GC-02.03 · Bùi Quang Tuân’s Side-and-Altitude Equilateral Triangles

Discoverer(s): Bùi Quang Tuân | Date: 2011 | Field: \(\mathbb{Q}(\sqrt{3}, \phi)\) | Loci: 2 (Lemoine \(S = 6\))

Let M be the midpoint of base BC of equilateral triangle ABC, and erect a second equilateral triangle AMN on altitude AM. Then circle cBA centered at B through A cuts side MN at P in the golden ratio \(|MN|/|MP| = |MP|/|PN| = \phi\).
GC-02.03 — Bùi Quang Tuân's Side-and-Altitude Equilateral Triangles cBA: y**2 + (x + 1)**2 - 4 = 0 MN: -sqrt(3)*x/2 + 3*y/2 = 0 Golden major MP = (-sqrt(3) + sqrt(15))/2 Golden minor PN = (-sqrt(15) + 3*sqrt(3))/2 A = (0, sqrt(3)) B = (-1, 0) C = (1, 0) M = (0, 0) N = (3/2, sqrt(3)/2) P = (-3/4 + 3*sqrt(5)/4, -sqrt(3)/4 + sqrt(15)/4) A B C M N P

Given & Constructive Steps

Given: Equilateral triangle ABC with B \(= (-1, 0)\), C \(= (1, 0)\), midpoint M \(= (0, 0)\), and apex A \(= (0, \sqrt{3})\).

step Construct equilateral triangle ABC of side \(2\) and bisect BC at M.
step On altitude AM of length \(\sqrt{3}\), construct equilateral triangle AMN with N \(= (\tfrac{3}{2}, \tfrac{\sqrt{3}}{2})\).
step Draw circle cBA centered at B through A (radius \(2\)), cutting MN at P.

Exact Symbolic Proof & Certificate

Ray MN makes angle \(30^\circ\) with \(MC\), so \(\angle BMP = 150^\circ\). Setting \(t = |MP|\) in \(\triangle BMP\) with \(|BM| = 1\) and \(|BP| = |BA| = 2\) yields \(t^2 - 2t\cos 150^\circ + 1 = 4 \implies t^2 + \sqrt{3}\,t - 3 = 0\), whose positive root is \(t = \frac{\sqrt{15} - \sqrt{3}}{2} = \frac{\sqrt{3}}{\phi}\). Since \(|MN| = \sqrt{3}\), we obtain \(|PN| = \sqrt{3} - \sqrt{3}/\phi = \sqrt{3}/\phi^2\) 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

\(|MP| = \frac{- \sqrt{3} + \sqrt{15}}{2}\), \(|PN| = \frac{- \sqrt{15} + 3 \sqrt{3}}{2}\)

Rational Trigonometry Quadrances

\(Q(M, P) = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\), \(Q(P, N) = \frac{21}{2} - \frac{9 \sqrt{5}}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

\(\mathbb{Q}(\sqrt{3}, \phi)\)

Géométrographie Loci

2 given points · 1 lines · 1 circles · 2 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 (M, P, N): long \(|MP|^2 = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\) vs. short \(|PN|^2 = \frac{21}{2} - \frac{9 \sqrt{5}}{2}\)

Historical Commentary & Attribution

Contributed by Bùi Quang Tuân (Cut-the-Knot), along with a companion variant (Cut-the-Knot II), this figure exploits the \(150^\circ\) cosine identity \(x^2 + \sqrt{3}\,x - 3 = 0\).