GC-06.01 · Symbolic Model & Source

Companion symbolic Python construction for GC-06.01 · Bùi Quang Tuân’s Infinite Family of Golden Circle Secants (GC-06.01).

Programmatic Usage

from geometor.divine.catalog import build_construction

        model, cert = build_construction("GC-06.01")
        assert cert.ratio_expr == sp.GoldenRatio
        

Exact Builder Implementation

def build_bui_circle_droves() -> ConstructionResult:
            """Bùi Quang Tuân's infinite rational secant family dividing a radius in phi."""
            m = new_catalog_model("bui-circle-droves")
            s3 = sp.sqrt(3)
            s5 = sp.sqrt(5)

            O = m.set_point(0, 0, classes=["given"], ID="O")
            B = m.set_point(1, 0, classes=["given"], ID="B")
            A = m.set_point(-1, 0, ID="A")

            c_o = m.construct_circle(O, B, classes=["primary"], ID="cO")
            _ = c_o
            ln_ab = m.construct_line(A, B, classes=["golden"], ID="AB")

            # First pair (x1 = 0, y1 = 2/3): C1=(0,1), D1=(2/3, -sqrt(5)/3)
            C1 = m.set_point(0, 1, ID="C1")
            D1 = m.set_point(sp.Rational(2, 3), -s5 / 3, ID="D1")
            ln_c1d1 = m.construct_line(C1, D1, classes=["golden"], ID="C1D1")

            # Second pair (x2 = -1/2, y2 = 7/8): C2=(-1/2, sqrt(3)/2), D2=(7/8, -sqrt(15)/8)
            C2 = m.set_point(-sp.Rational(1, 2), s3 / 2, ID="C2")
            D2 = m.set_point(sp.Rational(7, 8), -(s3 * s5) / 8, ID="D2")
            ln_c2d2 = m.construct_line(C2, D2, classes=["auxiliary"], ID="C2D2")
            _ = ln_c2d2

            m.intersect(ln_c1d1, ln_ab, classes=["golden"], ID="M")

            return make_result(m, ("M", "B"), ("O", "M"), ("O", "M", "B"))
        

Exact Symbolic Point Coordinates

All points in bui-circle-droves are certified free of floating-point approximations over \(\mathbb{Q}(\sqrt{3}, \phi)\):

Point

Exact \(x\)

Exact \(y\)

O

\(0\)

\(0\)

B

\(1\)

\(0\)

A

\(-1\)

\(0\)

C1

\(0\)

\(1\)

D1

\(\frac{2}{3}\)

\(- \frac{\sqrt{5}}{3}\)

C2

\(- \frac{1}{2}\)

\(\frac{\sqrt{3}}{2}\)

D2

\(\frac{7}{8}\)

\(- \frac{\sqrt{15}}{8}\)

M

\(\frac{3}{2} - \frac{\sqrt{5}}{2}\)

\(0\)

Certified Golden Ratio: MB / OM = GoldenRatio.