8. Let ABC be a triangle and P a point. The reflection of the line AP about BC intersects AB, AC in Ac, Ab respectively. Denote by (Ha) the hyperbola passing through the midpoints of BC, AAb, AAc and whose asymptotes are parallel to the sidelines AB, AC. Define (Hb) and (Hc) similarly. The centers of (Ha), (Hb), (Hc) are collinear if and only if P lies on K060. (Angel Montesdeoca, 20140606
Hechos Geométricos).
(K060 del catálogo de Bernard Gibert)
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 20131130. 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 20131130. Hechos Geométricos ).
(Q066 de Higher Degree Related Curves. Bernard Gibert)
3. See also http://amontes.webs.ull.es/otrashtm/HechosGeometricos.htm#HG091114.
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 20130806, 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 20130806). 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 circmrectangular hyperbolas passing through Ga, Gb, Gc (Angel Montesdeoca, Hyacinthos #21806).
(K211 del catálogo de Bernard Gibert)
In the reference triangle ABC, let P be a point and DEF the circumcevian triangle of P.
Ab is the center of the circle passing through D and tangent to AB at B. Ac is the center of the circle passing through D and tangent to AC at C. Similarly define Bc, Ba and Ca, Cb.
The triangle A'B'C' bounded by the lines AbAc, BcBa, CaCb is not degenerate and perspective to ABC if and only if P lies on the Jerabek strophoid (Angel M., private message 20140508).
Note that ABC and A'B'C' are also orthologic.
The locus of the orthology centers of ABC with respect to A'B'C', when P lies on the Jerabek strophoid (K039), is Ehrmann strophoid (K025).
These triangles are parallelogic if and only if P lies on the focal cubic nK(X571, X1994, X110) whose singular focus is the same as that of K039 i.e. F as above.
(K039 del catálogo de Bernard Gibert) (HGT)
The GK line (when defined) of a triangle is the line passing through its centroid and its symmedian point.
Let P be a point. Denote by Bc the intersection of AB with the parallel line to AC through P, by Cb the intersection of AC with the parallel line to AB through P. The other intersections Ca, Ac, Ab and Ba are defined cyclically. The GK lines of PBcCb, PCaAc and PAbBa concur if and only if P lies on the Napoleon sextic Q076 (Angel Montesdeoca 20140928, ADGEOM #1853).

Geometric properties:
The locus of P such that the nine point center of the cevian triangle of P lies on the line OP is a septic passing through the vertices of the triangle ABC (which are triple points) and those of the medial, antimedial and excentral triangles (Angel Montesdeoca, private message, 20141129 and also Anopolis #2003, Hatzipolakis).
See other locus properties of K044 in the page K646 and here (Angel Montesdeoca) and also K612 and K674 (César Lozada).
A generalization by Angel Montesdeoca
Let GaGbGc be the antimedial triangle and Q a fixed point. L(Q) is a variable line passing through Q.
The locus of the center of the hyperbola circumscribed to GaGbGc and having L(Q) as an asymptote is the cubic cK(#Q, X2).
This cubic is also the locus of the intersection S of L(Q) and the circumparabola whose axis is parallel to L(Q), S being the center of the hyperbola above.
nK(Q²,X_{2},Q)
Examples :
K015 = cK(#X2, X2) = nK(X2, X2, X2)
K052 = cK(#X99, X2) = nK(X4590, X2, X99)
K406 = cK(#X4, X2) = nK(X393, X2, X4)
QAP38 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 Ref13.
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 QuadriFigures )
A circumscribed QADTConic is a conic through the vertices of the QADiagonal Triangle QATr1.
There is a special property for these conics:
Let Sij be the intersection, other than the vertex of the QADiagonal Triangle, of the QADTConic 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 QADTConicPerspector.
This subject was being developed by the specific observation of Angel Montesdeoca in QACi1 and QAP38. It was generalized by observations of Eckart Schmidt, Randy Hutson and the author of EQF (July, 2012).
