Summary — Unit III: Inverse Reconstruction & Side Cubic (§§ 19–24)¶
Analytical Synopsis of §§ 19–24 (pp. 114–118)
Why the System Collapses to Rational Linear Equations (§§ 19–21): Introducing homogeneous dimension-2 invariants \(P = p^2\), \(Q = \frac{r}{p} = 2R_{\text{circ}}r_{\text{in}}\), and \(R = \frac{r^2}{4\mathcal{A}^2} = 4R_{\text{circ}}^2\) turns Equations I, II, III of § 21 into:
\(R = 4(f^2 + Q)\) from Equation I;
Eliminating \(\frac{Q^2}{R}\) between Equations II and III via \(5\,\text{II} - 2\,\text{III}\) gives a purely linear relation among \(P, Q, R\):
\[5g^2 - 2h^2 = \frac{5}{4}R - \frac{1}{3}P + 4Q.\]
Substituting \(R = 4(f^2 + Q)\) yields \(P = 27Q + 15f^2 - 15g^2 + 6h^2\), and substituting into \(\frac{Q^2}{R}\) cancels \(Q^2\) identically! Hence \((R, Q, P)\) are rational functions of the given quadrances \((f^2, g^2, h^2) = (GH^2, FH^2, FG^2)\).
Viète’s Side-Length Cubic (§ 22): A single square root \(p = \sqrt{P}\) recovers the perimeter \(p\), symmetric sum \(q\), and product \(r = Qp\), yielding \((z - a)(z - b)(z - c) = z^3 - pz^2 + qz - r = 0\).
Benchmark Reconstructions (:func:`~geometor.feuerbach.euler.solve_euler_reconstruction_cubic`):
Euler’s Example (§§ 23–24): \((f^2, g^2, h^2) = \left(\frac{35}{32}, \frac{155}{288}, \frac{1}{9}\right) \implies z^3 - 18z^2 + 107z - 210 = 0 \implies (5, 6, 7)\).
GEOMETOR Canonical Heronian Example: \((f^2, g^2, h^2) = \left(\frac{65}{64}, \frac{265}{576}, \frac{1}{9}\right) \implies z^3 - 42z^2 + 587z - 2730 = 0 \implies (13, 14, 15)\).