1. Let Q be a fixed point. The locus of P such that the line QP* (P* = isogonal of P) is perpendicular to the trilinear polar of P is a quartic C(Q). See Q066 obtained with Q = O. This quartic C(Q) contains Q if and only if Q lies on Q003. (Angel Montesdeoca, private message 2013-11-30. Hechos Geométricos).

(Q003 de Higher Degree Related Curves. Bernard Gibert)

1. Let P be a point and PaPbPc its cevian triangle, HaHbHcthe orthic triangle (cevian triangle of H). The parallel through the point Pb to the sideline HaHb and the parallel through the point Pc to the sideline HaHc intersect at A'. The points B' and C' defined likewise. The triangles ABC and A'B'C' are perspective if and nly if P lies on the Stammler quartic together with the circumcircle (Angel Motesdeoca, Hyacinthos #21817).

(Q066 de Higher Degree Related Curves. Bernard Gibert)

2. Let Q be a fixed point. The locus of P such that the line QP*(P* = isogonal of P) is perpendicular to the trilinear polar of P is a quartic C(Q). When Q = O, the quartic is Q066. (Angel Montesdeoca, private message 2013-11-30. Hechos Geométricos ).

(Q066 de Higher Degree Related Curves. Bernard Gibert)

19. Let Oa, Ob, Oc be the circumcenters of triangles PBC, PCA, PAB. The centroid of OaObOc lies on the line OP if and only if P lies on the Thomson cubic or on the circumcircle of ABC (Angel Montesdeoca, Anopolis #958). (Hechos Geométricos)

20. Let A'B'C' be the circumcevian triangle of P. The lines A'B' and A'C' meet BC at Ab and Ac. The points Bc, Ba and Ca, Cb are defined likewise and these six points lie on a same conic with center Q. The points X(6), P, Q are collinear if and only if P lies on the Thomson cubic (Angel Montesdeoca, ADGEOM #905, slightly rephrased). Similarly, the points X(3), P, Q are collinear if and only if P lies on the quartic Q098. (Hechos Geométricos)

Let A'B'C' the cevian triangle of a point P. Let Ab, Ac be the reflections of C, B in the perpendiculars dropped from A' onto AC, AB. Let Oa be the circumcenter of triangle AAbAc and define Ob, Oc likewise. Then

• OaObOc and A'B'C' are perspective if and only if P lies on K045 in which case the perspector lies on K044.

• OaObOc and ABC are perspective if and only if P lies on K646 in which case the perspector also lies on K044. (Angel Montesdeoca, Anopolis #845)

**K646** is the pivotal cubic pK(X97, X95). It is the isogonal transform of K350 and the isotomic transform of pK(X324, X264). It meets the circumcircle at the same points as pK(X6, X2979).

(Hechos Geométricos)

2. Let A' B' C' be the cevian triangle of a point P. The circumcircles (Oa), (Ob), (Oc) of AB'C', BC'A', CA'B' are concurrent at M. Denote by O* the circumcenter of OaObOc. K279 is the locus of point P such that O* lies on the Euler line (Angel Montesdeoca, private message 2013-08-06, see also Anopolis #758).

(K279 del catálogo de Bernard Gibert)

(Hechos Geométricos)

Let (C) be the inconic with perspector P and H(A), H(B), H(C) the Apollonius hyperbolas of A, B, C with respect to (C). The centers of these hyperbolas form a triangle perspective to ABC if and only if P lies on the Lucas cubic (Angel Montesdeoca, 2010/03/03).

(K007 del catálogo de Bernard Gibert)

(Triángulos)

Let A' B' C' be the cevian triangle of a point P. The circumcircles (Oa), (Ob), (Oc) of AB'C', BC'A', CA'B' are concurrent at M. Denote by O* the circumcenter of OaObOc. K007 is the locus of point P such that M, O*, O are collinear (or M is the reflection of O in O*). When P varies on the Lucas cubic, the locus of M is the Darboux cubic K004 (Angel Montesdeoca, private message 2013-08-06). See the related cubic K279.

