Trilinears           f(A,B,C) : f(B,C,A) : f(C,A,B), where f(A,B,C) = cos B + cos C - 2 cos A;
                                    = g(a,b,c) : g(b,c,a) : g(c,a,b), where g(a,b,c) = (b - c)² + a(b + c - 2a)

Barycentrics    (sin A)f(A,B,C) : (sin B)f(B,C,A) : (sin C)f(C,A,B)
                                    = ag(a,b,c) : bg(b,c,a) : cg(c,a,b)

Let XYZ be the intouch triangle of ABC; i.e., the pedal triangle of the incenter, I. The circles AIX, BIY, CIZ concur in two points. One of them is I; the other is X(1155). This result is obtain by inversion in

Heinz Schröder, "Die Inversion und ihre Anwendung im Unterricht der Oberstufe," Der Mathematikunterricht 1 (1957) 59-80.

Each vertex of the tangential triangle of any triangle T is the inverse-in-the-circumcircle-of-T of the midpoints of the sides of T. Applying this to triangle XYZ shows that X(1155) is the inverse-in-the-incircle of the centroid of XYZ; i.e., X(1155) is X(23)-of-the-intouch-triangle. (Darij Grinberg, #6319, 1/11/03; coordinates by Jean-Pierre Ehrmann, #6320, 1/11/03)

X(1155) lies on these lines:
1,3    10,535    11,516    37,750    44,513    47,582    63,210    88,105    89,1002    100,518    227,603    238,1054    243,653    244,902    404,960

X(1155) = midpoint of X(36) and X(484)
X(1155) = reflection of X(1319) in X(36)
X(1155) = isogonal conjugate of X(1156)
X(1155) = inverse-in-circumcircle of X(55)
X(1155) = inverse-in-incircle of X(354)
X(1155) = inverse-in-Bevan-circle of X(57)
X(1155) = crosspoint of X(I) and X(J) for these (I,J): (1,1156), (527,1323)
X(1155) = crosssum of X(1) and X(1155)
X(1155) = crossdifference of any two points on line X(1)X(650)