GC-05.05 · Symbolic Model & Source

Companion symbolic Python construction for GC-05.05 · Tran Quang Hung’s Six-Pentagon Reflection Star (GC-05.05).

Programmatic Usage

from geometor.divine.catalog import build_construction

        model, cert = build_construction("GC-05.05")
        assert cert.ratio_expr == sp.GoldenRatio
        

Exact Builder Implementation

def build_tran_star_six_pentagons() -> ConstructionResult:
            """Tran Quang Hung's star formed by reflecting a regular pentagon across its five sides."""
            m = new_catalog_model("tran-star-six-pentagons")
            phi, s36, _c36, _s72, _c72 = _pentagon_radicals()
            phi2 = phi + 1

            verts_xy = [
                (sp.Integer(-1), sp.Integer(0)),
                (sp.Integer(1), sp.Integer(0)),
                (phi, 2 * phi * s36),
                (sp.Integer(0), 2 * phi2 * s36),
                (-phi, 2 * phi * s36),
            ]
            names = ["A", "B", "C", "D", "E"]
            pts = [
                m.set_point(
                    x,
                    y,
                    classes=["given"] if nm in ("A", "B") else [],
                    ID=nm,
                )
                for nm, (x, y) in zip(names, verts_xy, strict=True)
            ]
            m.construct_polygon(pts, ID="ABCDE")

            # Reflect ABCDE across each of its 5 sides (each of exact length 2 => Q = 4)
            outer_pts = []
            for i in range(5):
                ux, uy = verts_xy[i]
                vx, vy = verts_xy[(i + 1) % 5]
                tri = []
                for idx_k, k in enumerate((4, 3, 2), start=1):
                    wx, wy = verts_xy[(i + k) % 5]
                    rx, ry = _reflect_across_side2(wx, wy, ux, uy, vx, vy)
                    p_out = m.set_point(rx, ry, guide=(idx_k == 2), ID=f"R{i + 1}{idx_k}")
                    tri.append(p_out)
                    outer_pts.append(p_out)
                m.construct_polygon([pts[i], pts[(i + 1) % 5], tri[2], tri[1], tri[0]], classes=["auxiliary"])

            ln_ab = m.construct_line(pts[0], pts[1], classes=["golden"], ID="AB")

            # Connect outer perimeter vertices with step 6 (the chiral {5/2} star on lateral vertices)
            # Indices in outer_pts (length 15): 0, 3, 6, 9, 12 (idx_k=1) and 2, 5, 8, 11, 14 (idx_k=3)
            s_a = m.construct_line(outer_pts[0], outer_pts[6], classes=["golden"], ID="S1")
            s_b = m.construct_line(outer_pts[3], outer_pts[9], classes=["auxiliary"], ID="S2")
            for idx in (0, 3, 6, 9, 12):
                m.construct_segment(outer_pts[idx], outer_pts[(idx + 6) % 15], classes=["auxiliary"])
            for idx in (2, 5, 8, 11, 14):
                m.construct_segment(outer_pts[idx], outer_pts[(idx + 6) % 15], classes=["auxiliary"])

            G = m.intersect(s_a, ln_ab, classes=["golden"], ID="G")
            H = m.intersect(s_b, ln_ab, classes=["golden"], ID="H")
            _ = H

            # Order long/short dynamically: G = (2 - sqrt(5), 0) or (sqrt(5) - 2, 0) on A=(-1,0)..B=(1,0)
            if float((G.x - pts[0].x).evalf()) >= float((pts[1].x - G.x).evalf()):
                return make_result(m, ("A", "G"), ("G", "B"), ("A", "G", "B"))
            return make_result(m, ("G", "B"), ("A", "G"), ("A", "G", "B"))
        

Exact Symbolic Point Coordinates

All points in tran-star-six-pentagons are certified free of floating-point approximations over \(\mathbb{Q}(\phi, \sqrt{10 - 2\sqrt{5}})\):

Point	Exact \(x\)	Exact \(y\)
A	\(-1\)	\(0\)
B	\(1\)	\(0\)
C	\(\frac{1}{2} + \frac{\sqrt{5}}{2}\)	\(\frac{\sqrt{2} \sqrt{5 - \sqrt{5}} + \sqrt{10} \sqrt{5 - \sqrt{5}}}{4}\)
D	\(0\)	\(\frac{\sqrt{10} \sqrt{5 - \sqrt{5}} + 3 \sqrt{2} \sqrt{5 - \sqrt{5}}}{4}\)
E	\(- \frac{\sqrt{5}}{2} - \frac{1}{2}\)	\(\frac{\sqrt{2} \sqrt{5 - \sqrt{5}} + \sqrt{10} \sqrt{5 - \sqrt{5}}}{4}\)
R11	\(- \frac{\sqrt{5}}{2} - \frac{1}{2}\)	\(\frac{- \sqrt{10} \sqrt{5 - \sqrt{5}} - \sqrt{2} \sqrt{5 - \sqrt{5}}}{4}\)
R13	\(\frac{1}{2} + \frac{\sqrt{5}}{2}\)	\(\frac{- \sqrt{10} \sqrt{5 - \sqrt{5}} - \sqrt{2} \sqrt{5 - \sqrt{5}}}{4}\)
R21	\(\frac{\sqrt{5}}{2} + \frac{3}{2}\)	\(- \frac{\sqrt{2} \sqrt{5 - \sqrt{5}}}{2}\)
R23	\(\frac{\sqrt{5}}{2} + \frac{5}{2}\)	\(\frac{\sqrt{2} \sqrt{5 - \sqrt{5}} + \sqrt{10} \sqrt{5 - \sqrt{5}}}{4}\)
R31	\(1 + \sqrt{5}\)	\(\frac{\sqrt{10} \sqrt{5 - \sqrt{5}} + 3 \sqrt{2} \sqrt{5 - \sqrt{5}}}{4}\)
R33	\(- \frac{1}{2} + \frac{\sqrt{5}}{2}\)	\(\frac{2 \sqrt{2} \sqrt{5 - \sqrt{5}} + \sqrt{10} \sqrt{5 - \sqrt{5}}}{2}\)
R41	\(\frac{1}{2} - \frac{\sqrt{5}}{2}\)	\(\frac{2 \sqrt{2} \sqrt{5 - \sqrt{5}} + \sqrt{10} \sqrt{5 - \sqrt{5}}}{2}\)
R43	\(- \sqrt{5} - 1\)	\(\frac{\sqrt{10} \sqrt{5 - \sqrt{5}} + 3 \sqrt{2} \sqrt{5 - \sqrt{5}}}{4}\)
R51	\(- \frac{5}{2} - \frac{\sqrt{5}}{2}\)	\(\frac{\sqrt{2} \sqrt{5 - \sqrt{5}} + \sqrt{10} \sqrt{5 - \sqrt{5}}}{4}\)
R53	\(- \frac{3}{2} - \frac{\sqrt{5}}{2}\)	\(- \frac{\sqrt{2} \sqrt{5 - \sqrt{5}}}{2}\)
G	\(-2 + \sqrt{5}\)	\(0\)
H	\(2\)	\(0\)

Certified Golden Ratio: AG / GB = GoldenRatio.