GC-01.02 · Right-Triangle Hypotenuse Transfer (1 : 2 : √5)

Discoverer(s): Euclid of Alexandria | Date: Classical | Field: \(\mathbb{Q}(\phi)\) | Loci: 4 (Lemoine \(S = 12\))

On segment AB of length \(2\), erect perpendicular leg BC of length \(1 = \tfrac{1}{2}|AB|\). Circle cCB cuts hypotenuse AC at D (and ray AC at E), and circle cAD transfers AD onto AB at G, dividing AB in the golden ratio \(|AG|/|GB| = \phi\).
GC-01.02 — Right-Triangle Hypotenuse Transfer (1 : 2 : √5) cBM: y**2 + (x - 2)**2 - 1 = 0 cCB: (x - 2)**2 + (y - 1)**2 - 1 = 0 cAD: x**2 + y**2 - (-2 + 2*sqrt(5)/5)**2 - (-1 + sqrt(5)/5)**2 = 0 AC: -x + 2*y = 0 Golden major AG = -1 + sqrt(5) Golden minor GB = 3 - sqrt(5) M = (1, 0) C = (2, 1) D = (2 - 2*sqrt(5)/5, 1 - sqrt(5)/5) E = (2*sqrt(5)/5 + 2, sqrt(5)/5 + 1) A = (0, 0) B = (2, 0) G = (-1 + sqrt(5), 0) M C D E A B G

Given & Constructive Steps

Given: Segment AB from A \(= (0, 0)\) to B \(= (2, 0)\) with midpoint M \(= (1, 0)\) and perpendicular leg endpoint C \(= (2, 1)\).

step Form the \(1 : 2 : \sqrt{5}\) right triangle ABC with right angle at B and line AC.
step Draw circle cCB centered at C through B (radius \(1\)), intersecting AC internally at D and externally at E.
step Draw circle cAD centered at A through D, meeting segment AB at G.

Exact Symbolic Proof & Certificate

Since \(|AB| = 2\) and \(|BC| = 1\), hypotenuse AC has length \(\sqrt{5}\). Circle cCB subtracts and adds unit radius \(1\) along line AC, yielding \(|AD| = \sqrt{5} - 1 = 2/\phi\) and \(|AE| = \sqrt{5} + 1 = 2\phi\). Along AC, \(|AE|/|DE| = (2\phi)/2 = \phi\), while transferring \(|AD|\) to AG on AB produces \(|AG|/|GB| = (\sqrt{5} - 1)/(3 - \sqrt{5}) = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

AG \(/\) GB \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (A, G, B)

Exact Segment Lengths

\(|AG| = -1 + \sqrt{5}\), \(|GB| = 3 - \sqrt{5}\)

Rational Trigonometry Quadrances

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

Algebraic Number Field

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

Géométrographie Loci

2 given points · 1 lines · 3 circles · 1 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 (A, D, E): long \(|DE|^2 = 4\) vs. short \(|AD|^2 = 6 - 2 \sqrt{5}\)

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

Historical Commentary & Attribution

Cataloged at the opening of Bogomolny's Golden Ratio overview and popularized by H. E. Huntley (Huntley 1970), this streamlined distillation of Euclid II.11 simultaneously exhibits both \(2/\phi\) (at D) and \(2\phi\) (at E) along a single secant line AC.