Contribuciones en otras web de Geometría del Triángulo

Angel Montesdeoca

(Última actualización: )

================= Laboratorio Virtual de Triángulos con Cabri ================

Problemas en el Laboratorio Virtual de Triángulos con Cabri. Ricardo Barroso.

( http://personal.us.es/rbarroso/trianguloscabri/ )

================= Catalogue of Triangle Cubics CTC Bernard Gibert =================

Locus properties of K001, pK(X6, X30), Neuberg Cubic

16. Let P be a finite point and denote by Ga, Ka the centroid and Lemoine point of triangle PBC. The line GaKa meets BC at A' and B', C' are defined similarly. ABC and A'B'C' are perspective (at Q) if and only if P lies on K001and then Q lies on the line X(2)X(6). (Angel Montesdeoca, 2018-02-12, see here).
(K001 del catálogo de Bernard Gibert)

Locus properties of K001, pK(X6, X30), Neuberg Cubic

21. P is a point (not on the circumcircle) and Ea is the radical axis of the circumcircle and the circle (A, AP). Eb and Ec are defined cyclically. The triangle formed by the three lines Ea, Eb, Ec is perspective to ABC if and only if P is on K001. The perspector, Q=Q(P), lies on the Jerabek hyperbola. If a line L, passing through X(3), intersects K001 at P1 and P2, then Q(P1) = Q(P2) is the second point of intersection of L and the Jerabek hyperbola (Angel Montesdeoca, 2021-05-18)

Locus properties of K002, pK(X6, X2), Thomson Cubic

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)

27. Given a triangle ABC, the parallel through a point P to the sideline BC cuts the perpendiculars to BC though B and C at Ba and Bc respectively. Let A' be the midpoint of segment BaBc. Points Cb, Ca, Ac, Ab, B' and C' are defined cyclically. The lines AA', BB', CC' are concurrent (at a point Q) if and only if P lies on the Thomson cubic K002. The point of concurrence lies on the Darboux cubic K004 (Angel Montesdeoca, 2020-10-25, further details here in Spanish).

( Mostrar/Ocultar figura )

28. Let P and Q be points isogonal conjugates with respect to the triangle ABC. Let A1B1C1 be the circumcevian triangle of P and let A2B2C2 be the cevian triangle of Q. The circles (AA1A2), (BB1B2), (CC1C2) are coaxal if and only if P lies on the Thomson cubic (together with the circumcircle)(Angel Montesdeoca, 2022-01-25). HG230122

Locus properties of K003 McCAY CUBIC, GRIFFITHS CUBIC, pK(X6, X3), a pK60+

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)

Locus properties related to Simson lines of K003 McCAY CUBIC, GRIFFITHS CUBIC, pK(X6, X3), a pK60+

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)

Locus properties related to Simson lines of K003 McCAY CUBIC, GRIFFITHS CUBIC, pK(X6, X3), a pK60+

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)

Locus properties related to Simson lines of K003 McCAY CUBIC, GRIFFITHS CUBIC, pK(X6, X3), a pK60+

Denote by PaPbPc the circumcevian triangle of a point P. The antipedal lines of Pa, Pb, Pc concur if and only if P lies on K003. The point of concurrence is the orthocenter of PaPbPc (Angel Montesdeoca, 2021-01-23). Recall that the antipedal line of a point X on the circumcircle is the line passing through X and its isogonal conjugate X*. It is the perpendicular at X to the Simson line of X.
(Property 33, K003 del catálogo de Bernard Gibert)

Locus properties of K004

Let (C) be any conic inscribed in ABC. Let P be a point with pedal triangle PaPbPc. The tangents (other than the sidelines of ABC) drawn from Pa, Pb, Pc to (C) form a triangle perspective to PaPbPc if and only if P lies on K004 (Angel Montesdeoca, private message, 2016-12-03).
(Hechos Geométricos en el Triángulo)

Locus properties of K005

10. Let PaPbPc be the anticevian triangle of a point P and denote by Qa, Qb, Qc the reflections of Pa, Pb, Pc in the sidelines of ABC respectively. ABC and QaQbQc are perspective (at Q) if and only if P lies on K005. The locus of Q is K060. (Angel Montesdeoca, 2021-08-02).
(Las cúbicas K005 y K060)

Locus properties of K005

13. Let P be a point and let PaPbPc be its anticevian triangle. Denote by Oa, Ob, Oc the circumcenters of the triangles PPbPc, PPcPa, PPaPb. The triangles PaPbPc and OaObOc are perspective (at Q) if and only if P lies on a circular sextic S (together with the polar circle, the line at infinity and the sidelines of ABC). When P traverses S, the locus of Q is K005. Note that S is the square root of K095. (Angel Montesdeoca, 2024-02-27)
(La cúbica de Napoleón-Feuerbach y una séxtica circunscrita a los triángulos anticevianos de sus puntos)

Locus properties of K007 LUCAS CUBIC, pK(X2, X69)

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)

Locus properties of K007 LUCAS CUBIC, pK(X2, X69)

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)

Locus properties of K007 LUCAS CUBIC, pK(X2, X69)

Let PaPbPc be the cevian triangle of a point P and MaMbMc the medial triangle of PaPbPc. ABC and MaMbMc are orthologic when P varies on the Lucas cubic. The loci of the centers of orthology are the Darboux cubic K004 and the Burek-Hutson central cubic K645. (Angel Montesdeoca, private message 2016-09-02). Other related properties to be found in the page K645.

Locus properties of K007 LUCAS CUBIC, pK(X2, X69)

20. Let PaPbPc be the cevian triangle of a point P. Let La be the radical axis of the circles (Pb, BPc) and (Pc, CPb), and define Lb and Lc similarly. The lines La, Lb, Lc concur if and only if P lies on K007 (Angel Montesdeoca, 2021-04-21).
Punto de Prasolov

Locus property of K009 Lemoine cubic

1. Denote by A' the intersection of the line AM with the polar of M in the circumcircle. Let La be the perpendicular at A' to BC, and Lb, Lc are defined similarly. The lines La, Lb, Lc concur (at Q) if and only if P lies on K009 (together with the line at infinity and the circumcircle). When M traverses K009, the locus of Q is a rather complicated circum-quintic passing through X(1), X(3), X(4), X(182), X(2574), X(2575), the excenters, the infinite points of the altitudes, the vertices of the CircumNormal triangle (Angel Montesdeoca, 2022-07-12, further details in Spanish here).

2. Given a triangle ABC, let L be a variable line through the circumcenter that cuts BC, CA, AB at D, E, F, respectively. The circles with diameters AD, BE, CF are coaxial and their radical axis R passes through the orthocenter. The locus of M = L ∩ R is the Lemoine cubic, K009.
Let Ma be the point of intersection of AM with the perpendicular bisector of BC, and Mb, Mc, are defined cyclically. The three points Ma, Mb, Mc are on a line L', which intersects L at P. The locus of P is the Stammler strophoid K038. The envelope of L' is the Kiepert parabola (HG201023 Angel Montesdeoca, 2023-10-21).

Locus properties of K010 SIMSON CUBIC, cK(#X2, X69), nK(X2, X69, X2)

4. The trilinear polar of a point P meets the sidelines of ABC at A', B', C'. Denote by Oa, Ob, Oc the centers of the circles AB'C', BC'A', CA'B' respectively. The triangles ABC and OaObOc are perspective and directly similar. They are homothetic if and only if P lies on K010 (Angel Montesdeoca, 2019-10-12).
(Una propiedad de la cúbica de Simson)

Locus properties of K018 Brocard (second) cubic

13. Let P be a point. The circles (Ba) and (Ca) pass through P and are tangent at B, C to AB and AC respectively. The circles (Cb), (Ab), (Ac) and (Bc) are defined cyclically. Then the points Ba, Ca, Cb, Ab, Ac and Bc lie on a same conic if and only if P lies on K018 (Angel Montesdeoca, 2016-04-28).
(Hechos Geométricos en el Triángulo)

Locus properties of K018

15. With the notations of 13, let A' be the intersection of BC with the radical axis of (Ba), (Ca) and define B', C' likewise. ABC and A'B'C' are perspective (at Q) if and only if P lies on the circumcircle (in which case X99 lies on the line PQ and Q lies on the Steiner ellipse) or P lies on K018 (in which case X2 lies on the line PQ and Q lies on K185). Angel Montesdeoca, 2017-06-01.
(Hechos Geométricos en el Triángulo)

Locus properties related to Simson lines of K024 Kjp, nK0+(X6, X6), a nK60+

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)

Other locus properties of K039 Jerabek strophoid

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 2014-05-08).

( Mostrar/Ocultar figura )

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)

Given a triangle ABC and a point P, the parabola with focus P and directrix BC intersects the A-altitude again at Pa. The points Pb and Pc are defined cyclically.
When P traverses the nine-point circle, ABC and PaPbPc are orthologic at H and another point P' that lies on K039 (Angel Montesdeoca, private message 2023-01-16).
(La estrofoide de Jerabek, lugar de centros ortológicos)

K044: Euler central cubic

See other locus properties of K044 in the page K646 and here (Angel Montesdeoca) and also K612 and K674 (César Lozada).

K052: A2(X115), an Allardice (second) cubic, cK(#X99, X2)

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 circum-parabola whose axis is parallel to L(Q), S being the center of the hyperbola above.
nK(Q²,X₂,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)

Other locus properties of the orthopivotal cubic of the orthopivot X(5), K060

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, 2014-06-06 Hechos Geométricos).
(K060 del catálogo de Bernard Gibert)

Locus properties of the K073=pK(X50, X3) K073

5. Let ABC be a triangle and P a point in its plane. The perpendicular through A to AP intersects the circumcircle at A'. Let (Da) be the directrix of the parabola, (Pa), with focus A and tangent to the lines A'B, A'C. The lines (Db), (Dc) and the parabolas (Pb), (Pc) are defined cyclically. The lines (Da), (Db), (Dc) are concurrent (at Q) if and only if P lies on the cubic K073, and then Q lies on K001. Q is the inverse in (O) of the X(50)-isoconjugate of P, hence O, P, Q are collinear. (Pa), (Pb), (Pc) are also tangent at A1, B1, C1 to the sidelines BC, CA, AB respectively. These points A1, B1, C1 are collinear on (L) if and only if P lies on K004, and then the trilinear pole of (L) lies on K002 (Angel Montesdeoca, 2023-03-17). (Angel Montesdeoca, 2014-06-06 Hechos Geométricos).
(K073 del catálogo de Bernard Gibert)

Locus property of K088 isotomic circular cK, nK(X2, X11054, X2)

Let G, K, S be the points X(2), X(6), X(99) in ETC and let M be a variable point on the line GK. The circumcircle of GSM meets the circum-conic of ABC passing through G and M at G (counted twice since the tangent is the same), M and then another point P. When M traverses GK, the locus of P is K088 (Angel Montesdeoca, 2020-04-06, HG060420).
(K088 del catálogo de Bernard Gibert)

A property by Angel Montesdeoca of K117 (2021-11-18), see here.

Let ABC be a triangle, P a point, C(P) the circum-conic with perspector P and A' the pole of BC with respect to the conic tangent at A to C(P) and passing through B, C, P. The points B' and C' are defined cyclically.
The triangles ABC and A'B'C' are orthologic if and only if P is on the cubic K117. The orthologic centers Q and Q', collinear with P, lie on the cubic K127.

Locus properties of K128

3. See another property here (in Spanish, 2017-11-19, Angel Montesdeoca).

K184 pK(X76, X76) parallel tripolars cubic

See here for another property (Angel Montesdeoca, in Spanish, 2022-03-26).

Locus properties of K191

2. Let A'B'C' be the pedal triangle of a point P. Let Ba, Ca be the orthogonal projections of A' onto lines CA, AB, resp. Define Cb, Ab, Ac, Bc cyclically. The locus of P such that the lines CbBc, AcCa and AbBa concur (at Q) is K191 (Angel Montesdeoca, private message, 2019-03-03).
HGT

Other locus properties of K211 nK(X4, X264, ?)

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)

K219 Allardice cubic A1(G)

See here for other related properties (Angel Montesdeoca, 2019-07-13).

Locus properties of K279 pK(X2, X3260), X(3260) = isotomic conjugate of X(74)

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)

K219 Allardice cubic A1(G)

See here for other related properties (Angel Montesdeoca, 2019-07-13).

K317 = pK(X81, X86)

Locus property (Angel Montesdeoca, 2022-04-15)

Denote by DEF the cevian triangle of P. IC cuts the perpendicular to IB though D at Ab, and IB cuts the perpendicular to IC though D at Ac. Points Bc, Ba and Ca, Cb are defined cyclically. Let A'B'C' be the triangle formed with lines AbAc, BcBa, CaCb.
ABC and A'B'C' are orthologic if and only if P lies on K317. The center of orthology of ABC with respect to A'B'C' lies on K033.

See Las cúbicas K033, K317, K365 (14/4/2022).

K327 Tucker-Poncelet cubic

See here for another property (Angel Montesdeoca, in Spanish, 2022-02-03)

K350 = pK(X3199, X4), the Thomson cubic of the orthic triangle

Locus property (Angel Montesdeoca, 2022-04-25)

Let DEF be the cevian triangle of a point P, the circumcircle of triangle ADB intersect AC again at Ab and the circumcircle of triangle ADC intersect AB again at Ab. Let Oa be the circumcenter of triangle AAbAc. Define Ob, Oc similarly. The lines AOa, BOb, COc concur (at Q) if and only if P lies on K350. The locus of Q is K919.

See La cúbica de Thomson del triángulo órtico ( (25/4/2022).

K365 = pK(X57, X7)

Locus property (Angel Montesdeoca, 2022-04-15)

Denote by DEF the cevian triangle of P. IC cuts the perpendicular to IB though D at Ab, and IB cuts the perpendicular to IC though D at Ac. Points Bc, Ba and Ca, Cb are defined cyclically. The triangle formed with lines AbAc, BcBa, CaCb is perspective to ABC if and only if P lies on K365.

See Las cúbicas K033, K317, K365 (14/4/2022).

K406 = nK(X393, X2, X4)

See here for a property by Angel Montesdeoca, in Spanish, 2023-04-13.

Let ABC be an triangle and L the tangent to circumcircle in a point P, and let La, Lb and Lc be the lines obtained by reflecting L in the lines BC, CA and AB, respectively. Then the circumcircle of the triangle A'B'C' determined by the lines La, Lb and Lc is tangent, at P', to the circumcircle. Let Q be the pole of the line PP' in the circumcircle, then K406 is the locus of isogonal conjugate of Q when P traverses the circumcircle.

Locus properties of K407 cK(#X1,X57), X(57) isogonal conjugate of Mittenpunkt.

3. Let M be a variable point on the incircle and T(M) its tangent at M. The perpendiculars at the incenter I to AI, BI, CI meet T(M) at A', B', C'. The lines AA', BB', CC' concur at P. When M varies on the incircle, the locus of M is K407 (Angel Montesdeoca, 2020-03-20).
See also La cúbica K407.

Locus properties of K634 nK(X6, X6, ?), orthoptic pedal cubic

( Mostrar/Ocultar figura )

( Mostrar/Ocultar figura )

Locus properties of K634 nK(X6, X6, ?), orthoptic pedal cubic

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)

K646 Montesdeoca cubic, pK(X97, X95)

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)

K647 pK(X2052, X264)

Let HaHbHc be the orthic triangle of ABC and a point P. Bap and Cap are the perpendicular feet of B on AP and C on AP respectively.
The triangles BHbBap and HcCCap are perspective at A (by construction) and the perpectrix meets BC at A'. The points B', C' are analogously defined.
The triangles ABC, A'B'C' are perspective if and only if P lies on the Darboux cubic K004. The perspector lies on K647. (Angel Montesdeoca, 2018-08-23, see here).

K656 Bataille acnodal cubic

Locus properties:

6. See here for other related properties (Angel Montesdeoca, 2019-07-13).

Generalization 2 (Angel Montesdeoca) :

Let U = u :v : w be a point with isotomic conjugate tU = 1/u : 1/v : 1/w.

Consider the mapping F which transforms U and tU into the barycentric product of their respective complements cU and ctU. Hence F is given by

V = F(U) = F(tU) = u (v + w)^2 : v (w + u)^2 : w (u + v)^2.

Now, let C(P) be the circum-conic with perspector P = p : q : r and equation p y z + q z x + r x y = 0 whose isotomic transform is the line L(P) with equation p x + q y + r z = 0, clearly the trilinear polar of tP.

F transforms C(P) and L(P) into a nodal cubic K(P) with node N which is the crossconjugate of P^2 and P.

N = p / (- p + q + r) : q / (p - q + r) : r / (p + q - r) = P x taP, where aP is the anticomplement of P.

K(P) meets the sidelines of ABC at the traces of the trilinear polars of P and (taP)^2, the tangents at these latter points being the sidelines of ABC. Note that these trilinear polars are parallel.

K(P) has only three inflexion points and one is always real. These points lie on the line Δ(L) with equation:

Σ (- p + q + r) [ p (- p + q + r) + 2 q r ] x = 0.

When P = X(2), the cubic K(P) is K656 and L(P), C(P) are the line at infinity, the Steiner ellipse respectively. Δ(L) is also the line at infinity.

K917 a nodal circum-cubic.

Locus properties :

See here (Angel Montesdeoca, 2021-02-15, in Spanish)

Denote by DEF the cevian triangle of point P.
Let Db, Dc be the reflections of D about B, C, respectively.
The circle (ADbDc) meets AB and AC again at Ab and Ac , respectively.
Let A2 be the pole of line AbDb in circle (BAbDb) and let A3 the pole of line AcDc in circle (CAcDc).
Define B3, B1 and C1, C2 similarly. A' is the intersection of the lines B3B1 and C1C2, and define B' and C' similarly.
The triangle A'B'C' is not degenerate and perspective to ABC if and only if P lies on the cubic K917.

K932 A cuspidal cubic

We meet K932 in Hyacinthos #26812 (A. P. Hatzipolakis) and here (Angel Montesdeoca, in Spanish).

K934 nK(X6,X3580,X3)

We meet K934 in Hyacinthos #26860 (A. P. Hatzipolakis) and more specifically here (Angel Montesdeoca, in Spanish).

K1155 pK(X6,X523)

Let P be a point and Pa, Pb, Pc its reflections in the sidelines BC, CA, AB of ABC respectively.
The circles with diameters APa, BPb, CPc concur (at X) if and only if P and P* lie on K1155 = pK(X6, X523). (Angel Montesdeoca, see here in Spanish). The locus of X is Q155 where further details are given.

================= Catalogue of Triangle Cubics CTC and Related Curves. Bernard Gibert =================

Locus properties of Q002 Euler-Morley quartic

Other properties:
3. Let P, P* be two isogonal conjugate points and A'B'C' the circumcevian triangle of P*. Let La be the tangent at A' to circle tangent at A to AP and passing through A'. Define Lb, Lc similarly. Q002 is the locus of point P such that the lines La, Lb, Lc concur (together with the line at infinity and the circumcircle). (Angel Montesdeoca, 2021-03-02, further details here in Spanish).

Other locus properties of the Euler-Morley quintic Q003

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).

6. Let P be a point and Na, Nb, Nc the nine-point centers of PBC, PCA, PAB respectively. If P lies on Q003 then the Euler lines of NaNbNc concur at the nine-point center X(5) of ABC (Angel Montesdeoca, 2017-06-29).

(Q003 de Higher Degree Related Curves. Bernard Gibert)

Locus properties of Q012

Locus property
5. Let M1 and M2 be the brocardians of a point M. The perpendicular bisector of M1M2 passes through M if and only if M lies on Q012 (Angel Montesdeoca, 2022-10-11). See here.

Locus properties of Q014

Locus property
Let P be a point and La, Lb, Lc the reflections of the sidelines of ABC in AP, BP, CP respectively. La, Lb, Lc concur if and only if P lies on Q014 (Angel Montesdeoca, 2019-12-05). See here.

Locus properties of Q023

4. Q023 is the locus of P such that the orthocenter H' of the pedal triangle of P lies on the line OP (Angel Montesdeoca, private message 2014-12-13). The locus of H' is a complicated circular septic. See here.
Sea ABC un triángulo, O su circuncentro y P un punto, el lugar geométrico de P tal que el ortocentro de su triángulo pedal queda en la recta OP es la cuártica Q023, del catálogo de Bernard Gibert).

Locus properties of Q024

Let P = p : q : r be a point with circumcevian triangle PaPbPc. Since the circumcircle (O) circumscribes both triangles ABC and PaPbPc, there must be a conic C(P) inscribed in these same triangles.
Note that C(P) is a parabola if and only if P lies on K024 (further comments in Spanish here -Triángulo circunceviano, porismo de Poncelet y la cúbica K024-).

Locus properties of Q026

Q026 is the locus of point P such that
5. the circumcevian triangle of P and the medial triangle are perspective, together with the circumcircle (Angel Montesdeoca, 2022-10-26). The locus of the perspector is a complicated 10th degree curve.
See "La cuártica Q026 y el triángulo medial".

Q030

Angel Montesdeoca gives another locus property related to orthologic triangles as follows. Let ABC be a triangle and P be a finite point. The circle of diameter AP cuts BP, CP at two other points. We denote the line that passes through these two points by La. Similarly, we have the lines Lb and Lc. The triangle A'B'C', formed by these lines, and ABC are orthologic if and only if P lies on Q030 (2024-02-12).
See "La bicircular séptica Q030 ".

Other locus properties of Q037 An inversible bicircular quintic. Bernard Gibert

See another property here (Angel Montesdeoca, 2020-08-17).
Denote by PaPbPc the pedal triangle of a point P that is neither on the circumcircle nor on the line at infinity. Let Pab, Pac be the second points of intersection of the lines PaPb, PaPc with the circumcircle of APbPc. Let Ha be the orthocenter of APabPac, and define Hb and Hc cyclically. Q037 is the locus of point P such that the lines AHa, BHb, CHc are concurrent.

Other locus properties of Q066 quartic

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)

3. See also http://amontes.webs.ull.es/otrashtm/HechosGeometricos.htm#HG091114.

4. The locus of P whose cevian circle passes through X(115) is Q066, together with the Kiepert hyperbola. (Angel Montesdeoca, private message 2016-01-27).

7. Let DEF be the cevian triangle of a point P. The trilinear polar of P meets the sidelines BC, CA, AB at D', E', F', respectively. Let A'B'C' be the triangle formed by the perpendicular bisectors of DD', EE', FF'. The triangles ABC and A'B'C' are directly similar and perspective. The center of similitude and the perspector, Q, lie on the circumcircle. Q066 is the locus of P such that Q is the Tarry point (Angel Montesdeoca, 2024-03-09, see here).

Other locus properties of Q068

• Let P be a point not lying on the line at infinity with pedal triangle A'B'C' and let Q be the isogonal conjugate of P. Denote by (Ca) the conic passing through A, P, Q, B', C' and define (Cb) and (Cc) similarly. (Ca), (Cb), (Cc) share the same tangent at P if and only if P lies on Q068. (Angel Montesdeoca, private message, 2018-02-11. See here).

Locus properties of the Napoleon sextic, Q076

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 2014-09-28, ADGEOM #1853).

( Mostrar/Ocultar figura )

Higher degree curves passing through the equilateral cevian points ( Table 10 Bernard Gibert)

curve	name	points	other points
Q033	X(370)-quartic	P, P	X(2), X(3), X(30)
Q034	X(370)-Fermat quintic	P, P	X(2), X(3), X(13), X(14)
Q035-A-B-C	X(370)-central quartics	P, P	Ga, Gb, Gc
Q036-A-B-C	X(370)-central quintics	P, P	Ga, Gb, Gc
Q074	X(370)-Soddy quintic	P, P	X(2), X(4), Soddy centers, X(616), X(617), etc
Q099	X(370)-quartic	P, P	X(3), X(6), X(1249)
Q100	Dergiades septic	P, P	X(2), X(7), X(369), etc
Q105	Montesdeoca septic	P, P	X(1), X(2), X(3), X(1138), etc

A generalization by Angel Montesdeoca Q098

It is known that the pedal triangles of two points P, Q are orthologic if and only if they are collinear with the circumcenter O of ABC. Obviously, the pedal triangles of O and P are always orthologic hence we will suppose P ≠ O in the sequel.
Let then P be a fixed point and Q a variable point on the line OP with respective pedal triangles PaPbPc and QaQbQc. Denote by O1 the center of orthology of QaQbQc with respect to PaPbPc and by O2 the other center of orthology. When Q traverses the line OP,
• the locus of O1 is a line (L) passing through the orthocenter Hp of PaPbPc,
• the locus of O2 is a rectangular hyperbola (H) passing through P, Pa, Pb, Pc and Hp.
Special cases :
• (L) contains O if and only if P lies on the cubic K389.
• (L) contains P if and only if P lies on Q098.

(Hechos Geométricos en el Triángulo)

Q105: Montesdeoca septic

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, 2014-11-29 and also Anopolis #2003, Hatzipolakis). See HG151114, Séptica que pasa por los "equilateral cevian points".

Q122: Montesdeoca bicircular central quintic

(contributed by Angel Montesdeoca, 2017-05-26, see here)

Q139: A bicircular quintic analogous to Q016

Q139 is mentioned by Angel Montesdeoca in Hyacinthos #26311.

Q169: An isogonal circum-sextic related to K024

A locus property (Angel Montesdeoca, 2022-01-23, HG210122)

PaPbPc denotes the pedal triangle of a point P, with respect to a triangle ABC. The perpendicular through P to PbPc intersects BC at A'. Similarly, points B' and C' are defined on CA and AB respectively. The lines AA', BB', CC' are concurrent (at Q) if and only if P lies on the sextic Q169.