(K007 del catálogo de Bernard Gibert)

Let P be a point, A*B*C* its circumcevian triangle, A'B'C' its pedal triangle and let A"B"C" be the antipodal triangle of A'B'C' in the pedal circle of P. A*B*C*and A"B"C" are perspective if P is on the circumcircle (the circumcevian triangle degenerates) or on K003 or on K634. If P is on K634, the perspector of the triangles A*B*C* and A"B"C" is a point in the circumcircle (Antreas P. Hatzipolakis, Hyacinthos #21738, Angel Montesdeoca, Hyacinthos #21743). See the two figures below (Angel Montesdeoca's work)

(K634 del catálogo de Bernard Gibert)

If P lies on the cubic K634 (together with a quadricircular octic), then the pedal and antipedal circles of P are tangent. The point of tangency of the two circles is on the circumcircle (Antreas P. Hatzipolakis, Angel Montesdeoca, Francisco García Capitán, Hyacinthos #21746 and sq.).

(K634 del catálogo de Bernard Gibert)

Let P be a point, A*B*C* its circumcevian triangle, A'B'C' its pedal triangle and let A"B"C" be the antipodal triangle of A'B'C' in the pedal circle of P. A*B*C*and A"B"C" are perspective if P is on the circumcircle (the circumcevian triangle degenerates) or on K003 or on K634. If P is on K634, the perspector of the triangles A*B*C* and A"B"C" is a point in the circumcircle (Antreas P. Hatzipolakis, Hyacinthos #21738, Angel Montesdeoca, Hyacinthos #21743).

(K003 del catálogo de Bernard Gibert)

Let P, P* be two isogonal conjugate points and A'B'C', A"B"C" the circumcevian triangles of P, P* resp. Denote sA', sB', sC' the Simson lines of A', B', C' wrt A"B"C" resp. These lines sA', sB', sC' are concurrent if and only if P lies on the McCay cubic (together with Kjp, the circumcircle). In other words, if P lies on the McCay cubic then the lines sA', sB', sC' are concurrent (Antreas P. Hatzipolakis, Angel Montesdeoca, Hyacinthos #21685).

(K003 del catálogo de Bernard Gibert)

Let P, P* be two isogonal conjugate points and A'B'C', A"B"C" the circumcevian triangles of P, P* resp. Denote sA', sB', sC' the Simson lines of A', B', C' wrt A"B"C" resp. These lines sA', sB', sC' are concurrent if and only if P lies on Kjp (together with the McCay cubic, the circumcircle). In other words, if P lies on Kjp then the lines sA', sB', sC' are concurrent (Hyacinthos #21685, Angel Montesdeoca).

(K024 del catálogo de Bernard Gibert)

Let P, P* be two isogonal conjugate points and A'B'C', A"B"C" the pedal triangles of P, P* resp. (P, P* share the same pedal circle). Denote sA', sB', sC' the Simson lines of A', B', C' wrt A"B"C" resp. These lines sA', sB', sC' are concurrent if and only if P lies on the McCay cubic (together with the circumcircle and the line at infinity (Antreas P. Hatzipolakis, Angel Montesdeoca, Hyacinthos #21686).

(K003 del catálogo de Bernard Gibert)

Let P be a point and PaPbPc its cevian triangle, HaHbHc the orthic triangle (cevian triangle of H). The parallel through the point Pb to the sideline HaHb and the parallel through the point Pc to the sideline HaHc intersect at A'. The points B' and C' defined likewise. The triangles PaPbPc and A'B'C' are perspective if and only if P lies on on K211, together with the three circm-rectangular hyperbolas passing through Ga, Gb, Gc (Angel Montesdeoca, Hyacinthos #21806).

(K211 del catálogo de Bernard Gibert)

QA-P38 is the Perspector of the Reference Quadrangle with the Cyclocevian Quadrangle.
The Cyclocevian Quadrangle CC1.CC2.CC3.CC4 is defined by:
CCi = Cyclocevian Conjugate of Pi wrt Pj.Pk.Pl for all permutations of (i,j,k,l) ? (1,2,3,4).
The definition of Cyclocevian Conjugate can be found at Ref-13.

