GC-05.02 · Dao Thanh Oai’s Regular Pentagon Edge Perpendicular

Discoverer(s): Dao Thanh Oai, Tran Quang Hung | Date: 2015 | Field: \(\mathbb{Q}(\phi, \sqrt{10 - 2\sqrt{5}})\) | Loci: 3 (Lemoine \(S = 9\))

In regular pentagon ABCDE, erect line EF perpendicular to side DE at vertex E. Line EF intersects opposite non-adjacent side AB internally at F such that \(|FB|/|AF| = |AB|/|FB| = \phi\).
GC-05.02 — Dao Thanh Oai's Regular Pentagon Edge Perpendicular AB: y = 0 DE: x*sqrt(10 - 2*sqrt(5))/2 + y*(-sqrt(5)/2 - 1/2) + sqrt(5 - sqrt(5))*(sqrt(2) + sqrt(10)/2) = 0 EF: x*(1/2 + sqrt(5)/2) + y*sqrt(10 - 2*sqrt(5))/2 - sqrt(5)/2 + 3/2 = 0 Golden major FB = -1 + sqrt(5) Golden minor AF = 3 - sqrt(5) C = (1/2 + sqrt(5)/2, sqrt(2)*sqrt(5 - sqrt(5))/4 + sqrt(10)*sqrt(5 - sqrt(5))/4) D = (0, sqrt(10)*sqrt(5 - sqrt(5))/4 + 3*sqrt(2)*sqrt(5 - sqrt(5))/4) E = (-sqrt(5)/2 - 1/2, sqrt(2)*sqrt(5 - sqrt(5))/4 + sqrt(10)*sqrt(5 - sqrt(5))/4) A = (-1, 0) B = (1, 0) F = (2 - sqrt(5), 0) C D E A B F

Given & Constructive Steps

Given: Regular pentagon ABCDE of side \(2\) on base A \(= (-1,0)\), B \(= (1,0)\).

step Construct regular pentagon ABCDE counterclockwise on base AB.
step Erect line EF perpendicular to upper-left side DE at vertex E.
step Mark intersection F of EF with base AB.

Exact Symbolic Proof & Certificate

Interior angle \(\angle DEA = 108^\circ\), while \(\angle DEF = 90^\circ\), leaving \(\angle AEF = 18^\circ\). Since \(\angle EAF = 108^\circ\), triangle \(\triangle EAF\) has \(\angle EFA = 54^\circ\). By the Law of Sines on \(\triangle EAF\) with \(|EA| = 2\), \(|AF| = \frac{2\sin 18^\circ}{\sin 54^\circ} = \frac{2(1/(2\phi))}{\phi/2} = \frac{2}{\phi^2}\), whence \(|FB| = 2 - \frac{2}{\phi^2} = \frac{2}{\phi}\) and \(|FB|/|AF| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

FB \(/\) AF \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (A, F, B)

Exact Segment Lengths

\(|FB| = -1 + \sqrt{5}\), \(|AF| = 3 - \sqrt{5}\)

Rational Trigonometry Quadrances

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

Algebraic Number Field

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

Géométrographie Loci

2 given points · 3 lines · 0 circles · 1 polygons (elementary loci \(= 3\), Lemoine \(S = 9\))

Automated Divine Scan

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

  • Section (A, F, B): long \(|FB|^2 = 6 - 2 \sqrt{5}\) vs. short \(|AF|^2 = 14 - 6 \sqrt{5}\)

Historical Commentary & Attribution

Tran Quang Hung discovered a dual regular-pentagon division via the incircle and radius midpoint (AnotherGoldenRatioInRegularPentagon.shtml), reflecting the ubiquitous \(18^\circ\text{-}54^\circ\text{-}72^\circ\) cosine ratios.