Properties:
When QADTConic = QACi1, then the QADTConicPerspector is QAP38.
(Encyclopedia of QuadriFigures )
• Let P1P2P3P4 be a Quadrangle. Let Qi (i=1,2,3,4) be the center of the circumconic to Diagonal Triangle with perspector Pi. QAP16 is the common intersection point of lines Pi.Qi (Angel Montesdeoca, January 18, 2015).
(Hechos Geométricos en el Triángulo)
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 21/08/13 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 ninepoint circle of AB'X(1), and define N_{BC} and N_{CA} cyclically. Let N_{AC} be the ninepoint 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 24/08/13 and Anopolis 860
f(a,b,c)= (b/a)^(1/3).
At X(8183), a degenerate conic is defined by using k = (a + b + c)a^(1/3)b^(1/3)c^(1/3). The conic consists of two lines whose trilinear polars are PU(124).. Contributed by Angel Montesdeoca, October 9, 2015.
See Hechos Geometricos 09/10/15.
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(b + c  a)(3b^{2} + 3c^{2}  2a^{2} + ab + ac  6bc)
X(5048) = (R  3r)*X(1) + r*X(3)
Let I be the incenter of a triangle ABC, let N_{A} be the ninepoint circle of the triangle IBC, and define N_{B} and N_{C} cyclically. Let R_{A} be the reflection of N_{A} in the line AI, and define R_{B} and R_{C} cyclically. Then X(5048) is the radical center of the circles R_{A}, R_{B}, R_{C}. (Angel Montesdeoca, Hyacinthos #22502, July 7, 2014.)
X(5048) lies on these lines:
{1, 3}, {8, 1392}, {11, 519}, {33, 1878}, {78, 3893}, {145, 1837}, {210, 3872}, {495, 4870}, {497, 3241}, {513, 4162}, {515, 1317}, {535, 3058}, {950, 3635}, {960, 4861}, {1318, 1320}, {1387, 1737}, {1391, 1870}, {1478, 3656}, {1836, 3476}, {2170, 2348}, {2269, 3723}, {3021, 3328}, {3318, 3319}, {3486, 3623}, {3655, 4302}, {3683, 3877}, {3693, 4919}, {3711, 4915}
(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_{B} be the reflection of A' in line BB', and define B_{C} and C_{A} cyclically. Let A_{C} be the reflection of A' in line CC', and define B_{A} and C_{B} cyclically. Let L be the Euler line of ABC, let L_{A} be the Euler line of AA_{B}A_{C}, and define L_{B} and L_{C} cyclically. Let M_{A} be the reflection of L_{A} in AA', and define M_{B} and M_{C} cyclically. The lines M_{A}, M_{B}, M_{C} concur in X(5494). Moreover, the four Euler lines L, L_{A}, L_{B}, L_{C} are parallel, concurring in X(30). (Antreas Hatzipolakis, May 25, 2013)
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 ninepoint center of the triangle BCO, where O = X(3), and define N_{B} and N_{C} cyclically. The ninepoint 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 Aexcenter of a triangle ABC and let N_{A} be the ninepoint 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 ninepoint center, N = X(5) of a triangle ABC. Let N_{A} be the ninepoint center of NB'C', and define N_{B} and N_{C} cyclically. The ninepoint 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^833c^10)+
a^4(b^2c^2)^2(21b^820b^6c^225b^4c^420b^2c^6+21c^8)
a^2(b^2c^2)^4(7b^613b^4c^213b^2c^4+7c^6)+
(b^2c^2)^6(b^44b^2c^2+c^4)a^12(13b^4+18b^2c^2+13c^4) (Angel Montesdeoca, June 2, 2013)
Let N be a the ninepoint center of triangle ABC. Let N_{A} be the ninepoint 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) = = 1/[(b^{2}  c^{2})(27^{1/2}b^{2}c^{2}S_{A}  S(S^{2} + 9S_{A}S_{A})
The negative Montesdeoca equilateral triangles for a point P are constructed as follows: in the construction of the positive Montesdeoca equilateral triangles atX(5618), replace the rotation angles (30, 60, 60) by (30, 60, 60). Barycentrics for X(5619) are obtained from those of X(5618) by replacing S by  S. (Peter Moses, November 8, 2013)
If you have The Geometer's Sketchpad, you can view Montesdeoca Equilateral Triangles.
