Critical & Symbolic Notes — Unit IV: Isosceles Locus, 60° Family & Trisection (§§ 25–34)

Foreshadowing X(5), the 60° Locus EG = GH, and Philological Audit of §§ 29–34

1. Euler’s Half-Segment d = ½ EH (§ 29) and the Nine-Point Center X(5)

In §§ 28–29 (pp. 119–120), Euler bisects the segment \(EH\) (\(= 2d\)) and classifies every isosceles triangle according as \(EG < d\) (Fig. 8, \(a > c\)) or \(EG > d\) (Fig. 9, \(a < c\)). Geometrically, the midpoint of \(EH\) is precisely Feuerbach’s nine-point center \(L = X(5)\). By Feuerbach’s Theorem (§ 57), \(LG = \tfrac{1}{2}R_{\text{circ}} - r_{\text{in}}\), which vanishes only when \(R_{\text{circ}} = 2r_{\text{in}}\) (the equilateral triangle). Since \(E, G, L, F, H\) all lie along the axis of symmetry of an isosceles triangle with \(EL = LH = d\), the non-vanishing distance \(LG > 0\) explains geometrically why Euler’s two regimes \(EG < d\) (\(a > c\), where \(G\) lies inside \(EL\)) and \(EG > d\) (\(a < c\), where \(G\) lies inside \(LF\)) are separated exclusively by the equilateral coalescence.

2. Why EG = GH (§§ 32–33) Characterizes Triangles with a 60° Angle

In §§ 32–33 (pp. 121–122), Euler discovers algebraically that whenever \(e = f\) (\(EG = GH\)), the side-length cubic factors rationally with root \(c = \frac{f^2\sqrt{3}}{\sqrt{4f^2 - k^2}} = R_{\text{circ}}\sqrt{3}\). Geometrically, \(c = 2R_{\text{circ}}\sin C = R_{\text{circ}}\sqrt{3}\) means \(C = 60^\circ\) (or \(120^\circ\), excluded here since \(Q > 0\) fixes \(C = 60^\circ\))! Indeed, at a \(60^\circ\) vertex \(C\), the distance from the circumcenter \(H\) to \(AB\) is \(R_{\text{circ}}\cos 60^\circ = \tfrac{1}{2}R_{\text{circ}}\), and reflection properties of \(X(1), X(3), X(4)\) force \(X(1)X(4) = X(1)X(3)\). For Euler’s numerical choice \(e^2 = f^2 = 2, k^2 = 7\) (§ 32), the sides \(a = \frac{3\sqrt{3}+\sqrt{7}}{2}\), \(b = \frac{3\sqrt{3}-\sqrt{7}}{2}\), \(c = 2\sqrt{3}\) satisfy \(a^2 + b^2 - ab = \frac{27 + 7}{2} - \frac{27 - 7}{4} = 17 - 5 = 12 = c^2\), confirming \(\angle C = 60^\circ\)!

3. Audit of Four Arithmetic/Printing Slips in §§ 29–34 (pp. 120–123)

  • p. 120, § 29: In the second case (\(EG = d + e\)), the folio prints OE = (d + e)^2 / (2e); substituting \(a, c\) into \(HO = \frac{2a^2 - c^2}{2\sqrt{4a^2 - c^2}}\) gives \(OE = \frac{(d + e)^2}{4e}\) (matching (d - e)^2 / (4e) in the first case).

  • p. 121, § 30, last line: In the normalized cubic \(y^3 - y^2 + \frac{q}{P}y - \frac{Q}{P} = 0\), the folio’s numerator prints + 3eekk for - 3eekk in \(q\) and omits the factor \((2e^2 + 2f^2 - k^2)\) multiplying \(f^2(k^2 - 2e^2 - f^2)\) in the constant term.

  • p. 121, § 31: Euler’s positivity conditions \(2e^2 + f^2 < k^2 < 2e^2 + 2f^2\) are exact (equivalent to \(Q > 0\) and \(R > 0\), i.e., \(2Rr_{\text{in}} > 0\) and \(4R^2 > 0\)), whereas his quick elimination for \(f^2 / e^2\) treats \(P > 0\) as binding; in fact, on any real triangle \(P = p^2 > 0\) holds automatically once the depressed cubic discriminant (Blundon’s inequality) is non-negative! Indeed, in Euler’s own valid example in § 34 (\(e^2 = 3, f^2 = 2, k^2 = 9\)), \(\frac{f^2}{e^2} = \frac{2}{3} = \frac{38}{57}\) satisfies \(\frac{8}{19} < \frac{2}{3} < \frac{19}{13}\), and even triangles with \(e = 0\) (\(X(1) = X(4)\), impossible for non-equilateral) are excluded by Blundon’s fundamental triangle inequality.

  • p. 123, § 34: For \(e^2 = 3, f^2 = 2, k^2 = 9\), we get \(2e^2 + 2f^2 - k^2 = 1\), \(R = 16\), \(Q = 2\), \(P = 71\), \(q = 22\), so Euler’s cubic \(z^3 - \sqrt{71}\,z^2 + 22z - 2\sqrt{71} = 0\) is 100% exact! Depressing via \(z = \frac{1}{3}\sqrt{71} + t\) gives

    \[t^3 - \frac{5}{3}t + \frac{2\sqrt{71}}{27} = 0.\]

    Comparing with \(\cos\alpha\) in \(t = -\frac{2}{3}\sqrt{5}\cos\frac{\theta}{3}\) gives \(\cos\alpha = \frac{\sqrt{71}}{5\sqrt{5}} = \sqrt{\frac{71}{125}}\) (Euler’s manuscript slipped \(5^3 = 125 \to 75\), and his angle \(11^\circ 32' 13'' = \arccos\sqrt{24/25}\) corresponds to \(72/75\)). With the exact \(\alpha = \arccos\sqrt{\frac{71}{125}} \approx 41^\circ 03' 16''\), Euler’s trisection formulas \(a = \frac{1}{3}\sqrt{71} + \frac{2}{3}\sqrt{5}\cos(60^\circ - \frac{1}{3}\alpha) \approx 3.84713\), \(b = \frac{1}{3}\sqrt{71} + \frac{2}{3}\sqrt{5}\cos(60^\circ + \frac{1}{3}\alpha) \approx 3.06379\), \(c = \frac{1}{3}\sqrt{71} - \frac{2}{3}\sqrt{5}\cos\frac{1}{3}\alpha \approx 1.51523\) satisfy \(a + b + c = \sqrt{71}\), \(ab + ac + bc = 22\), and \(abc = 2\sqrt{71}\) to full precision!