UNIT V — GEOMETRIC & STEREOMETRIC SUMMARY¶
Exact Symbolic Invariants of Tractatus Geometriae (geometor.pacioli.tractatus_geometriae)
1. Heron’s (13, 14, 15) Archetype (Part II, fol. 9v)¶
Coordinates: $G = (0, 0)$, $B = (14, 0)$, $A = (5, 12)$, Foot $D = (5, 0)$, Incenter $I = (6, 4)$, Circumcenter $O = \left(7, \frac{33}{8}\right)$.
Heron Product: $s = 21$, $(s - c, s - b, s - a) = (6, 7, 8)$, $\Delta = \sqrt{21 \cdot 6 \cdot 7 \cdot 8} = \sqrt{7056} = 84$.
Radii & Altitude: Cathetus $AD = 12$, Inradius $r = \Delta / s = 4$, Circumradius $R = \frac{abc}{4\Delta} = \frac{65}{8}$.
2. Proportional Division of Triangles (Part II, Dist. IV)¶
Vertex Cevian (`VI.1`): Dividing $GB$ at $K = \left(\frac{14p}{p+q}, 0\right)$ splits area $84$ into $\frac{84p}{p+q}$ and $\frac{84q}{p+q}$ (for $1:2$, $K = \left(\frac{14}{3}, 0\right)$, areas $28$ and $56$).
Parallel Cut (`VI.13` + `VI.19`): Constructing geometric mean $GU = \sqrt{14 \cdot \frac{14p}{p+q}}$ on the semicircle over $GB$ cuts off $\triangle GUV \sim \triangle GBA$ of exact rational area $\frac{84p}{p+q}$ despite radical vertex coordinates $U = (7\sqrt{2}, 0)$, $V = \left(\frac{5\sqrt{2}}{2}, 6\sqrt{2}\right)$ for $p:q = 1:1$.
3. Regular & Archimedean Solids in Sphere of Diameter $D = 6$ (fols. 68v–70r)¶
Inscribed Solid |
Possanza ($e^2$) |
Exact Edge ($e$) |
Surface Area |
Exact Volume |
|---|---|---|---|---|
Tetrahedron (XIII.13) |
$24$ |
$2\sqrt{6}$ |
$24\sqrt{3}$ |
$8\sqrt{3}$ |
Cube (XIII.15) |
$12$ |
$2\sqrt{3}$ |
$72$ |
$24\sqrt{3}$ |
Octahedron (XIII.14) |
$18$ |
$3\sqrt{2}$ |
$36\sqrt{3}$ |
$36$ |
Icosahedron (XIII.16) |
$\frac{18(5 - \sqrt{5})}{5}$ |
$3\sqrt{\frac{10 - 2\sqrt{5}}{5}}$ |
$18\sqrt{3}(5 - \sqrt{5})$ |
$9\sqrt{10 + 2\sqrt{5}}$ |
Dodecahedron (XIII.17) |
$18 - 6\sqrt{5}$ |
$\sqrt{15} - \sqrt{3}$ |
$18\sqrt{10(5 - \sqrt{5})}$ |
$30\sqrt{3} + 6\sqrt{15}$ |
Truncated Tetrahedron (fol. 69v) |
$\frac{72}{11} = 6\frac{6}{11}$ |
$\frac{6\sqrt{22}}{11}$ |
$\frac{504\sqrt{3}}{11}$ |
$\frac{1656\sqrt{11}}{121}$ |
Cuboctahedron (fol. 69v) |
$9$ |
$3$ |
$54 + 18\sqrt{3}$ |
$45\sqrt{2}$ |