X(5619) lies on the circumcircle and these lines: {14,74}, {115,2379}, {1989,2381}
(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_{A} be the ninepoint center of A'BC, and let O_{A} be the circumcircle of N_{A}BC. Define O_{B} and O_{C} cyclically. The circles O_{A}, O_{B}, O_{C} concur in X(5620). (Angel Montesdoca, Anapolis #1120, November 2013: see Concurrent Circumcircles)
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 )
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = S_{B}S_{C}(5a^{2}S_{A}  S_{B}S_{C})
Let ABC be a triangle, let P_{A} be the polar of A with respect to the circle with diameter BC, and define P_{B} and P_{C} cyclically. Let A_{B} = P_{A}∩AB and A_{C} = P_{A}∩AC, and define B_{C}, C_{A}, B_{A}, and C_{B} cyclically. The six points A_{B}, A_{C}, B_{C}, B_{A}, C_{A}, C_{B} lie on, and define, the Montesdeoca conic. (Angel Montesdeoca, June 23, 2014)
A barycentric equation for the Montesdeoca conic is found from A_{B} = S_{C} : 0 : 2S_{A} and A_{C} = S_{B} : 2S_{A} : 0 to be as follows:
2(S^{2}_{A}x^{2} + S^{2}_{B}y^{2} + S^{2}_{C}z^{2})  5(S_{B}S_{C}yz + S_{C}S_{A}zx + S_{A}S_{B}xy) = 0 (Peter Moses, June 23, 2014)
The Montesdeoca conic is the anticevianintersection conic when P = X(4); this conic is defined by Francisco J. Garcia Capitán ( The Anticevian Intersection Conic and Hyacinthos #20547 (December 19, 2011). Also, the perspector of the Montesdeoca conic is X(4).
X(5702) lies on these lines:
{4,6},{297,5032},{340,1992},{376,3284},{468,5304},{578,3183},{631,5158},{3163,3545}
(Encyclopedia of Triangle Centers )
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = (a^4+5 a^2 b^2+b^44 a^2 c^24 b^2 c^2+c^4) (a^44 a^2 b^2+b^4+5 a^2 c^24 b^2 c^2+c^4)
Suppose that P = p : q : r (barycentrics) and P* are a pair of isogonal conjugate points in the plane of a triangle ABC. Let A'B'C' be the pedal triangle of P and A''B''C'' the pedal triangle of P*. Let A* be the reflection of A' in line B''C'', and define B* and C* cyclically. In Hyacinthos (October 3, 2014), Antreas Hatzipolakis asks for the locus of P for which the circumcircles of A*BC, B*CA, C*AB concur. Angel Montesceoca responds that the three circumcircles concur for all choices of P. He further notes that if P is not on the circumcircle and not on the line at infinity, then the point Q of conurrence of the three circles has barycentrics Q(a,b,c,p,q,r) : Q(b,c,a,q,r,p) : Q(c,a,b,r,p,q) given by
Q(a,b,c,p,q,r) = (q + r)/[a^{4}qr(p + q)(p + r)  2a^{2}(q + r)(r + p)(p + q)(c^{2}q + b^{2}r) + p(q + r)(b^{4}r(p + q) + c^{4}q(q + r) + 2b^{2}c^{2}(q^{2} + r^{2} + pq + qr + rp))]
Writing Q as Q(P), the occurrence of (i,j) in the following list means that Q(X(i)) = X(j): (1,5620), (2,6094), (3, 1263), (8, 6095), (20, 265), (69, 6096), (1138, 1138). In particular, Q(X(2)) = X(6094).