================= Encyclopedia-of-quadri-figures EQF. Chris van Tienhoven =================

QA-P38: Montesdeoca-Hutson Point

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 )

QA-Co/1: QA-DT-Conic Perspector

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 )

QA-P16: QA-Harmonic Center

• Let P1P2P3P4 be a Quadrangle. Let Qi (i=1,2,3,4) be the center of the circum-conic to Diagonal Triangle with perspector Pi. QA-P16 is the common intersection point of lines Pi.Qi (Angel Montesdeoca, January 18, 2015).
(Hechos Geométricos en el Triángulo)

( Mostrar/Ocultar figura )

===================== Bicentric Pairs. Clark Kimberling =========================

P(44) = 1st LAEMMEL POINT

f(A,B,C) = sin B + (cos A sin C - sin A) cos B

Let I = X(1) and P = X(40), and let
DEF = pedal triangle of P,
A_b = AI∩PE, and define B_c and C_a cyclically,
A_c = AI∩PF, and define B_a and C_b cyclically.
The centers of the circles {{A_b, B_c, C_a}} and {{A_c, B_a, C_b}} are the bicentric pair PU(44). (Angel Montesdeoca, April 12, 2021)

Pares bicéntricos asociados a la recta IO

P(110) = 1st MONTESDEOCA RADICAL CENTER

f(a,b,c) = a²(b - 2c) - a(b² + bc - c²) - (b - c)c²

Let A'B'C' be the cevian triangle of X(1). Let A_B be the reflection of A' in BB', and define B_C and C_A cyclically. Let A_C be the reflection of A' in BC', and define B_A and C_B cyclically. Let O_AB be the circumcircle of AA'A_B, and define O_BC and O_CA cyclically. Let O_AC be the circumcircle of AA'A_C, and define O_BA and O_CB cyclically. Then P(110) is the radical center of O_AB, O_BC, O_CA, and the 2nd Montesdeoca circumcircles radical center, U(110) is defined symmetrically; i.e., as the radical center of O_AC, O_BA, O_CB. (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).

P(111) = 2nd MONTESDEOCA RADICAL CENTER

f(a,b,c)= (a + b - c) (a - b + c) (a³ b - a b³ + 2 a³ c - 2 a b² c - b³ c + a² c² - 3 a b c² - 3 b² c² - 2 a c³ - 3 b c³ - c⁴).

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 24/08/13 and Anopolis 860

P(124) = MONTESDEOCA TRILINEAR POLES

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.

P(160) = SIMILICENTER OF 1ST MONTESDEOCA BISECTOR TRIANGLE AND EXCENTRAL TRIANGLE

f(a,b,c)= a(-a^2c + ab(b + c) + bc(2b + c).

In the plane of a triangle ABC, let Ab be the point in which the internal bisector of angle A intersects the perpendicular bisector of segment AB, and define Bc and Ca cyclically.

The triangle AbBcCa, here named the 1st Montesdeoca bisector triangle, is inversely similar to the excentral triangle, A'B'C', with similicenter P(160).

Define AcBaCb symmetrically. This 2nd Montesdeoca bisector triangle is also inversely similar to the excentral triangle, with similicenter U(160).
The bicentric difference of PU(160) is X(1019), the center of the 1st Evans circle.

Let s₁ be the similarity mappping from A'B'C' to AbBcCa, and let s₂ be the similarity mapping from A'B'C' to AcBaCb.

There is a unique point X for which s₁(X) = s₂(X), specifically, X = X(5540), and s1(X(5540)) = X(1083).

Contributed by Angel Montesdeoca, April 25, 2017. See HGT2017 and AdvGeom3769

P(196) = 1ST MONTESDEOCA POINT

Barycentrics b^2 c^2 (b^2-c^2-3 a^2) : :

Contributed by Angel Montesdeoca, February 19, 2021

Let ℓ_ab be the axis of the parabola tangent to AC at A and also tangent to BG at B, and define ℓ_bc and ℓ_ca cyclically. Let
A_b = ℓ_bc∩ℓ_ca
B_c = ℓ_ca∩ℓ_ab
C_a = ℓ_ab∩ℓ_bc.

The triangle A_bB_cC_a is homothetic to ABC, and the center of homothety is P(196).

Let ℓ_ac be the axis of the parabola tangent to AB at A and also tangent to CG at C, and define ℓ_ba and ℓ_cb cyclically. Let
A_c = ℓ_ba∩ℓ_cb
B_a = ℓ_cb∩ℓ_ac
C_b = ℓ_ac∩ℓ_ba.

The triangle A_cB_aC_b is homothetic to ABC, and the center of homothety is the 2nd Montesdeoca point, U = b^2 c^2 (b^2-c^2+3 a^2) : :

The triangles A_bB_cC_a and A_cB_aC_b are homothetic, and their homothetic center is X(41300).
(See Un par bicéntrico sobre la tripolar de X(290))

P and U are a bicentric pair that lie on the line X(2)X(647).
X(850) = bicentric sum of PU
X(2) = bicentric difference of PU
X(237) = crossdifference of PU
X(290) = trilinear pole of the line PU
X(23878) = ideal point of line PU

P(210) = 3RD MONTESDEOCA POINT

Barycentrics b^2 (b^2-c^2) (a^2+b^2-c^2) (a^2-b^2+c^2)^2 : :

Contributed by Angel Montesdeoca, August 8, 2022

In the plane of a triangle ABC, let

A'B'C' = orthic triangle;
M = a variable point on line BC;
M₁ = point on B'C' such that ∠MAM₁ = ∠ A;
M₂ = point on B'C' such that ∠MAM₂ = - ∠ A;
p₁ = line through M₁ perpendicular to B'C';
p₂ = line through M₂ perpendicular to B'C';
h₁ = line through M₁ perpendicular to BC;
h₂ = line through M₂ perpendicular to BC;
H₁ = p₁ ? h₁;
H₂ = p₂ ? h₂;
ℋ₁ = locus of H₁ as M traverses BC;
ℋ₂ = locus of H₂ as M traverses BC;
A₁ = center of ℋ₁, and define B₁ and C₁ cyclically;
A₂ = center of ℋ₂, and define B₂ and C₂ cyclically;
σ₁ = the affine transformation that maps ABC onto A₁B₁C₁;
σ₂ = the affine transformation that maps ABC onto A₂B₂C₂;
F₁ = unique finite fixed point of σ₁;
F₂ = unique finite fixed point of σ₂.
Then F₁ and F₂ are a bicentric pair of points, and
P(210) = F₁= b^2(b^2-c^2)(a^2+b^2-c^2)(a^2-b^2+c^2)^2 : :
U(210) = F₂ = c^2(c^2-b^2)(a^2+c^2-b^2)(a^2-c^2+b^2)^2 : :

For a construction of P(210) and U(210), labeled as σ₁(F₁) and σ₂(F₂), see PU(210)

X(115) = bicentric sum of P(210) and U(210).

(See Un par bicéntrico asociado a puntos fijos propios de afinidades).

P(211) = 4TH MONTESDEOCA POINT

Barycentrics (a+b-3 c) (a+b-c) : :

Contributed by Angel Montesdeoca, November 29, 2022

In the plane of a triangle ABC, let

A'B'C' = medial triangle of ABC;
(I) = incircle;
Ta = line, other than BC, through A' tangent to (I);
Ba = Ta∩AC, and define Cb and Ac cyclically;
Ca = Ta∩AB, and define Ab and Bc cyclically;
Then the triangles CbCaAb and A'B'C' are perspective; let P denote their perspector, and triangles CbAcBa and A'B'C' are perspective; let U denote their perspector. The points P and U are a bicentric pair.

X(28818) = barycentric product P*U
X(4859) = bicentric sum of P and U
X(3667) = bicentric difference of P and U
X(3667) = ideal point of line PU
X(4859) = midpoint of P and U

(See Par bicéntrico asociado a hexágonos circunscritos a la circunferencia inscrita)

P(212) = 5TH MONTESDEOCA POINT

Barycentrics a ((a-c)^2-b^2) (a^3 b (b-c)-a^2 (b^3-4 b c^2+c^3)+a c (b^3-2 b^2 c+c^3)-b^2 (b-c)^2 c) : :

Contributed by Angel Montesdeoca, January 29, 2023

In the plane of a triangle ABC, let
I = X(1) = incenter
A2 = the point, other than B, where line BC meets circle (ABI)
A3 = the point, other than C, where line BC meets circle (ACI)
Ab = BI∩AA3, and define Bc and Ca cyclically
Ac = BI∩AA2, and define Ba and Cb
P = center of perspeconic of ABC and BcCbAc
U = center of perspeconic of ABC and CaAbBc
Then P and U are a bicentric pair given by 1st barycentric as shown above.
(See Un par bicéntrico formado con centros de cónicas)

================= Encyclopedia Triangle Centers ETC. Clark Kimberling ================

X(40) = BEVAN POINT

Barycentrics a (a^3 + a^2 (b + c) - (b - c)^2 (b + c) - a (b + c)^2) : :

Let I = X(1) and P = X(40), and let
DEF = pedal triangle of P,
A_b = AI∩PE, and define B_c and C_a cyclically,
A_c = AI∩PF, and define B_a and C_b cyclically.
The centers of the circles {{A_b, B_c, C_a}} and {{A_c, B_a, C_b}} are the bicentric pair PU(44). (Angel Montesdeoca, April 12, 2021)

X(57) ISOGONAL CONJUGATE OF X(9)
Barycentrics a/(b + c - a) : b/(c + a - b) : c/(a + b - c)

Let V_a, V_b, V_c be the antipodes of V=X(40) in the circles (VBC), (VCA), (VAB), respectively. The lines AV_a, BV_b, CV_c concur in X(57). (Angel Montesdeoca, October 14, 2019). See Antipodales del punto de Bevan y el centro X(57)

Let DEF be the intouch triangle. Let Ia be the internal bisector of angle BAC, and let D' be the point, other than D, where the line through D parallel to Ia meets the incircle. Let A' be the point, other than A, where AD' meets the incircle. Let La be the radical axis of the circumcircles of triangles A'BF and A'CE, and define Lb and Lc cyclically. The lines La, Lb, Lc concur in X(57). See X(57) and HG211219. (Angel Montesdeoca, December 21, 2019)
X(101) = Ψ(INCENTER, SYMMEDIAN POINT)
Barycentrics a²/(b - c) : b²/(c - a) : c²/(a - b)
X(101) = (r ² + 6rR + 8R² + s²)*X(1) - 6R(r + 4R)*X(2) - 2(r² + 4rR - s²)*X(3) (Peter Moses, April 2, 2013)

Let I_aI_bI_c be the excentral triangle. The Brocard axes of BCI_a, CAI_b, ABI_c concur in X(101). (Randy Hutson, February 10, 2016)

X(101) = intersection of Lemoine axes of 1st & 2nd Montesdeoca bisector triangles.
X(175) = ISOPERIMETRIC POINT and X(176) = EQUAL DETOUR POINT
Barycentrics (a + b - c)*(a - b + c)*(a*(a - b - c) + S) : :
and
Barycentrics (a + b - c)*(a - b + c)*(a*(a - b - c) - S) : :

In the plane of a triangle ABC, let

Oa = circle with diameter BC
C' = the point, other than B, where Oa meets AB
B' = the point, other than C, where Oa meets AC
Ub = B-mixtilinear incircle of BCC'
Uc = C-mixtilinear incircle of CBB'
Ab = touchpoint of Ub and BC; define Bc and Ca cyclically
Ac = touchpoint of Uc and BC; define Ba and Cb cyclically
The points Ab, Ac, Ba, Ba, Cb, Cb line on a conic, of which the center is X(176). A barycentric equation for this conic follows:

2 (a-b-c) (a^2-b^2-c^2) (a^2 (b+c)-(b-c)^2 (b+c)+2 a S) x^2 - ((a+b-c) (a-b+c) (3 a^4-2 a^3 (b+c)-3 (b^2-c^2)^2+2 a (b+c) (b^2+c^2))+2 (a^4-3 (b^2-c^2)^2+2 a^2 (b^2+c^2)) S) y z + (cyclic) = 0.

See X(175) and X(176). Continuing, let

Vb = B-mixtilinear excircle of BCC'
Vc = C-mixtilinear excircle of CBB'
A'b = touchpoint of Vb and BC; define B'c and C'a cyclically
A'c = touchpoint of Vc and BC; define B'a and C'b cyclically
The points A'b, A'c, B'a, B'c, C'a, C'b line on a conic, of which the center is X(175). A barycentric equation for this conic follows:

(a+b-c) (a-b+c) (2 (a-b-c) (2 a^3 b c+a^4 (b+c)-2 a b c (b+c)^2+(b-c)^2 (b+c)^3-2 a^2 (b^3+b^2 c+b c^2+c^3)+2 (a^3-2 b c (b+c)-a (b+c)^2) S) x^2 + (3 a^6-4 a^5 (b+c)-4 a (b-c)^2 (b+c)^3-3 (b-c)^2 (b+c)^4+a^2 (b+c)^2 (9 b^2+2 b c+9 c^2)-a^4 (9 b^2+14 b c+9 c^2)+8 a^3 (b^3+b^2 c+b c^2+c^3)-2 (a^4+6 a^3 (b+c)-4 a^2 (b+c)^2+3 (b^2-c^2)^2-2 a (3 b^3+7 b^2 c+7 b c^2+3 c^3)) S) y z) + (cyclic) = 0. (Angel Montesdeoca, September 9, 2022)

See HG050922. X(175) y X(176) centros de cónicas con los mismos puntos en el infinito que la cónica circunscrita de perspector el punto de Gergonne.
X(195) = X(5)-CEVA CONJUGATE OF X(3)
Barycentrics 4 cos 2A + cot²2A - cot A cot ω : :( (M. Iliev, 5/13/07)
a^2(a^8 + b^8 + c^8 - 4a^6(b^2 + c^2) + a^4(6b^4 + 6c^4 + 5b^2c^2) - a^2(4b^6 + 4c^6 - b^4c^2 - b^2c^4) - 2b^2c^2(b^4 + c^4 - b^2c^2)) : : : :

Hyacinthos #24180
Re: N, O, diameters, radical axes, reflections
Antreas Hatzipolakis Aug 28, 2016
[APH]:
Let ABC be a triangle.
Denote:
Ab, Ac = the orthogonal projections of A on NB, NC, resp.
(Oab), (Oac) = the circles with diameters AAb,AAc, resp.
Similarly (Obc), (Oba) and (Oca), (Ocb)
R1 = the radibal axis of (Oab), (Oac)
R2 = the radical axis of (Obc), (Oba)
R3 = the radical axis of (Oca), (Ocb)
The reflections of R1, R2, R3 in BC, CA, AB, resp. are concurrent.
[Angel Montesdeoca]:
The reflections of R1, R2, R3 in BC, CA, AB, resp. are concurrent. at X(195) = X(5)-CEVA CONJUGATE OF X(3)
X(221) = X(1)-CEVA CONJUGATE OF X(56)
Barycentrics a^2 (a^5+a^4 (b+c)-2 a^3 (b^2+c^2)-2 a^2 (b-c)^2 (b+c)+a (b^2-c^2)^2+(b-c)^4 (b+c)) : :

Let I = X(1) = incenter, and A'B'C' = medial triangle. Let A_b = AC∩IA', and define B_c and C_a cyclically. Let A_c = AB∩IA', and define B_a and C_b cyclically. Let O_a be the circumcircle of AA_bA_c, and define O_b and O cyclically. Then X(221) is the radical center of O_a, O_b, O_c. (Angel Montesdeoca, April 27, 2021)
X(221) como centro radical.
X(223) = X(2)-CEVA CONJUGATE OF X(57)
Barycentrics a (a + b - c) (a - b + c) (a^3 + a^2 (b + c) - (b - c)^2 (b + c) - a (b + c)^2) : :

Let V_a, V_b, V_c be the antipodes of V=X(40) in the circles (VBC), (VCA), (VAB), respectively. The triangle V_aV_bV_c is here named the Bevan antipodal triangle (Hyacinthos #29638). The point X(223) is the unique finite fixed point of the affine transformation that maps the reference triangle ABC onto V_aV_bV_c. (Angel Montesdeoca, October 15, 2019). See Antipodales del punto de Bevan y el centro X(57).

The construction of the triangle VaVbVc can be generalized for an arbitrary point P (instead of V=X(40)), and is, in fact, the antipedal triangle of P (Randy Hutson, message in Anopolis Group). Thus, "Bevan antipodal triangle" = antipedal triangle of X(40). (Angel Montesdeoca, October 16, 2019)
X(402) = ZEEMAN-GOSSARD PERSPECTOR
Barycentrics p(a,b,c)y(a,b,c) : :

p(a,b,c) = 2a⁴ - a²b² - a²c² - (b² - c²)²

y(a,b,c) = a⁸ - a⁶(b² + c²) + a⁴(2b² - c²)(2c² - b²) + [(b² - c²)²][3a²(b² + c²) - b⁴ - c⁴ - 3b²c²]

(From Angel Montesdeoca, November 25, 2020) Let P be a variable point on the Euler line and Q its isotomic conjugate. Let TP and TQ be the cevian triangles of P and Q, respectively. Let F be the finite fixed point of the affine transformation that carries TP onto TQ. The envelope of the line PF is a hyperbola (H) with center X(402), an asymptote of the Euler line. A barycentric equation for (H) follows:

cyclic sum [(-b^4+c^4+a^2 (b^2-c^2))^4 (-2 a^4+(b^2-c^2)^2+a^2 (b^2+c^2))^2x^2-2 (a^2-b^2)^2 (a^2-c^2)^2 (a^4-(b^2-c^2)^2)^2 (a^8-a^6 (b^2+c^2)+5 a^2 (b^2-c^2)^2 (b^2+c^2)+a^4 (-3 b^4+7 b^2 c^2-3 c^4)-(b^2-c^2)^2 (2 b^4+5 b^2 c^2+2 c^4))y z ] = 0.
X(427) = COMPLEMENT OF X(22)
Barycentrics tan A + tan ω : :
(a^2+b^2-c^2) (a^2-b^2+c^2) (b^2+c^2) : :

Let A'B'C' be the circummedial triangle. Let A" be the pole, wrt the polar circle, of line B'C', and define B" and C" cyclically. The lines AA", BB", CC" concur in X(427). Moreover, X(427) is the Euler line intercept of radical axis of nine-point circle and every circle with center on orthic axis that is orthogonal to nine-point circle, and X(427) is the point in which the extended trapezoid legs (P(4),P(4)-Ceva conjugate of U(4)) and (U(4),U(4)-Ceva conjugate of P(4)) meet. Also, X(427) is the QA-P38 center (Montesdeoca-Hutson Point) of quadrangle ABCX(2). (Randy Hutson, October 13, 2015)
X(845) = X(165)X(166) ∩ X(173)X(503)
Let Ia, Ib, Ic be the excenters of a triangle ABC. Let A' be the Clawson point of IaBC, and define B' and C' cyclically. The lines AA', BB', CC' concur in X(845). (Angel Montesdeoca, August 12, 2018)
X(900) = CROSSDIFFERENCE OF X(6) AND X(101)
Barycentrics (b - c)(b + c - 2a) : :

Let Ua be the line through X(80) perpendicular to the line AX(80), and define Ub and Uc cyclically. Let Va be the reflection of BC in Ua, and define Vb and Vc cyclically. The lines Va, Vb, Vc are parallel, and they concur in X(900). (Angel Montesdeoca, June 30, 2017)
(Una descripción del punto X(900) )
X(933) = X(4)-CROSS CONJUGATE OF X(250)
Barycentrics a^2((a^2-b^2)^2-(a^2+b^2) c^2) (a^4-b^2 c^2+c^4-a^2 (b^2+2 c^2))/((b^2-c^2) (-a^2+b^2+c^2)) : :

Let P be a point on the circumcircle, and let

(Oab) = circle through A tangent to line BC at B
(Oac) = circle through A tangent to line BC at C
Ba = point other than A where the line PB meets (Oab)
Ca = point other than A where the line PC meets (Oac)
Hab = orthocenter of APBa
Hac = orthocenter of APCa

Define Hbc and Hca cyclically, and define Hba and Hcb cyclically. The points Hab, Hac, Hbc, Hba, Hca, Hcb lies on a hyperbola with center = midpoint(P and H), where H = X(4), the orthocenter. This hyperbola is rectangular if and only if P = X(933). This note is related to a problem posed in 2017 Olympiad of Rusanovsky Lyceum in Kyiv, Ukraine , IX-X Team p9. For details, see El centro X(933) y una hipérbola rectangular. (Angel Montesdeoca, July 3, 2021)
X(953) = ISOGONAL CONJUGATE OF X(952)
Barycentrics a²/(2a⁴ - 2a³(b + c) - a²(b² - 4bc + c²) + (2a - b - c)(b - c)(b² - c²)) : :

X(953) = Miquel point of the sidelines of ABC and the Sherman line. (Angel Montesdeoca, July 24, 2019) See HG250719
X(962) = LONGUET-HIGGINS POINT
Barycentrics a^4 + 2a^3(b + c) - 4a^2bc - (b + c)(b - c)^2(2a + b + c) : :

Let I = X(1) and H = X(4) in a triangle ABC. Let L_b be the perpendicular to B at I, and let L_c be the perpendicular to CI at I. Let A_b = L_b∩AH and A_c = L_c∩AH. Let U be the circle (I, A_b, A_c). Let M_b be the point, other than I, where BI meets U, and let M_c be the point, other than I, where CI meets U. Let ℓ_a be the line M_bM_c, and define ℓ_b and ℓ_c cyclically. The lines ℓ_a, ℓ_b, ℓ_c concur in X(962). (Angel Montesdeoca, March 12, 2020). See Otra construcción del punto de Longuet-Higgins
X(974) = 2nd EHRMANN POINT
Barycentrics a^2 (a^2-b^2-c^2) (a^10 b^2-3 a^8 b^4+2 a^6 b^6+2 a^4 b^8-3 a^2 b^10+b^12+a^10 c^2+2 a^8 b^2 c^2-a^6 b^4 c^2-11 a^4 b^6 c^2+12 a^2 b^8 c^2-3 b^10 c^2-3 a^8 c^4-a^6 b^2 c^4+18 a^4 b^4 c^4-9 a^2 b^6 c^4+3 b^8 c^4+2 a^6 c^6-11 a^4 b^2 c^6-9 a^2 b^4 c^6-2 b^6 c^6+2 a^4 c^8+12 a^2 b^2 c^8+3 b^4 c^8-3 a^2 c^10-3 b^2 c^10+c^12) : :

The pedal triangles of circumcenter and orthocenter with respect to orthic and medial triangles, respectively, are perspective, with perspector X(974). (Angel Montesdeoca, August 29, 2019)

El centro X(974) y triángulos pedales.
X(1001) = MIDPOINT OF X(1) AND X(9)
Barycentrics a(a^2-a(b+c)-2bc) : :

Let A' be the line through X(1) parallel to line BC. Let A_B = A'∩AB and A_C = A'∩AC. Define B_C and C_A cyclically, and define B_A and C_B cyclically. The six points A_B, B_C, C_A, A_B, B_C, C_A lie on a conic whose center is X(1001). (Angel Montesdeoca, April 27, 2016)
(Hechos Geométricos en el Triángulo (2016) )
X(1083) = MIDPOINT OF X(105) AND X(644)
Trilinears a⁴ - a³(b + c) - a²bc + 2abc(b + c) - bc(b² + c²): :

X(1083) = X(112)-of-1st-Montesdeoca-bisector-triangle
X(1083) = X(112)-of-2nd-Montesdeoca-bisector-triangle
X(1083) = similicenter of 1st and 2nd Montesdeoca bisector triangles
X(1125) = COMPLEMENT OF X(10)
Barycentrics 2a+b+c : a+2b+c : a+b+2c

Hyacinthos #24185 Re: I, NPC, Concurrent Euler lines
Antreas Hatzipolakis Aug 28, 2016
[APH]:
Let ABC be a triangle.
Denote:
Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.
Aa, Ab, Ac = the orthogonal projections of Na on IA, IB, IC, resp.
Ba, Bb, Bc = the orthogonal projections of Nb on IA, IB, IC, resp.
Ca, Cb, Cc = the orthogonal projections of Nc on IA,IB, IC, resp.
The Euler lines of AaAbAc, BaBbBc, CaCbCc are concurrent.
[Angel Montesdeoca]:
The Euler lines of AaAbAc, BaBbBc, CaCbCc are concurrent. at X(1125)= the centroid of {A,B,C,X(1)} (Darij Grinberg, 12/28/02)
X(1141) = GIBERT POINT
Barycentrics ( (SA SB+S^2)(SA SC+S^2))/(b^2c^2-4SA^2) :...:...)

See Hyacinthos #24187 Re: O, NPC, Euler lines, parallelogic

Antreas Hatzipolakis Message 2 of 2 , Aug 29, 2016
[APH]:

Let ABC be a triangle.
Denote:
Na, Nb, Nc = the NPC centers of OBC, OCA, OAB, resp.
Aa, Ab, Ac = the orthogonal projections of Na on OA, OB, OC, resp.
Ba, Bb, Bc = the orthogonal projections of Nb on OA, OB, OC, resp.
Ca, Cb, Cc = the orthogonal projections of Nc on OA, OB, OC, resp.
The Euler lines of AaAbAc, BaBbBc, CaCbCc bound a triangle A*B*C*.

ABC, A*B*C* are parallelogic. The parallelogic center (ABC, A*B*C*) lies on the circumcircle.