This point was separately found by Angel Montesdeoca (Hyacinthos message 21075, see[11]) and Randy Hutson (private mail to author EQF) in the same week (June, 2012).

(Encyclopedia of Quadri-Figures )

A circumscribed QA-DT-Conic is a conic through the vertices of the QA-Diagonal Triangle QA-Tr1.
There is a special property for these conics:
Let Sij be the intersection, other than the vertex of the QA-Diagonal Triangle, of the QA-DT-Conic and line Pi.Pj, for all permutations of (i,j) ∈ (1,2,3,4). The lines Sij.Skl, for all permutations of (i,j,k,l) ∈ (1,2,3,4), concur in a new point which we shall call here the QA-DT-Conic-Perspector.
This subject was being developed by the specific observation of Angel Montesdeoca in QA-Ci1 and QA-P38. It was generalized by observations of Eckart Schmidt, Randy Hutson and the author of EQF (July, 2012).

Properties:
When QA-DT-Conic = QA-Ci1, then the QA-DT-Conic-Perspector is QA-P38.

(Encyclopedia of Quadri-Figures )

f(a,b,c) = a^{2}(b - 2c) - a(b^{2} + bc - c^{2}) - (b - c)c^{2}

Let A'B'C' be the cevian triangle of X(1). Let AB be the reflection of A' in BB', and define BC and CA cyclically. Let AC be the reflection of A' in BC', and define BA and CB cyclically. Let OAB be the circumcircle of AA'AB, and define OBC and OCA cyclically. Let OAC be the circumcircle of AA'AC, and define OBA and OCB cyclically. Then P(110) is the radical center of OAB, OBC, OCA, and U(110) is defined symmetrically; i.e., as the radical center of OAC, OBA, OCB. Angel Montesdeoca, August 26 2013

See Hechos Geometricos 240813 and Anopolis 885

P(110) and U(110) lie on the line X(942)X(1938); P(110)U(110) has ideal point X(513).

f(a,b,c)= (a + b - c) (a - b + c) (a^{3} b - a b^{3} + 2 a^{3} c - 2 a b^{2} c - b^{3} c + a^{2} c^{2} - 3 a b c^{2} - 3 b^{2} c^{2} - 2 a c^{3} - 3 b c^{3} - c^{4}).

Let A'B'C' be the cevian triangle of X(1).
Let N_{AB} be the nine-point circle of AB'X(1), and define N_{BC} and N_{CA} cyclically. Let N_{AC} be the nine-point circle of AC'X(1), and define N_{BA} and N_{CB} cyclically. Then P(111) is the radical center of N_{AB}, N_{BC}, N_{CA}, and U(111) is defined symmetrically; i.e., as the radical center of N_{AC}, N_{BA}, N_{CB}. Angel Montesdeoca, August 26 2013

See Hechos Geometricos 210813 and Anopolis 860

X(3628) is the centroid of the set {A', B', C', X(5)}, where A'B'C' is the medial triangle; more generally, H^{-1}(X; M, 2) is the centroid of the set {A', B', C', X}. (Angel Montesdeoca, 12/20/2011)

(Encyclopedia of Triangle Centers )

Let A'B'C' be the circumcevian triangle of the symmedian point (Lemoine point), X(6). The sidelines BC, CA, AB meet the sidelines of B'C', C'A', A'B' in 9 points, of which 6 do not lie on the trilinear polar of K; barycentrics for the 6 points are 0 : b^{2} : 2c^{2}, 0 : 2b^{2} : c^{2}, 2a^{2} : 0 : c^{2}, a^{2} : 0 : 2c^{2}, a^{2} : 2b^{2} : 0, 2a^{2} : b^{2} : 0. The 6 points lie on a conic with center X(5024) and equation

2(b^{4}c^{4}x^{2} + c^{4}a^{4}y^{2} + a^{4}b^{4}z^{2}) -5a^{2}b^{2}c^{2}(a^{2}yz + b^{2}zx + c^{2}xy) = 0.