X(6094) lies on these lines: {6,543},{263,2854}
X(6094) = isogonal conjugate of X(352)
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a (a^3a^2 ba b^2+b^32 a^2 c+7 a b c2 b^2 c2 a c^22 b c^2+c^3) (a^32 a^2 b2 a b^2+b^3a^2 c+7 a b c2 b^2 ca c^22 b c^2+c^3)
X(6095) = Q(X(8)); see X(6094).
X(6095) lies on these lines: {1,121}, {56,2802}, {106,1739}
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a (a^3a^2 ba b^2+b^32 a^2 c+7 a b c2 b^2 c2 a c^22 b c^2+c^3) (a^32 a^2 b2 a b^2+b^3a^2 c+7 a b c2 b^2 ca c^22 b c^2+c^3)
X(6096) = Q(X(69)); see X(6094).
X(6096) lies on these lines: (pending)
X(6096) = isogonal conjugate of X(5913)
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = (a^2 (a^2  b^2  b c  c^2) (a^5 (b + c)  2 a^3 (b^3 + c^3)  a^2 b c (b^2 + c^2) + a (b^5  b^4 c  b c^4 + c^5)+ b c (b^2  c^2)^2)
X(6097) = (r^{2} + 2rR  R^{2} + s^{2})*X(3) + R^{2}*X(4)
barycentrics, Peter Moses, October 3, 2014; combo, Angel Montesdoca, October 3, 2014
Let ABC be a triangle and A'B'C' the cevian triangle of X(1). Let O_{AB} be the circumcenter of ABA', and define O_{BC} and O_{CA} cyclically; let O_{AC} be the circumcenter of ACA', and define O_{BA} and O_{CB} cyclically. Let O_{A} be the circumcenter of triangle AO_{AB}O_{AC}, and define O_{B} and O_{C} cyclically. Hatzipolakis proposed, and Montesdeoca proved, that the Euler lines concur, in X(186), and that the orthocenter of triangle O_{A}O_{B}O_{C}, which is X(6097), lies on the Euler line. See Anthrakitis (October 3, 2014)
X(6097) lies on these lines: {2,3},{35,500},{55,5453},{511,5495},{3724,5492}
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a (a^2  b^2 + b c  c^2)
(a^2 (b + c)  2 a b c  b^3 + b^2 c + b c^2  c^3) / ((b^2 + c^2  a^2) (a^6  a^4 (b^2  b c + c^2)  a^3 b c (b + c)  a^2 (b^4  b^3 c  2 b^2 c^2  b c^3 + c^4) + a b (b  c)^2 c (b + c) + (b  c)^4 (b + c)^2))
Let A'B'C' be the antipedal triangle of the incenter. Let A'' = X(1986)ofA'BC, and define B'' and C'' cyclically. Let O_{A} be the circumcenter of A'BC, and define O_{B} and O_{C} cyclically. The circumcircles of the four triangles A''B''C'', A''BC, AB''C, ABC'' concur in X(6098). See Hyacinthos 22617, October 8, 2014.
X(6098) lies on these lines: (pending)
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a^2/((b  c) (a^3 b  a^2 b^2  a b^3 + b^4 + a^3 c + a b^2 c  a^2 c^2 + a b c^2  2 b^2 c^2  a c^3 + c^4))
Let A'B'C' be the antipedal triangle of the incenter. Let A'' = X(1986)ofA'BC, and define B'' and C'' cyclically. Let O_{A} be the circumcenter of A'BC, and define O_{B} and O_{C} cyclically. The circumcircles of the four triangles ABC, AB''C'', A''BC'', A"B''C concur in X(6099). The lines O_{A}A'', O_{B}B'', O_{C}C'' also concur in X(6099). See Hyacinthos 22617, October 8, 2014.
