GC-05.03 · Prasolov & Bùi Quang Tuân’s Equi-Area Irregular Pentagon

Discoverer(s): Viktor Vasil'evich Prasolov, Bùi Quang Tuân | Date: 2006 | Field: \(\mathbb{Q}(\phi)\) | Loci: 3 (Lemoine \(S = 9\))

Let ABCDE be any convex pentagon—regular or irregular—whose five peripheral triangles \(\triangle ABE, \triangle ABC, \triangle BCD, \triangle CDE, \triangle DEA\) have equal area. Then every diagonal is parallel to the opposite side and is divided by the crossing diagonals in the golden ratio, e.g. \(|BM|/|MD| = \phi\).
GC-05.03 — Prasolov & Bùi Quang Tuân's Equi-Area Irregular Pentagon BD: -2*x + y + 2 + 2*sqrt(5) = 0 CE: y - sqrt(5) + 1 = 0 AC: x*(1 - sqrt(5)) + y*(5/2 + 3*sqrt(5)/2) = 0 Golden major BM = 5/2 - sqrt(5)/2 Golden minor MD = -5/2 + 3*sqrt(5)/2 C = (5/2 + 3*sqrt(5)/2, -1 + sqrt(5)) E = (-1/2 + sqrt(5)/2, -1 + sqrt(5)) N = (sqrt(5)/2 + 5/2, 3 - sqrt(5)) A = (0, 0) B = (1 + sqrt(5), 0) D = (2 + sqrt(5), 2) M = (1/2 + 3*sqrt(5)/2, -1 + sqrt(5)) C E N A B D M

Given & Constructive Steps

Given: Sheared irregular convex pentagon ABCDE over \(\mathbb{Q}(\phi)\) with A \(= (0,0)\) and B \(= (2\phi, 0)\), each peripheral triangle having exact area \(2\).

step Place irregular vertices A \(= (0,0)\), B \(= (2\phi, 0)\), C \(= (3\phi+1, 2/\phi)\), D \(= (2\phi+1, 2)\), and E \(= (1/\phi, 2/\phi)\).
step Draw diagonals BD, CE, and AC.
step Mark intersections M \(= BD \cap CE\) and N \(= BD \cap AC\).

Exact Symbolic Proof & Certificate

Equal areas \([ABE] = [ABC]\) with shared base AB force \(CE \parallel AB\); cyclically, every diagonal is parallel to its opposite side, so \(ABME\) is a parallelogram with \(|EM| = |AB|\). Set \(x = |BM|/|MD|\). Then \([ABE] = [BM E]\) (half of parallelogram \(ABME\)), while \([CDE] = [BCD]\) implies \(|CE|/|EM| = 1 + 1/x\) and \(x = 1 + 1/x\), forcing \(x = \phi\) purely from affine area invariance!

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

BM \(/\) MD \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (B, M, D)

Exact Segment Lengths

\(|BM| = \frac{5}{2} - \frac{\sqrt{5}}{2}\), \(|MD| = - \frac{5}{2} + \frac{3 \sqrt{5}}{2}\)

Rational Trigonometry Quadrances

\(Q(B, M) = \frac{15}{2} - \frac{5 \sqrt{5}}{2}\), \(Q(M, D) = \frac{35}{2} - \frac{15 \sqrt{5}}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

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

Automated Divine Scan

geometor.divine.golden detected 6 exact collinear golden section(s) and 1 connected golden chain(s) in this model:

  • Section (B, N, M): long \(|BN|^2 = \frac{35}{2} - \frac{15 \sqrt{5}}{2}\) vs. short \(|NM|^2 = 45 - 20 \sqrt{5}\)

  • Section (B, N, D): long \(|ND|^2 = \frac{15}{2} - \frac{5 \sqrt{5}}{2}\) vs. short \(|BN|^2 = \frac{35}{2} - \frac{15 \sqrt{5}}{2}\)

  • Section (B, M, D): long \(|BM|^2 = \frac{15}{2} - \frac{5 \sqrt{5}}{2}\) vs. short \(|MD|^2 = \frac{35}{2} - \frac{15 \sqrt{5}}{2}\)

  • Section (N, M, D): long \(|MD|^2 = \frac{35}{2} - \frac{15 \sqrt{5}}{2}\) vs. short \(|NM|^2 = 45 - 20 \sqrt{5}\)

  • Section (E, M, C): long \(|EM|^2 = 2 \sqrt{5} + 6\) vs. short \(|MC|^2 = 4\)

  • Section (A, N, C): long \(|AN|^2 = \frac{43}{2} - \frac{7 \sqrt{5}}{2}\) vs. short \(|NC|^2 = 41 - 16 \sqrt{5}\)

Connected Golden Chains:

  • Chain (B — N — M — D) spanning 2 overlapping golden sections (progression > <)

Historical Commentary & Attribution

Because parallelisms and length ratios on parallel lines are affine invariants, every affine image of a regular pentagon retains all five golden pentagram chains while losing all metric symmetry—and lives entirely over \(\mathbb{Q}(\phi)\) without \(\sin 36^\circ\).