GC-01.05 · Mascheroni Five-Circle Compass-Only Construction

Discoverer(s): Kurt Hofstetter | Date: 2002 | Field: \(\mathbb{Q}(\sqrt{3}, \phi)\) | Loci: 5 (Lemoine \(S = 15\))

Using a compass alone and zero straightedge lines, five circles constructed from given points A and B generate four collinear points Y, C, D, X such that \(|CX|/|CD| = |CD|/|DX| = \phi\).
GC-01.05 — Mascheroni Five-Circle Compass-Only Construction c1: x**2 + y**2 - 1 = 0 c2: y**2 + (x - 1)**2 - 1 = 0 c3: (x - 1/2)**2 + (y - sqrt(3)/2)**2 - 3 = 0 c4: y**2 + (x - 1)**2 - 4 = 0 c5: x**2 + y**2 - 4 = 0 Golden major CD = sqrt(3) Golden minor DX = (-sqrt(3) + sqrt(15))/2 E = (-1, 0) F = (2, 0) Y = (1/2, sqrt(15)/2) A = (0, 0) B = (1, 0) C = (1/2, sqrt(3)/2) D = (1/2, -sqrt(3)/2) X = (1/2, -sqrt(15)/2) E F Y A B C D X

Given & Constructive Steps

Given: Two distinct points A \(= (0, 0)\) and B \(= (1, 0)\) with no lines drawn.

step Draw unit circles c1 \(= A(B)\) and c2 \(= B(A)\), intersecting at C and D.
step Draw circle c3 \(= C(D)\) of radius \(\sqrt{3}\); without a ruler, c3 meets c1 again at E \(= (-1, 0)\) and c2 again at F \(= (2, 0)\).
step Draw radius-\(2\) circles c4 \(= B(E)\) and c5 \(= A(F)\), intersecting at X \(= (\tfrac{1}{2}, -\tfrac{\sqrt{15}}{2})\) and Y \(= (\tfrac{1}{2}, \tfrac{\sqrt{15}}{2})\).

Exact Symbolic Proof & Certificate

For C \(= (\tfrac{1}{2}, \tfrac{\sqrt{3}}{2})\), the quadrance to \((-1, 0)\) is \((-\tfrac{3}{2})^2 + (\tfrac{\sqrt{3}}{2})^2 = 3 = Q(C, D)\), confirming that E and symmetrically F lie on c3 without drawing line \(AB\). Circles c4 and c5 both have radius \(2\) centered at B and A, so their intersections X and Y lie on the perpendicular bisector \(x = \tfrac{1}{2}\) alongside C and D, with ordinates \(\pm\tfrac{\sqrt{15}}{2}\). Thus \(|CD| = \sqrt{3}\) and \(|DX| = \tfrac{\sqrt{15} - \sqrt{3}}{2} = \sqrt{3}/\phi\), proving \(|CD|/|DX| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

CD \(/\) DX \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (C, D, X)

Exact Segment Lengths

\(|CD| = \sqrt{3}\), \(|DX| = \frac{- \sqrt{3} + \sqrt{15}}{2}\)

Rational Trigonometry Quadrances

\(Q(C, D) = 3\), \(Q(D, X) = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

2 given points · 0 lines · 5 circles · 0 polygons (elementary loci \(= 5\), Lemoine \(S = 15\))

Automated Divine Scan

geometor.divine.golden detected 2 exact collinear golden section(s) and 1 connected golden chain(s) in this model:

  • Section (X, D, C): long \(|DC|^2 = 3\) vs. short \(|XD|^2 = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\)

  • Section (D, C, Y): long \(|DC|^2 = 3\) vs. short \(|CY|^2 = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\)

Connected Golden Chains:

  • Chain (X — D — C — Y) spanning 2 overlapping golden sections (progression < >)

Historical Commentary & Attribution

In accordance with the Mohr–Mascheroni theorem, every ruler-and-compass point construction can be executed by compass alone. Kurt Hofstetter’s five-circle construction (Hofstetter 2002, Cut-the-Knot) achieves the golden section in just five arcs—and our geometor.divine.golden collinear point scanner detects the entire \((Y, C, D, X)\) golden chain automatically despite zero lines in the model.