[Angel Montesdeoca]:

The parallelogic center (ABC, A*B*C*) is X(1141) = Gibert point
X(1141) was first noted (Hyacinthos #1498, September 25, 2000) by Bernard Gibert as a point of intersection of the circumcircle and certain cubic.
X(1156) = ISOGONAL CONJUGATE OF X(1155)

Barycentrics af(a,b,c) : bf(b,c,a) : cf(c,a,b)
where f(a,b,c) = 1/[(b - c)² + a(b + c - 2a)]

Let B_c be the reflection of B in the internal angle bisector of angle C, and let C_b be the reflection of C in the internal angle bisector of angle B. Let Γ_a be the circle that passes through A, B_c, C_b. Let A' be the pole of BC, with respect to Γ_a and define B 'and C' cyclically. Then X(1156) is the perspector of A'B'C' and ABC, and X(35445) is the perspector of A'B'C' and the excentral triangle. (Angel Montesdeoca, January 13, 2021)
See HG120121.
X(1177) = 1st SARAGOSSA POINT OF X(67)
Trilinears f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a/(b⁶ + c⁶ - a⁴b² - a⁴c² + 2a²b²c² - b⁴c² - b²c⁴) (M. Iliev, 5/25/07)

See Angel Montesdeoca, Hyacinthos #21528, 2/12/2013

[Antreas Hatzipolakis]:
3. Let OaObOc be the medial triangle [pedal tr. of O], (N1),(N2),(N3) the reflections of the NPC (N) in the perp. bisectors
OOa,OOb,OOc, resp. and and A'B'C' the triangle bounded by the radical axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
The triangles ABC, A'B'C' are perspective.
Perspector?

[Angel Montesdeoca]:
The triangles ABC, A'B'C' are perspective.
Perspector: X(1177) = 1st SARAGOSSA POINT OF X(67) (M.Iliev, 5/25/07)
X(1320) = ISOGONAL CONJUGATE OF X(1319)
Barycentrics a(b + c - a)/ (b + c - 2a) : :

Let DEF be the intouch triangle. Let H_a be the orthocenter of IBC. Let A₁ be the point, other than A, where AI meets the circumcircle. Let t_a be the tangent to the circle (DH_aA₁) at H_a, and define t_b and t_c cyclically. The lines t_a, t_b , t_c concur in X(1320). (Angel Montesdeoca, September 3, 2020)
Conjugado isogonal del punto Bevan-Schröder
X(1380) = 2nd BROCARD-AXIS INTERCEPT OF CIRCUMCIRCLE
Barycentrics a^6-a^2 b^2 c^2-a^4 Sqrt[a^4+b^4-b^2 c^2+c^4-a^2 (b^2+c^2)] : :

In the plane of a triangle ABC, let
F₁ = X(39162) and F₂ = X(39163); these points are the foci of the Steiner inellipse;
L₁ = trilinear polar of F₁
L₂ = trilinear polar of F₁
A₁ = L₁∩BC, and define B₁ and C₁ cyclically
A₂ = L₂∩BC, and define B₂ and C₂ cyclically
The seven circle (ABC), (AB₁C₁), (BC₁A₁), (CA₁B₁), (AB₂C₂), (BC₂A₂), (CA₁B₂) concur in X(1380).
See X(1380) (Angel Montesdeoca, February 18, 2023).
X(1384) = 2nd GRINBERG HOMOTHETIC CENTER
Barycentrics a^2(b^2 + c^2 - 5a^2) :

Let DEF be the circumsymmedial triangle and let (Oa) be the circle tangent at D to circumcircle and with center Oa=AH∩OD. Define (Ob) and (Oc) cyclically. The radical center of the circles (Oa), (Ob), (Oc) is X(1384). (Angel Montesdeoca, October 2, 2018) HG300918
X(1441) = X(313)-BETH CONJUGATE OF X(313)
Barycentrics b c (b + c)/(a - b - c) : :

In the plane of a triangle ABC, let
I = incenter = X(1)
O = circumcenter = X(3)
DEF = cevian triangle of I
L_b = line through B perpendicular to AI
L_c = line through C perpendicular to AI
A_b = L_b∩AO
A_c = L_c∩AO
O_a = circumcircle of DA_bA_c
A' = the point of intersection, other than D of O_a and AI, and define B' and C' cyclically. Then X(1441) = finite fixed point of the affine transformation that carries ABC onto A'B'C'. (Angel Montesdeoca, May 30, 2023)
See HG300523
X(1493) = NAPOLEON CROSSSUM
Barycentrics sin A(3 sin²A - cos²A)(3 sin B sin C - cos B cos C) : :

Hyacinthos #24179
24179Re: N, NPC, radical axes, reflections
Antreas Hatzipolakis Aug 28,2016
[APH]:
Let ABC be a triangle.
Denote:
Ab, Ac = the orthogonal projections of A on NB, NC, resp.
(Nab), (Nac) = the NPCs of AAbN, AAcN, resp.
Similarly (Nbc), (Nba) and (Nca), (Ncb)
R1 = the radibal axis of (Nab), (Nac)
R2 = the radical axis of (Nbc), (Nba)
R3 = the radical axis of (Nca), (Ncb)
The reflections of R1, R2, R3 in BC, CA, AB, resp. are concurrent.
[Angel Montesdeoca]:
The reflections of R1, R2, R3 in BC, CA, AB, resp. are concurrent at X(1493) = NAPOLEON CROSSSUM
X(1576) = ISOGONAL CONJUGATE OF X(850)
Barycentrics a⁴/(b² - c²) : b⁴/(c² - a²) : c⁴/(a² - b²)

X(1576) is the center of the conic transform of the Stammler quartic (Q066 in Bernard Gibert' catalogue) by X(31)-isoconjugation. This conic is given by the barycentric equation b^4c^4(b^2-c^2)x^2+c^4a^4(c^2-a^2)y^2+a^4b^4(a^2-b^2)z^2 = 0, and it passes through the following triangle centers: X(6), X(31), X(48), X(154), X(1613), X(2578), X(2579), X(5638), X(5639). (Angel Montesdeoca, May 7, 2016)
X(1593) = POINT CEBALRAI
Barycentrics a^2 (a^4 + b^4 + c^4 - 2 a^2 b^2 - 2 a^2 c^2 + 6 b^2 c^2)/(a^2 - b^2 - c^2) : :

In the plane of a triangle ABC, let
Γ= circumcircle
DEF = medial triangle
(A) = circle that passes through E,F, and that intersects Γ in two antipodal points
A' = center of (A), and define B' and C' cyclically.
Then the triangles ABC and A'B'C' are homothetic, and their homothetic center is X(1593). (Angel Montesdeoca, February 14, 2023)
See X(1593) centro de homotecia
( Mostrar/Ocultar figura )
X(1650) = TRIPOLAR CENTROID OF X(525)
Barycentrics (-b^4 + c^4 + a^2 (b^2 - c^2))^2 (2 a^4 - (b^2 - c^2)^2 - a^2 (b^2 + c^2)) : :

Let W be the circumconic with center X(1650). One of the asymptotes of W is the Euler line. The other is in the direction of X(9033). Also, X(9033) is the infinity point of the isotomic line of the Euler line. For a sketch, click X(9033). (Angel Montesdeoca, April 19, 2016)
X(2052) = ISOGONAL CONJUGATE OF X(577)
Barycentrics sec²A : :

Let A'B'C' be the orthic triangle of a triangle ABC, and let O(A) = circle with center A and radius AA', and define O(B) and O(C) cyclically p(A) = polar of X(4) wrt O(A), and define p(B) and p(C) cyclically A'' = p(B)∩p(C), and define B'' and C'' cyclically. Then A''B''C'' is homothetic to ABC, and the center of homothety is X(2052). (Angel Montesdeoca, September 30, 2016)
X(2163) = X(2)-ISOCONJUGATE OF X(45)
Let IaIbIc be the excentral triangle of a triangle ABC. Let A2 and A3 be the points where the perpendicular bisector of BC meets the internal angle bisector of angle ABC and the internal angle bisector of angle BCA, respectively. Let A'2 and A'3 be the points where the perpendicular bisector of BC meets the external angle bisector of angle ABC and the external angle bisector of angle BCA, respectively. Let Oab, Oac be the circumcenters of triangles IbA2A'3, IcA3A'2, respectively. Define Obc, Oba, Oca, Ocb cyclically. Then the triangle having sidelines OabOac, ObcOba, OcaOcb is perspective to ABC, and the perspector is X(2163). See Angel Montesdeoca, HG090218: El centro del triángulo X(2163).
X(2185) = X(2)-ISOCONJUGATE OF X(181)
Barycentrics a (a + b)^2 (a - b - c) (a + c)^2 : :

Let DEF be the cevian triangle of the incenter. Let O_A be the circumcircle of AEF. Let T1 be the line tangent to O_A at E, and let T2 be the line tangent to O_A at F. Let A_B = T1∩BC and A_C = T2∩BC, and let A' be the radical center of the circumcircles of AEF, BFA_B, CAA_C.. Define B' and C' cyclically. The lines AA', BB', CC' concur in X(2185). (Angel Montesdeoca, January 4, 2021)

See Una construcción del X(2185)

Contribution by Angel Montesdeoca, September 10, 2017:
In the plane of a triangle ABC, let
O = X(3) = circumcenter
U(O,q) = circle with center O and radius q
A'B'C' = excentral triangle
A*B*C* = polar triangle of A'B'C' with respect to U(O,q).
There exists a unique q such that the lines AA*, BB*, CC* concur, and the point of concurrence is X(2217).
Moreover, the radius and a barycentric equation for U(O,q) are given as follows:

q^2 = R(r+R) = (a b c (2 a b c+(a+b-c) (a-b+c)(-a+b+c)))/(2 (a+b-c) (a-b+c) (-a+b+c) (a+b+c));

abc(x^2+y^2+z^2) + 2a (a+b) (a+c)y z + 2b(b+c)(b+a)z x + 2c(c+a)(c+b)x y = 0.
(Hechos Geométricos en el Triángulo (2017) )

X(2222) = X(2)-ISOCONJUGATE OF X(654)
Let A'B'C' be the excentral triangle in the plane of a triangle ABC. Let
A'' = X(110)-of-A'BC, and define B'' and C'' cyclically
(Oa) = circle with diameter AA'', and define (Ob) and (Oc) cyclically

The circles (Oa), (Ob), (Oc), and the circumcircle concur in X(2222). (Angel Montesdeoca, September 12, 2016)

(Hechos Geométricos en el Triángulo (2016) )
X(2292) = X(2)-ISOCONJUGATE OF X(1169)
Let FaFbFc be the Fuhrmann triangle, and AbBcCa and AcBaCb be the 1st and 2nd Montesdeoca bisector triangles. X(2292) is the radical center of the circumcircles of FaAbAc, FbBcBa, FcCaCb. (Randy Hutson, December 2, 2017)
X(2334) = X(2)-ISOCONJUGATE OF X(1449)
Barycentrics a^2/(3a + b + c) : b^2/(3b + c + a) : c^2/(3c + a + b)

Let A'B'C' = circumcevian triangle of X(1), and let
Oa = center of A-mixtilinear incircle
Ta = touchpoint of A-mixtilinear incircle and circumcircle
A1 = circumcenter of A'OaTa, and define B1 and C1 cyclically. Then AA1, BB1, CC1 concur in X(2334).
(Angel Montesdeoca, November 29, 2018) El centro del triángulo X(2334)

Let DEF be the cevian triangle of X(8), and let A5 is the QA-P5 center (Isotomic Center) of AEFX(8) (see Chris van Tienhoven, Quadrangle points and other objects). Let B5 be the QA-P5 center of BFDX(8) and C5 the QA-P5 center of CDEX(8). Then the triangles ABC and A5B5C5 are perspective, and X(2334) is their perspector. (Angel Montesdeoca, March 20, 2019).
El centro del triángulo X(2334), otra propiedad

In the plane of a triangle ABFC, let
DEF = cevian triangle of X(1);
Γ = circumcircle;
A_a = pole of EF wrt Γ
T₁ and T₂ = lines through A_a tangent to BC;
A_b and A_c = T₁∩BC and T₂∩BC;
Define B_c and C_a cyclically, and define B_a and C_b cyclically.
The six points A_b, A_c, B_c, B_a, C_a, C_b lie on a conic, 𝒞, given by the barycentric equation
b^2 c^2x^2+a^2 c^2y^2+a^2 b^2z^2+a^2 (a^2+b^2+c^2+2 a (b+c))yz+b^2 (a^2+2 a b+(b+c)^2)zx+c^2 (a^2+2 a c+(b+c)^2)xy= 0.
The polars of A_a, B_b, C_c with respect to 𝒞 form a perspective triangle to ABC, and the perspector is X(2334). (Angel Montesdeoca, November 27, 2022)
Triángulos con perspectriz el eje de Lemoine
X(2646) = SUM OF PU(80)
Barycentrics a (a - b - c) (2 a^2 - (b - c)^2 + a (b + c)) : :

In the place of a triangle ABC, let I = X(1) = incenter, and let
A'B'C' = intouch triangle;
Ma = midpoint of segment AI;
La = MaA', and define Lb and Lc cyclically;
Ra = reflection of La in AI, and define Bb and Rc cyclically.
The lines Ra, Rb, Rc concur in X(2646). (Angel Montesdeoca, October 9, 2021)
See Antreas Hatzipolakis and Angel Montesdeoca, Euclid 2662.
X(2883) = COMPLEMENTARY CONJUGATE OF X(4)
Barycentrics (3a^4 - 2a^2(b^2 + c^2) - (b^2 - c^2)^2)*(a^4(b^2 + c^2) - 2a^2(b^2 - c^2)^2 + (b^2 - c^2)^2(b^2 + c^2)) : :

In the plane of a triangle ABC, let H = X(4); L = X(20); Ab = AL∩BH; Ac = AL∩CH; U = line through Ab parallel to AC; V = line through Ac parallel to AB; A1 = U∩V; A2 = U∩BC; A3 = V∩BC; O = circle {{A1, A2, A3}}; A' = pole of BC wrt O, and define B' and C' cyclically. Then A'B'C' = reflection of ABC in X(2883). (Angel Montesdeoca, July 30, 2021)
El complemento del conjugado isogonal del punto de De Longchamps.
X(2904) = ORTHIC-ISOGONAL CONJUGATE OF X(24)
Let O be the circumcircle of a triangle ABC. Let
A' = perpendicular bisector of segment OA, and define B' and C' cyclically
A'' = B'∩C', and define B'' and C'' cyclically
A* = reflection of A* in BC, and define B* and C* cyclically.
Then A*B*C* is perspective to the orthic triangle of ABC, and the perspector is X(2904). (Angel Montesdeoca, November 12, 2016).

(Hechos Geométricos en el Triángulo (2016) )

Let P be an arbitrary point of the circumcircle , O, and let S(P) be the Simson-Wallace line of P, and P' the reflection of P in S(P). The locus of P' as P ranges through O is a bicircular circumquartic that meets the sidelines of ABC in 6 points other than A, B, C. The 6 points lie on a conic, here called the Montesdeoca conic, M. The center of M is X(2904), and the perspector of M is X(14111). Let u = S_A, v = S_B, w = S_C, f(a,b,c) = u²(S² - v²)(S² - w²), g(a,b,c) = 2vw(u² - S²)(S² + vw). A barycentric equation for M is

f(a,b,c)x² + f(b,c,a)y² + (f,c,a,b)z² + g(a,b,c)yz + g(b,c,a)zx + g(c,b,a)xy = 0.

See HG121017 and X(2904).
X(2963) = BARYCENTRIC PRODUCT X(17)*X(18)
Barycentrics 1/(a^4 + b^4 + c^4 - 2 a^2 b^2 - 2 a^2 c^2 - b^2 c^2) : :

Let Q be the pedal curve (a limaçon of Pascal) of the circle with center X(7728) and radius |OH|, with respect to X(265). Let A'B'C' be the triangle formed by the tangents at A,B,C to Q. Then the triangles ABC and A'B'C' are perspective, and their perspector is X(2963). Moreover, X(21975) is the only finite fixed point of the affine transformation that maps a triangle ABC onto A'B'C'. See X(21975). (Angel Montesdeoca, September 10, 2019). See HG080919.
X(3052) = INTERSECTION X(31)X(42)∩X(32)X(220)
Barycentrics a^2(3a - b - c) : :

Let A'B'C' be the tangential triangle of ABC, and let L be the line through X(1) parallel to BC. Let A'' = L∩B'C', and define B'' and C'' cyclically. Let A* = B'B''∩C'C'', and define B* and C* cyclically. The lines AA*, BB*, CC* concur in X(3445), and the lines A'A*, B'B*, C'C* concur in X(3052). (Angel Montesdeoca, April 29, 2016)

Hechos Geométricos en el Triángulo (2016). Los puntos X(3052) y X(3445) como centros de perspectividad.
X(3345) = ISOGONAL CONJUGATE OF X(1490)
Barycentrics a (a^12-6 a^10 (b-c)^2+(b-c)^8 (b+c)^4+5 a^8 (b-c)^2 (3 b^2+2 b c+3 c^2)+3 a^4 (b-c)^2 (b+c)^4 (5 b^2-2 b c+5 c^2)-2 a^2 (b-c)^2 (b+c)^4 (3 b^4-4 b^3 c+10 b^2 c^2-4 b c^3+3 c^4)-4 a^6 (b-c)^2 (5 b^4+12 b^3 c+18 b^2 c^2+12 b c^3+5 c^4)) : :

Let ABC be a triangle with incenter I = X(1), De Longchamps point L = X(20), and Bevan point Be = X(40). Let DEF = cevian triangle of L. Let A' = ABe∩ID, and define B' and C' cyclically. The triangles ABC and A'B'C' are orthologic. The center of orthology of ABC with respect to A'B'C' is X(3345) and the center of orthology of A'B'C' with respect to ABC is X(9121). (Angel Montesdeoca, October 3, 2023 *)
X(3346) = ISOGONAL CONJUGATE OF X(1498)
Let M be a point on the slideline BC of triangle ABC. Let (AM) be the circle with diameter AM. Let Mb be the point, other than A, in (AM)∩AC, and let Mc be the point, other than A, in (AM)∩AB. Let Tb be the line tangent to (AM) at Mb, and let Tc be the line tangent to (AM) at Mc. As M varies on BC, the locus of Tb∩Tc is a line, La. Define Lb and Lc cyclically. Let A' = Lb∩Lc, and define B' and C' cyclically. The triangle A'B'C' is perspective to ABCF, and the perspector is X(3346). (Angel Montesdeoca, November 11, 2017)
See also ejtr2224 (10 de Junio del 2008).
Una propiedad del centro X(3346) (11 de noviembre del 2017)
X(3445) = X(56)-VERTEX CONJUGATE OF X(56)
Barycentrics a^2/(3a - b - c) : b^2/(3b - c - a) : c^2/(3c - a - b)

Let A'B'C' be the tangential triangle of ABC, and let L be the line through X(1) parallel to BC. Let A'' = L∩B'C', and define B'' and C'' cyclically. Let A* = B'B''∩C'C'', and define B* and C* cyclically. The lines AA*, BB*, CC* concur in X(3445); also, the lines A'A*, B'B*, C'C* concur in X(3052). (Angel Montesdeoca, April 29, 2016)

Hechos Geométricos en el Triángulo (2016). Los puntos X(3052) y X(3445) como centros de perspectividad.
X(3447) = X(523)-VERTEX CONJUGATE OF X(523)
Let E be the Euler line and T_AT_BT_C the tangential triangle of ABC. Let D_A = E∩T_BT_C, and define D_B and D_C cyclically. Let A' = D_BT_B∩D_CT_C, and define B' and C' cyclically. Then A'B'C' is perspective to ABC, and the perspector is X(3447). For a sketch, click X(3447)andX(7669). (Angel Montesdeoca, April 22, 2016)

Hechos Geométricos en el Triángulo (2016). Una caracterización geométrica de X(3447).
X(3513) = 1st DILATION CENTER
Barycentrics a (a + b - c) (a - b + c) (a Sqrt[-a^2 - (b - c)^2 + 2 a (b + c)] - b Sqrt[-a^2 - (b - c)^2 + 2 a (b + c)] - c Sqrt[-a^2 - (b - c)^2 + 2 a (b + c)] - Sqrt[-a^4 - (b^2 - c^2)^2 + 2 a^2 (b^2 + c^2)]) : :

In the plane of a triangle ABC, let P be a point on the line X(1)X(3), and let DEF = intouch triangle. Let R_A = reflection of AP in EF, and define R_B and R_C cyclically. The lines R_A, R_B, R_C concur in a point, P', that lies on a rectangular hyperbola that circumscribes DEF. Let P'' = DEF-isogonal conjugate of P'. The mapping P -> P'' is a projective transformation of the line X(1)X(3), and its fixed points are X(3513) and X(3514). (Angel Montesdeoca, April 17, 2021).
Hechos Geométricos en el Triángulo (2021). Los centros X(3513) y X(3514) como puntos dobles de una proyectividad
X(3521) = ISOGONAL CONJUGATE OF X(3520)
Barycentrics a/(4 cos A + sec A) : :

Let Pa be the isotomic conjugate of X(1105) with respect to the triangle AX(3)X(4), and define Pb and Pc cyclically. Then PaPbPc and ABC are perspective at X(3521). (Angel Montesdeoca, September 23, 2018) HG220918
X(3532) = INTERSECTION OF LINES X(6,1204) AND X(64,1620)
Barycentrics a^2/(5 a^4 - 2 a^2 (b^2 + c^2)- 3 (b^2 - c^2)^2) : :

Let DEF be the cevian triangle of X(69) and A' the center of the circle that passes through D and through points where the line EF cuts to the circumcircle, and define B' and C' cyclically. The lines AA', BB', CC' concur in X(3532). (Angel Montesdeoca, October 30, 2019). Una propiedad del centro X(3532)
X(3627) = H^-1(X(3); M,-1/2)
Barycentrics a²(b² + c² - 4a²) + 3(b² - c²)² : :

Let ABC be a triangle and Γ_a a variable circle passing through A and the midpoint of side BC. Let Ab and Ac be the points, other than A, in which the circle Γ_a meets AC and AB, respectively. Let A' be the vertex of parallelogram with vertices A, A_b, A', A_c. When Γ_a varies, the locus of A' is a line ℓ_a. Define ℓ_b and ℓ_c cyclically. The triangle formed by the lines ℓ_a, ℓ_b, ℓ_c is orthologic to ABC with orthology center X(3627) . The reciprocal orthologic center of these triangles is X(3). See X(3627). (Angel Montesdeoca, July 13, 2022)
See HG130722
X(3628) = H^-1(X(5); M, 2)
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 )
X(3683) = X(3219)com(EXTOUCH TRIANGLE)
Barycentrics a(-a + b + c)(2a + b + c)

In the plane of a triangle ABC, let
A'B'C' = circumcevian triangle of X(1);
DEF = cevian triangle of X(10);
La = line tfhrough D perpendicular to BC;
A'' = La∩AA', and define B'' and C'' cyclically.
Then X(36;83) is the finite fixed point of the affine transformation that carries A'B'C' onto A''B''C'. (Angel Montesdeoca, November 21, 2022)
X(3683) punto fijo de una transformación afín (21/11/2022)
X(3817) = X(2)com(3rd EULER TRIANGLE)
Let A'B'C' be the circumcevian triangle of X(1). Let L1 be the Simson line of A', and define L2 and L3 cyclically. Let A'' = L2∩L3, and define B'' and C'' cyclically. Then X(3817) = X(2)-of-A''B''C''. (Angel Montesdeoca, June 28, 1017)
(Encyclopedia of Triangle Centers )
X(4012) = X(269)com(EXTOUCH TRIANGLE)
Barycentrics (b + c - a)3(a2 + b2 + c2 - 2bc) : :

Let DEF be the extouch triangle of ABC. Let Ha be the hyperbola with foci E and F, passing through A, and define Hb and Hc cyclically. These three hyperbolas have two common points, U and V. Let Pa be the pole of the line UV with respect to Ha, and define Pb and Pc cyclically. Then DEF and PaPBPc are perspective at X(4012). (Angel Montesdeoca, September 23, 2018) HG150918
X(4292) = INTERSECTION OF LINES X(1)X(7) AND X(4)X(57)
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = b⁴ + c⁴ - 2a⁴ - a³b - a³c + a²b² + a²c² - 2a²bc + ab³ + ac³ - ab²c - abc² - 2b²c²

A construction of X(4292) is given by Antreas Hatipolakis and Angel Montesdeoca at Hyacinthos 24046 (Aug 16, 2016).

[APH]:
Let ABC be a triangle and A'B'C' the pedal triangle of I.
Denote:
A"B"C" = the pedal triangle of the orthocenter H' of A'B'C' wrt the triangle A'B'C' (ie A"BC" = the orthic triangle of A'B'C')
Ab, Ac = the orthogonal projections of A" on AC, AB, resp.
(Nab), (Nac) = the NPCs of AbA"B', AcA"C', resp.
(Nbc), (Nba) = the NPCs of BcB"C', BaB"A', resp.
(Nca), (Ncb) = the NPCs of CaC"A', CbC"B', resp.
S1 = the radical axis of (Nba), (Nca).
S2 = the radical axis of (Ncb), (Nab)
S3 = the radical axis of (Nac), (Nbc)
1. S1, S2, S3 are concurrent.
2. the parallels to S1, S2, S3 through A, B, C, resp. are concurrent at H.
3. the parallels to S1, S2, S3 through A', B', C', resp. are concurrent at I. 4. the parallels to S1, S2, S3 through A",B",C", resp. are concurrent.
This is equivalent to:
The triangles ABC and the orthic triangle of the pedal triangle of I are orthologic.

