In triangle geometry, the sine-triple-angle circle is one of many circles that can be defined from a triangle. For triangle ABC, let A1 and A2 be points on side BC , with B1, B2, C1 and C2 defined similarly on CA and AB respectively. If
∠ A = ∠ A B 1 C 1 = A C 2 B 2 , {\displaystyle \angle A=\angle AB_{1}C_{1}=AC_{2}B_{2},}
∠ B = ∠ B C 1 A 1 = B A 2 C 2 , {\displaystyle \angle B=\angle BC_{1}A_{1}=BA_{2}C_{2},}
and
∠ C = ∠ C A 1 B 1 = C B 2 A 2 , {\displaystyle \angle C=\angle CA_{1}B_{1}=CB_{2}A_{2},}
then A1, A2, B1, B2, C1 and C2 lie on a circle called the sine-triple-angle circle, originally referred to by Tucker and Neuberg as the cercle triplicateur.
Properties
| A 1 A 2 | : | B 1 B 2 | : | C 1 C 2 | = sin 3 A : sin 3 B : sin 3 C {\displaystyle |A_{1}A_{2}|:|B_{1}B_{2}|:|C_{1}C_{2}|=\sin 3A:\sin 3B:\sin 3C} ., which gives the circle its name. However, there are an uncountably infinite amount of circles that also satisfy this identity. The centers of these circles are on the hyperbola through the incenter, three excenters, and X(49) (see below for X49). The homothetic centers of the Nine-point circle and the sine-triple-angle circle is the Kosnita point and the focus of the Kiepert parabola. The homothetic centers of the circumcircle and the sine-triple-angle circle is X(184), the inverse of Jerabek center in Brocard circle, and X(1147). Intersections of the Polar of A,B and C with the circle and BC,CA and AB are colinear. The radius of the sine-triple-angle circle is
R | 1 + 8 cos ( A ) cos ( B ) cos ( C ) | , {\displaystyle {\frac {R}{|1+8\cos(A)\cos(B)\cos(C)|}},}
where R is the circumradius of triangle ABC.
Center The center of sine-triple-angle circle is a triangle center designated as X(49) in Encyclopedia of Triangle Centers. with trilinear coordinates
cos ( 3 A ) : cos ( 3 B ) : cos ( 3 C ) {\displaystyle \cos(3A):\cos(3B):\cos(3C)} .
Generalization For a given natural number n>0, if
∠ A 1 C 1 A 2 = ( 2 n − 1 ) A − ( n − 1 ) π , {\displaystyle \angle A_{1}C_{1}A_{2}=(2n-1)A-(n-1)\pi ,}
∠ B 1 A 1 B 2 = ( 2 n − 1 ) B − ( n − 1 ) π , {\displaystyle \angle B_{1}A_{1}B_{2}=(2n-1)B-(n-1)\pi ,}
and
∠ C 1 B 1 C 2 = ( 2 n − 1 ) C − ( n − 1 ) π , {\displaystyle \angle C_{1}B_{1}C_{2}=(2n-1)C-(n-1)\pi ,}
then
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