X(94) = ISOGONAL CONJUGATE OF X(50)

Trilinears

\(csc 3A : csc 3B : csc 3C\)

Barycentrics

\(sin A csc 3A : sin B csc 3B : sin C csc 3C\)

\(b^2 c^2 / ((a^2 - b^2 - c^2)^2 - b^2 c^2) : : Let A1B1C1 and A2B2C2 be the 1st and 2nd Ehrmann circumscribing triangles. Let A' be the trilinear pole of line A1A2, and define B' and C' cyclically. The lines AA', BB', CC' concur in :ref:`X(94) <X(94)>\). Let A1’B1’C1’ and A2’B2’C2’ be the 1st and 2nd Ehrmann inscribed triangles. Then X(94) is the radical center of nine-point circles of AA1’A2’, BB1’B2’, CC1’C2’. (Randy Hutson, June 27, 2018) Let A’B’C’ be the medial or orthic triangle of ABC and A&Prime;, B&Prime;, C&Prime; the centers of circles {A’, X(3), X(4)}}, {B’, X(3), X(4)}} and {C’, X(3), X(4)}}, respectively. Then A&Prime;, B&Prime;, C&Prime; are collinear on the trilinear polar of X(94). (César Lozada, March 9, 2021). X(94) lies on the Kiepert hyperbola and these lines: 2,300 4,143 23,98 49,93 96,925 275,324 X(94) = isogonal conjugate of X(50) <X(50)>`

Notes

Let A1B1C1 and A2B2C2 be the 1st and 2nd Ehrmann circumscribing triangles. Let A’ be the trilinear pole of line A1A2, and define B’ and C’ cyclically. The lines AA’, BB’, CC’ concur in X(94). Let A1’B1’C1’ and A2’B2’C2’ be the 1st and 2nd Ehrmann inscribed triangles. Then X(94) is the radical center of nine-point circles of AA1’A2’, BB1’B2’, CC1’C2’. (Randy Hutson, June 27, 2018)

Let A’B’C’ be the medial or orthic triangle of ABC and A&Prime;, B&Prime;, C&Prime; the centers of circles {A’, X(3), X(4)}}, {B’, X(3), X(4)}} and {C’, X(3), X(4)}}, respectively. Then A&Prime;, B&Prime;, C&Prime; are collinear on the trilinear polar of X(94). (César Lozada, March 9, 2021).

X(94) lies on the Kiepert hyperbola and these lines: 2,300 4,143 23,98 49,93 96,925 275,324

X(94) = isogonal conjugate of X(50)

X(94) = isotomic conjugate of X(323)

X(94) = anticomplement of X(34834)

X(94) = cevapoint of X(49) and X(50)

X(94) = X(i)-cross conjugate of X(j) for these (i,j): (30,264), (50,93), (265,328)

X(94) = X(300)-Hirst inverse of X(301)

X(94) = trilinear pole of PU(5) (line X(5)X(523))

X(94) = pole wrt polar circle of trilinear polar of X(186)

X(94) = X(48)-isoconjugate (polar conjugate) of X(186)

X(94) = barycentric product X(476)

X(94) = trilinear pole of PU(173)

X(94) = trilinear product X(74) (circumcircle-X(4) antipodes)

X(94) = barycentric quotient X(476)/X(110)

X(94) = intersection of the tangent to hyperbola {A,B,C,:ref:X(6) <X(6)>,:ref:X(13) <X(13)>,:ref:X(16) <X(16)>}} at X(13) and the tangent to hyperbola {A,B,C,:ref:X(6) <X(6)>,:ref:X(14) <X(14)>,:ref:X(15) <X(15)>}} at X(14)