserved across our inspected scans: the Venetian colophon at the end of Part II (fol. 76r, scan n0618) is dated November 10, 1494, whereas the typographical register on the preliminary verso ([prelim. fol. 1v], scan n0005) is dated November 20, 1494—showing that Paganino de' Paganini completed printing the geometric body of the book on November 10 and struck the preliminary dedication, epigrams, and quire register ten days later on November 20, 1494!