[Angel Montesdeoca]:
1. S1, S2, S3 are concurrent at X(354)
4. the parallels to S1, S2, S3 through A",B",C", resp. are concurrent at X(4292)
X(4319) = INTERSECTION OF LINES X(1)X(7) AND X(19)X(25)
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a(b + c -a)²(a² + b² + c² - 2bc)

X(4319) is the point of concurrence of three lines associated with Soddy hyperbolas; see Angel Montesdeoca's construction at Hyacinthos 21290 (Nov 12, 2012)

(Paul Yiu, Introduction to the Geometry of the Triangle, 2002; 12.4 The Soddy hyperbolas, p. 143)
Given triangle ABC, consider the hyperbola passing through A, and with foci at B and C. We shall call this the A-Soddy hyperbola of the triangle.
The equation of A-Soddy hyperbola is
(Fa): (c+a-b)(a+b-c)(y^2+z^2)-2(a^2+(b-c)^2)y z+4(b-c)c x y -4b(b-c)z x=0.
The perspector of a (Fa): Pa = ( SB SC : - b c SC : -b c SB ).
The polar of Pa with respect to (Fa) is is the line "da":
2b(b-c)cx + (a^2+(b-c)^2)cy - b(a^2+(b-c)^2)z=0.
Similarly define the lines "db" and "dc"; then the lines da, db and dc are concurrent at the triangle center X(4319).
X(4350) = INTERSECTION OF LINES X(1)X(7) AND X(41)X(57)
Barycentrics a(a^2 + b^2 + c^2 - 2a b - 2a c)/(b + c - a)^2 ::

In the plane of a triangle ABC, let
I = X(1) = incenter;
DEF = intouch triangle;
(O) = circumcircle;
A' = the point, other than A, where the circle {{A,I,D}} meets (O), and define B' and C' cyclically;
Oa = center of the circle {{B',E,F,C'}}, and define Ob and Oc cyclically;
Oaa = A-extraversion of Oa, and define Obb and Occ cyclically;
T = affine transfomation that maps ABC onto OaObOc;
TT = affine transformation that maps ABC onto OaaObbOcc.
Then X(4350) = finite fixed point of T, and X(5250) = finite fixed point of TT. (Angel Montesdeoca, December 29, 2023)
See HG261223
X(4432) = X(44)com[T(a,c)]
Barycentrics (b + c - 2a)(a² - bc) : :

X(4432) is the intersection of the orthic axes of the 1st & 2nd Montesdeoca bisector triangles. (Randy Hutson, December 2, 2017)
X(4674) = X(3952)com[T(-a, a + c)]
Barycentrics a(b + c)(b + c - 2a) : :

Let A'B'C' be the cevian triangle, and let A"B"C" be the circumcevian triangle of X(1). The four points of intersection of the circumcircles of BA'I, CA'I with the side lines A"C", A"B", respectively, lie on a circle Oa; define Ob and Oc cyclically. Then the triangles ABC and OaObOc are orthologic, with orthology centers the X(3) and X(4674). (Angel Montesdeoca, April 14, 2019).
HG130519
X(4846) = ISOGONAL CONJUGATE OF X(378)
Barycentrics (b² + c² - a²)/(a⁴ + b⁴ + c⁴ - 2a²b² - 2a²c² + 4b²c²) : :

X(4846) is the Segovia point of the circumcevian triangle of the circumcenter. (Antreas Hatzipolakis, January 23 2012; Hyacinthos #20737; see Segovia Point, continued.)

Let A'B'C' be the translation of ABC by the vector X(4)X(3). Let M_A be the midpoint of the (possibly nonreal) points in which the circumcircle of ABC meets the line B'C', and define M_B and M_C cyclically. The triangle M_AM_BM_C is perspective to ABC, and the perspector is X(4846). If you have GeoGebra, you can view X(4846). (Angel Montesdeoca, May 20, 2018)
X(4882) = X(1058)com[INVERSE(nT(a, c - a)]
Barycentrics a(b + c - a)(b^2 + c^2 - a^2 + 6b c) :

Consider any triangle ABC. Let A-excircle ω_A of ABC is tangent to BC at A'. Similarly define ω_B, ω_C, B', C'. Consider two tangent lines ℓ_a, ℓ_a' from A' to ω_B, ω_C which are different from BC. Similarly define ℓ_b, ℓ_b', ℓ_c, ℓ_c'. Then ℓ_a, ℓ_a', ℓ_b, ℓ_b', ℓ_c, ℓ_c' form a hexagon which has an incircle.
The center of the incircle of the hexagon is X(4882). (Angel Montesdeoca, October 2, 2018). HG021018
X(5000) = WALSMITH POINT
Barycentrics 2 (a^6-a^4 (b^2+c^2)+(b^2-c^2)^2 (b^2+c^2)-a^2 (b^4-3 b^2 c^2+c^4)) (-2 a^4+2 a^2 b^2-b^4+2 a^2 c^2+2 b^2 c^2-c^4+(b^2-c^2)^2-((2 a^4-(b^2-c^2)^2-a^2 (b^2+c^2)) (-a^6+a^4 b^2+a^2 b^4-b^6+a^4 c^2-6 a^2 b^2 c^2+b^4 c^2+a^2 c^4+b^2 c^4-c^6+Sqrt[(a^2-b^2-c^2) (a^2+b^2-c^2) (a^2-b^2+c^2) (a^2+b^2+c^2) (a^4+(b^2-c^2)^2-2 a^2 (b^2+c^2))]))/(2 (a^6-a^4 (b^2+c^2)+(b^2-c^2)^2 (b^2+c^2)-a^2 (b^4-3 b^2 c^2+c^4)))) : :

In the plane of a triangle ABC, let P be a point on the Euler line and P' = isogonal conjugate of P, and let DEF = medial triangle. Let ℓ_a = reflection of AP' in EF, and define ℓ_b and ℓ_c cyclically. The lines ℓ_a, ℓ_b, ℓ_c concur in a point, P", that lies on the Euler line. The mapping P ↦ P'' is a projective transformation of the Euler line, and its fixed points are X(5000) and X(5001).
If P=X(3) + t X(4), then P"= 4 t SA SB SC X(3) + (a^2 b^2 c^2 + 4 t SA SB SC) X(4).
The fixed points are obtained for t = (2 SA SB SC ± S Sqrt[2 (a^2+b^2+c^2) SA SB SC])/(4 SA SB SC). (Angel Montesdeoca, April 19, 2021)
Proyectividad sobre la recta de Euler y los centros X(5000) y X(5001)
X(5024) = INVERSE-IN-1st-BROCARD-CIRCLE OF X(1384)
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 : b² : 2c², 0 : 2b² : c², 2a² : 0 : c², a² : 0 : 2c², a² : 2b² : 0, 2a² : b² : 0. The 6 points lie on a conic with center X(5024) and equation
2(b⁴c⁴x² + c⁴a⁴y² + a⁴b⁴z²) -5a²b²c²(a²yz + b²zx + c²xy) = 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
b⁴c⁴x² + c⁴a⁴y² + a⁴b⁴z² -2a²b²c²(a²yz + b²zx + c²xy) = 0.
(From Angel Montesdeoca, March 28, 2013)
(Encyclopedia of Triangle Centers )
X(5048) = INVERSE-IN-INCIRCLE OF X(3057)
Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a(b + c - a)(3b² + 3c² - 2a² + 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 nine-point 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 )

X(5173) = INVERSE-IN-INCIRCLE OF X(2078)

Barycentrics a(a + b - c)(a - b + c)(a2b + a2c - 2ab2 - 2ac2 - 2abc + b3 - b2c - bc2 + c3) : :

Let A_aB_aC_a be the intouch triangle, so that the lines BI and CI intersect B_bC_c at B_a and C_a, respectively. Let A₁A₂A₃ be the intouch triangle of A_aB_aC_a. Define B₁B₂B₃ and C₁C₂C₃ cyclically. Let d_a be the radical axis of (A₁B₃C₂) and (A₁A₂A₃). Let e_a be the radical axis of (A₁B₃C₂) and (A_aB_bC_c). Define d_b, d_c and e_b, e_c cyclically. Then the triangles A'B'C' and A"B"C" formed by the lines d_a, d_b, d_c and the lines e_a, e_b, e_c, respectively, are homothetic and the homothetic center is X(5173). (Angel Montesdeoca, April 1, 2020)

See X(1000) como centro de perspectividad.

X(5250) = INTERSECTION OF LINES X(1)X(21) AND X(2)X(40)

Barycentrics a(b + c - a)(a^2 + b^2 + c^2 + 2a b + 2a c) : :

In the plane of a triangle ABC, let
I = X(1) = incenter;
DEF = intouch triangle;
(O) = circumcircle;
A' = the point, other than A, where the circle {{A,I,D}} meets (O), and define B' and C' cyclically;
Oa = center of the circle {{B',E,F,C'}}, and define Ob and Oc cyclically;
Oaa = A-extraversion of Oa, and define Obb and Occ cyclically;
T = affine transfomation that maps ABC onto OaObOc;
TT = affine transformation that maps ABC onto OaaObbOcc.
Then X(4350) = finite fixed point of T, and X(5250) = finite fixed point of TT. (Angel Montesdeoca, December 29, 2023)
See HG261223

X(5450) = 3rd HATZIPOLAKIS-MOSES POINT

Barycentrics a (a^6-2 a^4 (b-c)^2-a^5 (b+c)-b c (b^2-c^2)^2+a^2 (b-c)^2 (b^2-b c+c^2)-a (b-c)^2 (b^3+c^3)+a^3 (2 b^3-b^2 c-b c^2+2 c^3)),b (-a^5 (b+c)+b (b-c)^3 (b+c)^2+a^4 b (b+2 c)-a^2 b (2 b^3+b^2 c-4 b c^2+c^3)+a^3 (2 b^3-3 b^2 c-b c^2+2 c^3)-a (b^5-4 b^4 c+b^3 c^2+3 b^2 c^3-2 b c^4+c^5)) : :

Let M be the isogonal conjugate of the trilinear polar of X(57) with respect to the circumcevian triangle of X(1). Then M is a conic, and its center is X(5450). (Angel Montesdeoca, August 9, 2019)

X(5466) = TRILINEAR POLE OF LINE X(115)X(523)

Barycentrics b² - c²)/(b² + c² - 2a² : :

In the plane of a triangle ABC, let
Γ= circumcircle;
D = line tangent to Γ at A;
Ta = D∩BC;
Oab = center of the circle through Ta that is tangent to Γ at B;
Oac = center of the circle through Ta that is tangent to Γ at C;
Ab = orthogonal projection of Oab onto AC;
Ac = orthogonal projection of Oac onto AB;
A' = AbAc∩BC, and define B' and C' cyclically.
The lines AA', BB', CC' concur in X(5466). See X(5466) (Angel Montesdeoca, August 21, 2022)
See Angel Montesdeoca, HG180822.

X(5482) = 1st HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics    f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a⁵b² + a⁵c² - 2a⁵b²c² + a³b³ + a³c³ + a³b²c + a³bc² + a²b⁴ + a²c⁴ - 3a²b³c - 3a²bc³ - ab⁵ - ac⁵ - ab⁴c - abc⁴ - bc(b² - c²)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 )

X(5494) = 2nd HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a[a⁹ - a⁸(b + c) - a⁷( b - c)² + a⁶(2b³ - b²c - bc² + 2c³) - a⁵(3b⁴ + b³c - 7b²c² + bc³ + 3c⁴) + 4a⁴bc(b - c)²(b + c) + a³(b² - c²)²(5b² - 4bc + 5c²) - a²(b - c)²(2b⁵ + 5b⁴c + b³c² + b²c³ + 5bc⁴ + 2c⁵) - a(b² - c²)²(2b⁴ - 3b³c + 5b²c² - 3bc³ + 2c⁴) + (b - c)⁴(b + c)³(b² + c²)] (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_BA_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 )

X(5495) = 3rd HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a²[a⁷(b + c) - a⁶(b² + c²) - a⁵(3b³ + 2b²c + 2bc² + 3c³) + a⁴(3b⁴ - b³c + 4b²c² - bc³ + 3c⁴) + a³(3b⁵ + b⁴c + 2b³c² + 2b²c³ + bc⁴ + 3c⁵) - a²(3b⁶ - 2b⁵c - 2bc⁵ + 3c⁶) - a(b⁷ - b⁴c³ - b³c⁴ + c⁷) + (b² - c²)²(b⁴ - b³c - bc³ - b²c² + c⁴)] (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 )

X(5496) = 4th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a(b + c)[a⁵ - 2a³(b² + c²) - a²bc(b+c) + a(b⁴ - b²c² + c⁴) + 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 )

X(5497) = 5th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a⁷ - a⁶(b + c) - a⁵(b + c)² + a⁴(2b³ + b²c + bc² + 2c³) - a²(b⁴ - b³c - 3b²c² - bc³ + c⁴) + abc(b² + c²)(b² - c²)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 )

X(5498) = 6th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = 2a¹⁰ - 5a⁸(b² + c²) + 2a⁶(b⁴ + 5b²c² + c⁴) + a⁴(4b⁶ - 5b⁴c² - 5b²c⁴ + 4c⁶) - a²(b² - c²)²(4b⁴ + 5b²c² + 4c⁴) + (b² - c²)sup>4(b² + c²) (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_AN_BN_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 )

X(5499) = 7th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a⁵(b² + 4bc + c²) - a⁴(b³ + b²c - bc² + c³) + a³(2b⁴ + 3b³c + 3bc³ + 2c⁴) + 2a²(b⁵ - b³c² - b²c³ + c⁵) + a(b² - c²)²(b² - bc + c²) - (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_ABC. Define N_B and N_C cyclically. The circumcenter of N_AN_BN_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 )

X(5500) = 8th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics    f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) =
2a²²
- 15a²⁰(b² + c²)
+ 6a¹⁸(8b⁴ + 13b²c² + 8c⁴)
- a¹⁶(81b⁶ + 52b⁴c² + 152b²c⁴ + 81c⁶)
+ a¹⁴(64b⁸ + 111b⁶c² + 128b⁴c⁴ + 111b²c⁶ + 64c⁸)
+ a¹²(14b¹⁰ + 29b⁸c² + 36b⁶c⁴ + 36b⁴c⁶ + 29b²c⁸ + 14c¹⁰)
- a¹⁰(84b^¹² + 67b¹⁰c² + 56b⁸c⁴ + 48b⁶c⁶ + 56b⁴c⁸ + 67b²c¹⁰ + 84c¹²)
+ a⁸(82b¹⁴ - 23b¹²c² - 31b¹⁰c⁴ - 19b⁸c⁶ - 19b⁶c⁸ - 31b⁴c¹⁰ - 23b²c¹²+ 82c¹⁴)
- a⁶(b² - c²)²(34b¹² + 11b¹⁰c² - 30b⁸c⁴ - 35b⁶c⁶ - 30b⁴c⁸ + 11b²c¹⁰ + 34c¹²)
+ a⁴(b²- c²)⁴(b¹⁰ - 2b⁸c² - 22b⁶c⁴ - 22b⁴c⁶ - 2b²c⁸+ c¹⁰)
+ a²(b² - c²)⁶(4b⁸ + 5b⁶c² + 8b⁴c⁴ + 5b²c⁶ + 4c⁸)
- (b² - c²)⁸(b⁶ + b⁴c² + b²c⁴ + c⁶)    (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_AN_BN_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 )

X(5501) = 9th HATZIPOLAKIS-MONTESDEOCA POINT

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_AN_BN_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}, {137,8254}
(Encyclopedia of Triangle Centers )

( Mostrar/Ocultar figura )

X(5502) = 10th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a²(a² - b²)[a² - c²)(a⁶ - a⁴(b² + c²) + a²(a² -b²)(a² - c²) + 3(b² - c²)²(b² + c²)] (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_AH_BH_C and M_AM_BM_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 )

X(5540) = GIBERT-BUREK-MOSES CONCURRENT CIRCLES IMAGE OF X(101)

The 1st and 2nd Montesdeoca bisector triangles and inversely similar to the excentral triangle. Let s1 and s2 be the similarity mappings. Then there is a unique point X such that s1(X) = s2(X), and X = X(5540). See HGT2017 and AdvGeom3769

X(5559) = GARCIA-FEUERBACH POINT GF(5/3)

Barycentrics 1/[(a + b + c)(b + c - a) - (b2 + c2 - a2)(5/3)] : :

Let Pa be the isotomic conjugate of X(1222) with respect to AX1)X(8), and define Pb and Pc cyclically. Then ABC and PaPbPc are perspective at X(5559). (Angel Montesdeoca, September 23, 2018). See HG220918

X(5618) = 1st MONTESDEOCA EQUILATERAL TRIANGLES POINT

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = 1/[(b² - c²)(27^1/2b²c²S_A + S(S² + 9S_AS_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_BA_C is equilateral. The lines A_BA_C, B_CB_A, C_BC_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 )

X(5619) = 2nd MONTESDEOCA EQUILATERAL TRIANGLES POINT

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = = 1/[(b² - c²)(27^1/2b²c²S_A - S(S² + 9S_AS_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 )

X(5620) = ISOGONAL CONJUGATE OF X(5127)

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = (b + c)[a⁶ - a⁴(b² + c²) - a²(b⁴ + c⁴ - 3b²c²) - 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 nine-point center of A'BC, and let O_A be the circumcircle of N_ABC. 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 )

X(5627) = YIU REFLECTION POINT

Barycentrics g(a,b,c) : g(b,c,a): g(c,a,b), where g(a,b,c) = 1/[(a²S_A - 2S_BS_C)(S² - 3S_AS_A)]

Let P = X(110) and A1A2, B1B2, C1C2 be the diameters of the circumcircle of ABC, parallel to AP, BP, CP, respectively. The Simson-Wallace lines of the points A1 and A2 intersect orthogonally in Pa, on the nine-point circle. The points Pb and Pc are defined similarly. The triangles ABC and PaPbPc are perspective and inversely similar; the perspector is X(5627). (Angel Montesdeoca, October 21, 2018)
HG201018

X(5643) = H(2) ON THE THOMSON-GIBERT-MOSES HYPERBOLA

Barycentrics f(A,B,C) : f(B,C,A) : f(C,A,B), where f(A,B,C) = a²(13S² + 7S²_A + 2S_BS_C) (Peter Moses, June 9, 2014)

X(5643) is the only point whose polar conic in the Napoleon cubic (K005) is a circle. (Bernard Gibert, June 22, 2014)

Let A' be the centroid of the A-altimedial triangle, and define B' and C' cyclically; then X(5643) is the center of similitude of ABC and A'B'C'. (Randy Hutson, July 7, 2014)

X(5643) is the only finite fixed point of the affine transformation that maps a triangle ABC onto the pedal triangle of X(5). (Angel Montesdeoca, August 19, 2016).

X(5691) = DE LONGCHAMPS POINT OF OUTER GARCIA TRIANGLE

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = 3 a^4- a^3 (b + c) - a^2 (b - c)^2 + a (b - c)^2 (b + c) - 2 (b^2 - c^2)^2
(Angel Montesdeoca, January 21, 2015)

Let I = X(1) and O = X(3). Let A'' be the reflection of I in line AO and let I_A be the reflection of A'' in line AI. Define I_B and I_C cyclically. Then ABC and I_AI_B I_C are orthologic triangles, and X(5691) is the ABC-orthology center of I_AI_B I_C. (Angel Montesdeoca, January 21, 2015)

X(5702) = CENTER OF MONTESDEOCA CONIC

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = S_BS_C(5a²S_A - S_BS_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(S²_Ax² + S²_By² + S²_Cz²) - 5(S_BS_Cyz + S_CS_Azx + S_AS_Bxy) = 0 (Peter Moses, June 23, 2014)

The Montesdeoca conic is the anticevian-intersection 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 )

X(5836) = INTERSECTION OF LINES X(5)X(10) AND X(7)X(8)

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a*(a^2*b - b^3 + a^2*c - 2*a*b*c + 3*b^2*c + 3*b*c^2 - c^3)

A construction of X(5836) is given by Antreas Hatipolakis and Angel Montesdeoca at Hyacinthos #24129. (Aug 23, 2016)

[APH]:
Let ABC be a triangle and A'B'C' the pedal triangle of I.
Denote:
A"B"C" = the orthic triangle of A'B'C'
Ab, Ac = the orthogonal projections of A on B'B", C'C", resp.
L1 = the Euler line of AAbAc. Similarly L2, L3
1. L1, L2, L3 are concurrent.
The parallels to L1, L2, L3 through:
2.1. A, B, C, resp.
2.2. A', B',C', resp.
2.3. A",B", C", resp.
are concurrent.

[Angel Montesdeoca]:
*** 1. L1, L2, L3 are concurrent. at X(5836)
*** 2.1. The parallels to L1, L2, L3 through A, B, C, resp.are concurrent. at X(8) = Nagel point
*** 2.2. The parallels to L1, L2, L3 through A', B', C', resp.are concurrent. at X(145) = anticomplement of Nagel point
*** 2.3. The parallels to L1, L2, L3 through A", B" C", resp.are concurrent. at X(10106) = 29th Hatzipolakis-Montesdeoca Point.

X(6042) = PERSPECTOR OF MONTESDEOCA-HUNG TRIANGLE AND ABC

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a(b + c)²(b² + c² + ab + ac)²

Let (Ap) be the Apollonius circle, and let (K_A), (K_B), (K_C) be the circles described at X(5973) in association with the Hung-Feuerbach circle at X(5974). Let L_A be the radical axis of (Ap) and (K_A), and define L_B and L_C cyclically. The lines L_A, L_B, (L_C form a triangle T (here named the Montesdeoca-Hung triangle) that is perspective to ABC, and the perspector is X(6042). (Tran Quang Hung, ADGEOM #1506; Angel Montesdeoca, ADGEOM #1525, August 24, 2014)

X(6043) = PERSPECTOR OF MONTESDEOCA-HUNG TRIANGLE AND EXCENTRAL TRIANGLE

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a(a + b)(a + c)(a³ + ab² + ac² + 3abc - b²c - bc²)

Continuing from X(6042), the triangle T is perspective to the excentral triangle, and the perspector is X(6043). (Peter Moses, August 24, 2014)

X(6048) = CENTER OF MOSES HULL CIRCLE

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a(a²b + a²c + ab² + ac² + abc - 3b²c - 3bc²)

Let (Ia), (Ib), (Ic) be the excircles of ABC, and let (AP) be the classical Apollonius circle (e.g., TCCT, p. 102). Let (Ja) be the circle internally tangent to (Ia) and externally tangent to (Ib) and (Ic), and define (Jb) and Jc) cyclically. Let (J) be the circle internally tangent to (Ja), (Jb), (Jc). Then X(6048) = center of (J). See ADGEOM #1541 and Problema de Apolonio relativo a las circunferencias exinscritas. (Angel Montesdeoca, February 25, 2017)

X(6094) = 11th HATZIPOLAKIS-MONTESDEOCA POINT

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^4-4 a^2 c^2-4 b^2 c^2+c^4) (a^4-4 a^2 b^2+b^4+5 a^2 c^2-4 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)/[a⁴qr(p + q)(p + r) - 2a²(q + r)(r + p)(p + q)(c²q + b²r) + p(q + r)(b⁴r(p + q) + c⁴q(q + r) + 2b²c²(q² + r² + 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)

X(6095) = 12th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a (a^3-a^2 b-a b^2+b^3-2 a^2 c+7 a b c-2 b^2 c-2 a c^2-2 b c^2+c^3) (a^3-2 a^2 b-2 a b^2+b^3-a^2 c+7 a b c-2 b^2 c-a c^2-2 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}

X(6096) = 13th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a (a^3-a^2 b-a b^2+b^3-2 a^2 c+7 a b c-2 b^2 c-2 a c^2-2 b c^2+c^3) (a^3-2 a^2 b-2 a b^2+b^3-a^2 c+7 a b c-2 b^2 c-a c^2-2 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)

X(6097) = 14th HATZIPOLAKIS-MONTESDEOCA POINT

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) = (r² + 2rR - R² + s²)*X(3) + R²*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_ABO_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_AO_BO_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}

X(6098) = 15th HATZIPOLAKIS-MONTESDEOCA POINT

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)-of-A'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)

X(6099) = 16th HATZIPOLAKIS-MONTESDEOCA POINT

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)-of-A'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_AA'', O_BB'', O_CC'' 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)

X(6100) = 17th HATZIPOLAKIS-MONTESDEOCA POINT

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)-of-A'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)

X(6101) = 18th HATZIPOLAKIS-MONTESDEOCA POINT

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 nine-point center of triangle AA_BA_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)

X(6102) = 19th HATZIPOLAKIS-MONTESDEOCA POINT

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^6-2b^4c^2-2b^2c^4+3c^6)-b^8+b^6c^2+b^2c^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 nine-point center of triangle AA_BA_C, and define B' and C' cyclically. Then X(6102) is the ABC-orthology center of A'B'C' on the circumcircle of A'B'C'. (Also, X(1141) is the ABC-orthology 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 A-altitude, and define N_B and N_C cyclically. Then X(6102) is the orthocenter of N_AN_BN_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)-of-circumorthic-triangle
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)

X(6126) = X(1)-CEVA CONJUGATE OF X(36)

Barycentrics f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = (a^2 (a^5 + a^4(b+c) - 2a^3(b^2+c^2) - a^2(2b^3-b*c(b+c)+2c^3)+ a(b^4+b^2c^2+c^4) + (b-c)^2(b^3+c^3))

X(6126) is the point QA-P41 ('Involutary Conjugate of QA-P4') of the quadrangle ABCX(1). X(6126)-of-orthocentroidal-triangle = X(1). These properties and others are presented in Hyacinthos messages 21651 and 22707-22710 by S. Topor, A. Montesdeoca, R. Hutson, and A. Hatzipolakis; see 22710)

X(6177) = CEVAPOINT OF THE REAL FOCI OF THE STEINER INELLIPSE

Barycentrics g(a,b,c) : g(b,c,a) : g(c,a,b), where g(a,b,c) = a⁴ + b⁴ + c⁴ - a²b² - a²c² + (b² + c²)(a⁴ + b⁴ + c⁴ - a²b² - a²c² - b²c²) ^1/2

In the plane of a triangle ABC, let
F1 and F2 be the real foci of the Steiner inellipse
A1 = AF1∩BC
A2 = AF2∩BC
Ga = conic tangent to AF1 at A1, to AF2 at A2, and to F1F2
Ta = Ga∩F1F2
Ao = center of Ga, and define Bo and Co cyclically
The lines AAo, BBo, CCo concur in X(14633), and the lines ATa, BTb, CTc concur in X(6177). See X(6177) and X(14633). (Angel Montesdeoca, July 21, 2023)

X(6188) = DAO (a,b,c,R) PERSPECTOR

Barycentrics 1 /[a^8 - 4 a^6 (b^2 + c^2) + (b^2 - c^2)^2 (b^4 + 6 b^2 c^2 + c^4) + a^4 (6 b^4 - 3 b^2 c^2 + 6 c^4) + a^2 (-4 b^6 + 3 b^4 c^2 + 3 b^2 c^4 - 4 c^6)]

Let r_A be a positive valued function of a,b,c, and define r_B and r_C cyclically. Let (A) denote the circle with center A and radius r_A, and define (B) and (C) cyclically. Let P be the radical center of (A), (B), (C). Let r_P be a positive valued function of a,b,c, let L_A be the radical axis of (A), (B), (P), and define L_B and L_C cyclically. Let A' = L_B∩L_C, and define B' and C' cyclically. Then the lines AA', BB', CC" concur in a point, D. Moreover, the six points A, B, C, X(4), P, D lie on a hyperbola. (Dao Thanh Oai, ADGEOM 893, November 26, 2013)

The hyperbola is here called the Dao (r_A,r_B,r_C,r_P) hyperbola. The triangle A'B'C' is the Dao (r_A,r_B,r_C,r_P) triangle, and the perspector D of ABC and A'B'C' is the Dao (r_A,r_B,r_C,r_P) perspector.