Moreover, the center of the conic tangent to the 6 lines BC, CA, AB, B'C', C'A', A'B' is X(39), and an equation for this conic is

b^{4}c^{4}x^{2} + c^{4}a^{4}y^{2} + a^{4}b^{4}z^{2} -2a^{2}b^{2}c^{2}(a^{2}yz + b^{2}zx + c^{2}xy) = 0.

(From Angel Montesdeoca, March 28, 2013)

(Encyclopedia of Triangle Centers )

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a^{5}b^{2} + a^{5}c^{2} - 2a^{5}b^{2}c^{2} + a^{3}b^{3} + a^{3}c^{3} + a^{3}b^{2}c + a^{3}bc^{2} + a^{2}b^{4} + a^{2}c^{4} - 3a^{2}b^{3}c - 3a^{2}bc^{3} - ab^{5} - ac^{5} - ab^{4}c - abc^{4} - bc(b^{2} - c^{2})^{2} (Angel Montesdeoca, May 13, 2013)

**X(5482) = 3*X(549) - X(970)
X(5482) = (R - 2r)*X(140) - R*X(143)
**

Let ABC be a triangle and let A'B'C' be the cevian triangle of the incenter, X(1). Let R be the radical center of the circles (A', |A'B|, {B',|B'C|), (C', |C'A|), and let S be the radical center of the circles (A',|A'C|), (B',|B'A|), (C',|C'B|). X(5482) is the midpoint of the segment RS. (Antreas Hatzipolakis, May 4, 2013)

X(5482) is the {X(3),X(1764)}-harmonic conjugate of X(3579) (Peter Moses, May 13, 2013)

For the construction and generalizations, see Hechos Geométricos en el Triángulo.

X(5482) lies on these lines: {1,3}, {140,143}, {549,970}

(Encyclopedia of Triangle Centers )

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a[a^{9} - a^{8}(b + c) - a^{7}( b - c)^{2} + a^{6}(2b^{3} - b^{2}c - bc^{2} + 2c^{3}) - a^{5}(3b^{4} + b^{3}c - 7b^{2}c^{2} + bc^{3} + 3c^{4}) + 4a^{4}bc(b - c)^{2}(b + c) + a^{3}(b^{2} - c^{2})^{2}(5b^{2} - 4bc + 5c^{2}) - a^{2}(b - c)^{2}(2b^{5} + 5b^{4}c + b^{3}c^{2} + b^{2}c^{3} + 5bc^{4} + 2c^{5}) - a(b^{2} - c^{2})^{2}(2b^{4} - 3b^{3}c + 5b^{2}c^{2} - 3bc^{3} + 2c^{4}) + (b - c)^{4}(b + c)^{3}(b^{2} + c^{2})] (Angel Montesdeoca, May 25, 2013)

**X(5494) = (2r + R)*X(110) - 4(r + R)X(1385)
X(5494) = 2R*X(65) + (2r + R)*X(74)
**

Let ABC be a triangle and let A'B'C' be the cevian triangle of the incenter, X(1). Let A

For the construction and discussion, see Hechos Geométricos en el Triángulo.

X(5494) lies on these lines: {1,2779},{21,104},{36,1725},{65,74},{125,860}

(Encyclopedia of Triangle Centers )

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a^{2}[a^{7}(b + c) - a^{6}(b^{2} + c^{2}) - a^{5}(3b^{3} + 2b^{2}c + 2bc^{2} + 3c^{3}) + a^{4}(3b^{4} - b^{3}c + 4b^{2}c^{2} - bc^{3} + 3c^{4}) + a^{3}(3b^{5} + b^{4}c + 2b^{3}c^{2} + 2b^{2}c^{3} + bc^{4} + 3c^{5}) - a^{2}(3b^{6} - 2b^{5}c - 2bc^{5} + 3c^{6}) - a(b^{7} - b^{4}c^{3} - b^{3}c^{4} + c^{7}) + (b^{2} - c^{2})^{2}(b^{4} - b^{3}c - bc^{3} - b^{2}c^{2} + c^{4})] (Angel Montesdeoca, May 28, 2013)

