Summary — Unit II: Six Mutual Quadrances & The Euler Line (§§ 10–18)

Analytical Synopsis of §§ 10–18 (pp. 107–114)

  1. Symmetric Basis (§ 11): Because the mutual distances among \(E, F, G, H\) cannot privilege any side of \(\triangle ABC\), Euler introduces \(p = a+b+c\), \(q = ab+ac+bc\), \(r = abc\), rewriting Heron’s formula as \(16\mathcal{A}^2 = -p^4 + 4p^2q - 8pr\). Since \(R_{\text{circ}} = \frac{abc}{4\mathcal{A}} = \frac{r}{4\mathcal{A}}\) and \(r_{\text{in}} = \frac{2\mathcal{A}}{p}\), the ubiquitous quantity \(\frac{r^2}{16\mathcal{A}^2}\) is simply \(R_{\text{circ}}^2\), and \(\frac{r}{p} = 2R_{\text{circ}}r_{\text{in}}\).

  2. Euler’s Six Squared-Distance Table (§ 18, Formulas I–VI) (alongside modern radius forms):

    Euler’s Six Mutual Quadrances (§ 18, p. 114)

    Segment

    Euler’s Symmetric Form (\(p, q, r, \mathcal{A}\))

    Radius / Side-Square Form (\(R, r_{\text{in}}, a, b, c\))

    \(EF^2\)

    \(\dfrac{r^2}{4\mathcal{A}^2} - \dfrac{4}{9}(p^2 - 2q)\)

    \(4R^2 - \dfrac{4}{9}(a^2 + b^2 + c^2)\)

    \(EG^2\)

    \(\dfrac{r^2}{4\mathcal{A}^2} - p^2 + 3q - \dfrac{4r}{p}\)

    \(4R^2 + 4Rr_{\text{in}} + 3r_{\text{in}}^2 - s^2 = 2r_{\text{in}}^2 - 4R^2\cos\alpha\cos\beta\cos\gamma\)

    \(EH^2\)

    \(\dfrac{9r^2}{16\mathcal{A}^2} - p^2 + 2q\)

    \(9R^2 - (a^2 + b^2 + c^2)\)

    \(FG^2\)

    \(-\dfrac{1}{9}p^2 + \dfrac{5}{9}q - \dfrac{2r}{p}\)

    \(\dfrac{1}{9}(s^2 + 5r_{\text{in}}^2 - 16Rr_{\text{in}})\)

    \(FH^2\)

    \(\dfrac{r^2}{16\mathcal{A}^2} - \dfrac{1}{9}(p^2 - 2q)\)

    \(R^2 - \dfrac{1}{9}(a^2 + b^2 + c^2)\)

    \(GH^2\)

    \(\dfrac{r^2}{16\mathcal{A}^2} - \dfrac{r}{p}\)

    \(R^2 - 2Rr_{\text{in}} = R(R - 2r_{\text{in}})\)

  3. Three Landmark Theorems on One Page (p. 114):

    • Euler’s Incenter–Circumcenter Theorem (§ 17, VI): \(GH^2 = R^2 - 2Rr_{\text{in}}\) (whence \(R \ge 2r_{\text{in}}\) with equality iff equilateral).

    • The Euler Line (§§ 16, 18): \(E, F, H\) are collinear in every triangle with \(F\) between \(E\) and \(H\) and \(EF : FH = 2 : 1\) (\(EH = \frac{3}{2}EF\)).

    • Stewart’s Collinearity Constraint (§ 18): Because \(F\) divides \(EH\) in the ratio \(2 : 1\), the four centers satisfy \(4GH^2 + 2EG^2 = 3EF^2 + 6FG^2\), meaning the four inverse problems of § 2 Collapse into a single three-parameter problem!