X3053

INTERSECTION X3X_6∩X4X_230

X3053 = inverse-in-circumcircle of X1692
< 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) circumcircle: Circumcircle of △ABC { 3, 1692 }: Brocard Axis { 3, 6 }: Brocard Axis { 24, 112 }: Central Line L(24, 112) { 4, 230 }: Central Line L(4, 230) { 64, 248 }: Central Line L(64, 248) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_3 = CIRCUMCENTER (0.5000, 0.1021) X_1692 = RADICAL TRACE OF CIRCUMCIRCLE AND 2nd LEMOINE CIRCLE (0.7995, 0.5968) X_6 = SYMMEDIAN POINT (LEMOINE POINT, GREBE POINT) (0.6000, 0.2672) X_24 = PERSPECTOR OF ABC AND ORTHIC-OF-ORTHIC TRIANGLE (0.6627, 0.5278) X_112 = Ψ(ORTHOCENTER, SYMMEDIAN POINT) (0.5162, 0.6121) X_4 = ORTHOCENTER (0.6122, 0.3958) X_230 = -CEVA CONJUGATE OF X(114) (0.9746, 0.6698) X_64 = ISOGONAL CONJUGATE OF X(20) (0.5579, 1.5943) X_248 = -LINE CONJUGATE OF X(132) (0.8229, -0.0790) X_3053 = INTERSECTION X(3)X(6)∩X(4)X(230) (0.7332, 0.4873) A B C X3 X1692 X6 X24 X112 X4 X230 X64 X248 X3053 (a, b, c) = (5, 6, 7) | AB = 1

X3053 (INTERSECTION X3X_6∩X4X_230) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(a(3a^{2} - b^{2} - c^{2}) : :\)
Search Value (6, 9, 13): \(-0.631285612775\)
▸ Alternate Coordinates5 forms
  • Trilinears: \(a(a^{2} - S_{A}) : :\)

  • Trilinears: \(\sin A - 2 \cos A \tan \omega : :\)

  • Trilinears: \(2 \cos A - \sin A \cot \omega : :\)

  • Trilinears: \(a - 4R \cos A \tan \omega : :\)

  • Trilinears: \(a (\cot A - \cot B \cot C) : :\)

▸ Incident Central Lines19 lines

{3, 6}, {4, 230}, {22, 1184}, {24, 112}, {25, 1611}, {35, 609}, {56, 1914}, {64, 248}, {76, 1003}, {115, 382}, {140, 2548}, {154, 237}, {218, 2251}, {550, 2549}, {988, 1100}, {999, 2241}, {1385, 1572}, {1498, 1971}, {2176, 2223}

▸ Functional Relations15 relations

Notes

Let Γ1 be the circumcircle, Γ2 the 2nd Lemoine circle, and Γ3 the circle {{X371,X372,PU(1),PU(39)}} (which has center X32). The three circles intersect in two points, at which let L and L' be the lines tangent to Γ1. Then X3035 = L∩L'. (Randy Hutson, January 29, 2015)