Summary — Unit III: The Nine-Point Circle & Feuerbach’s Tangency Theorem (§§ 49–60)¶
Logical Chain of Feuerbach’s Masterpiece (pp. 33–41)
Euler–Fuß Quadrances from Circumcenter K (§§ 49–50): \(\overline{KS}^2 = R^2 - 2rR\) (Euler E325 § 18 VI; Fuß 1797 § 32) and \(\overline{KS^{(k)}}^2 = R^2 + 2r^{(k)}R\), summing via \(r' + r'' + r''' - r = 4R\) to \(\sum \overline{KS_i}^2 = 12R^2\).
Orthocenter-Incenter & Orthocenter-Circumcenter Quadrances (§§ 51, 53): Direct coordinate-projection calculation on base \(AB = c\) yields \(\overline{OS}^2 = 2r^2 - 2\rho R\) and \(\overline{KO}^2 = R^2 - 4\rho R\).
Nine-Point Center L = X(5) Bisecting KO (§§ 54–56): Because \(O\) is the incenter of \(\triangle MNP\) (inradius \(\rho\)) and \(L\) is its circumcenter (circumradius \(\frac{1}{2}R\)), applying Euler’s \(R^2 - 2rR\) formula inside \(\triangle MNP\) gives \(\overline{OL}^2 = \frac{1}{4}R^2 - \rho R = \frac{1}{4}\overline{KO}^2\). Projection onto \(AB\) proves \(LJ = \frac{1}{2}(OP + Kc)\), establishing that \(L\) is the exact midpoint of \(KO\) and that circle \((L, \frac{1}{2}R)\) bisects all three sides \(BC, AC, AB\).
Apollonius’s Median Law & Feuerbach’s Theorem (§ 57): Applying \(2\overline{LS}^2 + 2\overline{OL}^2 = \overline{KS}^2 + \overline{OS}^2\) in \(\triangle KOS\) cancels \(-2\rho R\) identically, leaving
\[LS = \left|\frac{1}{2}R - r\right|, \qquad LS^{(k)} = \frac{1}{2}R + r^{(k)},\]proving internal tangency to the incircle at \(X(11)\) and external tangency to all three excircles!
Euler Line Harmonic Range (§ 60): On the Euler line, \((O, G, L, K) = (X_4, X_2, X_5, X_3)\) satisfy \(OL : LG : GK = 3 : 1 : 2\) and cross-ratio \((O, G; L, K) = -1\).