Demonstrations

100% Exact Symbolic SymPy Models, Möbius Inversion & Rational Trigonometry Proofs

In strict accordance with GEOMETOR’s Content / Code Separation, the historical Book Codex (De Docta Ignorantia et Scripta Mathematica — Coincidence of Opposites & Exact Geometry) embeds zero Python scripts. All executable geometric constructions live in the geometor.cusa package and the standalone demos/*.py scripts documented below, verified over \(\mathbb{Q}\), \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{6}, \sqrt{7}, \sqrt{15}, \sqrt{21})\), and \(\mathbb{Q}(n)\) using Model(..., auto_intersect=False) with zero floating-point approximations.


1. De Docta Ignorantia I.13 — Tangent Circle Pencil & Conformal Möbius Inversion

Codex Reference: Unit III — Coincidence of Figures (Critical Notes) · Script: demos/coincidentia_circle_line.py

De Docta Ignorantia I.13 — Coaxial Tangent Circle Pencil & Vanishing Sagitta ( C1 O ): (x - (0))^2 + (y - (1))^2 - (1) = 0 ( C2 O ): (x - (0))^2 + (y - (2))^2 - (4) = 0 ( C4 O ): (x - (0))^2 + (y - (4))^2 - (16) = 0 ( C8 O ): (x - (0))^2 + (y - (8))^2 - (64) = 0 [ O C8 ]: (1)*x + (0)*y + (0) = 0 [ X P1 ]: (1)*x + (0)*y + (-1) = 0 [ W Q1 ]: (1)*x + (0)*y + (1) = 0 [ A B ]: (0)*x + (1)*y + (0) = 0 [ Q1 P1 ]: (0)*x + (1)*y + (-1) = 0 [ Q2 P2 ]: (0)*x + (1)*y + (-2 + sqrt(3)) = 0 [ Q4 P4 ]: (0)*x + (1)*y + (-4 + sqrt(15)) = 0 [ Q8 P8 ]: (0)*x + (1)*y + (-8 + 3*sqrt(7)) = 0 A = (-2, 0) B = (2, 0) W = (-1, 0) X = (1, 0) C1 = (0, 1) C2 = (0, 2) O = (0, 0) P1 = (1, 1) Q1 = (-1, 1) P2 = (1, 2 - sqrt(3)) Q2 = (-1, 2 - sqrt(3)) P4 = (1, 4 - sqrt(15)) Q4 = (-1, 4 - sqrt(15)) P8 = (1, 8 - 3*sqrt(7)) Q8 = (-1, 8 - 3*sqrt(7)) A B C1 C2 O P1 Q1 P2 Q2 P4 Q4 P8 Q8
Tangent circle pencil C_1, C_2, C_4, C_8 above straight horizon AB (y = 0) with exact radical sagittae 1, 2 - sqrt(3), 4 - sqrt(15), 8 - 3*sqrt(7) along vertical secant x = 1.
De Docta Ignorantia I.13 — Mobius Inversion Unifying Tangent Circles & Straight Line ( S O ): (x - (0))^2 + (y - (-1))^2 - (1) = 0 ( C1 O ): (x - (0))^2 + (y - (1))^2 - (1) = 0 ( M1 O ): (x - (0))^2 + (y - (-1/3))^2 - (1/9) = 0 ( M2 O ): (x - (0))^2 + (y - (-2/5))^2 - (4/25) = 0 ( M4 O ): (x - (0))^2 + (y - (-4/9))^2 - (16/81) = 0 ( Minf O ): (x - (0))^2 + (y - (-1/2))^2 - (1/4) = 0 [ S Z ]: (1)*x + (0)*y + (0) = 0 [ A B ]: (0)*x + (1)*y + (0) = 0 [ S X ]: (-1)*x + (1)*y + (1) = 0 A = (-3/2, 0) B = (3/2, 0) Z = (0, 3/2) C1 = (0, 1) N1 = (0, -2/3) N2 = (0, -4/5) N4 = (0, -8/9) X = (1, 0) O = (0, 0) M1 = (0, -1/3) M2 = (0, -2/5) M4 = (0, -4/9) Minf = (0, -1/2) Xp = (1/2, -1/2) S = (0, -1) A B Z C1 N1 N2 N4 O M1 M2 M4 M∞ S
Bounded conformal inversion in unit circle Omega at pole S = (0, -1), mapping C_R to finite rational circles C'_R centered at (0, -R/(2R+1)) converging smoothly to C'_inf = inverted line AB.
  • Exact Symbolic Verification: Rationalizing the sagitta \(h(x, R) = R - \sqrt{R^2 - x^2} = \frac{x^2}{R + \sqrt{R^2 - x^2}}\) proves \(\lim_{R \to \infty} h(x, R) = 0\), \(\lim_{R \to \infty} R\,h(x, R) = \frac{x^2}{2}\), and curvature \(\kappa(R) = \frac{1}{R} \to 0\). Under unit-circle inversion at S \(= (0, -1)\), every circle \(C_R\) maps to the exact rational circle centered at \(\left(0, -\frac{R}{2R+1}\right)\) of radius \(\frac{R}{2R+1}\), whose \(R \to \infty\) limit at Minf \(= \left(0, -\frac{1}{2}\right)\) is identically the inverted straight line AB.