Let T be the triangle whose sidelines are the reflections of the line X(3)X(11) in the sidelines of ABC. Then T is perspective to ABC, and X(6099) is the perpsector. (Randy Hutson, October 16, 2014)
X(6099) lies on these lines: (pending)
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a (a^14 (b + c)  2 a^13 (b + c)^2  (b  c)^6 (b + c)^5 (b^2 + c^2)^2 + a^12 (3 b^3 + 5 b^2 c + 5 b c^2  3 c^3) + 2 a^11 (b + c)^2 (4 b^2  5 b c + 4 c^2) + a^10 (b^5  21 b^4 c + b^3 c^2 + b^2 c^3  21 b c^4 + c^5)  10 a^9 (b^2  c^2)^2 (b^2  b c + c^2) + a^8 (5 b^7 + 15 b^6 c  21 b^5 c^2 + 21 b^4 c^3 + 21 b^3 c^4  21 b^2 c^5 + 15 b c^6 + 5 c^7) + 2 a^7 b c (10 b^6 + 7 b^5 c + 4 b^4 c^2  18 b^3 c^3 + 4 b^2 c^4 + 7 b c^5  10 c^6) + a^6 (5 b^9 + 15 b^8 c + 4 b^7 c^2  32 b^6 c^3 + 22 b^5 c^4 + 22 b^4 c^5  32 b^3 c^6 + 4 b^2 c^7 + 15 b c^8  5 c^9) + 2 a^5 (b  c)^2 (5 b^8 + 10 b^7 c  b^6 c^2 + 4 b^5 c^3 + 14 b^4 c^4 + 4 b^3 c^5  b^2 c^6 + 10 b c^7 + 5 c^8)  a^4 (b  c)^2 (b^9 + 23 b^8 c + 16 b^7 c^2 + 28 b^5 c^4 + 28 b^4 c^5 + 16 b^2 c^7 + 23 b c^8 + c^9)  2 a^3 (b^2  c^2)^2 (4 b^8  7 b^7 c + b^6 c^2 + 5 b^5 c^3  12 b^4 c^4 + 5 b^3 c^5 + b^2 c^6  7 b c^7 + 4 c^8) + a^2 (b  c)^4 (b + c)^3 (3 b^6 + 8 b^5 c  4 b^4 c^2 +
18 b^3 c^3  4 b^2 c^4 + 8 b c^5 + 3 c^6) + 2 a (b^2  c^2)^4 (b^6  3 b^5 c + 4 b^4 c^2  6 b^3 c^3 + 4 b^2 c^4  3 b c^5 + c^6))
Let A'B'C' be the antipedal triangle of the incenter. Let A'' = X(1986)ofA'BC, and define B'' and C'' cyclically. Let O_{A} be the circumcenter of A'BC, and define O_{B} and O_{C} cyclically. The points A'', B'', C'', X(6098) lie on a circle, of which the center is X(6100). See Hyacinthos 22617, October 8, 2014.
X(6100) lies on these lines: (pending)
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a^2 (a^6 (b^2 + c^2)  a^4 (3b^4 + 4b^2 c^2 + 3c^4) + a^2(3b^6 + 2b^4 c^2 + 2b^2 c^4 + 3c^6)  b^8 + b^6 c^2 + b^2 c^6  c^8)
Let A_{B} be the reflection of A in line OB, where O = circumcenter of ABC, and define
B_{C} and C_{A} cyclically. Let A_{C} be the reflection of A in line OC, and define B_{A} and C_{B} cyclically. Let A' be the ninepoint center of triangle AA_{B}A_{C}, and define B' and C' cyclically. The circumcircles of the four triangles A'B'C', A'BC, AB'C, ABC' concur in X(6101). (Also, the circumcircles of the four triangles ABC, AB'C', A'BC', A'B'C concur in X(930)). See Hyacinthos 22624, October 10, 2014.
