X2671

1st GOLDEN ARBELOS POINT

Suppose UV is a segment and 1/2 < s < 1. Let P be the point on UV satisfying |UP|/|VP| = s. Let O be a semicircle having diameter UV. Let H be the semicircle having diameter VP, on the same side of UV as O. Let K be the semicircle having diameter UP, on the same side of UP as O. The three circular arcs form an arbelos, R.
< A B C >: Reference Triangle △ABC [ A B ]: Sideline AB (Base c = 1) [ B C ]: Sideline BC (Side a) [ C A ]: Sideline CA (Side b) { 2, 2674 }: Central Line L(2, 2674) { 6, 2672 }: Central Line L(6, 2672) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_2673 = ISOGONAL CONJUGATE OF X(2671) (0.5710, 0.2193) X_2 = CENTROID (0.5374, 0.2000) X_2674 = ISOGONAL CONJUGATE OF X(2672) (0.6690, 0.3812) X_6 = SYMMEDIAN POINT (LEMOINE POINT, GREBE POINT) (0.6000, 0.2672) X_2672 = 2nd GOLDEN ARBELOS POINT (0.3508, 0.0923) X_2671 = 1st GOLDEN ARBELOS POINT (0.5720, 0.2475) A B C X2673 X2 X2674 X6 X2672 X2671 (a, b, c) = (5, 6, 7) | AB = 1

X2671 (1st GOLDEN ARBELOS POINT) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(1/(2 \cos A + τ^{2}\sin A) : :\), where τ = (1 + \(\sqrt{5}\))/2 = golden ratio
Barycentrics: \((S-(\sqrt{5}-3) S_{B}) \cdot (S-(\sqrt{5}-3) S_{C}) : :\) (C. Lozada, 7/13/19)
Search Value (6, 9, 13): \(1.116945892931\)
▸ Alternate Coordinates2 forms

  • Trilinears: \(\sec[A - arccot(3 - 5^{1/2})] : :\) (M. Iliev, 5/13/07)

  • Trilinears: found by Peter J. C. Moses, July, 2004. For a general account of the arbelos, visit

▸ Linear Combinations1 combination

  • At the Eleventh International Conference on Fibonacci Numbers and Their Applications (TU Braunschweig, Germany, July 2004), Zvonko Cerin established that the lines AA', BB', CC' concur if and only if s = τ. The point of concurrence is X2671\(.\)

X2671 lies on the Kiepert hyperbola.

▸ Incident Central Lines2 lines

{2, 2674}, {6, 2672}

▸ Functional Relations1 relation

Notes

Let L be the line through P perpendicular to BC. The circle tangent to semicircles O and K and line L is the s-Archimedean circle of the arbelos R. Now continuing with the variable s, suppose ABC is an arbitrary triangle. Let A' be the center of the s-Archimedean circle of the outward arbelos on segment BC, and define B' and C' cyclically.

MathWorld.