UNIT IV — DISTINCTIO V–VI: CUBE ROOTS, SOLIDS & THE FIVE REGULAR POLYHEDRA

Cube-Root Extraction, Stereometry of Pyramids & Spheres, and the Five Platonic Solids in Sphere Diameter Six

Source Text: fibonacci:pg-5, fibonacci:pg-6 (Boncompagni Vol. II, pp. 148–202 · fol. 92v–130v) · Latin (1220 / Boncompagni 1862) · Summary · GEOMETOR Notes

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[Folio fol. 92v–99r | p. 148–159]

Distinctio V — On the Finding of Cube Roots (De Radicibus Cubicis)

Latin Locus: Distinctio V (pp. 148–159) · Commentary: Notes on Cubic Binomial Extraction

Here begins the Fifth Distinction, on finding cube roots (de radicibus cubicis inveniendis), needful for the duplication and comparison of corporeal solids.

Just as square-root extraction rests upon the gnomon $(a + b)^2 = a^2 + 2ab + b^2$, cube-root extraction rests upon the solid dissection of a cube cut into two parts $a$ and $b$:

$$(a + b)^3 = a^3 + 3a^2 b + 3a b^2 + b^3 = a^3 + 3ab(a + b) + b^3.$$

Grouping Hindu-Arabic digits in triads ($10^3 = 1{,}000$) from the right, Leonardo subtracts the greatest cube $a^3$ from the leading triad, forms the trial divisor $300 a^2$, determines the next digit $b$ so that $300 a^2 b + 30 a b^2 + b^3$ does not exceed the remainder, and for non-cubic integers $N = a^3 + R$ supplies the rational fractional bound:

$$\sqrt[3]{a^3 + R} \approx a + \frac{R}{3a(a + 1) + 1},$$

followed by multiplication and division of cubic radicals ($\sqrt[3]{p}\cdot\sqrt[3]{q} = \sqrt[3]{pq}$).

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[Folio fol. 99v–121v | p. 159–190]

Distinctio VI (Part 1) — Columns, Pyramids, Frustums & the Sphere

Latin Locus: De Dimensione Corporum (pp. 159–190) · Commentary: Notes on Stereometry & Archimedes’ Sphere

A solid pertica (pertica corporalis) contains $6 \times 6 \times 6 = 216$ cubic pedes. Proceeding from Euclid Books XI–XII (euclid:elements) and Archimedes’ On the Sphere and Cylinder (archimedes:sphere), Leonardo demonstrates the fundamental stereometric laws:

1. Prisms & Circular Columns (Cylinders): Solid measure equals the planar embadum of the base multiplied into the perpendicular height (cathetus). [XI. 31, XII. 11]

2. Pyramids & Cones: Every pyramid (triangular, quadrangular, or multi-angular) and every circular cone is one-third of the prism or cylinder sharing the same base and altitude: $V = \frac{1}{3} B h$. [XII. 7, 10]

3. Truncated Pyramids & Conical Frustums (Pyramis caput abscisa): If a pyramid or cone is cut parallel to its base, leaving lower base area $B_1$, upper top area $B_2$, and perpendicular height $h$, its exact volume is one-third the height multiplied into the sum of the two bases and their geometric mean:

   $$V = \frac{h}{3}\!\left(B_1 + \sqrt{B_1 B_2} + B_2\right).$$

4. The Sphere: Following Archimedes (archimedes:sphere) and the Banū Mūsā (banu-musa:verba), every sphere is two-thirds of its circumscribed cylinder (whose base diameter and height both equal the sphere’s diameter $d$), and its surface area equals four times its greatest circle ($4 \pi r^2 = \pi d^2$). Hence $V_{\text{sphere}} = \frac{1}{6}\pi d^3 \approx \frac{11}{21} d^3$.

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[Folio fol. 122r–130v | p. 190–202]

Distinctio VI (Part 2) — The Five Regular Polyhedra in Sphere Diameter Six

Latin Locus: De Quinque Corporibus Regularibus in Spera (pp. 190–202) · Commentary: Notes on Euclid XIII, Hypsicles XIV & Pacioli

To find the sides, surfaces, altitudes, and solid volumes of the five regular bodies of Euclid’s thirteenth book—namely the tetrahedron (4 triangular bases), cube (6 square bases), octahedron (8 triangular bases), dodecahedron (12 pentagonal bases), and icosahedron (20 triangular bases)—inscribed within a sphere of given rational diameter $d = 6$. [XIII. 13–18]

Euclid XIII.18 master semicircle ( O A ) on sphere diameter AI = 6 (center O, trisection AK = 2): tetrahedron edge / B I / (Q = 24), octahedron edge / A C / (Q = 18), cube edge / A B / (Q = 12), golden cut N giving dodecahedron edge / A N / (Q = 18 - sqrt(180)), and / M I / giving icosahedron edge (Q = 18 - sqrt(64 4/5)).

