Trilinears f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = bc(a2 - bc)(b3 + c3 - a3 - abc)
Barycentrics af(a,b,c) : bf(b,c,a) : cf(c,a,b)
The Sharygin points are described in
Darij Grinberg, Sharygin Points Report, Hyacinthos #6293 (1/8/03) and #6315 (1/10/03)
The first of ten sections is an Introduction quoted, in part, here:
We will treat two remarkable triangles: the triangle bounded by the perpendicular bisectors of the internal angle bisectors of a triangle ABC, and the triangle bounded by the perpendicular bisectors of the external angle bisectors of triangle ABC. These two triangles and the triangle ABC are three perspective triangles, having a common perspectrix: the Lemoine axis of ABC. The mutual perspectors of the three triangles will be called the first, second and third Sharygin points of ABC (after a problem of Igor Sharygin - see Section 10).
The report introduces fifteen Sharygin points, of which the 1st, 2nd, 4th, and 6th are X(256), X(291), X(846),
and X(1054), respectively. X(1281) is the 3rd Sharygin point. See also Hyacinthos #6293 and #6315.
Let A' be the point where the internal angle bisector of angle CAB meets line BC, and let A" be the point where the external angle bisector of angle CAB meets line BC. Let x be the perpendicular bisector of segment AA', and let x' be the perpendicular bisector of segment AA". Define y, z, y', z' cyclically. Let D be the point where lines y and z meet, and let D' be the point where lines y' and z' meet. Define E, F, E', F' cyclically. Then
X(1281) = points of concurrence of lines DD', EE', FF'
X(846) = homothetic center of the excentral triangle and triangle DEF
X(1054) = center of similitude of the excentral triangle and triangle D'E'F'.
X(1281) lies on these lines:
2,846 21,99 63,147 98,100 256,291 350,1284 385,740 659,804
, select the center name from the list and, then, click on the vertices A, B and C successively.