Let ABC be a triangle and let A'B'C' be the cevian triangle of the incenter, X(1). Let L_{A} be the line through A' perpendicular to line AA', and define L_{B} add L_{C} cyclically. Let

U_{A} = reflection of L_{A} in AA'

U_{B} = reflection of L_{A} in BB'

U_{C} = reflection of L_{A} in CC'

V_{A} = reflection of L_{B} in AA'

V_{B} = reflection of L_{B} in BB'

V_{C} = reflection of L_{B} in CC'

W_{A} = reflection of L_{C} in AA'

W_{B} = reflection of L_{C} in BB'

W_{C} = reflection of L_{C} in CC'

T_{A} = triangle formed by the lines in U_{A}, U_{B}, U_{C}

T_{B} = triangle formed by the lines in V_{A}, V_{B}, V_{C}

T_{C} = triangle formed by the lines in W_{A}, W_{B}, W_{C}

O_{A} = circumcenter of T_{A}, O_{B} = circumcenter of T_{A}, O_{C} = circumcenter of T_{A}, O = X(3) = circumcenter of ABC. The points O, O_{A}, O_{B}, O_{C} are concyclic. The center of their circle is X(5495). (Antreas Hatzipolakis, May 28, 2013)

For the construction and discussion, see Concyclic Circumcenters.

X(5495) lies on these lines: (pending)

(Encyclopedia of Triangle Centers )

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a(b + c)[a^{5} - 2a^{3}(b^{2} + c^{2}) - a^{2}bc(b+c) + a(b^{4} - b^{2}c^{2} + c^{4}) + bc(b + c)(b - c)^{2} (Angel Montesdeoca, May 29, 2013)

Let ABC be a triangle and let A'B'C' be the cevian triangle of the incenter, X(1). Let L_{A} be the line through A' perpendicular to line AA', and define L_{B} add L_{C} cyclically. Using the notation at X(5495), let M_{A} be the line parallel to U_{A} through B', and define M_{B} and M_{C} cyclically. Let A'' = M_{B}∩M_{C}, and define B'' and C'' cyclically. Let O_{A} = circumcenter of A''B'C', and define O_{B} and O_{C} cyclically. Then the points X(1), O_{A}, O_{B}, O_{C} are concyclic, and the center of their circle is X(5496). (Antreas Hatzipolakis, May 29, 2013)

For a discussion, see Concurrent Circles.

X(5496) lies on these lines: (pending)

(Encyclopedia of Triangle Centers )

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a^{7} - a^{6}(b + c) - a^{5}(b + c)^{2} + a^{4}(2b^{3} + b^{2}c + bc^{2} + 2c^{3}) - a^{2}(b^{4} - b^{3}c - 3b^{2}c^{2} - bc^{3} + c^{4}) + abc(b^{2} + c^{2})(b^{2} - c^{2})^{2} (Angel Montesdeoca, May 29, 2013)

Let ABC be a triangle and let A'B'C' be the cevian triangle of the incenter, X(1). The circles O_{A}, O_{B}, O_{C} defined at X(5496) concur in X(5497). (Antreas Hatzipolakis, May 29, 2013)

For a discussion, see Hechos Geométricos en el Triángulo.

X(5497) lies on these lines: (pending)

(Encyclopedia of Triangle Centers )

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = 2a^{10} - 5a^{8}(b^{2} + c^{2}) + 2a^{6}(b^{4} + 5b^{2}c^{2} + c^{4}) + a^{4}(4b^{6} - 5b^{4}c^{2} - 5b^{2}c^{4} + 4c^{6}) - a^{2}(b^{2} - c^{2})^{2}(4b^{4} + 5b^{2}c^{2} + 4c^{4}) + (b^{2} - c^{2})sup>4(b^{2} + c^{2})
(Angel Montesdeoca, May 30, 2013)