X(6101) lies on these lines:
{2,143},{3,54},{4,2889},{5,141},{20,5663},{22,156},{26,394},{30,5562},{49,323},{51,3628},{52,140},{67,68},{110,2937},{155,1350},{185,548},{389,549},{546,5891},{568,631},{632,3819},{1092,1511},{1147,5944},{1656,3060},{2392,5694},{2781,5609},{3313,3564},{3526,3567},{3627,5907},{5070,5640}
X(6101) = reflection of X(i) in X(j) for these (i,j): (5,1216), (52,140), (185,548), (389,5447), (3627,5907), (5946,3917)
X(6101) = anticomplement of X(143)
X(6101) = {X(i),X(j)}harmonic conjugate of X(k) for these (i,j,k): (3,195,5012), (52,140,5946), (52,3917,140), (389,5447,549), (1092,1658,1511), (3819,5462,632)
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a^2(a^6(b^2+c^2)
3a^4(b^4+c^4)+a^2(3b^62b^4c^22b^2c^4+3c^6)b^8+b^6c^2+b^2c^6c^8
Let A_{B} be the reflection of A in line OB, where O = circumcenter of ABC, and define
B_{C} and C_{A} cyclically. Let A_{C} be the reflection of A in line OC, and define B_{A} and C_{B} cyclically. Let A' be the ninepoint center of triangle AA_{B}A_{C}, and define B' and C' cyclically. Then X(6102) is the ABCorthology center of A'B'C' on the circumcircle of A'B'C'. (Also, X(1141) is the ABCorthology center of ABC on the circumcircle of ABC.) See Hyacinthos 22628 and Hechos Geometricos 121014, October 12, 2014.
Let N_{A} be the reflection of X(5) in the Aaltitude, and define N_{B} and N_{C} cyclically. Then X(6102) is the orthocenter of N_{A}N_{B}N_{C}. Let A'B'C' be the reflection triangle. Let A'' be the trilinear pole, with respect to A'B'C', of the line BC, and define B'' and C'' cyclically. Let A* be the trilinear pole, with respect to A'B'C', of line B''C'', and define B* and C* cyclically. The lines A'A*, B'B*,C'C* concur in X(6102), and the lines A'A'', B'B'', C'C'' concur in X(382). (Randy Hutson, October 16, 2014)
X(6102) lies on these lines:
{3,54},{4,94},{5,389},{24,156},{26,1181},{30,52},{49,186},{51,546},{140,5562},{155,2929},{184,1658},{381,3567},{382,3060},{511,550},{549,1216},{567,1199},{576,2781},{632,5892},{974,1204},{1147,1511},{1614,2070},{1994,3520},{3530,3917},{3627,5446},{3628,5891},{3851,5640}
X(6102) = midpoint of X(i) and X(j) for these (i,j): (52,185), (3,5889)
X(6102) = reflection of X(i) in X(j) for these (i,j): (4,143), (5,389), (5562,140), (3627,5446), (5876,5), (5907,5462)
X(6102) = X(5)ofcircumorthictriangle
X(6102) = {X(i),X(j)}harmonic conjugate of X(k) for these (i,j,k): (4,568,143), (5,389,5946), (184,1658,5944), (389,5907,5462), (5462,5907,5), (5876,5946,5), (5889,5890,3)
Barycentrics a^2(a^6(b^2+c^2)+2a^4(b^2+c^2)^2a^2(4b^6+7b^4c^2+7b^2c^4+4c^6)+b^83b^6c^22b^4c^43b^2c^6+c^8 + 4S^3(2a^2b^2c^2) csc ω) : :
X(8160) = 3 X[3]  X[1671] = 3 X[1670] + X[1671] = 2 X[1671]  3 X[8161] = 2 X[1670] + X[8161]
Let ABC be a triangle, and let A'B'C' be the cevian triangle of X(6). Let U be the line through A' parallel to AB, and let V be the line through A' parallel to AC. Let U = U∩BC and V' = V∩AB. The 4 points B, C, U', V' lie on a circle, (O)_{A}. Define (O)_{B} and (O)_{C} cyclically. Let M be the circle tangent to (O)_{A}, (O)_{B}, (O)_{C} that encompasses them; call M the outer MontesdeocaLemoine circle. Let M' be the circle tangent to (O)_{A}, (O)_{B}, (O)_{C} that is encompassed by each of them; call M' the inner MontesdeocaLemoine circle. Then X(8160) is the center of M, and X(8161) is the center of M'. The contact points of M with (O)_{A}, (O)_{B}, (O)_{C} are the vertices of a triangle perspective to ABC, with perspector X(1343). Likewise, the contact points of M' with (O)_{A}, (O)_{B}, (O)_{C} are the vertices of a triangle perspective to ABC, with perspector X(1342). (Based on notes from Angel Montesdeoca, October 2, 2015)
The points X(8160) and X(8161) are labeled Z_{1} and Z_{2}, respectively, in this sketch:
Hechos Geométricos 02/10/2015.