For details and examples, including coordinates and equations, see Angel Montesdeoca's presentation at Hechos Geometricos.

X(6240) = CIRCUMORTHIC-TRIANGLE-ORTHOLOGIC CENTER OF MACBEATH TRIANGLE

Barycentrics (a^2+b^2-c^2) (a^2-b^2+c^2) (2 a^6-3 a^4 b^2+b^6-3 a^4 c^2+2 a^2 b^2 c^2-b^4 c^2-b^2 c^4+c^6) : :

Let MaMbMc = medial triangle and DEF = circumorthic triangle. Let (Oa) be the circle with center A that passes through E and F (and X(4)). Let E' be the point, other than E, in which (Oa) meets EMa, and let F' be the point, other than F, in which (Oa) meets FMa. Let A' = EF'∩FE', and define B' and C' cyclically. The lines A'DF, B'E, C'F concur in X(6240). (Angel Montesdeoca, September 25, 2020)
El centro del triángulo X(6340) como centro de perspectividad

X(6288) = JOHNSON-TRIANGLE-ORTHOLOGIC CENTER OF REFLECTION TRIANGLE

Barycentrics a^10-2 a^8 b^2+a^6 b^4-a^4 b^6+2 a^2 b^8-b^10-2 a^8 c^2+3 a^6 b^2 c^2-2 a^4 b^4 c^2-2 a^2 b^6 c^2+3 b^8 c^2+a^6 c^4-2 a^4 b^2 c^4-2 b^6 c^4-a^4 c^6-2 a^2 b^2 c^6-2 b^4 c^6+2 a^2 c^8+3 b^2 c^8-c^10 : :

Let (Ha) be the hyperbola having diameter X(3)X(4), passing through A, with an asymptote parallel to the internal bisector of angle A. Let A' be the reflecttion of A in X(5), and let Ta be the tangent to (Ha) at A'. Define Tb and Tc cyclically. The lines Ta, Tb, Tc concur in X(6288). See X(6288). (Angel Montesdeoca, December 9, 2019) HGT

X(6525) = SS(a → SBSC) of X(55)

Barycentrics [3a^4 - 2a^2b^2 - 2a^2c^2 - (b^2 - c^2)^2]/(b^2 + c^2 - a^2)^2 : :

Let DEF be the orthic triangle and D' the midpoint of AD . Let A' be the reflection of A in D'X(3), Let La be the perpendicular bisector of A'D, and define Lb and Lc cyclically. Let A" = Lb∩Lc, B"= Lc∩La, C" = La∩Lb. The lines DA", EB", FC" concur in X(6525). (Angel Montesdeoca, August 21, 2019). HG210819.

X(6595) = (ABC-1st-SCHIFFLER) CYCLOLOGIC CENTER

Barycentrics a (a^9+a^7 (-6 b^2+7 b c-6 c^2)-(b-c)^4 (b+c)^3 (2 b^2+5 b c+2 c^2)+a^6 (2 b^3-3 b^2 c-3 b c^2+2 c^3)+a (b^2-c^2)^2 (3 b^4-b^3 c+2 b^2 c^2-b c^3+3 c^4)+a^5 (12 b^4-15 b^3 c+13 b^2 c^2-15 b c^3+12 c^4)+a^4 (-6 b^5+3 b^4 c+5 b^3 c^2+5 b^2 c^3+3 b c^4-6 c^5)+a^3 (-10 b^6+9 b^5 c-3 b^4 c^2+5 b^3 c^3-3 b^2 c^4+9 b c^5-10 c^6)+a^2 (6 b^7+3 b^6 c-11 b^5 c^2+b^4 c^3+b^3 c^4-11 b^2 c^5+3 b c^6+6 c^7)) : :

Let I = X(1), the incenter of a triangle ABC. Let L_A be the Euler line of triangle IBC, and define L_B and L_C cyclically. These lines concur in S = X(21), the Schiffler point of ABC. Let C₂ be the point, other than S, of intersection of the line L_B and the circle (C, |CS|). Let B₃ be the point, other than S, of intersection of the line L_C and the circle (B, |BS|). Let O₁ be the circumcenter of SC₂B₃, and define O₂ and O₃ cyclically. The triangle O₁O₂O₃ is here named the 1st Schiffler triangle. The triangles ABC and O₁O₂O₃ are cyclologic, and X(6595) is the (ABC, O₁O₂O₃)-cyclologic center. See X(6596)-X(6599) and Hyacinthos #23098 and Hyacinthos #23101. (S. Kirikami, A. Hatzipolakis and A.Montesdeoca, February 4-5, 2015)

X(6596) = (1st SCHIFFLER, 2nd SCHIFFLER) CYCLOLOGIC CENTER

Barycentrics a (a^9 + 10 a^7 b c - 3 a^8 (b + c) + (b - c)^6 (b + c)^3 - a (b - c)^4 (b + c)^2 (3 b^2 - 2 b c + 3 c^2) + a^6 (8 b^3 - 6 b^2 c - 6 b c^2 + 8 c^3) - a^5 (6 b^4 + 12 b^3 c - 13 b^2 c^2 + 12 b c^3 + 6 c^4) - a^4 (6 b^5 - 18 b^4 c + 7 b^3 c^2 + 7 b^2 c^3 - 18 b c^4 + 6 c^5) - a^2 b c (6 b^5 - 13 b^4 c + 11 b^3 c^2 + 11 b^2 c^3 - 13 b c^4 + 6 c^5) + a^3 (8 b^6 - 6 b^5 c - 9 b^4 c^2 + 26 b^3 c^3 - 9 b^2 c^4 - 6 b c^5 + 8 c^6)) : :

Continuing from X(6595), let W = X(3065). Let C'₂ be the point, other than S, of intersection of the Euler line of the triangle WCA and the circle (C, |CS|). Let B'₃ be the point, other than S, of intersection of the Euler line of the triangle WAB and the circle (B, |BS|). Let O'₁ be the circumcenter of SC₂B₃, and define O'₂ and O'₃ cyclically. The triangle O'₁O'₂O'₃ is here named the 2nd Schiffler triangle.The triangles ABC, O₁O₂O₃, and O'₁O'₂O'₃ and pairwise cyclologic:

X(6595) = (ABC, O₁O₂O₃)-cyclologic center
X(3065) = (O₁O₂O₃, ABC)-cyclologic center
X(1320) = (ABC, O'₁O'₂O'₃)-cyclologic center
X(1) = (O'₁O'₂O'₃, ABC)-cyclologic center
X(6596) = (O₁O₂O₃, O'₁O'₂O'₃)-cyclologic center
X(6597) = (O'₁O'₂O'₃, O₁O₂O₃)-cyclologic center

The points O'₁, O'₂, O'₃ lie on the Feuerbach hyperbola. (Randy Hutson, September 14, 2016)

See Hyacinthos #23101. (S. Kirikami, A. Hatzipolakis and A.Montesdeoca, February 4-5, 2015)

X(6599) = (2nd SCHIFFLER, 1st SCHIFFLER)-ORTHOLOGIC CENTER

Barycentrics a^13 - 2 a^12 (b + c) - (b - c)^8 (b + c)^5 + a^11 (-3 b^2 + 5 b c - 3 c^2) + 2 a (b - c)^6 (b + c)^4 (b^2 - b c + c^2) - a^8 (b + c) (3 b^2 - 2 b c + 3 c^2)^2 + a^10 (7 b^3 + 2 b^2 c + 2 b c^2 + 7 c^3) + a^2 (b - c)^4 (b + c)^3 (3 b^4 - 3 b^3 c - 5 b^2 c^2 - 3 b c^3 + 3 c^4) + a^9 (4 b^4 - 14 b^3 c + 15 b^2 c^2 - 14 b c^3 + 4 c^4) - 3 a^7 (2 b^6 - 6 b^5 c + 6 b^4 c^2 - 7 b^3 c^3 + 6 b^2 c^4 - 6 b c^5 + 2 c^6) - a^4 (b + c)^3 (4 b^6 - 16 b^5 c + 26 b^4 c^2 - 27 b^3 c^3 + 26 b^2 c^4 - 16 b c^5 + 4 c^6) - a^3 (b^2 - c^2)^2 (7 b^6 - 17 b^5 c + 11 b^4 c^2 - 6 b^3 c^3 + 11 b^2 c^4 - 17 b c^5 + 7 c^6) + a^6 (6 b^7 - 4 b^6 c + 7 b^5 c^2 + 4 b^4 c^3 + 4 b^3 c^4 + 7 b^2 c^5 - 4 b c^6 + 6 c^7) + a^5 (9 b^8 - 20 b^7 c + 3 b^6 c^2 + 3 b^5 c^3 + 6 b^4 c^4 + 3 b^3 c^5 + 3 b^2 c^6 - 20 b c^7 + 9 c^8) : :

In addition to the notes at X(6596), Angel Montesdeoca notes in Hyacinthos 23101 that the following 15 points lie on the Feuerbach hyperbola: O₁, O₂, O₃, O'₁, O'₂, O'₃, the cyclologic ceners X(1), X(1320), X(3065), X(6595)-X(6599), and S.

X(7004) = PERSPECTOR OF ABC AND THE EXTRA-TRIANGLE OF X(201)

Barycentrics a(b - c)2(b + c - a)(b2 + c2 - a2) : :

In the plane of a triangle ABC, let
P = X(80) = reflection of X(1) in X(11);
R1 = reflection of AB in CP;
R2 = reflection of AC in BP;
A' = R1∩R2, and define B' and C' cyclically;
T = the affine transformation that carries ABC onto A'B'C'.
Then X(7004) = the finite fixed point of T. The fixed lines of T are parallel to the asymptotes of the Jerabek hyperbola. (Angel Montesdeoca, March 14, 2024)

See Afinidad con puntos fijos X(7004) y los del infinito de la hipérbola de Jerabek

X(7100) = PERSPECTOR OF ABC AND THE EXTRA-TRIANGLE OF X(1807)

Barycentrics sin(A) / (2 + sec(A)) : :

Let I be the incenter of triangle ABC. Let L_B be the line through I perpendiculat to AC, and let A_B = L_B∩BC and B_A = L_B∩BA. Define B_C and C_A cyclically, and define C_B and A_C cyclically. Let A' be the circumcenter of IA_BA_C, and define B' and C' cyclically. The triangle A'B'C' is perspective to ABC, and the perspector is X(7100). (Angel Montesdeoca, June 11, 2016)

X(7319) = PERSPECTOR OF ABC AND THE EXTRA-TRIANGLE OF X(5556)

Barycentrics 1 / [3(b^2 + c^2 - a^2) - 2bc] : : : :

Let DEF be the extouch triangle of ABC. Let D' be the point, other than A, in which the line AD meets the A-excircle, and define E' and F' cyclically. Let La be the line tangent to the A-excircle at D', and define Lb and Lc cyclically. Let A' = Lb∩Lc, and define B' and C' cyclically. Then A'B'C' is perspective to ABC, and the perspector is X(7319). (Angel Montesdeoca, July 14, 2018)

X(7494) = {X(2),X(22)}-HARMONIC CONJUGATE OF X(4)

Barycentrics (b^2 + c^2 - a^2)(3a^4 - b^4 - c^4 + 2a^2b^2 + 2a^2c^2 + 2b^2c^2) : :

Let DEF = medial triangle and D'E'F' = circummedial triangle. Let Γ = circumcircle, and let Ab be the point, other than E', in which the line E'F intersects Γ. Define Bc and Ca cyclically. Let Ac be the point, other than F', in which the line F'E intersects Γ. Let Oa be the center of the conic tangent to the five lines BC, AE', AF', BAb, CAc, and define Ob and Oc cyclically. The finite fixed point of the affine transformation that carries ABC onto OaObOC is X(7494). (Angel Montesdeoca, May 3, 2020)
HG020520.

X(7619) = X(5)-OF-McCAY-TRIANGLE

Barycentrics 8 a^4-14 a^2 b^2+5 b^4-14 a^2 c^2-8 b^2 c^2+5 c^4 : :
X(7619) = (1 + 20 sin²ω)*X(2) + (-1 + 4 sin²ω)*X(99)

Let G be the centroid of a triangle ABC, and
Oa = circumcenter of GBC, and define Ob and Oc cyclically
Na = nine-point center of GObOc, and define Nb and Nc cyclically
L = Euler line of NaNbNc
L' = Euler line of McCay triangle
Then X(7619) = L∩L'. See Angel Montesdeoca, X(7619) and Hyacinthos #24690. (October 25, 2016)

X(7666) = GIUGIUC CENTER OF SIMILITUDE

Trilinears (4 + 6 cos 2A) cos(B - C) - 16 cos A - 9 cos 3A : : (César Lozada)
Barycentrics a^2 (9 a^8 - 24 a^6 (b^2 + c^2) + a^4 (18 b^4 + 37 b^2 c^2 + 18 c^4) - 15 a^2 b^2 c^2 (b^2 + c^2) - (b^2 - c^2)^2 (3 b^4 + 4 b^2 c^2 + 3 c^4)) : : (Angel Montesdeoca)

Let O_A = reflection of X(3) in line AX(4), let H_A = reflection of X(4) in O_A, let M_A = midpoint of segment O_AH_A, and define M_B and M_C cyclically. Then M_AM_BM_C is similar to ABC, and the center of similitude is X(7666). See Hyacinthos 23263 and 23277.

(Triángulos inversamente semejantes)

X(7669) = POLE OF X(115)X(125) WITH RESPECT TO THE CIRCUMCIRCLE

For a sketch, click X(3447)andX(7669). (Angel Montesdeoca, April 22, 2016)

X(7740) = MIDPOINT OF X(3) AND X(5502)

Barycentrics ((-a^2+b^2+c^2)^2-b^2*c^2)*(2*a^10-2*(b^2+c^2)*a^8-(5*b^4-12*b^2*c^2+5*c^4)*a^6+7*(b^4-c^4)*(b^2-c^2)*a^4-(b^2-c^2)^2*(b^4+8*b^2*c^2+c^4)*a^2-(b^4-c^4)*(b^2-c^2)^3)*a^2 : :

X(7740) = center of the circle D = {{X(3), X(110), X(4240}}. Dao Thanh Oai finds seven other points on D, as follows. Let A' = (Euler line)∩BC, and define B' and C' cyclically. The points X(3)-of-AB'C', X(3)-of-A'BC', X(3)-of-A'B'C, X(110)-of-AB'C', X(110)-of-A'BC', X(110)-of-A'B'C all lie on D. Let P be the paralogic triangle whose perspectrix is the Euler line of ABC. Then X(3)-of-P lies on D. See ADGEOM 2342 (February 2015) and Centro X(5502) (A. Montesdeoca. Hechos Geométricos en el Triángulo).

X(7971) = ORTHOLOGIC CENTER OF THESE TRIANGLES: 5th MIXTILINEAR AND EXTOUCH

Barycentrics a (a^6-4 a^5 (b+c)+a^4 (b^2+6 b c+c^2)+8 a^3 (b-c)^2 (b+c)-a^2 (b-c)^2 (5 b^2+14 b c+5 c^2-4 a (b-c)^4 (b+c)+(b^2-c^2)^2 (3 b^2-2 b c+3 c^2))) : :

Let A' = reflection of A in X(1), and define B' and C' cyclically. Let A" be the orthogonal projection of A' on BC. Let Ta be the tangent at A' to the circle (IA'A"), and define Tb and Tc cyclically. The lines Ta, Tb, Tc concur in X(7971). (Angel Montedeoca, April 20, 2020) HG200420

X(8041) = DANNEELS POINT OF X(4576)

Barycentrics a^2(b^2 + c^2)^2 : :

In the plane of a triangle ABC, let
   G = centroid = X(2);
   DEF = cevian triangle of X(6);
   T_A = line tangent to circumcircle at A;
   L_AB = line through B parallel to AG;
   L_AC = line through C parallel to AG;
   A₂ = T_A∩L_AB;
   A₃ = T_A∩L_AC;
   A_b = DA₂∩AG;
   A_c = DA₃∩AG;
   A' = {A_b,A_c}-harmonic conjugate of A, and define B' and C' cyclically.
The finite fixed point of the affine transformation that carries ABC onto A'B'C' is X(8041). See X(8041) (Angel Montesdeoca, February 23, 2021) and HG200221

X(8051) = CYCLOCEVIAN CONJUGATE OF X(7319)

Barycentrics 1/((-a+b+c) (a^2+2 a b+b^2+2 a c-6 b c+c^2)) : :

Let Na be the Nagel point. The incircle intersects ANa at two points, the closer of which to the vertex A is denoted by A1 and the other by A2. Define B1, C1 and B2, C2 cyclically. X(8051) is the trilinear pole of the perspectrix of the triangles A1B1C1 and A2B2C2, line X(3667)X(3669). (Angel Montesdeoca, November 17, 2021)

X(8105) = X(2)-CEVA CONJUGATE OF X(1313)

Barycentrics (b^2-c^2) (2 a^4-a^2 b^2-b^4-a^2 c^2+2 b^2 c^2-c^4 + a^2 (a^2-b^2-c^2) J) : : , where J = |OH|/R

Let MaMbMc = medial triangle and HaHbHc = orthic triangle. Let A' = reflection of A in X(3), and let A'' be the point, other than A', where the line A'Ma meets the circumcircle. Define B'' and C'' cyclically. The points D=BC∩HbHc, E=CA∩HcHa, F=AB∩HaHb, D'=B"C"∩MbMc, E'=C"A"∩McMa, F'=A"B"∩MaMb lie on orthic axis (common perspectrix of ABC and HaHbHc, MaMbMc and A"B"C" ). The fixed points of the projectivity that maps D, E, F onto D', E', F', respectively, are X(8105) and X(8106). (Angel Montesdeoca, May 17, 2019)
Proyectividad sobre el eje órtico.

X(8160) = CENTER OF THE OUTER MONTESDEOCA-LEMOINE CIRCLE

Barycentrics a^2(a^6(b^2+c^2)+2a^4(b^2+c^2)^2-a^2(4b^6+7b^4c^2+7b^2c^4+4c^6)+b^8-3b^6c^2-2b^4c^4-3b^2c^6+c^8 + 4S^3(2a^2-b^2-c^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 Montesdeoca-Lemoine 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 Montesdeoca-Lemoine 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 Z₁ and Z₂, 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)

X(8161) = CENTER OF THE INNER MONTESDEOCA-LEMOINE CIRCLE

Barycentrics a^2(a^6(b^2+c^2)+2a^4(b^2+c^2)^2-a^2(4b^6+7b^4c^2+7b^2c^4+4c^6)+b^8-3b^6c^2-2b^4c^4-3b^2c^6-c^8 - 4S^3(2a^2-b^2-c^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)

X(8183) = MONTESDEOCA DEGENERATE CONICS POINT

Barycentrics: a - a^1/3b^1/3c^1/3 : b - a^1/3b^1/3c^1/3 : c - a^1/3b^1/3c^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_AC_A is parallel to BC at distance |kr(A)|, where r(A) is the inradius of triangle AB_AC_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)x² - a(a² + b² + c² + 2bc + 2ca + 2ab + bck²)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^-1/3b^-1/3c^-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 - a² : ba - bc : ca - cb
ab - ac : ac - b² : cb - ca
ac - ab : bc - ba : ba - c²

These last three points are collinear on the trilinear pole of X(86).

X(8770) = BARYCENTRIC PRODUCT OF PU(128)

Barycentrics a^2 (a^2 + b^2 - 3 c^2) (a^2 - 3 b^2 + c^2) : :

Let A'B'C' be the tangential triangle and L the orthic axis. Let A'' = L∩B'C', and define B'' and C'' cyclically. Let Ta be the line, other than B'C', through A'' tangent to the circumcircle, and define Tb and Tc cyclically. Let A* = Tb∩Tc, and define B* and C* cyclically. Then A*B*C* is perspective to ABC, and the perspector is X(8770). (Angel Montesdeoca, July 17, 2019).

X(9033) = CROSSDIFFERENCE OF X(6) AND X(2781)

Barycentrics (b^2 - c^2)*(-a^2 + b^2 + c^2)*(-2*a^4 + a^2*b^2 + b^4 + a^2*c^2 - 2*b^2*c^2 + c^4) : :

Let W be the circumconic with center X(1650). One of the asymptotes of W is the Euler line. The other is in the direction of X(9033). For a sketch, click X(9033). (Angel Montesdeoca, April 19, 2016)

X(9121) = (1,84,1498)-(4,3,20)-(1490,40,64)-LOUD POINT OF THE DARBOUX CUBIC

Barycentrics a (a^9+a^8 (b+c)+(b-c)^6 (b+c)^3+a (b-c)^2 (b+c)^6-4 a^7 (b^2-b c+c^2)-4 a^2 (b-c)^2 (b+c)^3 (b^2-b c+c^2)-4 a^3 (b^2-c^2)^2 (b^2+b c+c^2)+2 a^4 (b+c)^3 (3 b^2-4 b c+3 c^2)+2 a^5 (b-c)^2 (3 b^2+4 b c+3 c^2)-4 a^6 (b^3+2 b^2 c+2 b c^2+c^3)) : :

Let ABC be a triangle with incenter I = X(1), De Longchamps point L = X(20), and Bevan point Be = X(40). Let DEF = cevian triangle of L. Let A' = ABe∩ID, and define B' and C' cyclically. The triangles ABC and A'B'C' are orthologic. The center of orthology of ABC with respect to A'B'C' is X(3345) and the center of orthology of A'B'C' with respect to ABC is X(9121). (Angel Montesdeoca, October 3, 2023 *)

X(9311) = CEVAPOINT OF PU(47)

Barycentrics 1/(a^2-(b+c)*a+2*b*c) : :

X(9311) is the trilinear pole of line X(2254)X(3667), which is the radical axis of incircle and excircles radical circle, and also the polar of X(1) wrt {circumcircle, nine-point circle}-inverter. (Randy Hutson, February 10, 2016)

Let DEF be the intouch triangle of triangle ABC. Let L be the line through D parallel to CA, and let C_A = L∩AB, and define A_B and B_C cyclically. Let L' be the line through D parallel to AB, and let B_A = L'∩CA, and define C_B and A_C cyclically. Let A'B'C' be the triangle having sidelines B_AC_A, C_BA_B, A_CB_C. Then A'B'C' is perspective to ABC, and the perspector is X(9311). Click here for a sketch showing X(9311). (Angel Montesdeoca, March 20, 2016.)

X(9581) = EXSIMILCENTER OF THESE CIRCLES: BEVAN AND 2ND JOHNSON-YFF

Barycentrics (a^3+(b-c)^2*a+2*(b^2-c^2)*(b-c))*(a-b-c) : :

Let
(I) = incircle;
T = a point on (I);
t = line tangent to (I) at T;
T₁ and T₂ = points of intersection of t and the nine-point circle;
t₁ = line through T₁ tangent to (I);
t₂ = line through T₂ tangent to (I).
The locus of t₁∩t₂ as T varies on (I) is a conic, Γ, with center X(50443).
The center of the reciprocal polar of Γ wrt (I) is X(9581).
The center of the reciprocal polar of (I) wrt Γ is X(50444).
If you have Geogebra, you can view X(9581) (Angel Montesdeoca, June 8, 2022)
See Angel Montesdeoca, HG060622.

X(9969) = ORTHOLOGIC CENTER OF THESE TRIANGLES: MIDHEIGHT TO 1ST EHRMANN

Barycentrics a^2*((b^2+c^2)*(a^4-(b^2-c^2)^2)+2*a^2*b^2*c^2) : :

In the plane of a triangle ABC, let
Γ = circumcircle of ABC
T_aT_bT_c = tangential triangle of ABC
DEF = cevian triangle of X(1176)
A_b = the point, other than B, of intersection of circles Γ and (BDT)
A_c = the point, other than C, of intersection of circles Γ and (CDT)
L_a = A_bA_c, and define L_b and L_c cyclically
A' = L_b∩L_c, and define B' and C' cyclically
Then A'B'C' is perspective to ABC, and the perspector is X(9969). (Angel Montesdeoca, February 10, 2021)

See El centro X(9969) y el triángulo ceviano del 1º punto Saragossa del conjugado isogonal del punto de Exeter

X(10018) = 20th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics (a^2+b^2-c^2) (a^2-b^2+c^2) (2 a^6-5 a^4 b^2+4 a^2 b^4-b^6-5 a^4 c^2+2 a^2 b^2 c^2+b^4 c^2+4 a^2 c^4+b^2 c^4-c^6) : :

Centers X(10018)-X(10021) lie on the Euler line. For constructions and properties, see Hyacinthos messages beginning with 23422 (May 30, 2016). See also X(6102).

X(10019) = 21st HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics (a^2+b^2-c^2) (a^2-b^2+c^2) (2 a^6+a^4 b^2-8 a^2 b^4+5 b^6+a^4 c^2+8 a^2 b^2 c^2-5 b^4 c^2-8 a^2 c^4-5 b^2 c^4+5 c^6) : :

Centers X(10018)-X(10021) lie on the Euler line. For constructions and properties, see Hyacinthos messages beginning with 23422 (May 30, 2016).

