GC-06.05 · Jerzy Kocik’s Six-Circle Apollonian Golden Window (2010)

Discoverer(s): Jerzy Kocik | Date: 2010 | Field: \(\mathbb{Q}(\phi)\) | Loci: 8 (Lemoine \(S = 24\))

Place two unit-diameter circles cC1, cC2 tangent at O and two flanking congruent circles cC3, cC4 inside outer circle cO. Requiring symmetric upper and lower circles cU1, cU2 to be simultaneously tangent to all five circles uniquely forces the flanking circles to have exact radius \(\phi\), outer radius \(\phi^3\), and \(|OC4|/|C4A4| = \phi\).
GC-06.05 — Jerzy Kocik's Six-Circle Apollonian Golden Window (2010) cC1: y**2 + (x + 1/2)**2 - 1/4 = 0 cC2: y**2 + (x - 1/2)**2 - 1/4 = 0 cC3: y**2 + (x + sqrt(5)/2 + 3/2)**2 - (1/2 + sqrt(5)/2)**2 = 0 cC4: y**2 + (x - 3/2 - sqrt(5)/2)**2 - (1/2 + sqrt(5)/2)**2 = 0 cO: x**2 + y**2 - (2 + sqrt(5))**2 = 0 cU1: x**2 + (y - 3*sqrt(5)/5 - 1)**2 - (2*sqrt(5)/5 + 1)**2 = 0 cU2: x**2 + (y + 1 + 3*sqrt(5)/5)**2 - (2*sqrt(5)/5 + 1)**2 = 0 A3A4: y = 0 Golden major OC4 = sqrt(5)/2 + 3/2 Golden minor C4A4 = 1/2 + sqrt(5)/2 P1 = (-1, 0) C3 = (-3/2 - sqrt(5)/2, 0) A3 = (-sqrt(5) - 2, 0) U1 = (0, 1 + 3*sqrt(5)/5) U2 = (0, -3*sqrt(5)/5 - 1) T1 = (0, 2 + sqrt(5)) T2 = (0, -sqrt(5) - 2) P2 = (1, 0) O = (0, 0) C4 = (sqrt(5)/2 + 3/2, 0) A4 = (2 + sqrt(5), 0) P1 C3 A3 U1 U2 T1 T2 P2 O C4 A4

Given & Constructive Steps

Given: Origin O \(= (0,0)\) and tangency point P2 \(= (1,0)\) defining central unit-diameter circles cC1 and cC2.

step Construct central circles cC1 and cC2 of radius \(\tfrac{1}{2}\) touching at O and reaching P1, P2 \(= (\pm 1, 0)\).
step Construct flanking circles cC3 and cC4 of radius \(\phi\) centered at C3, C4 \(= (\pm \phi^2, 0)\) inside outer circle cO of radius \(\phi^3\).
step Inscribe symmetric circles cU1 and cU2 of radius \(\phi^3/\sqrt{5}\) centered at \((0, \pm 2\phi^2/\sqrt{5})\), tangent to all five circles.

Exact Symbolic Proof & Certificate

Let flanking circles have unknown radius \(R\), so outer circle cO has radius \(1 + 2R\). For upper circle cU1 of radius \(r_u\) at \((0, 1 + 2R - r_u)\), tangency to cC2 gives \((1/2)^2 + (1 + 2R - r_u)^2 = (r_u + 1/2)^2\), while tangency to cC4 gives \((1 + R)^2 + (1 + 2R - r_u)^2 = (r_u + R)^2\). Eliminating \(r_u\) yields \(R^2 - R - 1 = 0\), uniquely forcing \(R = \phi\)!

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

OC4 \(/\) C4A4 \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (O, C4, A4)

Exact Segment Lengths

\(|OC4| = \frac{\sqrt{5}}{2} + \frac{3}{2}\), \(|C4A4| = \frac{1}{2} + \frac{\sqrt{5}}{2}\)

Rational Trigonometry Quadrances

\(Q(O, C4) = \frac{3 \sqrt{5}}{2} + \frac{7}{2}\), \(Q(C4, A4) = \frac{\sqrt{5}}{2} + \frac{3}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

2 given points · 1 lines · 7 circles · 0 polygons (elementary loci \(= 8\), Lemoine \(S = 24\))

Automated Divine Scan

geometor.divine.golden detected 10 exact collinear golden section(s) and 3 connected golden chain(s) in this model:

  • Section (A3, C3, O): long \(|C3O|^2 = \frac{3 \sqrt{5}}{2} + \frac{7}{2}\) vs. short \(|A3C3|^2 = \frac{\sqrt{5}}{2} + \frac{3}{2}\)

  • Section (A3, P1, P2): long \(|A3P1|^2 = 2 \sqrt{5} + 6\) vs. short \(|P1P2|^2 = 4\)

  • Section (A3, P1, A4): long \(|P1A4|^2 = 6 \sqrt{5} + 14\) vs. short \(|A3P1|^2 = 2 \sqrt{5} + 6\)

  • Section (A3, O, C4): long \(|A3O|^2 = 4 \sqrt{5} + 9\) vs. short \(|OC4|^2 = \frac{3 \sqrt{5}}{2} + \frac{7}{2}\)

  • Section (A3, P2, A4): long \(|A3P2|^2 = 6 \sqrt{5} + 14\) vs. short \(|P2A4|^2 = 2 \sqrt{5} + 6\)

  • Section (C3, P1, O): long \(|C3P1|^2 = \frac{\sqrt{5}}{2} + \frac{3}{2}\) vs. short \(|P1O|^2 = 1\)

  • Section (C3, O, A4): long \(|OA4|^2 = 4 \sqrt{5} + 9\) vs. short \(|C3O|^2 = \frac{3 \sqrt{5}}{2} + \frac{7}{2}\)

  • Section (P1, P2, A4): long \(|P2A4|^2 = 2 \sqrt{5} + 6\) vs. short \(|P1P2|^2 = 4\)

  • Section (O, P2, C4): long \(|P2C4|^2 = \frac{\sqrt{5}}{2} + \frac{3}{2}\) vs. short \(|OP2|^2 = 1\)

  • Section (O, C4, A4): long \(|OC4|^2 = \frac{3 \sqrt{5}}{2} + \frac{7}{2}\) vs. short \(|C4A4|^2 = \frac{\sqrt{5}}{2} + \frac{3}{2}\)

Connected Golden Chains:

  • Chain (A3 — C3 — O — A4) spanning 2 overlapping golden sections (progression < <)

  • Chain (A3 — P1 — P2 — A4) spanning 2 overlapping golden sections (progression > <)

  • Chain (A3 — O — C4 — A4) spanning 2 overlapping golden sections (progression > >)

Historical Commentary & Attribution

Published by Jerzy Kocik in Mathematics Magazine (2010), the Golden Window embeds powers \(1, \phi, \phi^2, \phi^3\), golden rectangles, and Kepler’s right triangle in a single rigid Apollonian circle gasket.