X(49) = CENTER OF SINE-TRIPLE-ANGLE CIRCLE

Trilinears

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

Barycentrics

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

\(a^4 (a^2 - b^2 - c^2) (a^4 + b^4 + c^4 - 2 a^2 b^2 - 2 a^2 c^2 - b^2 c^2) : :\)

Notes

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

X(49) lies on these lines: 1,215 3,155 4,156 5,54 24,568 52,195 93,94 381,578

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,B’,C’}}, {C,A,C’,A’}}, {A,B,A’,B’}}, and the circles {B,C,A’}}, {C,A,B’}}, {A,B,C’}}, where A’,B’,C’ are the excenters of ABC. Moreover, if P = X(3) and Q = X(4), then MAMBMC is not only perspective, but homothetic, to ABC, and the center of homothety is X(49). See Hyacinthos 23265, June 1, 2015.

X(49) is the {X(54),:ref:X(110) <X(110)>}-harmonic conjugate of X(5). For a list of other harmonic conjugates of X(49), click Tables at the top of this page.

X(49) = isogonal conjugate of X(93)

X(49) = isotomic conjugate of X(20572)

X(49) = anticomplement of X(34826)

X(49) = X(4)-isoconjugate of X(2962)

X(49) = X(92)-isoconjugate of X(2963)

X(49) = eigencenter of cevian triangle of X(94)

X(49) = eigencenter of anticevian triangle of X(50)

X(49) = X(94)-Ceva conjugate of X(50)