X(10020) = 22nd HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics 2 a^10-5 a^8 b^2+2 a^6 b^4+4 a^4 b^6-4 a^2 b^8+b^10-5 a^8 c^2+4 a^6 b^2 c^2-2 a^4 b^4 c^2+6 a^2 b^6 c^2-3 b^8 c^2+2 a^6 c^4-2 a^4 b^2 c^4-4 a^2 b^4 c^4+2 b^6 c^4+4 a^4 c^6+6 a^2 b^2 c^6+2 b^4 c^6-4 a^2 c^8-3 b^2 c^8+c^10 : :

Centers X(10018)-X(10021) lie on the Euler line. For constructions and properties, see Hyacinthos messages beginning with 23422, especially 23429

X(10021) = 23rd HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics 2 a^7-2 a^6 b-5 a^5 b^2+5 a^4 b^3+4 a^3 b^4-4 a^2 b^5-a b^6+b^7-2 a^6 c+2 a^5 b c+a^4 b^2 c+a^3 b^3 c+2 a^2 b^4 c-3 a b^5 c-b^6 c-5 a^5 c^2+a^4 b c^2+4 a^3 b^2 c^2+2 a^2 b^3 c^2+a b^4 c^2-3 b^5 c^2+5 a^4 c^3+a^3 b c^3+2 a^2 b^2 c^3+6 a b^3 c^3+3 b^4 c^3+4 a^3 c^4+2 a^2 b c^4+a b^2 c^4+3 b^3 c^4-4 a^2 c^5-3 a b c^5-3 b^2 c^5-a c^6-b c^6+c^7 : :

Centers X(10018)-X(10021) lie on the Euler line. For constructions and properties, see Hyacinthos messages beginning with 23422, especially 23436

X(10032) = X(8)X(30)∩X(21)X(551)

Barycentrics (7a^3-a^2(b+c)-a(4b^2+b c+4c^2)-2(b-c)^2(b+c) : :
X(10032) = 5 X[21] - 4 X[551] = 5 X[79] - 8 X[3634] = 7 X[3624] - 10 X[3647] = X[8] + 5 X[3648] = X[8] - 10 X[3650] = X[3648] + 2 X[3650]

Let Ia be the A-excenter of a triangle ABC, and define Ib and Ic cyclically. Let A' = reflection of Ia in BC, and define B' and C' cyclically. Let A'' = reflection of IA in A', and define B'' and C'' cyclically. The Euler lines of A''BC, B''CA,C''AB concur in X(10032). (Tran Quang Hung and Angel Montesdeoca, July 19, 2016: Hyacinthos 23831)
X(10032) lies on these lines:
{8,30}, {21,551}, {79,3634}, {191,6175}, {527,2346}, {553,5284}, {1281,6054}, {2796,4921}, {2975,3656}, {3624,3647}
X(10032) = reflection of X(6175) in X(9)

X(10033) = 24th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics 4 a^8+a^4 (4 b^4+7 b^2 c^2+4 c^4)-2 a^2 (3 b^6-5 b^4 c^2-5 b^2 c^4+3 c^6)-(b^2-c^2)^2 (2 b^4+7 b^2 c^2+2 c^4) : :

Let ABC be a triangle and A'B'C' the cevian triangle of the centroid, G, and let
Oa = circumcenter of GB'C'; define Ob and Oc cyclically
Oab = circumcenter of AB'G; define Obc and Oca cyclically
Oac = circumcenter of AC'G; define Oba and Ocb cyclically
Ga = centroid of OaObOc; define Gb and Gc cyclically
Na = nine-point center of GB'C'; define Nb and Nc cyclically
Nab = nine-point center of AB'G; define Nbc and Nca cyclically
Nac = nine-point center of AC'G; define Nba and Ncb cyclically

The triangles GaGbGc, G1G2G3 are perspective, and their perspector is X(10033). Let Ea be the Euler line of OaOabOac, and define Eb and Ec cyclically; then Ea, Eb, Ec are parallel, and they concur in X(524). Let Fa be the Euler line of NaNabNac, and define Fb and Fc cyclically. Let A'' = Fb∩Fc, and define B'' and C'' cyclically. Then the triangles ABC and A''B''C'' are parallelogic, and the parallelogic center of ABC with respect to A''B''C'' is X(6094), the 11th Hatzipolakis-Montesdeoca point, and the parallelogic center of A''B''C'' with respect to ABC is X(10034). (Antreas Hatzipolakis and Angel Montesdeoca, August 1, 2016; see Hyacinthos 23907)

X(10033) lies on these lines:
{2,1495}, {4,3849}, {30,7697}, {98,381}, {114,8592}, {183,3830}, {262,542}, {3545,7694}, {3839,9753}, {3845,9993}, {5066,7792}, {6054,9830}, {8370,9873}

X(10034) = 25th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics 10 a^12-33 a^10 (b^2+c^2)-6 a^8 (11 b^4-34 b^2 c^2+11 c^4)+a^6 (221 b^6-108 b^4 c^2-108 b^2 c^4+221 c^6)-3 a^4 (41 b^8-112 b^6 c^2+288 b^4 c^4-112 b^2 c^6+41 c^8)-6 a^2 (5 b^10-28 b^8 c^2+8 b^6 c^4+8 b^4 c^6-28 b^2 c^8+5 c^10)+13 b^12-81 b^10 c^2+153 b^8 c^4-154 b^6 c^6+153 b^4 c^8-81 b^2 c^10+13 c^12 : :

Let A''B''C'' be as at X(10035). The parallelogic center of A''B''C'' with respect to ABC is X(10034). (Antreas Hatzipolakis and Angel Montesdeoca, August 1, 2016; see Hyacinthos 23907)

X(10035) = 26th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics 2 a^9 (b+c)+b c (b^2-c^2)^4-2 a^8 (b^2+c^2)+a (b-c)^4 (b+c)^3 (b^2+b c+c^2)-a^7 (7 b^3+b^2 c+b c^2+7 c^3)+2 a^2 (b^2-c^2)^2 (b^4-2 b^3 c-2 b c^3+c^4)+a^6 (6 b^4-2 b^3 c+8 b^2 c^2-2 b c^3+6 c^4)-a^3 (b-c)^2 (5 b^5+7 b^4 c+6 b^3 c^2+6 b^2 c^3+7 b c^4+5 c^5)+a^5 (9 b^5-4 b^4 c+b^3 c^2+b^2 c^3-4 b c^4+9 c^5)-a^4 (6 b^6-5 b^5 c+2 b^4 c^2+6 b^3 c^3+2 b^2 c^4-5 b c^5+6 c^6) : :

Let ABC be a triangle and A'B'C' the cevian triangle of the incenter, I, and let

Oa = circumcenter of IB'C'; define Ob and Oc cyclically
Oab = circumcenter of AB'I; define Obc and Oca cyclically
Oac = circumcenter of AC'I; define Oba and Ocb cyclically
Ga = centroid of OaObOc; define Gb and Gc cyclically
Na = nine-point center of IB'C'; define Nb and Nc cyclically
Nab = nine-point center of AB'I; define Nbc and Nca cyclically
Nac = nine-point center of AC'I; define Nba and Ncb cyclically

Let Ea be the Euler line of NaNabNac, and define Eb and Ec cyclically; then Ea, Eb, Ec concur in X(10035). Let Fa be the Euler line of OaOabOac, and define Fb and Fc cyclically; the lines Fa, Fb, Fc are parallel, and they meet in X(517). (Antreas Hatzipolakis and Angel Montesdeoca, August 1, 2016; see Hyacinthos 23914)

X(10035) lies on these lines:
{11,500}, {30,1319}, {496,5495}, {511,6713}, {549,4271}, {952,5453}

X(10036) = 27th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics 4 a^8 b c-2 a^9 (b+c)+b c (b^2-c^2)^4-12 a^6 b c (b^2+c^2)-6 a^2 b c (b^2-c^2)^2 (b^2+c^2)-a (b-c)^4 (b+c)^3 (b^2+3 b c+c^2)+a^7 (7 b^3+5 b^2 c+5 b c^2+7 c^3)+5 a^3 (b-c)^2 (b^5+3 b^4 c+2 b^3 c^2+2 b^2 c^3+3 b c^4+c^5)-a^5 (9 b^5+6 b^4 c-5 b^3 c^2-5 b^2 c^3+6 b c^4+9 c^5)-a^4 (-13 b^5 c+6 b^3 c^3-13 b c^5) : :

Let ABC be a triangle and A'B'C' the cevian triangle of the incenter, I. Continuing from X(10035), let Pa be the line through A' parallel to Ea, and define Pb and Pc cyclically. Then Pa,Pb,Pc concur in X(10036). (Antreas Hatzipolakis and Angel Montesdeoca, August 1, 2016; see Hyacinthos 23914)

X(10036) lies on these lines:
{11,8143}, {115,119}, {952,5492}, {1317,2771}

X(10105) = 28th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics a (a^7 b^2-3 a^5 b^4+3 a^3 b^6-a b^8-2 a^7 b c-6 a^6 b^2 c+3 a^5 b^3 c+13 a^4 b^4 c-8 a^2 b^6 c-a b^7 c+b^8 c+a^7 c^2-6 a^6 b c^2-16 a^5 b^2 c^2+9 a^4 b^3 c^2+15 a^3 b^4 c^2-2 a^2 b^5 c^2-b^7 c^2+3 a^5 b c^3+9 a^4 b^2 c^3+12 a^3 b^3 c^3+10 a^2 b^4 c^3+a b^5 c^3-3 b^6 c^3-3 a^5 c^4+13 a^4 b c^4+15 a^3 b^2 c^4+10 a^2 b^3 c^4+2 a b^4 c^4+3 b^5 c^4-2 a^2 b^2 c^5+a b^3 c^5+3 b^4 c^5+3 a^3 c^6-8 a^2 b c^6-3 b^3 c^6-a b c^7-b^2 c^7-a c^8+b c^8) : :

Let I be the incenter of a triangle ABC, and
A'B'C' = intouch triangle (the pedal triangle of I)
A''B''C'' = cevian triangle of I
Ab = orthogonal projection of A'' on IB, and define Bc and Ca cyclically
Ac = orthogonal projection of A'' on IC, and define Ba and Cb cyclically
A'b = orthogonal projection of A'' on IB', and define B'c and C'a cyclically
A'c = orthogonal projection of A'' on IC', and define B'a and C'b cyclically
(Nab) = nine-point cricle of A''AbA'b, and define (Nbc) and (Nca) cyclically
(Nac) = nine-point cricle of A''AcA'c, and define (Nba) and (Ncb) cyclically
Ra = radical axis of (Nab) and (Nac) = perpendicular bisector of segment NabNac.

The lines Ra, Rb, Rc concur in X(10105). Also, the parallels to Ra, Rb, Rc through A', B', C, respectively, concur in X(942); and the parallels to Ra, Rb, Rc through A'', B'', C'', respectively, concur in X(500). (Antreas Hatzipolakis and Angel Montesdeoca, August 8, 2016; see Hyacinthos 23972)

X(10106) = 29th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics 2 a^4-a^3 (b+c)+a (b-c)^2 (b+c)-(b^2-c^2)^2-a^2 (b^2-6 b c+c^2) : :
X(10106) = (2R - r)*X(1) + r*X(4)

Let I be the incenter of a triangle ABC, and
A'B'C' = intouch triangle (the pedal triangle of I)
H' = X(4)-of-A'B'C'
Ab = orthogonal projection of A on H'B', and define Bc and Ca cyclically
Ac = orthogonal projection of A on H'C', and define Ba and Cb cyclically
La = Euler line of AAbAc, and define Lb and Lc cyclically
Pa = line through A'' parallel to La, and define Pb and Pc cyclically

The lines La, Lb, Lc concur in X(10106). Let

Qa = line through A parallel to La, and define Qb and Qc cyclically
Ra = line through A' parallel to La, and define Rb and Rc cyclically

The lines La, Lb, Lc concur in X(5836), the midpoint of X(8) and X(65). The lines Qa, Qb, Qc concur in X(8), and the lines Ra, Rb, Rc concur in X(145). (Antreas Hatzipolakis and Angel Montesdeoca, August 9, 2016; see Hyacinthos 23990)

X(10107) = 30th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics a(3a^2(b+c)-2a b c-3b^3+5b c(b+c)-3c^3) : :
X(10107) = 3(4R + r)*X(7) + (4R-3r)*X(8)

Let I be the incenter of a triangle ABC, and
A'B'C' = intouch triangle (the pedal triangle of I)
H' = X(4)-of-A'B'C'
Ab = orthogonal projection of A on H'B', and define Bc and Ca cyclically
Ac = orthogonal projection of A on H'C', and define Ba and Cb cyclically
La = Euler line of AAbAc, and define Lb and Lc cyclically
Ia = excenter of ABC, and define Ib and Ic cyclically
Pa = orthogonal projection of Ia on BC, and define Pb and Pc cyclically
A* = midpoint of AbAc, and define B* and C* cyclically
Qa = line through A* parallel to La, and define Qb and Qc cyclically

The lines Qa, Qb, Qc concur in X(10107). Let

Qa = line through Ia parallel to La, and define Qb and Qc cyclically

The lines Qa, Qb, Qc concur in X(2136), which is the X(145)-Ceva conjugate of X(1). (Antreas Hatzipolakis and Angel Montesdeoca, August 10, 2016; see Hyacinthos 23998)

X(10108) = 31st HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics a (a^4 (b-c)^2-a^2 (b^4+7 b^3 c+16 b^2 c^2+7 b c^3+c^4)-a (b^5+b^4 c+8 b^3 c^2+8 b^2 c^3+b c^4+c^5)+b c (b^2-c^2)^2+a^3 (b^3-7 b^2 c-7 b c^2+c^3)) : :

Let I be the incenter of a triangle ABC, and
A'B'C' = cevian triangle of X(I)
Ab = orthogonal projection of A' on BB', and define Bc and Ca cyclically
Ac = orthogonal projection of A' on CC', and define Ba and Cb cyclically
Abc = orthogonal projection of Ab on CC', and define Bca and Cab cyclically
Acb = orthogonal projection of Ac on BB', and define Bac and Cba cyclically
La = Euler line of IAbcAcb, and define Lb and Lc cyclically

The lines La, Lb, Lc concur in X(10108). (Antreas Hatzipolakis and Angel Montesdeoca, August 10, 2016; see Hyacinthos 23949)

X(10122) = 32nd HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics a (a^4 (b-c)^2-a^5 (b+c)+(b^2-c^2)^2 (b^2-b c+c^2)-a (b-c)^2 (b^3+4 b^2 c+4 b c^2+c^3)+a^3 (2 b^3+3 b^2 c+3 b c^2+2 c^3)+a^2 (-2 b^4+3 b^3 c+6 b^2 c^2+3 b c^3-2 c^4)) : :
X(10122) = (r+2R)*X(1) - r*X(21) = (r+4R)*X(7) - (r+2R)*X(79)

Let ABC be a triangle, and let
A'b = orthogonal projection of A on the external bisector of angle ABC
A'c = orthogonal projection of A on the external bisector of angle ACB
Ea = Euler line of AA'bA'c, and define Eb and Ec cyclically
Ia = A-excenter of ABC, and define Ib and Ic cyclically
A'B'C' = intouch triangle (the pedal triangle of the incenter)
A'' = orthogonal projection of IA on B'C', and define B'' and C'' cyclically
Pa = line through A'' parallel to Ea, and define Pb and Pc cyclically
A''' = orthogonal projections of A' on IbIc, and define B''' and C''' cyclically
Qa = line through A''' parallel to Ea, and define Qb and Qc cyclically

The lines Pa, Pb, Pc concur in X(10122). The lines Qa, Qb, Qc concur in X(10123). The lines Ea, Eb, Ec concur in X(442). (Antreas Hatzipolakis and Angel Montesdeoca, August 14, 2016. See Hyacinthos 24025 and Hyacinthos 24015.

X(10123) = 33rd HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics 2 a^7-a^6 (b+c)-(b-c)^4 (b+c)^3+5 a b c (b^2-c^2)^2+a^3 (b+c)^2 (2 b^2-3 b c+2 c^2)-a^5 (4 b^2+6 b c+4 c^2)+a^4 (b^3-4 b^2 c-4 b c^2+c^3)+a^2 (b-c)^2 (b^3+6 b^2 c+6 b c^2+c^3) : :
X(10123) = (r+3R)*X(21) - (r+4R)*X(142) = (2r+R)*X(35) - (2r+5R)*X(79)

For a construction and references, see X(10122).

X(10124) = 34th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics 10 a^4-17 a^2 (b^2+c^2)+7 (b^2-c^2)^2 : :
X(10124) = 5X(3) + 7X(5) = 7X(2) + X(3)

Let O be the circumcenter and N the nine-point center of a triangle ABC. Let
Ma = midpoint of OA, and define Mb and Mc cyclically
Aa = orthogonal projection of Ma on NA, and define Bb and Cc cyclically
Ba = orthogonal projection of Mb on NA, and define Cb and Ac cyclically
Ca = orthogonal projection of Mc on NA, and define Cb and Ac cyclically
M1 = midpoint of NA, and define M2 and M3 cyclically
A1 = orthogonal projection of M1 on OA, and define A2 and A3 cyclically
B1 = orthogonal projection of M2 on OA, and define B2 and B3 cyclically
C1 = orthogonal projection of M3 on OA, and define C2 and C3 cyclically
Oa = circumcenter of AaAbAc, and define Ob and Oc cyclically
O1 = circumcenter of A1A2A3, and define O2 and O3 cyclically
La = Euler line of OaO2O3, and define Lb and Lc cyclically
L1 = Euler line of O1ObOC, and define L2 and L3 cyclically

Then L1, L2, L3 concur in X(10124), which lies on the Euler line, and La, Lb, Lc concur in X(140). (Antreas Hatzipolakis and Angel Montesdeoca, August 14, 2016. See Hyacinthos 24036.

X(10189) = CENTER OF THE CENTROIDAL CONIC OF THE SCHRÖTER TRIANGLE

Barycentrics (3*a^4-3*(b^2+c^2)*a^2-b^4+5*b^2*c^2-c^4)*(b^2-c^2) : :

Let ABC be a triangle, P a point, DEF the cevian triangle of P, and A'B'C' the Schröeter triangle. As P moves on GK, the centroids of the triangles DB'C', EC'A', FA'C' are aligned' and their line envelops a conic having center X(10189). This conic is given by the barycentric equation
(a^2-b^2) (a^2-c^2)((a^2-b^2) (a^2-c^2) (4 a^4-4 a^2 (b^2+c^2)+9 b^4-14 b^2 c^2+9 c^4) x^2-6 (b^2-c^2)^2 (3 a^4-3 a^2 (b^2+c^2)+2 b^4-b^2 c^2+2 c^4) y z) + cyclic sum =0.
See Angel Montesdeoca HG131223. (December 15, 2023.)

X(10203) = X(5)-OF-ANTIPEDAL-TRIANGLE OF X(54)

Barycentrics a^2 (a^14-4 a^12 (b^2+c^2)-4 a^8 b^2 c^2 (b^2+c^2)-b^2 c^2 (b^2-c^2)^4 (b^2+c^2)+a^10 (5 b^4+8 b^2 c^2+5 c^4)-a^2 (b^2-c^2)^2 (b^8-5 b^6 c^2-7 b^4 c^4-5 b^2 c^6+c^8)-5 a^6 (b^8-b^6 c^2-b^4 c^4-b^2 c^6+c^8)+a^4 (4 b^10-11 b^8 c^2-5 b^6 c^4-5 b^4 c^6-11 b^2 c^8+4 c^10)) : :
X(10203) = 4R²X(3) - |OH²|2X(54)
X(10203) is the nine-point center of the antipedal triangle of the Kosnita point. (Tran Quang Hung and Angel Montesdeoca, August 31, 2016; see Hyacinthos #24193):

[Tran Quang Hung]:
1. The centroid of antipedal triangle of symmedian point of a triangle lies on Euler line of this triangle.
2. The NPC center of the antipedal triangle of Kosnita point of a triangle lies on the line connecting this Kosnita point and the circumcenter of this triangle.
3. The Euler line of triangle ABC bisects the segment connecting symmedian point L of ABC and symmedian point of antipedal triangle of L with respect to ABC.
4. The line connecting the NPC center of the antipedal triangle of N of a triangle passes through circumcenter of this triangle.

[Antreas Hatzipolakis]:
Questions:
Which are the points 1,2,3 and which is the NPC of the antipedal triangle of N of 4?

[Angel montesdeoca]:
*** 1. The centroid of antipedal triangle of symmedian point of a triangle is the circumcenter
*** 2. The NPC center of the antipedal triangle of Kosnita point of a triangle is W2 = 4R^2 X(3)- |OH|^2 X(54)

W2 = ( a^2 (a^14-4 a^12 (b^2+c^2)-4 a^8 b^2 c^2 (b^2+c^2)-b^2 c^2 (b^2-c^2)^4 (b^2+c^2)+a^10 (5 b^4+8 b^2 c^2+5 c^4)-a^2 (b^2-c^2)^2 (b^8-5 b^6 c^2-7 b^4 c^4-5 b^2 c^6+c^8)-5 a^6 (b^8-b^6 c^2-b^4 c^4-b^2 c^6+c^8)+a^4 (4 b^10-11 b^8 c^2-5 b^6 c^4-5 b^4 c^6-11 b^2 c^8+4 c^10)) : ... : ...),

with (6-9-13)-search numbers (253.269435013061,-86. 4545432539761,-53. 3997755790604).
*** 3. The Euler line of triangle ABC bisects the segment connecting symmedian point L of ABC and symmedian point of antipedal triangle of L with respect to ABC at

W3 = (a^2 (a^8-b^8+12 a^4 b^2 c^2-7 b^6 c^2+24 b^4 c^4-7 b^2 c^6-c^8-4 a^6 (b^2+c^2)+a^2 (4 b^6+3 b^4 c^2+3 b^2 c^4+4 c^6)) : ... : ... ),

with (6-9-13)-search numbers (3.22662598230530,2. 34859413859511,0. 525502701816087 )
On lines: {2,3}, {6235,9871}, {8546,9830}
*** The NPC of the antipedal triangle of N of 4. is W4 on Euler line of ABC:

W4 = (2 a^16-13 a^14 (b^2+c^2)-(b^2-c^2)^6 (b^4+c^4)+a^12 (37 b^4+50 b^2 c^2+37 c^4)+3 a^2 (b^2-c^2)^4 (b^6+b^4 c^2+b^2 c^4+c^6)-a^10 (59 b^6+71 b^4 c^2+71 b^2 c^4+59 c^6)+a^4 (b^2-c^2)^2 (3 b^8-4 b^6 c^2-7 b^4 c^4-4 b^2 c^6+3 c^8)+a^8 (55 b^8+34 b^6 c^2+32 b^4 c^4+34 b^2 c^6+55 c^8)+a^6 (-27 b^10+13 b^8 c^2+5 b^6 c^4+5 b^4 c^6+13 b^2 c^8-27 c^10) : ... :... ),

with (6-9-13)-search numbers (14.263493810476143237204984, )
On lines: {2,3}, {252,1263}

X(10204) = MIDPOINT OF X(6) AND X(6)-OF-ANTIPEDAL-TRIANGLE OF X(6)

Barycentrics a^2 (a^8-b^8+12 a^4 b^2 c^2-7 b^6 c^2+24 b^4 c^4-7 b^2 c^6-c^8-4 a^6 (b^2+c^2)+a^2 (4 b^6+3 b^4 c^2+3 b^2 c^4+4 c^6)) : :

In the plane of a triangle ABC, let K = X(6) K' = X(6)-of-antipedial-triangle of X(6) X(10204) = midpoint of K and K'. X(10204) lies on the Euler line of ABC. (Tran Quang Hung and Angel Montesdeoca, August 31, 2016; see Hyacinthos #24193).

X(10205) = (Name pending)

Barycentrics (2 a^16-13 a^14 (b^2+c^2)-(b^2-c^2)^6 (b^4+c^4)+a^12 (37 b^4+50 b^2 c^2+37 c^4)+3 a^2 (b^2-c^2)^4 (b^6+b^4 c^2+b^2 c^4+c^6)-a^10 (59 b^6+71 b^4 c^2+71 b^2 c^4+59 c^6)+a^4 (b^2-c^2)^2 (3 b^8-4 b^6 c^2-7 b^4 c^4-4 b^2 c^6+3 c^8)+a^8 (55 b^8+34 b^6 c^2+32 b^4 c^4+34 b^2 c^6+55 c^8)+a^6 (-27 b^10+13 b^8 c^2+5 b^6 c^4+5 b^4 c^6+13 b^2 c^8-27 c^10) : :

In the plane of a triangle ABC, let N = X(5) X(10205) = X(5)-of-antipedial-triangle of X(5)-of-[triangle, pending]; X(10205) lies on the Euler line of ABC. (Tran Quang Hung and Angel Montesdeoca, August 31, 2016; see Hyacinthos #24193).
X(10205) lies on these lines: {2,3}, {252,1263}
X(10205) = anticomplement of X(5501).

X(10206) = HUNG-MONTESDEOCA RADICAL CENTER

Barycentrics a (a^2 (b+c)+2 a b c-(b-c)^2 (b+c)) (a^6-2 a^5 (b+c)-a^4 (b^2+3 b c+c^2)+4 a^3 (b^3+b^2 c+b c^2+c^3)-a^2 (b+c)^2 (b^2-6 b c+c^2)-2 a (b-c)^2 (b+c)^3+(b^2-c^2)^2 (b^2-b c+c^2)) : :

In the plane of a triangle ABC, let HaHbHc = orthic triangle.
Ja = incenter of AHbHc, and define Jb and Jc cyclically
Ab = JbJc∩CA, and define Bc and Ca cyclically
Ac = JcJa∩BA, and define Ba and Cb cyclically

The points Jb,Jc,Ca,Ba lie on a circle, (Oa); define (Ob) and (Oc) cyclically.

