GC-07.02 · John Molokach’s Golden Rhombus from Shifted 1×3 Rectangles

Discoverer(s): John Molokach | Date: 2014 | Field: \(\mathbb{Q}(\phi)\) | Loci: 2 (Lemoine \(S = 6\))

Stack two \(3 \times 1\) rectangles RECT1 and RECT2 shifted by one unit around center O, draw unit circle cO, and draw the two tangents to cO perpendicular to outer diagonal R1R6. Together with the horizontal outer edges, they bound a golden rhombus ABCD (\(|AC|/|BD| = \phi\)) with \(|T1K|/|KA| = \phi\).
GC-07.02 — John Molokach's Golden Rhombus from Shifted 1×3 Rectangles cO: x**2 + y**2 - 1 = 0 BA: y - 1 = 0 Golden major T1K = 1 Golden minor KA = -1/2 + sqrt(5)/2 T2 = (0, -1) R1 = (-2, 1) R6 = (2, -1) R7 = (-1, -1) B = (1/2 - sqrt(5)/2, 1) C = (-sqrt(5)/2 - 1/2, -1) D = (-1/2 + sqrt(5)/2, -1) O = (0, 0) T1 = (0, 1) K = (1, 1) A = (1/2 + sqrt(5)/2, 1) T2 R1 R6 R7 B C D O T1 K A

Given & Constructive Steps

Given: Center O \(= (0,0)\) and rectangle corner K \(= (1,1)\) of shifted \(3 \times 1\) rectangles RECT1 and RECT2.

step Place \(3 \times 1\) rectangles \([-2,1]\times[0,1]\) and \([-1,2]\times[-1,0]\) and draw unit circle cO tangent to \(y = \pm 1\) at T1 and T2.
step Draw long diagonal R1R6 from \((-2,1)\) to \((2,-1)\) and erect tangents to cO where R1R6 crosses cO.
step Intersect the four tangents to obtain golden rhombus ABCD with top vertices B \(= (-1/\phi, 1)\) and A \(= (\phi, 1)\).

Exact Symbolic Proof & Certificate

Diagonal R1R6 has direction \((2, -1)\), meeting unit circle cO at \(\pm(\tfrac{2}{\sqrt{5}}, -\tfrac{1}{\sqrt{5}})\). The tangents there are \(2x - y = \pm\sqrt{5}\), which intersect \(y = 1\) at \(x_A = \frac{\sqrt{5}+1}{2} = \phi\) and \(x_B = -\frac{\sqrt{5}-1}{2} = -1/\phi\). Along \(y = 1\), points \(B(-1/\phi,1), T1(0,1), K(1,1), A(\phi,1)\) form a 4-point golden chain, while \(Q(A,C)/Q(B,D) = (4\phi^2 + 4)/(4\phi^{-2} + 4) = \phi^2\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

T1K \(/\) KA \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (T1, K, A)

Exact Segment Lengths

\(|T1K| = 1\), \(|KA| = - \frac{1}{2} + \frac{\sqrt{5}}{2}\)

Rational Trigonometry Quadrances

\(Q(T1, K) = 1\), \(Q(K, A) = \frac{3}{2} - \frac{\sqrt{5}}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

\(\mathbb{Q}(\phi)\)

Géométrographie Loci

2 given points · 1 lines · 1 circles · 3 polygons (elementary loci \(= 2\), Lemoine \(S = 6\))

Automated Divine Scan

geometor.divine.golden detected 6 exact collinear golden section(s) and 2 connected golden chain(s) in this model:

  • Section (R1, B, A): long \(|BA|^2 = 5\) vs. short \(|R1B|^2 = \frac{15}{2} - \frac{5 \sqrt{5}}{2}\)

  • Section (B, T1, K): long \(|T1K|^2 = 1\) vs. short \(|BT1|^2 = \frac{3}{2} - \frac{\sqrt{5}}{2}\)

  • Section (T1, K, A): long \(|T1K|^2 = 1\) vs. short \(|KA|^2 = \frac{3}{2} - \frac{\sqrt{5}}{2}\)

  • Section (C, R7, T2): long \(|R7T2|^2 = 1\) vs. short \(|CR7|^2 = \frac{3}{2} - \frac{\sqrt{5}}{2}\)

  • Section (C, D, R6): long \(|CD|^2 = 5\) vs. short \(|DR6|^2 = \frac{15}{2} - \frac{5 \sqrt{5}}{2}\)

  • Section (R7, T2, D): long \(|R7T2|^2 = 1\) vs. short \(|T2D|^2 = \frac{3}{2} - \frac{\sqrt{5}}{2}\)

Connected Golden Chains:

  • Chain (B — T1 — K — A) spanning 2 overlapping golden sections (progression < >)

  • Chain (C — R7 — T2 — D) spanning 2 overlapping golden sections (progression < >)

Historical Commentary & Attribution

A golden rhombus is the planar cross-section of Kepler’s rhombic triacontahedron; Molokach’s construction produces both \(\phi\), \(1/\phi\), \(1/\phi^2\), and the golden rhombus itself in a single stroke.