GC-01.04 · Hofstetter’s 2004 Five-Step Secant Division

Discoverer(s): Kurt Hofstetter | Date: 2004 | Field: \(\mathbb{Q}(\sqrt{3}, \phi)\) | Loci: 5 (Lemoine \(S = 15\))

Given points A and B, draw line AB and unit circles cAB, cBA meeting at C and D, with cAB meeting ray BA beyond A at E. Circle cEB cuts cBA above AB at F, and secant DF divides AB at G in the golden ratio \(|AG|/|GB| = \phi\) in exactly five Euclidean loci.
GC-01.04 — Hofstetter's 2004 Five-Step Secant Division cAB: x**2 + y**2 - 1 = 0 cBA: y**2 + (x - 1)**2 - 1 = 0 cEB: y**2 + (x + 1)**2 - 4 = 0 AB: y = 0 DF: x*(-sqrt(15)/4 - sqrt(3)/2) + y/4 + sqrt(15)/8 + 3*sqrt(3)/8 = 0 Golden major AG = -1/2 + sqrt(5)/2 Golden minor GB = 3/2 - sqrt(5)/2 E = (-1, 0) C = (1/2, sqrt(3)/2) D = (1/2, -sqrt(3)/2) F = (3/4, sqrt(15)/4) A = (0, 0) B = (1, 0) G = (-1/2 + sqrt(5)/2, 0) E C D F A B G

Given & Constructive Steps

Given: Points A \(= (0, 0)\) and B \(= (1, 0)\) in the Euclidean plane.

step Draw carrier line AB (Locus 1).
step Draw circle cAB centered at A through B (Locus 2), meeting line AB beyond A at E \(= (-1, 0)\).
step Draw circle cBA centered at B through A (Locus 3), intersecting cAB at C and D \(= (\tfrac{1}{2}, -\tfrac{\sqrt{3}}{2})\).
step Draw circle cEB centered at E through B of radius \(2\) (Locus 4), meeting cBA above AB at F \(= (\tfrac{3}{4}, \tfrac{\sqrt{15}}{4})\).
step Draw line DF (Locus 5), cutting AB at G \(= (1/\phi, 0)\).

Exact Symbolic Proof & Certificate

Subtracting \((x - 1)^2 + y^2 = 1\) from \((x + 1)^2 + y^2 = 4\) gives \(4x = 3\), so F has exact coordinates \((\tfrac{3}{4}, \tfrac{\sqrt{15}}{4})\). Along secant DF joining \((\tfrac{1}{2}, -\tfrac{\sqrt{3}}{2})\) to \((\tfrac{3}{4}, \tfrac{\sqrt{15}}{4})\), the \(x\)-intercept G is \(x_G = \tfrac{1}{2} + \tfrac{1}{4} \cdot \frac{\sqrt{3}/2}{\sqrt{3}/2 + \sqrt{15}/4} = \tfrac{1}{2} + \frac{1}{2(\sqrt{5} + 2)} = \frac{\sqrt{5} - 1}{2} = \frac{1}{\phi}\). Therefore \(|AG| = 1/\phi\), \(|GB| = 1/\phi^2\), and \(|AG|/|GB| = \phi\).

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 · 3 circles · 0 polygons (elementary loci \(= 5\), Lemoine \(S = 15\))

Automated Divine Scan

geometor.divine.golden detected 2 exact collinear golden section(s) and 1 connected golden chain(s) in this model:

  • Section (E, A, G): long \(|EA|^2 = 1\) vs. short \(|AG|^2 = \frac{3}{2} - \frac{\sqrt{5}}{2}\)

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

Connected Golden Chains:

  • Chain (E — A — G — B) spanning 2 overlapping golden sections (progression > >)

Historical Commentary & Attribution

Published in Forum Geometricorum (Hofstetter 2004), this construction requires only two lines and three circles even when the line through A and B is not drawn in advance. Notice that its abscissae \(x_D = \tfrac{1}{2}\) and \(x_F = \tfrac{3}{4}\) (or under reflection \(\tfrac{1}{4}, \tfrac{1}{2}\)) are a special case of Bùi Quang Tuân’s general bilinear circle theorem (Cut-the-Knot).