X(10206) = radical center of (Oa), (Ob), (Oc); X(10206) lies on the Euler line of OaObOc.

Let Go = X(5902) = X(2)-of-OaObOc and Ho = X(4)-of-OaObOc.
X(10206) = 3[(r + 2R)2 - s2]Go + 2s2Ho.
(Tran Quang Hung and Angel Montesdeoca, August 31, 2016; see Hyacinthos #24219).

X(10207) = HUNG-MONTESDEOCA PERSPECTOR

Barycentrics a (2 a^7 (b^2+b c+c^2) +7 a^6 b c (b+c)-2 a^5 (3 b^4+3 b^3 c-b^2 c^2+3 b c^3+3 c^4) -a^4 b c (15 b^3+17 b^2 c+17 b c^2+15 c^3)+2 a^3 (b+c)^2 (3 b^4-3 b^3 c-4 b^2 c^2-3 bc^3+3 c^4)+9 a^2 b (b-c)^2 c (b+c)^3-2 a (b^2-c^2)^2 (b^4+b^3 c-3 b^2 c^2+bc^3+c^4)-b (b-c)^4 c (b+c)^3) : :

In the plane of a triangle ABC, let HaHbHc = orthic triangle
Ja = incenter of AHbHc, and define Jb and Jc cyclically
Ab = JbJc∩CA, and define Bc and Ca cyclically
Ac = JcJa∩BA, and define Ba and Cb cyclically

The points Jb,Jc,Ca,Ba lie on a circle, (Oa); define (Ob) and (Oc) cyclically.

The triangles JaJbJc and OaObOc are perspective, and their perspector is X(10207); see X(10206). (Tran Quang Hung and Angel Montesdeoca, August 31, 2016; see Hyacinthos #24219).

X(10208) = 1st HUNG-MONTESDEOCA-MOSES

Barycentrics (2 a^7 (b+c)^3-(b-c)^4 (b+c)^6-2 a (b-c)^4 (b+c)^3 (b^2+3 b c+c^2)+a^8 (b^2+6 b c+c^2)-2 a^5 (b+c)^3 (3 b^2-b c+3 c^2)+2 a^3 (b-c)^2 (b+c)^3 (3 b^2+4 b c+3 c^2)-a^4 b^2 c^2 (11 b^2+18 b c+11 c^2)-2 a^6 (b^4+5 b^3 c+4 b^2 c^2+5 b c^3+c^4)+a^2 (b^2-c^2)^2 (2 b^4+6 b^3 c+19 b^2 c^2+6 b c^3+2 c^4) : :
X(10208) = 3 X[5947] - 2 X[5953]

In the plane of a triangle ABC, let FaFbFc = Feuerbach triangle.
U = projection of A on line FbFc, and define V and W cyclically

The lines UFa, VFb, WFc concur in X(10208); see Hyacinthos #23529).

X(10209) = 2nd HUNG-MONTESDEOCA-MOSES POINT

Barycentrics a^11 (b-c)^2-(b-c)^6 (b+c)^7-a (b-c)^4 (b+c)^6 (b^2-7 b c+c^2)+a^10 (b^3-5 b^2 c-5 b c^2+c^3)-a^9 (5 b^4+5 b^3 c+12 b^2 c^2+5 b c^3+5 c^4)+a^8 (-5 b^5+b^4 c+b c^4-5 c^5)+a^3 (b^2-c^2)^2 (5 b^6-4 b^5 c-42 b^4 c^2-59 b^3 c^3-42 b^2 c^4-4 b c^5+5 c^6)+a^2 (b-c)^2 (b+c)^3 (5 b^6+12 b^5 c-8 b^4 c^2-26 b^3 c^3-8 b^2 c^4+12 b c^5+5 c^6)+a^7 (10 b^6+22 b^5 c+25 b^4 c^2+14 b^3 c^3+25 b^2 c^4+22 b c^5+10 c^6)+a^6 (10 b^7+28 b^6 c+45 b^5 c^2+33 b^4 c^3+33 b^3 c^4+45 b^2 c^5+28 b c^6+10 c^7)+a^5 (-10 b^8-16 b^7 c+22 b^6 c^2+57 b^5 c^3+54 b^4 c^4+57 b^3 c^5+22 b^2 c^6-16 b c^7-10 c^8)-2 a^4 (5 b^9+20 b^8 c+20 b^7 c^2-14 b^6 c^3-41 b^5 c^4-41 b^4 c^5-14 b^3 c^6+20 b^2 c^7+20 b c^8+5 c^9) : :
X(10209) = 2 X[442] - 3 X[5947]

In the plane of a triangle ABC, let FaFbFc = Feuerbach triangle.
U = AFbFc-isogonal conjugates of Fa, and define V and W cyclically

The lines UFa, VFb, WFc concur in X(10209); see Hyacinthos #23530. The construction was originally posted in ADGEOM #1550 by Tran Quang Hung, 9/1/2014.

X(10212) = 35th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics -6 a^10 + 13 a^8 (b^2 + c^2) - (b^2 - c^2)^4 (b^2 + c^2) - 2 a^6 (b^4 + 13 b^2 c^2 + c^4) + a^2 (b^2 - c^2)^2 (8 b^4 + 13 b^2 c^2 + 8 c^4) + a^4 (-12 b^6 + 13 b^4 c^2 + 13 b^2 c^4 - 12 c^6) : :

Let O be the circumcenter of a triangle ABC, and let
Na = X(5)-of-OBC, and define Nb and Nc cyclically
Aa = orthogonal projection of Na on OA, and define Ab and Ac cyclically
Ba = orthogonal projection of Nb on OA, and define OB and OC cyclically
Ca = orthogonal projection of Nc on OA, and define OB and OC cyclically
Oa = circumcenter of AaAbAc, and define Ob and Oc cyclically.

X(10212) = X(5)-of-OaObOc; X(10212) lies on the Euler line of ABC. (Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos #24189)

X(10213) = 36th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics (a^2-b^2-c^2) (2 a^20-11 a^18 (b^2+c^2)-(b^2-c^2)^8 (b^4+b^2 c^2+c^4)+a^16 (25 b^4+42 b^2 c^2+25 c^4)+a^2 (b^2-c^2)^6 (8 b^6+7 b^4 c^2+7 b^2 c^4+8 c^6)-a^14 (29 b^6+61 b^4 c^2+61 b^2 c^4+29 c^6)+6 a^12 (2 b^8+7 b^6 c^2+9 b^4 c^4+7 b^2 c^6+2 c^8)-a^4 (b^2-c^2)^4 (26 b^8+12 b^6 c^2+13 b^4 c^4+12 b^2 c^6+26 c^8)-a^8 (b^2-c^2)^2 (44 b^8+31 b^6 c^2+35 b^4 c^4+31 b^2 c^6+44 c^8)+a^10 (19 b^10-26 b^8 c^2-20 b^6 c^4-20 b^4 c^6-26 b^2 c^8+19 c^10)+a^6 (b^2-c^2)^2 (45 b^10-11 b^8 c^2+9 b^6 c^4+9 b^4 c^6-11 b^2 c^8+45 c^10)) : :

In the plane of a triangle ABC, let
O = circumcenter, X(3)
N = nine-point center, X(5)
Na = N-of-OBC, and define Nb and Nc cyclically
Aa = orthogonal projection of Na on OA, and define Ab and Ac cyclically
Ba = orthogonal projection of Nb on OA, and define OB and OC cyclically
Ca = orthogonal projection of Nc on OA, and define OB and OC cyclically
Ea = Euler line of AaAbAc, and define Eb and Ec cyclically
A' = Eb∩Ec, and define B' and C' cyclically

Then ABC and A'B'C' are parallelogic, and
X(10213) = (A'B'C',ABC)-parallelogic center
X(1141) = (ABC,A'B'C')-parallelogic center.
(Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos #24183).

X(10221) = HATZIPOLAKIS-MONTESDEOCA-EULER-PEDAL POINT

Barycentrics a^2/3b^2/3c^2/3S_BS_C - a²S_A(S_AS_BS_C)^1/3 : :
X(10221) = -2(S_AS_BS_C)^1/3*X(3) + a^2/3b^2/3c^2/3*X(4)
Suppose that W is a triangle center on the Euler line of a triangle ABC. Let A'B'C' be the pedal triangle of W. Then W(ABC) = W(A'B'C') if and only if W = X(10221). (Regarding the notation, recall that a triangle center is a function defined on a set of triangles, so that the notation W(T) is analogous to the notation f(x); i.e., W-of-T.) (Antreas Hatzipolakis and Angel Montesdeoca, September 13, 2016.) See 24354 and HG100916
X(10221) lies on these lines: {2,3}

X(10223) = 37th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics 2 a^14 b^2-9 a^12 b^4+15 a^10 b^6-10 a^8 b^8+3 a^4 b^12-a^2 b^14+2 a^14 c^2-6 a^12 b^2 c^2+5 a^10 b^4 c^2-3 a^8 b^6 c^2+10 a^6 b^8 c^2-14 a^4 b^10 c^2+7 a^2 b^12 c^2-b^14 c^2-9 a^12 c^4+5 a^10 b^2 c^4+2 a^8 b^4 c^4-8 a^6 b^6 c^4+19 a^4 b^8 c^4-15 a^2 b^10 c^4+6 b^12 c^4+15 a^10 c^6-3 a^8 b^2 c^6-8 a^6 b^4 c^6-16 a^4 b^6 c^6+9 a^2 b^8 c^6-15 b^10 c^6-10 a^8 c^8+10 a^6 b^2 c^8+19 a^4 b^4 c^8+9 a^2 b^6 c^8+20 b^8 c^8-14 a^4 b^2 c^10-15 a^2 b^4 c^10-15 b^6 c^10+3 a^4 c^12+7 a^2 b^2 c^12+6 b^4 c^12-a^2 c^14-b^2 c^14 : :

Let A'B'C' be the pedal triangle of a point P in the plane of a triangle ABC. Let A'' = reflection of A' in the Euler line, and define B'' and C'' cyclically Na = X(5)-of-A''B''C'', and define Nb and Nc cyclically

The locus of P such that Na, Nb, Nc are collinear is the union of the cubic K187 and a circum-quintic that passes through X(74) and X(1304). If P = X(4), the line NaNbNc meets the Euler line in X(10223). (Antreas Hatzipolakis and Angel Montesdeoca, September 14, 2016). See Hyacinthos #24377).

X(10227) = HUNG-MONTESDEOCA-EULER POINT

Barycentrics 2 a^28 -19 a^26 (b^2+c^2)+a^24 (77 b^4+142 b^2 c^2+77 c^4) -2 a^22 (83 b^6+215 b^4 c^2+215 b^2 c^4+83 c^6)+4 a^20 (44 b^8+161 b^6 c^2+221 b^4 c^4+161 b^2 c^6+44 c^8)+a^18 (11 b^10-421 b^8 c^2-744 b^6 c^4-744 b^4 c^6-421 b^2 c^8+11 c^10)+a^16 (-297 b^12-22 b^10 c^2+91 b^8 c^4+144 b^6 c^6+91 b^4 c^8-22 b^2 c^10-297 c^12)+2 a^14 (198 b^14+54 b^12 c^2+95 b^10 c^4+75 b^8 c^6+75 b^6 c^8+95 b^4 c^10+54 b^2 c^12+198 c^14)-2 a^12 (99 b^16-20 b^14 c^2+41 b^12 c^4-6 b^10 c^6+3 b^8 c^8-6 b^6 c^10+41 b^4 c^12-20 b^2 c^14+99 c^16)-a^10 (77 b^18-131 b^16 c^2+82 b^14 c^4+42 b^12 c^6+29 b^10 c^8+29 b^8 c^10+42 b^6 c^12+82 b^4 c^14-131 b^2 c^16+77 c^18)+a^8 (b^2-c^2)^2 (187 b^16-120 b^14 c^2+82 b^12 c^4+56 b^10 c^6+87 b^8 c^8+56 b^6 c^10+82 b^4 c^12-120 b^2 c^14+187 c^16)-2 a^6 (b^2-c^2)^4 (67 b^14-5 b^12 c^2+26 b^10 c^4+22 b^8 c^6+22 b^6 c^8+26 b^4 c^10-5 b^2 c^12+67 c^14)+2 a^4 (b^2-c^2)^6 (26 b^12+6 b^10 c^2+5 b^8 c^4+5 b^4 c^8+6 b^2 c^10+26 c^12)-a^2 (b^2-c^2)^8 (11 b^10+3 b^8 c^2-6 b^6 c^4-6 b^4 c^6+3 b^2 c^8+11 c^10)+(b^2-c^2)^12 (b^2+c^2)^2 ) ) : :

Let N be the nine-point center of a triangle ABC. Let A'B'C' be the circumcevian triangle of N, and let A''B''C'' be the pedal triangle of N with respect to A'B'C'. The Euler lines of AB''C'', B'C''A'', C'A''B'' concur in X(10227). (Tran Quang Hung and Angel Montesdeoca, September 14, 2016). See Hyacinthos #24387).

X(10228) = HUNG-MONTESDEOCA CENTER OF SIMILITUDE

Barycentrics
a^2 (a^26-8 a^24 (b^2+c^2)+28 a^22 (b^2+c^2)^2-6 a^20 (9 b^6+28 b^4 c^2+28 b^2c^4+9 c^6)+a^18 (53 b^8+277 b^6 c^2+406 b^4 c^4+277 b^2 c^6+53 c^8)+a^16 (6 b^10-273 b^8 c^2-499 b^6 c^4-499 b^4 c^6-273 b^2 c^8+6 c^10)+a^14 (-96 b^12+184 b^10 c^2+307 b^8 c^4+386 b^6 c^6+307 b^4 c^8+184 b^2 c^10-96 c^12)+2 a^12 (66 b^14-56 b^12 c^2-9 b^10 c^4-38 b^8 c^6-38 b^6 c^8-9 b^4 c^10-56 b^2 c^12+66 c^14) -a^10 (69 b^16+6 b^14 c^2+97 b^12 c^4+43 b^10 c^6+5 b^8 c^8+43 b^6 c^10+97 b^4 c^12+6 b^2 c^14+69 c^16)-a^8 (28 b^18-238 b^16 c^2+176 b^14 c^4-11 b^12c^6+63b^10 c^8+63 b^8 c^10-11 b^6 c^12+176 b^4 c^14-238 b^2 c^16+28 c^18)+a^6 (b^2-c^2)^2 (68 b^16-256 b^14 c^2+89 b^12 c^4+65 b^10 c^6+97 b^8 c^8+65 b^6 c^10+89 b^4 c^12-256 b^2 c^14+68 c^16)-a^4 (b^2-c^2)^4 (46 b^14-120 b^12 c^2+26 b^10 c^4+73 b^8 c^6+73 b^6 c^8+26 b^4 c^10-120 b^2 c^12+46 c^14) +a^2 (b^2-c^2)^6 (15 b^12-29 b^10 c^2+16 b^8 c^4+32 b^6 c^6+16 b^4 c^8-29 b^2 c^10+15 c^12)-(b^2-c^2)^8 (2 b^10-3 b^8 c^2+5 b^6 c^4+5 b^4 c^6-3 b^2 c^8+2 c^10) ) : :

Let N be the nine-point center of a triangle ABC. Let A'B'C' be the circumcevian triangle of N, and let A''B''C'' be the pedal triangle of N with respect to A'B'C'. The triangle A''B''C'' is similar to ABC, and the center of similitude is X(10228). (Tran Quang Hung and Angel Montesdeoca, September 14, 2016). See Hyacinthos #24387).

X(10282) = X(3)X(64)∩X(51)X(54)

Barycentrics a^2 (2 a^8-5 a^6 (b^2+c^2)+a^4 (3 b^4+4 b^2 c^2+3 c^4)+a^2 (b^2-c^2)^2 (b^2+c^2)-(b^2-c^2)^2 (b^4+c^4)) : :
X(10282) = 5 X(3) - X(64)

Let O be the circumcenter of a triangle ABC, and let
Oa = circumcenter of OBC, and define Ob and OC cyclically
N1 = nine-point center of OObOC, and define N2 and N3 cyclically.
Then ABC and N1N2N3 are orthologic triangles, and X(10282) = (N1N2N3,ABC)-orthologic center, and X(74) = (ABC,N1N2N3)-othologic center. X(10282) lies on the circumcircle of N1N2N3. See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos #24665.

X(10284) = X(1)X(3)∩X(5)X(2802)

Barycentrics a (a^5 (b+c) -a^4 (b^2+6 b c+c^2) -a^3 (2 (b^3+c^3)-7 bc(b+c)) +2 a^2 (b^4+2 b^3 c-7 b^2 c^2+2 b c^3+c^4)+a (b-c)^2 (b^3-6 b c(b+c)+c^3)-(b-c)^4 (b+c)^2) : :

Let A1B1C1 be the intouch triangle of a triangle ABC, and let A2 = reflection of A1 in X(1), and define B2 and C2 cyclically A3 = reflection of A in A2, and define B3 and C3 cyclically. Then X(10284) = nine-point center of A3B3C3. See Tran Quang Hung and Angel Montesdeoca, Hyacinthos #24438.

X(10287) = X(3)X(2575)∩X(5)X(523)

Barycentrics a^2 (b^4 (a^2-b^2) (-a^2+b^2-a c-c^2) (-a^2+b^2+a c-c^2) (-a^4+2 a^2 b^2-b^4-a^2 c^2-b^2 c^2+2 c^4+c^2 (-a^2-b^2+c^2) J)-c^4 (-a^2+c^2) (-a^2-a b-b^2+c^2) (-a^2+a b-b^2+c^2) (-a^4-a^2 b^2+2 b^4+2 a^2 c^2-b^2 c^2-c^4+b^2 (-a^2+b^2-c^2) J)) : : , where J = |OH|/R (Peter Moses, October 23, 2016)

Let H be the orthocenter of a triangle ABC. Let La be the Euler line of AHX(1113), and define Lb and Lc cyclically. The lines La, Lb, Lc concur in X(10287). See X(10288) and Seiichi Kirikami and Angel Montesdeoca, 24541 and 24545.

X(10288) = X(3)X(2574)∩X(5)X(523)

Barycentrics a^2 (b^4 (a^2-b^2) (-a^2+b^2-a c-c^2) (-a^2+b^2+a c-c^2) (-a^4+2 a^2 b^2-b^4-a^2 c^2-b^2 c^2+2 c^4-c^2 (-a^2-b^2+c^2) J)-c^4 (-a^2+c^2) (-a^2-a b-b^2+c^2) (-a^2+a b-b^2+c^2) (-a^4-a^2 b^2+2 b^4+2 a^2 c^2-b^2 c^2-c^4-b^2 (-a^2+b^2-c^2) J)) : : , where J = |OH|/R (Peter Moses, October 23, 2016)

Let H be the orthocenter of a triangle ABC. Let La be the Euler line of AHX(1114), and define Lb and Lc cyclically. The lines La, Lb, Lc concur in X(10288).

See X(10287) and Seiichi Kirikami and Angel Montesdeoca, 24541 and 24545.

X(10293) = PERSPECTOR OF ABC AND MID-TRIANGLE OF ORTHOCENTROIDAL AND ANTI-ORTHOCENTROIDAL TRIANGLES

Barycentrics 1/[a^8 - 2a^6(b^2 + c^2) + 11a^4b^2c^2 + 2a^2(b^2 + c^2)(b^4 - 4b^2c^2 + c^4) - (b^2 - c^2)^2(b^4 + 5b^2c^2 + c^4)] : :

Let ABC be a triangle and HaHbHc the orthic triangle. The segment AHa has two trisectors; let A' be the trisector closer to A. Let (Ha) be the (rectangular) hyperbola through A, A', X(6), with asymptotes parallel to those of the Jerabek hyperbola. Let Ao be the center of (Ha), and define Bo and Co cyclically. The lines AAo, BBo, CCo concur in X(10293). Let Ta be the line through A tangent to (Ha), and define Tb and Tc cyclically. The lines Ta, Tb, Tc concur in X(3426). (Angel Montesdeoca, January 1, 2017)

An equation for the hyperbola (Ha), in barycentrics: (2S^2-3SB SC) (c^2y^2-b^2z^2) + S^2(b^2-c^2)y z - b^2(2S^2-3SA SB)z x + c^2(2S^2-3SA SC)x y = 0. (Angel Montesdeoca, January 1, 2017)

X(10618) = (name pending)

Barycentrics (a (7a^5(b+c)+a^4(b^2+4b c+c^2)-a^3(14b^3+9b^2c+9b c^2+14c^3)-2a^2(b^4+5b^3c+7b^2c^2+5b c^3+c^4)+a(b-c)^2(7b^3+16b^2c+16b c^2+7c^3)+(b^2-c^2)^2(b^2+6b c+c^2)) : :
X(10618) = (2r^2+5rR+4s^2)*X(1) + r(2r+3R)*X(3)

Let I be the incenter of a triangle ABC, and let A'B'C' be the cevian triangle of I.
Let
Na = nine-point center of IB'C', and define Nb and Nc cyclically
N1 = nine-point center of INbNc, and define N2 and N3 cyclically.
X(10618) = nine-point center of N1N2N3. See Tran Quang Hung and Angel Montesdeoca, July 19, 2016: Hyacinthos 24692. )

X(10628) = INFINITY POINT OF X(54)X(74)

Barycentrics a^2 (a^12 (b^2+c^2) - 2 a^10 (2 b^4+b^2 c^2+2 c^4) + a^8 (5 b^6+2 b^4 c^2+2 b^2 c^4+5 c^6) + a^6 (-5 b^6 c^2+4 b^4 c^4-5 b^2 c^6) - a^4 (b^2-c^2)^2 (5 b^6+2 b^4 c^2+2 b^2 c^4+5 c^6) + a^2 (b^2-c^2)^2 (4 b^8+3 b^6 c^2+3 b^2 c^6+4 c^8) - (b^2-c^2)^4 (b^2+c^2)^3) : :

Let H = X(4) be the orthocenter of a triangle ABC, and suppose that t > 0. Let
A' = the point on line AH such that |A'A|/|A'H| = t, and define B' and C' cyclically
Ab = orthogonal projection of A' on AB, and define Bc and Ca cyclically
Ac = orthogonal project of A' on AC, and define Ba and Cb cyclically
Ea = Euler line of A'AbAc, and define Eb and Ec cyclically
E'a = Euler line of AAbAc, and define E'b and E'c cyclically.

The lines Ea, Eb, Ec concur in a point whose locus as t varies is the line W =X(4)X(54). The lines E'a, E'b, E'c concur in a point whose locuse as t varies is the line W' = X(4)X(74). The lines W and W' are parallel, and X(10628) is their point of intersection on the line at infinity. See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 24703.

X(10683) = (name pending)

Barycentrics 2 a^10 + 4 a^9 (b+c) + a^8 (-9 b^2+2 b c-9 c^2) - 2 a^7 (8 b^3+7 b^2 c+7 b c^2+8 c^3) + a^6 (10 b^4-7 b^3 c-7 b c^3+10 c^4) + a^5 (24 b^5+15 b^4 c+11 b^3 c^2+11 b^2 c^3+15 b c^4+24 c^5) + 2 a^4 (2 b^6+6 b^5 c+3 b^4 c^2+6b^3 c^3+3 b^2 c^4+6 b c^5+2 c^6) - 2 a^3 (b-c)^2 (8 b^5+18 b^4 c+21 b^3 c^2+21 b^2 c^3+18 b c^4+8 c^5) - a^2 (b^2-c^2)^2 (12 b^4+11 b^3 c+6 b^2 c^2+11 b c^3+12 c^4) + a (b-c)^4 (b+c)^3 (4 b^2+3 b c+4 c^2) + (b^2-c^2)^4 (5 b^2+4 b c+5 c^2) : :
X(10683) = (3r^2+17rR+24R^2-7s^2)*X[2] - (r^2+5rR+6R^2-s^2)*X[3]

Let ABC be a triangle, and let
Fa = A-Feuerbach point, and define Fb and Fc cyclically
Oa = circumcenter of AFbFc, and define Ob and Oc cyclically.
Then X(10683) = centroid of OaObOc, and X(10683) lies on Euler line of triangle ABC. See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 24711 and Hyacinthos 24719.

X(10684) = (name pending)

Barycentrics a^10 (b^2+c^2) - a^8 (2 b^4+b^2 c^2+2 c^4) + a^6 (b^6+b^4c^2+b^2 c^4+c^6) - a^4 b^2 c^2 (b^4-b^2 c^2+c^4) + b^4 c^4 (b^2-c^2)^2 : :

Let ABC be a triangle with two Brocard points Ω₁ and Ω₂ A1B1C1 is cevian triangle of Ω₁ and Ω₁* is isogonal conjugate of Br1 wrt A1B1C1. A2B2C2 is cevian triangle of Ω₂ and Ω₂* is isogonal conjugate of Ω₂ wrt A2B2C2. Then the lines Ω₁Ω₁* and Ω₂Ω₂* intersect on Euler of ABC.
See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 24712.

X(10685) = (name pending)

Barycentrics a^10 (b^2+c^2) - a^8 (3 b^4+b^2 c^2+3 c^4) + 2 a^6 (b^6+b^4 c^2+b^2 c^4+c^6) + a^4 (-2 b^6 c^2+b^4 c^4-2 b^2 c^6) - a^2 b^2 c^2 (b^2-c^2)^2 (b^2+c^2)b^4 c^4 (b^2-c^2)^2 : :

Let ABC be a triangle with two Brocard points Ω₁ and Ω₂
A1B1C1 is anticevian triangle of Ω₁ and Ω₁* is isogonal conjugate of Ω₁ wrt A1B1C1.
A2B2C2 is anticevian triangle of Ω₂ and Ω₂* is isogonal conjugate of Ω₁2 wrt A2B2C2.
Then the lines Ω₁Ω₁* andΩ₂Ω₂* intersect on Euler of ABC.
See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 24717.