Let ABC be a triangle, let N_{A} be the nine-point center of the triangle BCO, where O = X(3), and define N_{B} and N_{C} cyclically. The nine-point center of the triangle N_{A}N_{B}N_{C} is X(5498), which lies on the Euler line of ABC. (Antreas Hatzipolakis, May 30, 2013)

X(5498) lies on these lines: (2,3}, (more pending)

(Encyclopedia of Triangle Centers )

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a^{5}(b^{2} + 4bc + c^{2}) - a^{4}(b^{3} + b^{2}c - bc^{2} + c^{3}) + a^{3}(2b^{4} + 3b^{3}c + 3bc^{3} + 2c^{4}) + 2a^{2}(b^{5} - b^{3}c^{2} - b^{2}c^{3} + c^{5}) + a(b^{2} - c^{2})^{2}(b^{2} - bc + c^{2}) - (b - c)^{4}(b + c)^{3} (Angel Montesdeoca, May 30, 2013)

Let I_{A} be the A-excenter of a triangle ABC and let N_{A} be the nine-point center of I_{A}BC. Define N_{B} and N_{C} cyclically. The circumcenter of N_{A}N_{B}N_{C} is X(5499), which lies on the Euler line of ABC. (Antreas Hatzipolakis, May 30, 2013)

X(5499) lies on these lines: (2,3}, (more pending)

(Encyclopedia of Triangle Centers )

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) =

2a^{22}

- 15a^{20}(b^{2} + c^{2})

+ 6a^{18}(8b^{4} + 13b^{2}c^{2} + 8c^{4})

- a^{16}(81b^{6} + 52b^{4}c^{2} + 152b^{2}c^{4} + 81c^{6})

+ a^{14}(64b^{8} + 111b^{6}c^{2} + 128b^{4}c^{4} + 111b^{2}c^{6} + 64c^{8})

+ a^^{12}(14b^{10} + 29b^{8}c^{2} + 36b^{6}c^{4} + 36b^{4}c^{6} + 29b^{2}c^{8} + 14c^{10})

- a^{10}(84b^^{12} + 67b^{10}c^{2} + 56b^{8}c^{4} + 48b^{6}c^{6} + 56b^{4}c^{8} + 67b^{2}c^{10} + 84c^{12})

+ a^{8}(82b^{14} - 23b^{12}c^{2} - 31b^{10}c^{4} - 19b^{8}c^{6} - 19b^{6}c^{8} - 31b^{4}c^{10} - 23b^{2}c^^{12}+ 82c^{14})

- a^{6}(b^{2} - c^{2})^{2}(34b^{12} + 11b^{10}c^{2} - 30b^{8}c^{4} - 35b^{6}c^{6} - 30b^{4}c^{8} + 11b^{2}c^{10} + 34c^{12})

+ a^{4}(b^{2}- c^{2})^{4}(b^{10} - 2b^{8}c^{2} - 22b^{6}c^{4} - 22b^{4}c^{6} - 2b^{2}c^{8}+ c^{10})

+ a^{2}(b^{2} - c^{2})^{6}(4b^{8} + 5b^{6}c^{2} + 8b^{4}c^{4} + 5b^{2}c^{6} + 4c^{8})

- (b^{2} - c^{2})^{8}(b^{6} + b^{4}c^{2} + b^{2}c^{4} + c^{6}) (Angel Montesdeoca, May 30, 2013)

Let A'B'C' be the antipedal triangle of the nine-point center, N = X(5) of a triangle ABC. Let N_{A} be the nine-point center of NB'C', and define N_{B} and N_{C} cyclically. The nine-point center of N_{A}N_{B}N_{C} is X(5500), which lies on the Euler line of ABC. (Antreas Hatzipolakis, May 30, 2013)

X(5500) lies on these lines: (2,3}, (more pending)

(Encyclopedia of Triangle Centers )

