Summary — Unit I: The Four Classical Centers (§§ 1–9)¶
Analytical Synopsis of §§ 1–9 (pp. 103–107)
Problem Statement (§§ 1–3): Euler isolates the four classical concurrency centers of a non-equilateral plane triangle \(\triangle ABC\) with side lengths \(BC = a\), \(AC = b\), \(AB = c\), and area \(\mathcal{A}\):
\(E = X(4)\) — orthocenter (intersection of altitudes),
\(F = X(2)\) — centroid (intersection of medians),
\(G = X(1)\) — incenter (intersection of internal angle bisectors),
\(H = X(3)\) — circumcenter (intersection of perpendicular bisectors).
Given any three of these four points in position, the triangle is determined (save in the equilateral case where all four coalesce).
Orthogonal Coordinate Table on Base AB (§§ 5–9): Taking vertex \(A\) as origin and base \(AB = c\) along the positive abscissa axis, Euler expresses every abscissa and ordinate purely as rational functions of \((a, b, c)\) and \(\mathcal{A}\):
Euler’s Base Coordinates (§§ 6–9)¶ Euler Point
Kimberling ID
Abscissa along \(AB\)
Ordinate normal to \(AB\)
\(E\) (Orthocenter)
\(X(4)\)
\(AP = \dfrac{c^2 + b^2 - a^2}{2c}\)
\(PE = \dfrac{(c^2 + b^2 - a^2)(a^2 + c^2 - b^2)}{8c\mathcal{A}}\)
\(F\) (Centroid)
\(X(2)\)
\(AQ = \dfrac{3c^2 + b^2 - a^2}{6c}\)
\(QF = \dfrac{2\mathcal{A}}{3c}\)
\(G\) (Incenter)
\(X(1)\)
\(AR = \dfrac{c + b - a}{2}\)
\(RG = \dfrac{2\mathcal{A}}{a + b + c}\)
\(H\) (Circumcenter)
\(X(3)\)
\(AS = \dfrac{c}{2}\)
\(SH = \dfrac{c(a^2 + b^2 - c^2)}{8\mathcal{A}}\)
Immediate Vectorial Consequence: Even before computing squared distances in Unit II, comparing \((AP, PE)\), \((AQ, QF)\), and \((AS, SH)\) already reveals \(AQ - AS = \frac{b^2 - a^2}{6c} = \frac{1}{3}(AP - AS)\) and \(QF - SH = \frac{1}{3}(PE - SH)\) via Heron’s expansion \(16\mathcal{A}^2 = 2a^2b^2 + 2a^2c^2 + 2b^2c^2 - a^4 - b^4 - c^4\), foreshadowing the Euler Line vector law \(\overrightarrow{HF} = \frac{1}{3}\overrightarrow{HE}\).