GC-05.04 · Tran Quang Hung’s Shared-Base Regular Pentagon & Hexagon

Discoverer(s): Tran Quang Hung | Date: 2015 | Field: \(\mathbb{Q}(\sqrt{3}, \phi, \sqrt{10 - 2\sqrt{5}})\) | Loci: 4 (Lemoine \(S = 12\))

Erect regular hexagon ABCDEF and regular pentagon AMNPF on the same side AF. The perpendiculars MQ and PR to diagonal MP at M and P intersect upper hexagon sides BC and DE at Q and R such that \(|CQ|/|QB| = |DR|/|RE| = \phi\).
GC-05.04 — Tran Quang Hung's Shared-Base Regular Pentagon & Hexagon BC: -sqrt(3)*x + y - 3*sqrt(3) = 0 DE: sqrt(3)*x + y - 3*sqrt(3) = 0 MQ: x + 1/2 + sqrt(5)/2 = 0 PR: x - sqrt(5)/2 - 1/2 = 0 Golden major CQ = -1 + sqrt(5) Golden minor QB = 3 - sqrt(5) E = (2, sqrt(3)) D = (1, 2*sqrt(3)) M = (-sqrt(5)/2 - 1/2, sqrt(2)*sqrt(5 - sqrt(5))/4 + sqrt(10)*sqrt(5 - sqrt(5))/4) N = (0, sqrt(10)*sqrt(5 - sqrt(5))/4 + 3*sqrt(2)*sqrt(5 - sqrt(5))/4) P = (1/2 + sqrt(5)/2, sqrt(2)*sqrt(5 - sqrt(5))/4 + sqrt(10)*sqrt(5 - sqrt(5))/4) R = (1/2 + sqrt(5)/2, -sqrt(15)/2 + 5*sqrt(3)/2) A = (-1, 0) F = (1, 0) C = (-1, 2*sqrt(3)) B = (-2, sqrt(3)) Q = (-sqrt(5)/2 - 1/2, -sqrt(15)/2 + 5*sqrt(3)/2) E D M N P R A F C B Q

Given & Constructive Steps

Given: Shared horizontal base A \(= (-1,0)\), F \(= (1,0)\) of length \(2\).

step Construct regular hexagon ABCDEF of side \(2\) above AF, with upper-left side B \(= (-2, \sqrt{3})\) to C \(= (-1, 2\sqrt{3})\).
step Construct regular pentagon AMNPF of side \(2\) on the same side of AF, with horizontal diagonal MP.
step Erect perpendiculars MQ and PR to MP at M and P, cutting BC at Q and DE at R.

Exact Symbolic Proof & Certificate

For base \(|AF| = 2\), diagonal MP of regular pentagon AMNPF has length \(2\phi\), so M has abscissa \(x_M = -\phi\) while B and C have abscissas \(x_B = -2\) and \(x_C = -1\). Because MQ is vertical (\(x = -\phi\)), linear projection onto the \(x\)-axis yields \(|CQ|/|QB| = (-1 - (-\phi))/(-\phi - (-2)) = (\phi - 1)/(2 - \phi) = (1/\phi)/(1/\phi^2) = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

CQ \(/\) QB \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (C, Q, B)

Exact Segment Lengths

\(|CQ| = -1 + \sqrt{5}\), \(|QB| = 3 - \sqrt{5}\)

Rational Trigonometry Quadrances

\(Q(C, Q) = 6 - 2 \sqrt{5}\), \(Q(Q, B) = 14 - 6 \sqrt{5}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

2 given points · 4 lines · 0 circles · 2 polygons (elementary loci \(= 4\), Lemoine \(S = 12\))

Automated Divine Scan

geometor.divine.golden detected 2 exact collinear golden section(s) and 0 connected golden chain(s) in this model:

  • Section (B, Q, C): long \(|QC|^2 = 6 - 2 \sqrt{5}\) vs. short \(|BQ|^2 = 14 - 6 \sqrt{5}\)

  • Section (D, R, E): long \(|DR|^2 = 6 - 2 \sqrt{5}\) vs. short \(|RE|^2 = 14 - 6 \sqrt{5}\)

Historical Commentary & Attribution

Projected onto the horizontal axis, the hexagon’s lateral overhang has width \(1\) while the pentagon’s lateral overhang has width \(\phi - 1 = 1/\phi\), immediately sectioning the overhang in extreme and mean ratio.