GC-01.07 · Bùi Quang Tuân’s Five-Step Three-Circle Secant

Discoverer(s): Bùi Quang Tuân | Date: 2014 | Field: \(\mathbb{Q}(\sqrt{5}, \sqrt{7})\) | Loci: 5 (Lemoine \(S = 15\))

On line L through O, draw circle cO meeting L at A and B, circle cBA meeting L at C, and circle cCO meeting cO above L at M and cBA below L at N. Then chord MN cuts cO at P with \(|MN|/|MP| = |MP|/|PN| = \phi\).
GC-01.07 — Bùi Quang Tuân's Five-Step Three-Circle Secant cO: x**2 + y**2 - 1 = 0 cBA: y**2 + (x + 1)**2 - 4 = 0 cCO: y**2 + (x + 3)**2 - 9 = 0 L: y = 0 MN: x*(sqrt(35)/6 + 3*sqrt(7)/4) - 7*y/12 + sqrt(7)/8 + sqrt(35)/8 = 0 Golden major MP = sqrt(7)*sqrt(-sqrt(2)*sqrt(sqrt(5) + 3) + 4 + sqrt(10)*sqrt(sqrt(5) + 3))/4 Golden minor PN = sqrt(7)*sqrt(-5*sqrt(2)*sqrt(sqrt(5) + 3) + sqrt(10)*sqrt(sqrt(5) + 3) + 12)/4 A = (1, 0) C = (-3, 0) O = (0, 0) B = (-1, 0) M = (-1/6, sqrt(35)/6) N = (-3/4, -3*sqrt(7)/4) P = (-7*sqrt(2)*sqrt(sqrt(5) + 3)/32 - 7*sqrt(5)/48 - 1/48 + 7*sqrt(10)*sqrt(sqrt(5) + 3)/96, -17*sqrt(14)*sqrt(sqrt(5) + 3)/96 - sqrt(7)/48 + sqrt(35)/48 + sqrt(70)*sqrt(sqrt(5) + 3)/32) A C O B M N P

Given & Constructive Steps

Given: Center O \(= (0, 0)\) and unit point B \(= (-1, 0)\) defining carrier line L.

step Draw line L through O and B, and unit circle cO meeting L again at A \(= (1, 0)\).
step Draw circle cBA of radius \(2\) centered at B through A, meeting L beyond B at C \(= (-3, 0)\).
step Draw circle cCO of radius \(3\) centered at C through O, cutting cO above L at M \(= (-\tfrac{1}{6}, \tfrac{\sqrt{35}}{6})\) and cBA below L at N \(= (-\tfrac{3}{4}, -\tfrac{3\sqrt{7}}{4})\).
step Join MN, intersecting unit circle cO a second time at P.

Exact Symbolic Proof & Certificate

Direct quadrance computation between M and N yields \(Q(M, N) = (\tfrac{7}{12})^2 + (\tfrac{\sqrt{35}}{6} + \tfrac{3\sqrt{7}}{4})^2 = \tfrac{21}{4} + \tfrac{7\sqrt{5}}{4} = \tfrac{7}{2}\phi^2\). Because O, M, and N all lie on radius-\(3\) circle cCO while M and P lie on unit circle cO, inscribed-angle similarity with diameter endpoint \((-6, 0)\) of cCO gives \(Q(M, P) = \tfrac{7}{2}\) and \(Q(P, N) = \tfrac{7}{2}\phi^{-2}\). Hence \(Q(M, P) / Q(P, N) = \phi^2\) and \(|MP|/|PN| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

MP \(/\) PN \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (M, P, N)

Exact Segment Lengths

\(|MP| = \frac{\sqrt{7} \sqrt{- \sqrt{2} \sqrt{\sqrt{5} + 3} + 4 + \sqrt{10} \sqrt{\sqrt{5} + 3}}}{4}\), \(|PN| = \frac{\sqrt{7} \sqrt{- 5 \sqrt{2} \sqrt{\sqrt{5} + 3} + \sqrt{10} \sqrt{\sqrt{5} + 3} + 12}}{4}\)

Rational Trigonometry Quadrances

\(Q(M, P) = \frac{7}{4} + \frac{- 7 \sqrt{2} \sqrt{\sqrt{5} + 3} + 7 \sqrt{10} \sqrt{\sqrt{5} + 3}}{16}\), \(Q(P, N) = \frac{- 35 \sqrt{2} \sqrt{\sqrt{5} + 3} + 7 \sqrt{10} \sqrt{\sqrt{5} + 3}}{16} + \frac{21}{4}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

\(\mathbb{Q}(\sqrt{5}, \sqrt{7})\)

Géométrographie Loci

2 given points · 2 lines · 3 circles · 0 polygons (elementary loci \(= 5\), Lemoine \(S = 15\))

Automated Divine Scan

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

  • Section (N, P, M): long \(|PM|^2 = \frac{7}{4} + \frac{- 7 \sqrt{2} \sqrt{\sqrt{5} + 3} + 7 \sqrt{10} \sqrt{\sqrt{5} + 3}}{16}\) vs. short \(|NP|^2 = \frac{- 35 \sqrt{2} \sqrt{\sqrt{5} + 3} + 7 \sqrt{10} \sqrt{\sqrt{5} + 3}}{16} + \frac{21}{4}\)

Historical Commentary & Attribution

Devised by Bùi Quang Tuân (Cut-the-Knot), this five-step construction starts from an arbitrary radius on line L rather than a pre-specified unit segment, generating \(\phi\) inside the biquadratic field \(\mathbb{Q}(\sqrt{5}, \sqrt{7})\).