Critical & Symbolic Notes — Unit III: Inverse Reconstruction Cubic (§§ 19–24)

Algebraic Mechanism of Q² Cancellation & 1767 Folio Notes

1. Two Typographical Details in §§ 19–21 (pp. 114–116)

  • p. 114, § 19, penult. line: The folio prints Quoniam igitur hinc est -s/p = QQ/R; since \(R = \frac{r^2}{ps}\) and \(Q = \frac{r}{p}\), we have \(\frac{Q^2}{R} = \frac{r^2/p^2}{r^2/(ps)} = +\frac{s}{p}\) (as Euler immediately uses in \(q = \frac{1}{4}P + 2Q + \frac{Q^2}{R}\)).

  • p. 116, § 21, Eq. III: The folio prints hh = 1/36 P - 8/9 Q + 6 QQ / (9 R) (6 pro 5), whereas on the facing page 115 (§ 19 IV) Euler correctly printed + 5 QQ / (9 R). Euler’s solved rational expressions for \(R, Q, P\) immediately below Eq. III use the true coefficient 5: only 5 · II - 2 · III eliminates \(\frac{Q^2}{R}\) to produce the denominator \(3g^2 + 6h^2 - 2f^2\).

2. Perfect-Square Factorization of 3g² + 6h² − 2f² Under Collinearity

Consider the denominator governing all three invariants \(R, Q, P\) in § 21:

\[\mathcal{D} = 3g^2 + 6h^2 - 2f^2 = 3FH^2 + 6FG^2 - 2GH^2.\]

Whenever the incenter \(G\) lies on the Euler line so that \(g = f - h\) (Euler’s § 25), substituting \(g = f - h\) into \(\mathcal{D}\) yields an exact square:

\[3(f - h)^2 + 6h^2 - 2f^2 = f^2 - 6fh + 9h^2 = (f - 3h)^2 \ge 0.\]

Thus \(\mathcal{D}\) vanishes only at the equilateral coalescence \(f = 3h\) (where \(E, F, G, H\) coincide), and for every non-equilateral collinear configuration \(P\) factors up to the single quadratic radical \(\sqrt{3h(4f - 3h)}\), forcing a double root \(a = b\) in Unit IV (§§ 25–26).

3. Constructibility Barrier: Why Compass and Straightedge Cannot Construct (A, B, C)

In § 22, Euler remarks that once \(z^3 - pz^2 + qz - r = 0\) is formed, its three roots give the sides \((a, b, c)\). Over a general rational configuration of centers \((E, F, G, H)\), this cubic is irreducible over \(\mathbb{Q}(f, g, h)\) with three distinct positive real roots (casus irreducibilis). By Wantzel’s theorem (1837), reconstructing \(\triangle ABC\) from \((E, F, G, H)\) in general requires angle trisection (or conic intersections) and is impossible by Euclidean ruler and compass alone—a fact already recognized empirically by Euler in § 34 (p. 123)!