GC-05.06 · Tran Quang Hung’s Fivefold Wu Xing Equilateral Star

Discoverer(s): Tran Quang Hung | Date: 2015 | Field: \(\mathbb{Q}(\sqrt{3}, \phi, \sqrt{10 - 2\sqrt{5}})\) | Loci: 2 (Lemoine \(S = 6\))

Arrange five congruent equilateral triangles OV1W1 … OV5W5 of unit side with fivefold rotational symmetry around shared vertex O. The pentagram star joining outer vertices V1 … V5 cuts the radial sides of the equilateral triangles in the golden ratio, e.g. chord V2V5 cuts OW1 at F with \(|OF|/|FW1| = \phi\).
GC-05.06 — Tran Quang Hung's Fivefold Wu Xing Equilateral Star V2V5: y - sqrt(5)/4 + 1/4 = 0 OW1: -x/2 + sqrt(3)*y/2 = 0 Golden major OF = -1/2 + sqrt(5)/2 Golden minor FW1 = 3/2 - sqrt(5)/2 V2 = (-sqrt(10)*sqrt(5 - sqrt(5))/8 - sqrt(2)*sqrt(5 - sqrt(5))/8, -1/4 + sqrt(5)/4) W2 = (-sqrt(10)*sqrt(5 - sqrt(5))/16 - sqrt(3)/8 - sqrt(2)*sqrt(5 - sqrt(5))/16 + sqrt(15)/8, -1/8 + sqrt(6)*sqrt(5 - sqrt(5))/16 + sqrt(5)/8 + sqrt(30)*sqrt(5 - sqrt(5))/16) V3 = (-sqrt(2)*sqrt(5 - sqrt(5))/4, -sqrt(5)/4 - 1/4) W3 = (-sqrt(15)/8 - sqrt(2)*sqrt(5 - sqrt(5))/8 - sqrt(3)/8, -sqrt(5)/8 - 1/8 + sqrt(6)*sqrt(5 - sqrt(5))/8) V4 = (sqrt(2)*sqrt(5 - sqrt(5))/4, -sqrt(5)/4 - 1/4) W4 = (-sqrt(15)/8 - sqrt(3)/8 + sqrt(2)*sqrt(5 - sqrt(5))/8, -sqrt(6)*sqrt(5 - sqrt(5))/8 - sqrt(5)/8 - 1/8) V5 = (sqrt(2)*sqrt(5 - sqrt(5))/8 + sqrt(10)*sqrt(5 - sqrt(5))/8, -1/4 + sqrt(5)/4) W5 = (-sqrt(3)/8 + sqrt(2)*sqrt(5 - sqrt(5))/16 + sqrt(10)*sqrt(5 - sqrt(5))/16 + sqrt(15)/8, -sqrt(30)*sqrt(5 - sqrt(5))/16 - sqrt(6)*sqrt(5 - sqrt(5))/16 - 1/8 + sqrt(5)/8) V1 = (0, 1) O = (0, 0) W1 = (sqrt(3)/2, 1/2) F = (-sqrt(3)/4 + sqrt(15)/4, -1/4 + sqrt(5)/4) V2 W2 V3 W3 V4 W4 V5 W5 V1 O W1 F

Given & Constructive Steps

Given: Center O \(= (0,0)\) and leading apex V1 \(= (0,1)\) of five \(72^\circ\)-spaced unit equilateral triangles.

step Construct the five \(72^\circ\)-rotated unit equilateral triangles OV1W1 … OV5W5 sharing vertex O.
step Draw pentagram chord V2V5 (along with the remaining four chords of star \(\{5/2\}\)).
step Mark intersection F of horizontal pentagram chord V2V5 with equilateral side OW1.

Exact Symbolic Proof & Certificate

Vertices V2 and V5 lie at angles \(90^\circ \pm 72^\circ\), so chord V2V5 is horizontal at height \(y = \sin 18^\circ = \frac{\sqrt{5}-1}{4} = \frac{1}{2\phi}\). Meanwhile, rotating V1 \(= (0,1)\) clockwise by \(60^\circ\) places W1 at \((\tfrac{\sqrt{3}}{2}, \tfrac{1}{2})\) (\(30^\circ\) above the horizontal). Hence OW1 reaches height \(y_F = \frac{1}{2\phi}\) at distance \(|OF| = \frac{y_F}{\sin 30^\circ} = \frac{1}{\phi}\), leaving \(|FW1| = 1 - \frac{1}{\phi} = \frac{1}{\phi^2}\) and proving \(|OF|/|FW1| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

OF \(/\) FW1 \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (O, F, W1)

Exact Segment Lengths

\(|OF| = - \frac{1}{2} + \frac{\sqrt{5}}{2}\), \(|FW1| = \frac{3}{2} - \frac{\sqrt{5}}{2}\)

Rational Trigonometry Quadrances

\(Q(O, F) = \frac{3}{2} - \frac{\sqrt{5}}{2}\), \(Q(F, W1) = \frac{7}{2} - \frac{3 \sqrt{5}}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

\(\mathbb{Q}(\sqrt{3}, \phi, \sqrt{10 - 2\sqrt{5}})\)

Géométrographie Loci

2 given points · 2 lines · 0 circles · 5 polygons (elementary loci \(= 2\), Lemoine \(S = 6\))

Automated Divine Scan

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

  • Section (O, F, W1): long \(|OF|^2 = \frac{3}{2} - \frac{\sqrt{5}}{2}\) vs. short \(|FW1|^2 = \frac{7}{2} - \frac{3 \sqrt{5}}{2}\)

Historical Commentary & Attribution

This construction fuses \(C_5\) (\(72^\circ\)) and \(C_3\) (\(60^\circ\)) rotational symmetries: their \(12^\circ\) phase difference aligns \(\sin 18^\circ\) with \(\cos 60^\circ = 1/2\).