X49

CENTER OF SINE-TRIPLE-ANGLE CIRCLE

Suppose that P and Q are distinct points in the plane of a triangle ABC . Let PA = reflection of P in line AQ, let QA = reflection of Q in line AP, and let MA = midpoint of segment PAQA. Define MB and MC cyclically. César Lozada found that if Q = isogonal conjugate of P, then the locus of P for which MAMBMC is perspective to ABC is the union of a cubic and 6 circles: specifically, the McCay cubic (K003), the circles {B,C,Jb,Jc}}, {C,A,Jc,Ja}}, {A,B,Ja,Jb}}, and the circles {B,C,Ja}}, {C,A,Jb}}, {A,B,Jc}}, where Ja,Jb,Jc are the excenters of ABC. Moreover, if P = X3 and Q = X4, then MAMBMC is not only perspective, but homothetic, to ABC, and the center of homothety is X49. See Hyacinthos 23265, June 1, 2015.
< 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) { 5, 54 }: Central Line L(5, 54) { 3, 155 }: Central Line L(3, 155) { 4, 156 }: Central Line L(4, 156) { 93, 94 }: Central Line L(93, 94) { 1, 215 }: Central Line L(1, 215) { 52, 195 }: Central Line L(52, 195) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_93 = ISOGONAL CONJUGATE OF X(49) (0.5299, 0.2878) X_5 = NINE-POINT CENTER (0.5561, 0.2489) X_54 = KOSNITA POINT (0.5880, 0.2176) X_3 = CIRCUMCENTER (0.5000, 0.1021) X_155 = EIGENCENTER OF ORTHIC TRIANGLE (1.1658, 0.5033) X_4 = ORTHOCENTER (0.6122, 0.3958) X_156 = -OF-TANGENTIAL-TRIANGLE (0.6520, -0.1707) X_94 = ISOGONAL CONJUGATE OF X(50) (0.9069, -0.1331) X_1 = INCENTER (0.5714, 0.2333) X_215 = -CEVA CONJUGATE OF X(50) (0.4113, 0.3887) X_52 = ORTHOCENTER OF ORTHIC TRIANGLE (0.7155, 0.4580) X_195 = -CEVA CONJUGATE OF X(3) (0.6761, 0.3331) X_49 = CENTER OF SINE-TRIPLE-ANGLE CIRCLE (0.6275, 0.1789) A B C X93 X5 X54 X3 X155 X4 X156 X94 X1 X215 X52 X195 X49 (a, b, c) = (5, 6, 7) | AB = 1

X49 (CENTER OF SINE-TRIPLE-ANGLE CIRCLE) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(\cos 3A : \cos 3B : \cos 3C\)
Barycentrics: \(\sin A \cos 3A : \sin B \cos 3B : \sin C \cos 3C\)
Search Value (6, 9, 13): \(1.301655102607\)
▸ Alternate Coordinates1 form
  • Barycentrics: \(a^{4} (a^{2} - b^{2} - c^{2}) (a^{4} + b^{4} + c^{4} - 2 a^{2} b^{2} - 2 a^{2} c^{2} - b^{2} c^{2}) : :\)

▸ Linear Combinations1 combination
  • X49 = \((r^{2} + 2 r R+ s^{2})\) X1 - \(6 R (R + r)\) X2 - \(3(r^{2} + 4 r R + R^{2} - s^{2})\) X3 (Peter Moses, April 2, 2013)

▸ Incident Central Lines8 lines

{1, 215}, {3, 155}, {4, 156}, {5, 54}, {24, 568}, {52, 195}, {93, 94}, {381, 578}

▸ Functional Relations8 relations

Notes

V. Thebault, “Sine-triple-angle-circle,” Mathesis 65 (1956) 282-284.

X49 is the {X54,X110}-harmonic conjugate of X5.