Let the sphere diameter $AI = 6$ lie along AK with center O ($AO = OI = 3$) and great circle (OA). Cut $AI$ at K so that $AK = \frac{1}{3} AI = 2$ and $KI = 4$; erect perpendiculars KB and OC to meet the semicircle at B and C. [XIII. 18]

1. Regular Tetrahedron (Pyramis quatuor basium triangularium, [XIII. 13]): Since $BI^2 = AI \cdot KI = \frac{2}{3} d^2$, the tetrahedron edge is BI with $e_4^2 = 6 \times 4 = 24$ ($e_4 = \sqrt{24} = 2\sqrt{6}$). Its altitude is $h_4 = 4$, base area $\sqrt{108} = 6\sqrt{3}$, and solid volume $V_4 = \frac{1}{3} \times 4 \times 6\sqrt{3} = 8\sqrt{3} = \sqrt{192}$.

2. Cube (Cubus sex basium quadratarum, [XIII. 15]): Since $AB^2 = AI \cdot AK = \frac{1}{3} d^2$, the cube edge is AB with $e_6^2 = 6 \times 2 = 12$ ($e_6 = \sqrt{12} = 2\sqrt{3}$), and its solid volume is $V_6 = 12\sqrt{12} = 24\sqrt{3} = \sqrt{1{,}728} = 3\,V_4$.

3. Regular Octahedron (Octaedrum octo basium triangularium, [XIII. 14]): Since $AC^2 = AO^2 + OC^2 = \frac{1}{2} d^2$, the octahedron edge is AC with $e_8^2 = 9 + 9 = 18$ ($e_8 = \sqrt{18} = 3\sqrt{2}$). Composed of two square pyramids of base $e_8^2 = 18$ and combined height $d = 6$, its volume is the exact rational integer $V_8 = \frac{1}{3} \times 18 \times 6 = 36$.

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[Folio fol. 125v–130v | p. 196–202]

Extreme and Mean Ratio: The Dodecahedron & Icosahedron

Latin Locus: Modus Dividendi Lineam Media et Extrema Proportione (pp. 196–202) · Commentary: Notes on the Golden Chord & Hypsicles Book XIV

To construct the dodecahedron, each pentagonal face is crowned upon an inscribed cube whose edge ($e_6 = \sqrt{12}$) serves as the pentagonal chord (diagonal) BE. Since in every regular pentagon the diagonal and side stand in extreme and mean ratio [XIII. 8, 17], Leonardo inserts an explicit constructive chapter (Modus dividendi lineam media et extrema proportione, p. 196):

Cube edge BE = 2*sqrt(3) divided in extreme and mean ratio: erecting / E C / = 1/2 BE = sqrt(3) via midpoint M and circle ( E M ), cutting hypotenuse [ B D ] (BC = sqrt(15)) with circle ( C E ) at D, and transferring ( B D ) to G yields dodecahedron edge / B G / = / A B / = / A E / = sqrt(15) - sqrt(3).

• Golden-Section Construction: Let the given chord $BE = \sqrt{12} = 2\sqrt{3}$ be bisected at M, and by circle (EM) erect at E the perpendicular EC $= EM = \sqrt{3}$. [II. 11, VI. 30] Join the hypotenuse BC, whose square is $BE^2 + EC^2 = 12 + 3 = 15$, so $BC = \sqrt{15}$. With center C and radius $CE = \sqrt{3}$, describe circle (CE) cutting BC at D, leaving $BD = \sqrt{15} - \sqrt{3}$. With center B and radius $BD$, describe arc (BD) cutting $BE$ at G.

• Dodecahedron Edge (*Apotome*): Both BG (and AN on AB in the sphere diagram) and the pentagonal sides AB and AE of ABE are equal to the greater golden segment, an irrational recisum (apotome):

  $$e_{12} = \sqrt{15} - \sqrt{3}, \qquad e_{12}^2 = 18 - 6\sqrt{5} = 18 - \sqrt{180}.$$

  Leonardo computes the pentagon circumradius squared $af^2 = 6 - \sqrt{7\frac{1}{5}}$ and perpendicular in-sphere squared from the sphere center to each pentagonal face $hf^2 = 9 - af^2 = 3 + \sqrt{7\frac{1}{5}}$, summing the $12$ pentagonal pyramids into $V_{12} = 30\sqrt{3} + 6\sqrt{15} = \sqrt{2{,}700} + \sqrt{540}$.

• Regular Icosahedron & Hypsicles’ Book XIV Ratio: Erecting $IL = AO = 3$ tangent at $I$ so ray $OL$ cuts the lower semicircle at M along OM [XIII. 16], chord MI is the icosahedron edge (a minor irrational line):

  $$e_{20}^2 = 18 - \frac{18}{5}\sqrt{5} = 18 - \sqrt{64\frac{4}{5}}, \qquad V_{20} = 18\sqrt{10 + 2\sqrt{5}}.$$

  Invoking Hypsicles’ Book XIV theorem that the dodecahedron and icosahedron inscribed in the same sphere stand to one another as the cube edge $e_6$ to the icosahedron edge $e_{20}$:

  $$\frac{V_{12}}{V_{20}} = \frac{e_6}{e_{20}} = \sqrt{\frac{12}{18 - \frac{18}{5}\sqrt{5}}} = \sqrt{\frac{5 + \sqrt{5}}{6}}.$$

Here ends the Sixth Distinction on the dimension of solid bodies.