X(8160) lies on these lines: {3,6}, {30,5404}, {35,3238}, {36,3237}, {140,5403}, {1676,6683}, {2546,7786}
X(8160) = midpoint of X(3) and X(1670)
X(8160) = reflection of X(8161) in X(3)
X(8160) = {X(i),X(j)}harmonic conjugate of X(k) for these (i,j,k): (3,1343,5092), (39,5092,8161)
Barycentrics a^2(a^6(b^2+c^2)+2a^4(b^2+c^2)^2a^2(4b^6+7b^4c^2+7b^2c^4+4c^6)+b^83b^6c^22b^4c^43b^2c^6c^8  4S^3(2a^2b^2c^2) csc ω) : :
X(8161) = 3 X[3]  X[1670] = X[1670] + 3 X[1671] = 2 X[1670]  3 X[8160] = 2 X[1671] + X[8160]
See X(8160).
X(8161) lies on these lines: {3,6}, {30,5403}, {35,3237}, {36,3238}, {140,5404}, {1677,6683}, {2547,7786}
X(8161) = midpoint of X(3) and X(1671)
X(8161) = reflection of X(8160) in X(3)
X(8161) = {X(i),X(j)}harmonic conjugate of X(k) for these (i,j,k): (3,1342,5092), (39,5092,8160)
Barycentrics: a  a^{1/3}b^{1/3}c^{1/3} : b  a^{1/3}b^{1/3}c^{1/3} : c  a^{1/3}b^{1/3}c^{1/3}
(Contributed by Angel Montesceoca, Ocftober 9, 2015)
Hechos Geométricos 09/10/2015.
Let ABC be a triangle and k a real number. Let B_{A} on line AB and C_{A} on line AC be points such that B_{A}C_{A} is parallel to BC at distance kr(A), where r(A) is the inradius of triangle AB_{A}C_{A}. Points C_{B}, A_{B}, A_{C}, B_{C} are defined cyclically. The six points B_{A}, C_{A}, C_{B}, A_{B}, A_{C}, B_{C} lie on a conic, with barycentric equation
0 = cyclic sum of kbc(a + b + c)x^{2}  a(a^{2} + b^{2} + c^{2} + 2bc + 2ca + 2ab + bck^{2})yz
The 4 degenerate real conics are given by these values of k: (a + b + c)/a, (a + b + c)/b, (a + b + c)/c, and (a + b + c)a^{1/3}b^{1/3}c^{1/3}. The 4 singular points of degenerate conics (i.e., points of intersection of the pairs of lines comprising each degenerate conic) are X(8183) and
bc  a^{2} : ba  bc : ca  cb
ab  ac : ac  b^{2} : cb  ca
ac  ab : bc  ba : ba  c^{2}
These last three points are collinear on the trilinear pole of X(86).