X(10688) = X(30)X(6344)∩X(265)X(6000)

Barycentrics (b^10-3 b^8 c^2+2 b^6 c^4+2 b^4 c^6-3 b^2 c^8+c^10+(-2 b^8+6 b^6 c^2-8 b^4 c^4+6 b^2 c^6-2 c^8) a^2+(-b^6-2 b^4 c^2-2 b^2 c^4-c^6) a^4+(5 b^4+3 b^2 c^2+5 c^4) a^6+(-4 b^2-4 c^2) a^8+a^10) / (a^2(4SA^2-b^2c^2) : :

Let ABC be a triangle with orthocenter H and circumcenter O, and let
da = reflection of Euler line of triangle AOH in line OH, and define db and dc cyclically
A' = db∩dc, and define B' and C' cyclically.
The triangle A'B'C' is perspective to ABC, and the perspector is X(10688). See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 24745.

X(10694) = EULER INTERCEPT OF X(880)X(1502)

Barycentrics b^4 c^4 (b^2-c^2)^2+2 b^2 c^2 (b^2-c^2)^2 (b^2+c^2) a^2+b^2 c^2 (b^4+b^2 c^2+c^4) a^4-(b^6+b^4 c^2+b^2 c^4+c^6)a^6-b^2 c^2 a^8+(b^2+c^2) a^10 : :

In the plane of a triangle ABC, let
Ω₁ = 1st Brocard point
Ω₂ = 2nd Brocard point
Ω₁* = Orion transform of &Omega1
Ω₂* = Orion transform of &Omega2.
X(10694) = Ω₁Ω₁*∩Ω₂Ω₂*, and X(10694) lies on the Euler line of ABC. (Orion transform is define just before X(2055).) See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 24769.

X(10980) = {X(10655),X(10656)}-HARMONIC CONJUGATE OF X(3062)

Barycentrics a( a^2 + 2a(b + c) - 3(b - c)^2) : :

Let M_aM_bM_c be the medial triangle of the intouch triangle DEF. Let Γ_a be the circle, other than the circle with diameter ID, tangent to the incircle and passing through M_b and M_c. Define Γ_b and Γ_c cyclically. X(10980) is the radical center of Γ_a, Γ_b and Γ_c. (Angel Montesdeoca, October 2, 2020).
Ver HG031020

X(11008) = REFLECTION OF X(69) IN X(193)

Barycentrics 7 a^2-3 (b^2+c^2) : :
X(11008) = 9 X[2] - 10 X[6], 6 X[2] - 5 X[69], 4 X[6] - 3 X[69], 7 X[69] - 8 X[141], 7 X[6] - 6 X[141], 4 X[141] - 7 X[193], 3 X[2] - 5 X[193], 2 X[6] - 3 X[193], 19 X[6] - 18 X[597], 19 X[193] - 12 X[597], 11 X[69] - 12 X[599], 11 X[2] - 10 X[599], 11 X[6] - 9 X[599], 11 X[193] - 6 X[599]

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 24817 (November 16, 2016).

X(11016) = (name pending)

Barycentrics a^16-3 a^14 b^2-a^12 b^4+14 a^10 b^6-20 a^8 b^8+9 a^6 b^10+3 a^4 b^12-4 a^2 b^14+b^16-3 a^14 c^2+16 a^10 b^4 c^2-10 a^8 b^6 c^2-18 a^6 b^8 c^2+17 a^4 b^10 c^2+a^2 b^12 c^2-3 b^14 c^2-a^12 c^4+16 a^10 b^2 c^4-3 a^8 b^4 c^4-18 a^6 b^6 c^4-13 a^4 b^8 c^4+21 a^2 b^10 c^4-2 b^12 c^4+14 a^10 c^6-10 a^8 b^2 c^6-18 a^6 b^4 c^6-14 a^4 b^6 c^6-18 a^2 b^8 c^6+19 b^10 c^6-20 a^8 c^8-18 a^6 b^2 c^8-13 a^4 b^4 c^8-18 a^2 b^6 c^8-30 b^8 c^8+9 a^6 c^10+17 a^4 b^2 c^10+21 a^2 b^4 c^10+19 b^6 c^10+3 a^4 c^12+a^2 b^2 c^12-2 b^4 c^12-4 a^2 c^14-3 b^2 c^14+c^16 : :

In the plane of a triangle ABC, let N = X(5) = nine-point center, ) = X(3) = circumcenter, and
Na = nine-point center of NBC, and define Nb and Nc cyclically
Oa = reflection of O in BC, and define Ob and Oc cyclically
Then NaNbNc are OaObOc are perspective, and X(11016) is their perspector. See Antreas Hatzipolakis, Peter Moses, and Angel Montesdeoca, Hyacinthos 24853 (November 21, 2016).)

X(11538) = X(324)X(9381)∩X(1370,10155)

Barycentrics (b^2-c^2)^6-4 (b^2-c^2)^4 (b^2+c^2) a^2+(5 b^8-2 b^6 c^2-7 b^4 c^4-2 b^2 c^6+5 c^8) a^4-4 b^2 c^2 (b^2+c^2) a^6+(-5 b^4-4 b^2 c^2-5 c^4) a^8+4 (b^2+c^2)a^10-a^12 : :

In the plane of a triangle ABC, let O = X(3) = circumcenter, N = X(5) = nine-point center, and
A' = reflection of O in BC, and define B' and C' cyclically
Nab = N(AOB'), and define Nbc and Nca cyclically
Nac = N(AOC'), and define Nba and Ncb cyclically
Oa = O(ANabNac), and define Ob and Oc cyclically
The triangles ABC and OaObOc are orthologic, and
X(11538) = ABC-to-OaObOc orthologic center; and X(5) = OaObOc-to-ABC orthologic center. See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 24908 (Nov. 29, 2016)

X(11539) = MIDPOINT OF X(2) AND X(5054)

Barycentrics 8a^4-13a^2 (b^2+c^2)+5 (b^2-c^2)^2 : :

In the plane of a triangle ABC, let G = X(2) = centroid, and
A'B'C' = cevian triangle of G; i,e., A'B'C' = medial triangle
A''B''C'' = pedal triangle of G
Ma = midpoint of AA', and define Mb and Mc cyclically
P = a point (as a function) on the Euler line of ABC
Pa = P(A''MbMc), and define Pb and Pc cyclically
The triangles ABC and PaPbPc are orthologic.
X(11539) = PaPbPc-to-ABC orthologic center, for P = X(2)
X(11540) = PaPbPc-to-ABC orthologic center, for P = X(3)
X(4) = PaPbPc-to-ABC orthologic center, for P = X(11541)
See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 24914 (Nov 30, 2016).

X(11540) = MIDPOINT OF X(140) AND X(10124)

Barycentrics 22 a^4-35 a^2 (b^2+c^2)+13 (b^2-c^2)^2 : :

See X(11539) and Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 24914 Hyacinthos 24914 (Nov 30, 2016).

X(11541) = (name pending)

Barycentrics 17 a^4 - 8 a^2 (b^2 + c^2)- 9 (b^2 - c^2)^2 : :
X(11541) = 8 X(3) - 9 X(4)

See X(11539) and Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 24914 (Nov 30, 2016).

X(11548) = EULER INTERCEPT OF X(230)X(233)

Barycentrics (2 a^6-3 a^4 (b^2+c^2)-2 a^2 (b^4+4 b^2 c^2+c^4)+3 (b^2-c^2)^2 (b^2+c^2) : :

In the plane of a triangle ABC, let H = X(4) = orthocenter, and let
A'B'C' = pedal triangle of H; i.e., A'B'C' = orthic triangle
Ab = orthogonal projection of A' on HB, and define Bc and Ca cyclically
Ac = orthogonal projection of A' on HC, and define Ba and Cb cyclically
Na = nine-point cneter of A'AbAc, and define Nb and Nc cyclically
Oa = nine-point circle of A'AbAc, and define Ob and Oc cyclically
Qa = reflection of Qa in BC, and define Qb and Qc cyclically
X(11548) = radical center of Qa, Qb, Qc, a point on the Euler line of ABC. See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 24935 (Dec 3, 2016).

X(11567) = X(1)X(3)∩X(3244)X(10265)

Barycentrics (2 a^6-3 a^4 (b^2+c^2)-2 a^2 (b^4+4 b^2 c^2+c^4)+3 (b^2-c^2)^2 (b^2+c^2) : :
X(11567) = (R+4r) X[1] - R X[3]

In the plane of a triangle ABC, let I = X(1) = incenter, and let
A'B'C' = pedal triangle of I
Na = X(5)-of-IBC, and define Nb and Nc cyclically,
Then X(11567) = radical center of the circles (A', A'Na), (B', B'Nb), (C', C'Nc). See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 25041 (Dec 24, 2016).

X(11567) = X(11576) = X(4)X(93)∩X(25)X(54)

Trilinears a^2 (a^12 (b^2+c^2)-4 a^10 (b^2+c^2)^2+5 a^8 (b^6+2 b^4 c^2+2 b^2 c^4+c^6)+2 a^6 b^2 c^2 (b^2-c^2)^2+-a^4 (b^2-c^2)^2 (5 b^6+11 b^4 c^2+11 b^2 c^4+5 c^6)+2 a^2 (b^2-c^2)^4 (2 b^4+3 b^2 c^2+2 c^4)-(b^2-c^2)^4 (b^6-2 b^4 c^2-2 b^2 c^4+c^6)) : :

In the plane of a triangle ABC, let H = X(4) = orthocenter, and let
HaHbHc = orthic triangle = pedal triangle of H
Ab = orthogonal projection of Ha on HHb, and define Bc and Ca cyclically
Ac = orthogonal projection of Ha on HHc, and define Ba and Cb cyclically
A2 = orthogonal projection of A on HaAb, and define B2 and C2 cyclically
A3 = orthogonal projection of A on HaAc, and define B3 and C3 cyclically
A'B'C' = pedal triangle of H with respect to the triangle HaHbHc
A'' = orthogonal projection of A on HbHc, and define B'' and C'' cyclically
La = Euler line of AA2A3, and define Lb and Lc cyclically
Pa = line through A' parallel to La, and define Pb and Pc cyclically
Qa = line through A'' parallel to La, and define Qb and Qc cyclically.

The lines Pa, Pb, Pc concur in X(11576). The lines Qa, Qb, Qc concur in A(11577). See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 25068 (Dec 29, 2016).

X(11577) = X6)X(24)∩X(185)X(550)(name pending)

Barycentrics (a^2 (-a^12 (b^2+c^2) +4 a^10 (b^4+3 b^2 c^2+c^4)-a^8 (5 b^6+26 b^4 c^2+26 b^2 c^4+5 c^6)+20 a^6 b^2 c^2 (b^4+b^2 c^2+c^4)+a^4 (b^2-c^2)^2 (5 b^6+b^4 c^2+b^2 c^4+5 c^6)-4 a^2 (b^2-c^2)^2 (b^8+b^4 c^4+c^8)+(b^2-c^2)^4 (b^6+c^6)) : :

See X(11576) and Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 25068 (Dec 29, 2016).

X(11578) = (name pending)

Barycentrics (a-b-c)/(a^4-2 a^2 (b^2-b c+c^2)+b^4-2 b^3 c+10 b^2 c^2-2 b c^3+c^4) : :

In the plane of a triangle ABC, let
(Ia) = A-excircle, and define (Ib) and (Ic) cyclically
A' = (Ia)∩BC, and define B' and C' cyclically, so that A'B'C' = excentral triangle
Ta = line B'Ab, tangent to (Ia), and define Tb and Tc cyclically
{Ua, Va} = lines through B' tangent to (Ia), and define {Ub, Vb} and {Uc,Vc} cyclically
{Ab,Ac} = {Ua,Va}∩(Ia), and define {Bc, Ba} and {Ca, Cb} cyclically
Wa = BcBa∩CaAb, and define Wb and Wc cyclically.

The triangle WaWbWc is perspective to A'B'C' at X(11578), and the perpendicular bisectors of AbAc, BcBa, CaCb concur in X(4882). See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 25070 (Dec 30, 2016).

X(12048) = X(3)X(6)∩X(237)X(8881)

Barycentrics a^2 (SA+S Cot[w]^2 (Cot[w]-Csc[w])) : :

See Angel Montedeoca, Hechos Geométricos en el Triángulo (2015) and Circunferencias de Apolonio.

X(12049) = X(3)X(6)∩X(237)X(8880)

Barycentrics a^2 (SA+S Cot[w]^2 (Cot[w]+Csc[w])) : :

See Angel Montedeoca, Hechos Geométricos en el Triángulo (2015) and Circunferencias de Apolonio.

X(12050) = X(3)X(6)∩X(1501)X(8880)

Barycentrics = a^2 (SA+S (Csc[w]+Tan[w])) : :

See Angel Montedeoca, Hechos Geométricos en el Triángulo (2015) and Circunferencias de Apolonio.

X(12051) = X(3)X(6)∩(1501)X(8881)

Barycentrics = a^2 (SA-S (Csc[w]+Tan[w])) : :

See Angel Montedeoca, Hechos Geométricos en el Triángulo (2015) and Circunferencias de Apolonio.

X(12053) = X(1)X(4)∩X(10)X(11)

Barycentrics a^3 (b+c)+a^2 (b^2-6 b c+c^2)-a (b-c)^2 (b+c)-(b^2-c^2)^2: :

The line AX(8) meets the incircle in two points, A' and A'', where A' is the point closer to A. Let σ be the affine transformation that carries A'B'C' onto A''B''C''. The finite fixed point of σ is X(12053). (Angel Montesdeoca, July 5, 2021). Ver Hechos Geométrico del Triángulo

X(12053) = X(1)X(4)∩X(10)X(11)

Barycentrics a^3 (b+c)+a^2 (b^2-6 b c+c^2)-a (b-c)^2 (b+c)-(b^2-c^2)^2 : :

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 25502 .

Let ABC be a triangle and A'B'C' the pedal triangle of I.
Denote:
A", B", C" = the reflections of I in BC, CA, AB, resp.
A'''B'''C''' = the orthic triangle of A"B"C"
Ma, Mb, Mc = the midpoints of the altitudes A"A''', B"B''', C"C''', resp.

R1 = the radical axis of the circles (Mb, MbB'), (Mc, McC')
R2 = the radical axis of the circles (Mc, McC'), (Ma, MaA')
R3 = the radical axis of the circles (Ma, MaA'), (Mb, MaB')
[the radical center of the circles is the point they concur at = the NPC center of A"B"C"]

Ra, Rb, Rc = the reflections of R1, R2, R3 in IA, IB, IC, resp.
A*B*C* = the triangle bounded by Ra,Rb,Rc
ABC, A*B*C* are parallelogic. The parallelogic center (ABC, A*B*C*) lies on the OI line of ABC [ = Euler line of A"B"C"]

The parallelogic center (ABC, A*B*C*) is X(56)
The parallelogic center (A*B*C*, ABC) is X(12053) = (r-2R) X(1) + r X(4)

X(12054) = X(3)X(6)∩X(30)X(83)

Barycentrics a^2 (a^6+a^4 b^2-2 a^2 b^4+a^4 c^2-5 a^2 b^2 c^2-3 b^4 c^2-2 a^2 c^4-3 b^2 c^4) : :

See Angel Montedeoca, Hechos Geométricos en el Triángulo (2015) , where circles Oa, Ob, Oc are defined. Let A' be the point, other than A, in which the circles Ob and Oc intersect. Define B' and C' cyclically. Then X(12054) = X(3)-of-A'B'C'. (Peter Moses, February 25, 2017)

X(12055) = X(3)X(6)∩X(99)X(3589)

Barycentrics a^2 (a^4-2 a^2 b^2-2 b^4-2 a^2 c^2-5 b^2 c^2-2 c^4) : :

See Angel Montedeoca, Hechos Geométricos en el Triángulo (2015), where circles Oa, Ob, Oc are defined. Let A' be the point, other than A, in which the circles Ob and Oc intersect. Define B' and C' cyclically. Then X(12055) = X(6)-of-A'B'C'. (Peter Moses, February 25, 2017)

X(12079) = X(30)X(74)∩X(125)X(523)

Barycentrics (b^2-c^2)^2/(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2) : :

Let M_aM_bM_c = medial triangle. Let (𝒫_a) be the parabola, tangent to Euler line, to NM_a, and to the line BC at its vertex, so that its directrix, d_a, is parallel to BC. Define the lines d_b and d_c cyclically. Let T be the triangle bounded by the lines d_a, d_b, d_c. Then T is homothetic to ABC, and the center of homothety is X(12079). For a construction, see Paris Pamfilos, A Gallery of Conics by Five Elements, Forum Geometricorum 14 (2014) 295-348, paragraph 13.3, page 346: construct a conic tangent to the line at infinity, i.e. a parabola, tangent to three lines a, b, c and passing through [D], i.e. with given axis-direction. (Angel Montesdeoca, March 1, 2022) See HG010322

X(13160) = POINT BEID 42

Barycentrics (a^8 (b^2+c^2)-4 a^4 b^2 c^2 (b^2+c^2)-(b^2-c^2)^4 (b^2+c^2)-2 a^6 (b^4-b^2 c^2+c^4)+2 a^2 (b^8-b^6 c^2-b^2 c^6+c^8) : ... : ..

In the plane of a triangle ABC, let
DEF = circummedial triangle,
H_aH_bH_c = orthic triangle,
Γ_a = circumcircle of EFH_a,
Γ_b = circumcircle of FDH_b,
Γ_c = circumcircle of DEH_b,
U = circle tangent to and encompassing Γ_a, Γ_b, Γ_c.
Then X(13160) = center of U. See X(13160) (Angel Montesdeoca, March 25, 2020)
X(13160) como centro de una circunferencia de Apolonio.

https://faculty.evansville.edu/ck6/encyclopedia/X13160.png

X(13291) = PARALLELOGIC CENTER OF THESE TRIANGLES: 2nd PARRY TO 1st HYACINTH

Barycentrics (2*a^10-4*(b^2+c^2)*a^8+2*(b^4+4*b^2*c^2+c^4)*a^6-(b^2+c^2)^3*a^4+2*(b^4-b^2*c^2+c^4)^2*a^2-(b^6+c^6)*(b^2-c^2)^2)*(b^2-c^2) : :

Let A* be the intersection of the lies through X(110) perpendicular to BC, and define B* and C* cyclically. Then X(13291) is the centroid of the degenerate triangle A*B*C*. (Angel Montesdeoca, November 1, 2021). See X(13291).

See Antreas Hatzipolakis and Angel Montesdeoca, euclid 2948.

X(13351) = CROSSSUM OF X(6) AND X(1656)

Barycentrics a^2 (a^4 b^2-2 a^2 b^4+b^6+a^4 c^2-3 a^2 b^2 c^2-b^4 c^2-2 a^2 c^4-b^2 c^4+c^6) : :

In the plane of a triangle ABC, let
DEF = circum-orthic triangle;
Ab = center of circle (BFO), and define Bc and Ca cyclically;
Ac = center of circle (CEO), and define Ba and Cb cyclically;
A' = BAc∩CAb, and define B' and C' cyclically;
T = the affine transformation that carries ABC onto A'B'C'.
Then X(13351) = the finite fixed point of T. (Angel Montesdeoca, March 4, 2024)

See El centro X(13351) como punto fijo de una transformación afín.

X(13419) = POINT BEID 75

Barycentrics 2 a^10-4 a^8 b^2+a^6 b^4+a^4 b^6+a^2 b^8-b^10-4 a^8 c^2-a^4 b^4 c^2+2 a^2 b^6 c^2+3 b^8 c^2+a^6 c^4-a^4 b^2 c^4-6 a^2 b^4 c^4-2 b^6 c^4+a^4 c^6+2 a^2 b^2 c^6-2 b^4 c^6+a^2 c^8+3 b^2 c^8-c^10 : :

Let P be a point on the circumcircle and D a point on the line BC. Let U and V be the reflections of D in PB and PC respectively. The envelope of lines UV when D moves on BC is a parabola with focus P and directrix d_a. The envelope of d_a when P moves on circumcircle is a circle (O_a). Define (O_b), (O_c) cyclically. The radical center of the circles (O_a), (O_b), (O_c) is X(13419); see Euclid #611 (Angel Montesdeoca, February 6, 2020)

X(13624) = X(1)X(3)∩X(30)X(1125)

Barycentrics a(4*a^3-(b+c)*a^2-2*(2*b^2-b*c+ 2*c^2)*a+(b^2-c^2)*(b-c)) : :

Let DEF be the cevian triangle of I=X(1) and A"B"C" the 2nd circumperp triangle. Let D_b, D_c be the circumcenters of ABD, ACD. The lines through X(3) parallel to DD_b, DD_c intersects the lines through I perpendicular to IB, IC at A_b, A_c respectively. Let (A") be the circle with center centered at A" and tangent to line A_bA_c, and define (B"), (C") cyclically. X(13624) is the radical center of (A"), (B"), (C"). (Angel Montesdeoca, November 7, 2019) El centro X(13624) como centro radical

X(13867) = POINT BEID 149

Barycentrics a (a^5 (b+c) -a^4 (b^2+14 b c+c^2) +a (b-c)^2 (b^3-9 b^2 c-9 b c^2+c^3) -2 a^3 (b^3-5 b^2 c-5 b c^2+c^3) +2 a^2 (b^4+4 b^3 c+14 b^2 c^2+4 b c^3+c^4) -(b^2-c^2)^2 (b^2-6 b c+c^2)) : :

Let ABC be a triangle and A'B'C',IaIbIc the pedal, antipedal triangles of I, resp.
Denote:
A",B",C" = the antipodes of A', B', C', in the incircle, resp.
If Ra is the the radical center of incircle and circumcircle of IA"Ia, and define Rb, Rc cyclically.
Ra, Rb , Rc concur in a point in X(13867).
See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 26354 (Jul 14, 2017).

X(13881) = ORTHOLOGIC CENTER OF THESE TRIANGLES: 3rd TRI-SQUARES TO 4th TRI-SQUARES

Barycentrics a^4-(b^2+c^2)*a^2+2*(b^2-c^2)^2 : :

Let DEF be the orthic triangle. Let Ab be the point in which the line through B perpendicular to AB meets the perpendicular bisector of segment BD . Let Ac be the point in which the line through C perpendicular to AC meets the perpendicular bisector of segment CD. Define Bc, Ba, Ca, Cb cyclically. The six ponts Ab, Ac, Bc, Ba, Ca, Cb lie on a conic with center X(13881). A barycentric equation for this conic follows:
0 = cyclic sum of (b^2+c^2-a^2)(5 a^4-12 a^2 (b^2+c^2)+7 b^4+2 b^2 c^2+7 c^4)x^2- 2 (11a^6-13 a^4 (b^2+c^2)+a^2 (b^2-c^2)^2+(b^2-c^2)^2 (b^2+c^2))y z. (Angel Montesdeoca, December 23, 2018)

El punto X(13881) como centro de una cónica.

X(14221) = REFLECTION OF X(316) IN X(3260)

Barycentrics (a^2-b^2) (a^2-c^2) (a^6 b^2-a^4 b^4-a^2 b^6+b^8+a^6 c^2-2 a^4 b^2 c^2+2 a^2 b^4 c^2-3 b^6 c^2-a^4 c^4+2 a^2 b^2 c^4+4 b^4 c^4-a^2 c^6-3 b^2 c^6+c^8) : :

Let P=X(110) be the focus of Kiepert parabola, and ℓ₁. ℓ₂ perpendicular lines through P, that intersect the sidelines BC, CA, AB at points A₁ and A₂, B₁ and B₂, C₁ and C₂, respectively . Let A_bc = A₁C₂ ∩ A₂B₁ and A_cb = A₁B₂ ∩ A₂C₁. When the lines ℓ₁ and ℓ₂ rotate around P, the points A_bc and A_cb lie on the same conic C(A), through B and C. Let A_o be its center. The conics C(B), C(C) and and their centers B_o and C_o are defined cyclically. Then the lines AA_o, BB_o and CC_o concur in X(14221). (See HG060721). (Angel Montesdeoca, July 13, 2021)

X(14249) = X(3)X(107)∩X(4)X(51)

Barycentrics b^2*c^2*(-a^2 + b^2 - c^2)^2*(a^2 + b^2 - c^2)^2*(-3*a^4 + 2*a^2*b^2 + b^4 + 2*a^2*c^2 - 2*b^2*c^2 + c^4) : :

Let DEF be the Euler triangle of ABC. Let La be the line through D perpendicular to OD, and define Lb, Lc, cyclically. Let A'B'C' be the triangle formed by the lines La, Lb, Lc. The point X(14249) is the unique finite fixed point of the affine transformation that maps the reference triangle ABC onto A'B'C'. (Angel Montesdeoca, July 5, 2022)

See Angel Montesdeoca, HG040722.

X(14356) = X(2)X(476)∩X(5)X(523)

Barycentrics (a^2 - a*b + b^2 - c^2)*(a^2 + a*b + b^2 - c^2)*(a^2 - b^2 - a*c + c^2)*(a^2 - b^2 + a*c + c^2)*(a^2*b^2 - b^4 + a^2*c^2 - c^4) : :

Let ABC be a triangle with anticomplementary triangle A'B'C'. Let DEF and D'E'F' be the cevian triangles of Steiner point and its isotomic conjugate, respectively. Let Ao, Bo, Co be the circumcenters of triangles A'DD', B'EE', C'FF', respectively. Then the triangles ABC and AoBoCo are perspective and the perspector is X(14356). (Angel Montesdeoca, July 18, 2022)
HG160722.

X(14491) = NGUYEN-MOSES IMAGE OF X(381)

Barycentrics a^2*(2*a^4 - 7*a^2*b^2 + 5*b^4 - 4*a^2*c^2 - 7*b^2*c^2 + 2*c^4)*(2*a^4 - 4*a^2*b^2 + 2*b^4 - 7*a^2*c^2 - 7*b^2*c^2 + 5*c^4) : :

From Angel Montesdeoca, February 28, 2020:

Let ℰ_a be the ellipse with major axis the segment BC and length of the minor axis a/Sqrt[3]. Let L_a be the polar of A wrt ℰ_a, and define L_b and L_c cyclically. