GC-04.03 · Jo Niemeyer’s Three Equal Leaning Segments

Discoverer(s): Jo Niemeyer | Date: 2011 | Field: \(\mathbb{Q}(\sqrt{3}, \phi)\) | Loci: 3 (Lemoine \(S = 9\))

Place segment A1B1 perpendicular to line L at A1, lean equal segment B2A2 from midpoint B2 of A1B1 onto L, and lean a third equal segment B3A3 from midpoint B3 of B2A2 onto L. Then A2 divides A1A3 in the golden ratio: \(|A1A2|/|A2A3| = \phi\).
GC-04.03 — Jo Niemeyer's Three Equal Leaning Segments cB2: y**2 + (x - 1)**2 - 4 = 0 cB3: (x - 1/2)**2 + (y - sqrt(3)/2)**2 - 4 = 0 L: x = 0 Golden major A1A2 = sqrt(3) Golden minor A2A3 = (-sqrt(3) + sqrt(15))/2 B2 = (1, 0) B3 = (1/2, sqrt(3)/2) B1 = (2, 0) A1 = (0, 0) A2 = (0, sqrt(3)) A3 = (0, sqrt(3)/2 + sqrt(15)/2) B2 B3 B1 A1 A2 A3

Given & Constructive Steps

Given: Segment A1B1 of length \(2\) from A1 \(= (0,0)\) to B1 \(= (2,0)\) perpendicular to vertical line L (\(x = 0\)).

step Bisect A1B1 at B2 \(= (1,0)\) and swing circle cB2 of radius \(2\) to meet L at A2 \(= (0, \sqrt{3})\).
step Bisect B2A2 at B3 \(= (\tfrac{1}{2}, \tfrac{\sqrt{3}}{2})\).
step Swing circle cB3 of radius \(2\) centered at B3 to meet L above A2 at A3.

Exact Symbolic Proof & Certificate

Right triangle \(\triangle A1B2A2\) has hypotenuse \(2\) and leg \(1\), so \(|A1A2| = \sqrt{3}\) and B3 \(= (\tfrac{1}{2}, \tfrac{\sqrt{3}}{2})\). Circle cB3 satisfies \((x - \tfrac{1}{2})^2 + (y - \tfrac{\sqrt{3}}{2})^2 = 4\); at \(x = 0\), \((y - \tfrac{\sqrt{3}}{2})^2 = \frac{15}{4}\), giving \(y_{A3} = \frac{\sqrt{3} + \sqrt{15}}{2} = \sqrt{3}\,\phi\). Hence \(|A1A2| = \sqrt{3}\) and \(|A2A3| = \sqrt{3}(\phi - 1) = \sqrt{3}/\phi\), proving \(|A1A2|/|A2A3| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

A1A2 \(/\) A2A3 \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (A1, A2, A3)

Exact Segment Lengths

\(|A1A2| = \sqrt{3}\), \(|A2A3| = \frac{- \sqrt{3} + \sqrt{15}}{2}\)

Rational Trigonometry Quadrances

\(Q(A1, A2) = 3\), \(Q(A2, A3) = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

2 given points · 1 lines · 2 circles · 0 polygons (elementary loci \(= 3\), Lemoine \(S = 9\))

Automated Divine Scan

geometor.divine.golden detected 1 exact collinear golden section(s) and 0 connected golden chain(s) in this model:

  • Section (A1, A2, A3): long \(|A1A2|^2 = 3\) vs. short \(|A2A3|^2 = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\)

Historical Commentary & Attribution

Concrete artist Jo Niemeyer designed minimal sculptures and reliefs built from this three-bar linkage, realizing \(\phi\) mechanically with three identical rods.