GC-01.03 · Lemoine–Reusch–Hofstetter 5-Step Division

Discoverer(s): Émile Michel Hyacinthe Lemoine, Kurt Hofstetter | Date: 1902 / 2003 | Field: \(\mathbb{Q}(\sqrt{3}, \phi)\) | Loci: 6 (Lemoine \(S = 18\))

Given points A and B on line AB, let unit vesica circles cAB and cBA meet at C and D. Circle cCA cuts cAB at E and chord CD at F; then circle cEF cuts AB internally at G and externally at G1 so that \(|AG|/|GB| = |G_1A|/|AB| = \phi\).
GC-01.03 — Lemoine–Reusch–Hofstetter 5-Step Division cAB: x**2 + y**2 - 1 = 0 cBA: y**2 + (x - 1)**2 - 1 = 0 cCA: (x - 1/2)**2 + (y - sqrt(3)/2)**2 - 1 = 0 cEF: (x + 1/2)**2 + (y - sqrt(3)/2)**2 - 2 = 0 AB: y = 0 CD: x - 1/2 = 0 Golden major AG = -1/2 + sqrt(5)/2 Golden minor GB = 3/2 - sqrt(5)/2 C = (1/2, sqrt(3)/2) D = (1/2, -sqrt(3)/2) E = (-1/2, sqrt(3)/2) F = (1/2, -1 + sqrt(3)/2) G1 = (-sqrt(5)/2 - 1/2, 0) A = (0, 0) B = (1, 0) G = (-1/2 + sqrt(5)/2, 0) C D E F G1 A B G

Given & Constructive Steps

Given: Unit base points A \(= (0, 0)\) and B \(= (1, 0)\) on carrier line AB.

step Draw circles cAB and cBA of unit radius, intersecting at C \(= (\tfrac{1}{2}, \tfrac{\sqrt{3}}{2})\) and D \(= (\tfrac{1}{2}, -\tfrac{\sqrt{3}}{2})\).
step Join perpendicular bisector line CD.
step Draw circle cCA centered at C through A, intersecting cAB at E \(= (-\tfrac{1}{2}, \tfrac{\sqrt{3}}{2})\) and segment CD at F \(= (\tfrac{1}{2}, \tfrac{\sqrt{3}}{2} - 1)\).
step Draw circle cEF centered at E through F, cutting AB internally at G and externally at G1.

Exact Symbolic Proof & Certificate

Because E and C share ordinate \(\tfrac{\sqrt{3}}{2}\) at horizontal separation \(1\), while F lies \(1\) unit directly below C, triangle \(\triangle ECF\) is right isosceles with \(Q(E, F) = 1^2 + 1^2 = 2\). Circle cEF therefore satisfies \((x + \tfrac{1}{2})^2 + y^2 = 2\); setting \(y = 0\) yields \((x + \tfrac{1}{2})^2 = 2 - \tfrac{3}{4} = \tfrac{5}{4}\), giving G \(= (1/\phi, 0)\) and G1 \(= (-\phi, 0)\). Thus \(|AG|/|GB| = \phi\) and \((G_1, A, G, B)\) forms a connected golden chain along AB.

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

AG \(/\) GB \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (A, G, B)

Exact Segment Lengths

\(|AG| = - \frac{1}{2} + \frac{\sqrt{5}}{2}\), \(|GB| = \frac{3}{2} - \frac{\sqrt{5}}{2}\)

Rational Trigonometry Quadrances

\(Q(A, G) = \frac{3}{2} - \frac{\sqrt{5}}{2}\), \(Q(G, B) = \frac{7}{2} - \frac{3 \sqrt{5}}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

\(\mathbb{Q}(\sqrt{3}, \phi)\)

Géométrographie Loci

2 given points · 2 lines · 4 circles · 0 polygons (elementary loci \(= 6\), Lemoine \(S = 18\))

Automated Divine Scan

geometor.divine.golden detected 2 exact collinear golden section(s) and 0 connected golden chain(s) in this model:

  • Section (G1, A, B): long \(|G1A|^2 = \frac{\sqrt{5}}{2} + \frac{3}{2}\) vs. short \(|AB|^2 = 1\)

  • Section (A, G, B): long \(|AG|^2 = \frac{3}{2} - \frac{\sqrt{5}}{2}\) vs. short \(|GB|^2 = \frac{7}{2} - \frac{3 \sqrt{5}}{2}\)

Historical Commentary & Attribution

Rediscovered independently by Kurt Hofstetter (Hofstetter 2003), this five-locus division of a pre-drawn line segment had appeared a century earlier in Émile Lemoine’s Géométrographie (Lemoine 1902) and J. Reusch’s Planymetrische Konstruktionen in geometrographischer Ausführung (1904).