import sympy as sp
        from geometor.cusa import (
            construct_tangent_circle_pencil,
            verify_mobius_inversion_coincidence,
            verify_sagitta_and_curvature_limits,
        )

        model = construct_tangent_circle_pencil()
        assert not any(pt.has(sp.Float) for pt in model.points)

        limits = verify_sagitta_and_curvature_limits()
        assert limits["identity_holds"] and limits["sagitta_limit"] == 0

        mobius = verify_mobius_inversion_coincidence()
        assert mobius["limit_center_y"] == sp.Rational(-1, 2)
        assert mobius["limit_radius"] == sp.Rational(1, 2)
        

2. De Docta Ignorantia I.14–15 — Flattening Triangles in Rational Trigonometry

Codex Reference: Unit III — Coincidence of Figures (Critical Notes) · Script: demos/coincidentia_triangle_line.py

De Docta Ignorantia I.14–15 — Flattening Triangles & Vanishing Rational Spread [ M A1 ]: (1)*x + (0)*y + (0) = 0 [ B C ]: (0)*x + (1)*y + (0) = 0 [ B A1 ]: (-1)*x + (1)*y + (-1) = 0 [ A1 C ]: (1)*x + (1)*y + (-1) = 0 [ B A2 ]: (-1/2)*x + (1)*y + (-1/2) = 0 [ A2 C ]: (1/2)*x + (1)*y + (-1/2) = 0 [ B A4 ]: (-1/4)*x + (1)*y + (-1/4) = 0 [ A4 C ]: (1/4)*x + (1)*y + (-1/4) = 0 [ B A8 ]: (-1/8)*x + (1)*y + (-1/8) = 0 [ A8 C ]: (1/8)*x + (1)*y + (-1/8) = 0 B = (-1, 0) C = (1, 0) M = (0, 0) A1 = (0, 1) A2 = (0, 1/2) A4 = (0, 1/4) A8 = (0, 1/8) B C M A1 A2 A4 A8
Nested isosceles triangles B A_n C over fixed base BC = 2 with altitudes 1/n: pure rational apex spreads s_A = 4n^2/(n^2+1)^2 -> 0 and Triple Quad collinearity residual 16/n^2 -> 0.
  • Exact Rational Trigonometry Proof: For B \(= (-1, 0)\), C \(= (1, 0)\), and apex \(A_n = (0, 1/n)\) (A1, A2, A4, A8 descending toward M), Wildberger's invariants belong purely to \(\mathbb{Q}(n)\): \(Q(A_n, B) = 1 + \frac{1}{n^2}\), \(s_B = s_C = \frac{1}{n^2+1} \to 0\), \(s_{A_n} = \frac{4n^2}{(n^2+1)^2} \to 0\), the Triple Spread Formula vanishes identically, and the Triple Quad Formula residual equals \(\frac{16}{n^2} \to 0\).

import sympy as sp
        from geometor.cusa import (
            compute_flattening_triangle_metrics,
            verify_triangle_line_coincidence,
        )

        m4 = compute_flattening_triangle_metrics(4)
        assert m4.spread_apex == sp.Rational(64, 289)
        assert m4.triple_quad_residual == sp.Integer(1)

        proof = verify_triangle_line_coincidence()
        assert proof["triple_spread_identity"] is True
        assert proof["limit_triple_quad"] == 0
        

