Critical & Symbolic Notes — Unit II: Orthocenter & Orthic Inradius ρ (§§ 19–48)¶
Parity of \(\rho R\), Translation to Euler’s \((p, q, r)\), and (15, 13, 14) Values
Exact Bridge Between Feuerbach’s ρ and Euler’s (p, q, r): Comparing § 27 (\(a^2 + b^2 + c^2 = 8R^2 + 4\rho R\)) with Euler’s § 14 (\(EH^2 = 9R^2 - (a^2 + b^2 + c^2)\)) immediately yields
\[\overline{KO}^2 = EH^2 = R^2 - 4\rho R = R(R - 4\rho),\]which is precisely § 53 of Feuerbach’s monograph! Expressing \(\rho R\) in terms of the side squares (§ 24 and \(R = \frac{abc}{4\Delta}\)):
\[\rho R = \frac{(-a^2 + b^2 + c^2)(a^2 - b^2 + c^2)(a^2 + b^2 - c^2)}{64\Delta^2}.\]Observe that every side \(a, b, c\) enters \(\rho R\) exclusively through its square \(a^2, b^2, c^2\)! Consequently, \(\rho R\) is strictly invariant under \(a \mapsto -a\) (because \(R \mapsto -R\) and \(\rho \mapsto -\rho\) simultaneously).
Duality Between Sections I and II (§ 1 vs. § 24): Feuerbach systematically exploits the self-duality of orthocentric quadrangles:
For the excentral quadrangle \((S, S', S'', S''')\), the pedal triangle is \(\triangle ABC\) (circumradius \(R = \frac{1}{2}R_{\text{ex}}\)).
For the orthocentric quadrangle \((O, A, B, C)\), the pedal triangle is \(\triangle MNP\) (circumradius \(R_{MNP} = \frac{1}{2}R\)), so every identity of Section I (§§ 2–7) carries over instantaneously to \((\rho, \rho^{(1)}, \rho^{(2)}, \rho^{(3)})\) in § 30 (p. 21)!
Exact Rational Values on the Canonical (15, 13, 14) Triangle: On \(A = (0, 0), B = (14, 0), C = (5, 12)\) (\(a = 15, b = 13, c = 14\), \(\Delta = 84\), \(R = \frac{65}{8}\)):
Cosine product: \(\cos\alpha = \frac{33}{65}\), \(\cos\beta = \frac{3}{5}\), \(\cos\gamma = \frac{5}{13}\), so \(\cos\alpha\cos\beta\cos\gamma = \frac{99}{845}\), whence \(\rho = 2 \cdot \frac{65}{8} \cdot \frac{99}{845} = \frac{99}{52}\).
Check § 27: \(8R^2 + 4\rho R = 8 \cdot \frac{4225}{64} + 4 \cdot \frac{99}{52} \cdot \frac{65}{8} = \frac{4225}{8} + \frac{495}{8} = \frac{4720}{8} = 590 = 15^2 + 13^2 + 14^2\)!
Check § 28: \(4R(2R - \rho) = \frac{65}{2}\left(\frac{65}{4} - \frac{99}{52}\right) = \frac{65}{2} \cdot \frac{746}{52} = \frac{1865}{4} = 4^2 + 14^2 + \left(\frac{21}{2}\right)^2 + 12^2 = 16 + 196 + \frac{441}{4} + 144 = \frac{1865}{4}\)!