Unit II Commentary & Notes: Euclid, Mascheroni, and Steiner on Book I Primitives

Minimal-Instrument Counterparts & Deductive Order vs. Constructive Economy

1. Why Euclid Could Not Use Thales’s Circle (\(S = 8\)) in Proposition I.11

In Problem XX.2, Lemoine erects the perpendicular at \(A \in BC\) in a single circle (\(4R_1 + 2R_2 + C_1 + C_3, S = 8\)) via Thales’s semicircle theorem (Euclid, Elements III.31). Why did Euclid instead use three circles (\(S = 9\) or \(10\)) in Proposition I.11?

Because Euclid III.31 depends deductively on the parallel postulate (I.29, I.32: angle sum of a triangle equals two right angles)! In absolute (neutral) geometry—the first twenty-eight propositions of Book I, valid identically in Euclidean and Bolyai–Lobachevsky hyperbolic planes—the angle inscribed in a semicircle is strictly less than a right angle in hyperbolic space! Euclid’s three-circle equilateral construction in I.11 holds in absolute geometry without the parallel postulate, whereas Lemoine’s \(S = 8\) Thales construction exploits Euclidean flatness to save two circles and one operation.

2. Dual Minimal-Instrument Comparison on the Parallel (XVII) and Perpendicular (XX)

Comparing Lemoine’s Book I benchmarks with Mascheroni (mascheroni-1797) and Steiner (steiner-1833) reveals an extraordinary structural symmetry (see XVII — Parallel Through an External Point to a Given Line and XX.1 — Perpendicular to a Traced Line from an External Point):

• Parallel Through an External Point: * Lemoine / Tarry Rhombus (Ruler + Compass): \(2R_1 + R_2 + 3C_1 + 3C_3\) (\(S = 9, E = 5\)). * Mascheroni Libro IV, § 80 (Compass Alone): Given \(A, B\) on the line and external point \(C\), sweep \(B(AC)\) (\(3C_1 + C_3\)) and \(C(AB)\) (\(2C_1 + C_3\) if \(AB\) was spanned first) meeting at \(D\). Point \(D\) completes parallelogram \(ABCD\) so pair \((C, D)\) represents the parallel in \(5C_1 + 2C_3\) (\(S = 7, E = 5\), zero lines!). Notice that Tarry’s rhombus (\(S = 9\)) is simply Mascheroni’s parallelogram specialized to equal sides \(R\) plus the final straightedge join \(2R_1 + R_2\)! * Steiner § 18, Erste Aufgabe (Straightedge + Fixed Circle :math:`M`): When the given line passes through the center \(M\) of the fixed circle, cutting it at diameter endpoints \(a, b\), M is the known midpoint of \(ab\). A complete quadrangle over \(a, M, b\) constructs the harmonic conjugate of the point at infinity—hence the parallel through \(P\)—in \(11R_1 + 6R_2\) (\(S = 17, E = 11\), zero circles).

• Perpendicular From an External Point: * Lemoine XX.1 (Ruler + Compass): \(2R_1 + R_2 + 3C_1 + C_2 + 2C_3\) (\(S = 9\), 1 line, 2 circles). * Mascheroni Libro IV, § 78 (Compass Alone): Reflect \(A\) across \(B, C\) to \(A'\) in \(4C_1 + 2C_3\) (\(S = 6\), 0 lines, 2 circles), determining the perpendicular line implicitly as \((A, A')\)! * Steiner § 18, Dritte Aufgabe (Straightedge + Fixed Circle :math:`M`): Join external \(P\) to diameter endpoints \(a, b\) cutting the fixed circle again at \(c, d\); by Thales’s theorem \(ad \perp Pb\) and \(bc \perp Pa\), so \(H = ad \cap bc\) is the orthocenter of \(\triangle Pab\), and line \(PH\) is perpendicular to \(ab\) in \(10R_1 + 5R_2\) (\(S = 15\), 5 lines, 0 circles)!