Trilinears f(A,B,C) : f(B,C,A) : f(C,A,B), where f(A,B,C) = 1/(2 cos A + τ2sin A), where τ = (1 + sqrt(5))/2 = golden ratio
Trilinears g(A,B,C) : g(B,C,A) : g(C,A,B), where g(A,B,C) = sec[A - arccot(3 - 51/2)] (M. Iliev, 5/13/07)
Barycentrics (sin A)f(A,B,C) : (sin B)f(B,C,A) : (sin C)f(C,A,B)
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.
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.
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 X(2671).
X(2671) lies on the Kiepert hyperbola and these lines: 2,2674 6,2672
, select the center name from the list and, then, click on the vertices A, B and C successively.