UNIT V — TRACTATUS GEOMETRIAE: HERON, SURFACE DIVISION & REGULAR SOLIDS

Part II, Distinctiones I–VIII: Plane Mensuration, Archimedes’s Circle, De Divisione Superficierum, Inscribed Polyhedra & the 1494 Colophon (fols. 1r–76r)

[Folio Part II, fols. 1r–9v]

Euclid Digest & Heron’s (13, 14, 15) Triangle Mensuration (Part II, fol. 9v)

Opening the separately foliated second tome (Tractatus Geometriae, fols. 1r–2r), Fra Luca Pacioli condenses the foundational definitions and constructions of Euclid Books I, II, and VI before turning in Distinctio I, Cap. 8 (fol. 9v) to the crown jewel of practical surveyor’s mensuration inherited from Heron of Alexandria and Leonardo Pisano's Practica Geometriae: determining the exact area, perpendicular cathetus (catetto), and incircle of a scalene triangle from its three sides alone.

G D T B A I AG = 13 AB = 15 AD = 12 r = 4 s = 21 | ℞(6 · 7 · 8 · 21) = ℞.7056 = 84 braccia²
[Heron's Formula & Euclid II.13 / IV.4]

Let scalene triangle GBA have sides GA $= 13$, base GB $= 14$, and BA $= 15$ braccia ([I.22]). Sum the three sides ($13 + 14 + 15 = 42$) and take the semiperimeter (mità dela summa) $s = 21$. Subtract each side in turn to obtain the three excesses (differentie):

$$21 - 15 = 6, \qquad 21 - 14 = 7, \qquad 21 - 13 = 8$$

Multiply $6 \times 7 \times 8 = 336$, and multiply $336$ by the semiperimeter $21$ to obtain $7056$, whose exact square root gives the area:

$$\text{Area}(\triangle GBA) = \sqrt{7056} = 84 \text{ square braccia}$$

Dividing $2 \times 84 = 168$ by base GB ($14$) yields the perpendicular cathetus AD $= 12$, splitting $GB$ at D into right-triangle base segments GD $= 5$ (inside $(5, 12, 13)$ right triangle GDA) and DB $= 9$ (inside $(9, 12, 15)$). Dividing area $84$ by semiperimeter $21$ locates the incircle (IT) centered at I $= (6, 4)$ of exact integer inradius IT $= 4$ ([IV.4]).

[Folio Part II, fol. 31r & fols. 34r–43v]

Archimedes’s Circle (fol. 31r) & De Divisione Superficierum (Dist. IV)

In Distinctio III, Cap. 6 (fol. 31r), Pacioli expounds Archimedes's Dimensio Circuli, bounding the circumference-to-diameter ratio by inscribed and circumscribed $96$-gons between $3\frac{10}{71}$ and $3\frac{1}{7}$ ($22 : 7$). In Distinctio IV (De divisione superficierum), he solves estate-partition problems dividing triangles, quadrilaterals, and circles in prescribed ratios $p : q$ both through vertices ([VI.1]) and parallel to a side ([VI.13], [VI.19]).

G M U B A V S GU = GS = ℞(14·7) = ℞.98 ⇒ Area(GUV) = Area(UBAV) = 42
[Euclid VI.13 & VI.19]To bisect the area ($84$) of GBA by a cut UV parallel to BA: erect perpendicular MS at midpoint M $(7, 0)$ of semicircle (MB) to meet the arc at S, and swing arc (GS) onto GB at U, so GU is equal in chord length to GS $= \sqrt{14 \cdot 7} = \sqrt{98} = 7\sqrt{2}$. Drawing UV $\parallel$ BA cuts off similar sub-triangle GUV of exact area $\frac{GU^2}{GB^2} \cdot 84 = \frac{98}{196} \cdot 84 = 42$.

In Distinctio V–VI, Pacioli gauges wine casks (botte) and solves algebraic-geometric circle packings via Regola dela Cosa—culminating on fol. 57r in inscribing three mutually tangent equal circles inside the $(10, 10, 12)$ triangle via the Cossic quadratic $5\frac{3}{4}.\text{ce.} \; \bar{p} \; 16.\text{co.} = 48$ ($r = \frac{8\sqrt{85} - 32}{23}$), three centuries before Malfatti (1803).

[Folio Part II, fols. 68v–70r & 76r]

De Corporibus Regularibus (Dist. VIII) & the November 10, 1494 Venetian Colophon

O P Q R S T U W X Y Z Spera D = 6 (R = 3) | Cuboctaedro lato = 3 | Rettangolo aureo WXYZ

Closing the entire work in Distinctio VIII (fols. 68v–70v), Pacioli inscribes the five regular Platonic solids ([XIII.13]–[XIII.17]) and dependent Archimedean truncations (following Piero della Francesca's Libellus de quinque corporibus regularibus) inside a sphere (OP) of diameter $D = 6$ (radius $R = 3$):

  • Tetrahedron (`fol. 69r`): Possanza $e^2 = \frac{2}{3} D^2 = 24$ ($e = \sqrt{24}$); Cube: $e^2 = \frac{1}{3} D^2 = 12$ ($e = \sqrt{12}$); Octahedron (`fol. 68v`): $e^2 = \frac{1}{2} D^2 = 18$ ($e = \sqrt{18}$).

  • Truncated Tetrahedron (`fol. 69v, prob. 15–16`, 4 triangles + 4 hexagons): Exact squared edge $e^2 = \frac{2}{11} D^2 = \frac{72}{11} = 6\frac{6}{11}$, matching Pacioli’s printed radical $\mathcal{R}.6\frac{6}{11}$!

  • Cuboctahedron (`fol. 69v–70r, prob. 17–19`, 6 squares + 8 triangles): Its equatorial great-circle section PQRSTU across diameters PS, QT, and RU is a regular hexagon whose side PQ is equal to the sphere radius $R = 3$!

  • Icosahedron & Dodecahedron (`fol. 70r, prob. 21–22`): Spanned internally by golden rectangle WXYZ (WX : XY $= \phi : 1$). In Problem 21 (fol. 70r), for a sphere whose axis has possanza $48$ ($D = \sqrt{48}$), the inscribed cube has edge $\sqrt{48/3} = 4$, and dividing $4$ in extreme and mean ratio yields the exact dodecahedron edge $\sqrt{20} - 2$ and minor segment $6 - \sqrt{20}$, right beside Problem 20’s hydrostatic method of Archimedes measuring irregular sculptural bodies by displaced water.

On Part II, fol. 76r, Paganino de' Paganini sets his colophon in Venice on November 10, 1494, under the ten-year privilege of Doge Agostino Barbarigo (Con spesa e diligentia: e opifitio del prudente homo Paganino de Paganini da Brescia. Nella excelsa cita de vinegia… Negli anni de nostra Salute M.cccc.lxliiii. adi .10. de nouembre)—launching the mathematical Renaissance in print.