3. Polygon Exhaustion (I.3) & Scripta Mathematica (1445–1459)

Codex Reference: Unit I & Unit V · Script: demos/polygon_exhaustion_and_complementa.py

De Geometricis Transmutationibus (1445) — Quarter-Side Sesquiquarta Rectification ( A D ): (x - (0))^2 + (y - (0))^2 - (1) = 0 ( A F ): (x - (0))^2 + (y - (0))^2 - (1/4) = 0 ( A H ): (x - (0))^2 + (y - (0))^2 - (175/256) = 0 [ A F ]: (1)*x + (0)*y + (0) = 0 [ B C ]: (0)*x + (1)*y + (1/2) = 0 [ C D ]: (-3/2)*x + (-sqrt(3)/2)*y + (sqrt(3)/2) = 0 [ D B ]: (3/2)*x + (-sqrt(3)/2)*y + (sqrt(3)/2) = 0 [ A H ]: (5/8)*x + (-5*sqrt(3)/16)*y + (0) = 0 D = (0, 1) B = (-sqrt(3)/2, -1/2) C = (sqrt(3)/2, -1/2) A = (0, 0) E = (-sqrt(3)/4, -1/2) F = (0, -1/2) G = (sqrt(3)/4, -1/2) H = (-5*sqrt(3)/16, -5/8) D B C A E F G H
De Geometricis Transmutationibus (1445, §§9–11): quarter-side ray AE = sqrt(7)/4 extended in 5:4 sesquiquartal ratio to AH = 5*sqrt(7)/16, implying pi ~= 24*sqrt(21)/35 ~= 3.14233762.
De Mathematica Perfectione (1458) & Aurea Propositio (1459) — Three-Ray Arc Rectification ( A D ): (x - (0))^2 + (y - (0))^2 - (1) = 0 [ E G ]: (1)*x + (0)*y + (-1) = 0 [ A E ]: (-sqrt(3)/3)*x + (1)*y + (0) = 0 [ A D ]: (0)*x + (1)*y + (0) = 0 [ A G ]: (sqrt(3)/3)*x + (1)*y + (0) = 0 [ B C ]: (1)*x + (0)*y + (-sqrt(3)/2) = 0 [ I L ]: (1)*x + (0)*y + (9/13 - 12*sqrt(3)/13) = 0 D = (1, 0) B = (sqrt(3)/2, 1/2) C = (sqrt(3)/2, -1/2) A = (0, 0) H = (sqrt(3)/2, 0) E = (1, sqrt(3)/3) G = (1, -sqrt(3)/3) K = (-9/13 + 12*sqrt(3)/13, 0) I = (-9/13 + 12*sqrt(3)/13, 12/13 - 3*sqrt(3)/13) L = (-9/13 + 12*sqrt(3)/13, -12/13 + 3*sqrt(3)/13) D B C A H E G K I L
Aurea Propositio in Mathematicis (1459, §§1–2) over Q(sqrt(3)) at theta = pi/6: balanced rectifying segment IKL satisfying |AI| + |AK| + |AL| = 3 and fifth-order Maclaurin accuracy theta^5/180.
  • Exact Symbolic Verification: Verifies (a) monotone radical apothems \(r_n = \cos(\pi/n)\) across \(n \in \{3, 4, 6, 8, 12\}\), (b) Cusa’s 1445 sesquiquartal equilateral-triangle rectification \(\pi_{\text{Cusa},1445} = \frac{24\sqrt{21}}{35}\), and (c) Cusa’s 1458–1459 three-ray law \(L_{\text{Cusa}}(\theta) = \frac{6\sin\theta}{2 + \cos\theta}\) with exact Maclaurin series error \(\theta - \frac{3\sin\theta}{2+\cos\theta} = \frac{\theta^5}{180} + \frac{\theta^7}{1512} + \frac{\theta^9}{25920} + O(\theta^{11})\).

import sympy as sp
        from geometor.cusa import (
            verify_cusa_snellius_approximation,
            verify_transmutatio_1445,
        )

        tr = verify_transmutatio_1445()
        assert sp.simplify(tr["pi_cusa_1445"] - 24 * sp.sqrt(21) / 35) == 0

        au = verify_cusa_snellius_approximation()
        assert au["tangent_ratio_equal"] is True
        assert au["coeff_theta_5"] == sp.Rational(1, 180)