Unit I · Executive Geometric Summary

Core Theses of Mysterium Cosmographicum (1596 / 1621)

  1. Dimensional Shift from Infinite Plane Polygons to Five Convex Solids: While teaching the zodiacal progression of Saturn–Jupiter great conjunctions at Graz on July 19, 1595, Kepler observed that the incircle and circumcircle of an equilateral triangle stand in ratio $1 : 2$ ($Q_{\text{in}}/Q_{\text{out}} = 1/4$). Realizing that regular 2D polygons form an infinite family whereas three-dimensional Euclidean space admits strictly five regular convex polyhedra ([XIII.18]), he deduced an a priori geometric explanation for why Copernicus's heliocentric system possesses exactly six mobile planets separated by five interplanetary intervals.

  2. The Two Great Treasures of Geometry: In Chapter XII, Kepler formulates his enduring aphorism pairing the Pythagorean Theorem (measure of gold, governing the Cube, Tetrahedron, and Octahedron over $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{3})$) with the Extreme and Mean Ratio (sectio divina $\phi = \frac{1+\sqrt{5}}{2}$, precious jewel, governing the Dodecahedron and Icosahedron over $\mathbb{Q}(\phi)$).

  3. Polar-Duality Insphere/Circumsphere Invariance: Dual regular solids exchange vertices ($V$) and faces ($F$) while preserving edge count ($E$) and exact insphere-to-circumsphere quadrance ratios:

Exact Platonic Sphere Ratios (geometor.kepler.mysterium_spheres)

Solid

Schläfli

Planetary Pair

Exact Quadrance Ratio $Q_{\text{in}} / Q_{\text{out}}$

Exact Radius Ratio $r_{\text{in}} / R_{\text{out}}$

Cube

$\{4, 3\}$

Saturn $\to$ Jupiter

$\frac{1}{3}$

$\frac{\sqrt{3}}{3} \approx 0.57735$

Tetrahedron

$\{3, 3\}$

Jupiter $\to$ Mars

$\frac{1}{9}$

$\frac{1}{3} \approx 0.33333$

Dodecahedron

$\{5, 3\}$

Mars $\to$ Earth

$\frac{5 + 2\sqrt{5}}{15} = \frac{4\phi + 3}{15}$

$\sqrt{\frac{5 + 2\sqrt{5}}{15}} \approx 0.79465$

Icosahedron

$\{3, 5\}$

Earth $\to$ Venus

$\frac{5 + 2\sqrt{5}}{15} = \frac{4\phi + 3}{15}$

$\sqrt{\frac{5 + 2\sqrt{5}}{15}} \approx 0.79465$

Octahedron

$\{3, 4\}$

Venus $\to$ Mercury

$\frac{1}{3}$

$\frac{\sqrt{3}}{3} \approx 0.57735$

  1. Earth as Measuring Boundary (*Circulus mensor omnium*): Placing Earth ($R_\oplus = 1$) between the three outer trihedral primary solids (Dodecahedron, Tetrahedron, Cube) and the two inner secondary solids (Icosahedron, Octahedron) reproduces the qualitative spacing of the six classical planets—especially the vast Jovian gap ($1/3$ tetrahedral ratio between Mars and Jupiter) and tight terrestrial cluster ($0.795$ golden ratios around Earth and Venus).