X1138

ISOGONAL CONJUGATE OF X399

X1138 is the isogonal conjugate of X399.
< 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) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_399 = PARRY REFLECTION POINT (0.0032, 0.9936) X_1138 = ISOGONAL CONJUGATE OF X(399) (-0.2714, 0.2752) A B C X399 X1138 (a, b, c) = (5, 6, 7) | AB = 1

X1138 (ISOGONAL CONJUGATE OF X399) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(f(A,B,C) : f(B,C,A) : f(C,A,B)\), where \(f(A,B,C) = 1/(5 \cos A - 4 \cos B \cos C - 8 \cos^{2} A \sin B \sin C)\)
Barycentrics: \(1/(a^{8} - 4 a^{6} (b^{2} + c^{2}) + a^{4} (6 b^{4} + b^{2} c^{2} + 6 c^{4}) - a^{2} (4 b^{6} - b^{4} c^{2} - b^{2} c^{4} + 4 c^{6}) + (b^{2} - c^{2})^{2} (b^{4} + 4 b^{2} c^{2} + c^{4})) : :\)
Search Value (6, 9, 13): \(7.791344667477\)

X1138 lies on the following curves and lines: K001 (Neuberg Cubic), K279, K449, K490, K515, K528, K614, Q066, Q105, {1,5677}, {3,3471}, {4,2132}, {15,5624}, {16,5623}, {30,146}, {74,5670}, {186,1990}, {484,3464}, {616,5675}, {617,5674}, {1157,5667}, {1272,3260}, {3258,5627}, {3465,5685}, {3479,5679}, {3480,5678}, {3484,5684}, {5672,5673}

▸ Functional Relations8 relations

Notes

There are only two points X such that the pedal triangle of X is similar to the cevian triangle of X. They are X4 and X1138. (Jean-Pierre Ehrmann, January 4, 2003)

Let KaKbKc be the anticomplementary triangle of a triangle ABC, and let EA be the line through A parallel to the Euler line. Let A″ be the point of intersection, other than A, of EA and the circumcircle. Define EB and EC cyclically. The locus of a point P such that the Euler line of PBC is parallel to the Euler line of ABC is a conic ABCA’A″ having the midpoint of segment BC as center. The conics, ABCA’A″, ABCB’B″, ABCC’C″ pass through X1138. (Francisco Javier García Capitán, April 3, 2015: ADGEOM 2458)