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) =

-2a^16+ 9a^14(b^2+c^2)-
a^10(b^6+b^4c^2+b^2c^4+c^6)+
a^8(25b^8+10b^6c^2+8b^4c^4+10b^2c^6+25c^8) +
a^6(-33b^10+31b^8c^2+11b^6c^4+11b^4c^6+31b^2c^8-33c^10)+
a^4(b^2-c^2)^2(21b^8-20b^6c^2-25b^4c^4-20b^2c^6+21c^8)-
a^2(b^2-c^2)^4(7b^6-13b^4c^2-13b^2c^4+7c^6)+
(b^2-c^2)^6(b^4-4b^2c^2+c^4)-a^12(13b^4+18b^2c^2+13c^4) (Angel Montesdeoca, June 2, 2013)

Let N be a the nine-point center of triangle ABC. Let N_{A} be the nine-point center of NBC, and define N_{B} and N_{C} cyclically. The circumcenter of N_{A}N_{B}N_{C} is X(5501), which lies on the Euler line of ABC. (Antreas Hatzipolakis, June 2, 2013)

See For a discussion, see Hechos Geométricos en el Triángulo.

X(5501) lies on these lines: (2,3}, (more pending)

(Encyclopedia of Triangle Centers )

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) =
a^{2}(a^{2} - b^{2})[a^{2} - c^{2})(a^{6} - a^{4}(b^{2} + c^{2}) + a^{2}(a^{2} -b^{2})(a^{2} - c^{2}) + 3(b^{2} - c^{2})^{2}(b^{2} + c^{2})] (Angel Montesdeoca, June 3, 2013)

Let L be the Euler line of a triangle ABC. Let L_{A} be the reflection of L in line BC, and define L_{B} and L_{C} cyclically. Let A' = L∩BC, and define B' and C' cyclically. The circles whose diameters are the segments AA', BB', CC' are coaxial. Let D be their coaxial axis (the line X(4)X(74)); let D_{A} be the reflection of D in line BC, and define D_{B} and D_{C} cyclically. Let H_{A} = L_{B}∩D_{C}, and define H_{B} and H_{C} cyclically. Let M_{A} = L_{C}∩D_{B}, and define M_{B} and M_{C} cyclically. The triangles H_{A}H_{B}H_{C} and M_{A}M_{B}M_{C} are perspective, and their perspector is X(5502). (Antreas Hatzipolakis, June 3, 2013)

See For a discussion, see Hechos Geométricos en el Triángulo.

X(5502) lies on these lines: {3,64}, {110, 351}

(Encyclopedia of Triangle Centers )

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = 1/[(b^{2} - c^{2})(27^{1/2}b^{2}c^{2}S_{A} + S(S^{2} + 9S_{A}S_{A})

Let AP, BP, CP be the cevians of a point P in the plane of a triangle ABC. Let A_{B} be a point on BP and A_{C} a point on CP such that the triangle AA_{B}A_{C} is equilateral. The lines A_{B}A_{C}, B_{C}B_{A}, C_{B}C_{A} concur in X(5618). Moreover, the centers of the three equilateral triangles are collinear with P; denote their line by L(P). If P is on the circumcircle of ABC, then L(P) passes through X(110). (Angel Montesdeoca, November 3, 2013)

For the construction and generalizations, see Hechos Geométricos en el Triángulo.

X(5618) lies on the circumcircle and these lines: {13,74}, {115,2378}, {1989,2380}

(Encyclopedia of Triangle Centers )

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = (b + c)[a^{6} - a^{4}(b^{2} + c^{2}) - a^{2}(b^{4} + c^{4} - 3b^{2}c^{2}) - 2abc(b + c)(b - c)^{2} + (b + c)^{2}(b - c)^{4}]

**X(5620) = R*X(65) - (2r + R)*X(1365)
**

Let A'B'C' be the excentral triangle of ABC. Let N

X(5620) lies on these lines:

{1,149},{10,1109},{36,759},{37,115},{65,1365},{162,1838},{267,3336},{897,1738},{1054,1247},{1737,2166},{2218,2915}

X(5620) = isogonal conjugate of X(5127)

(Encyclopedia of Triangle Centers )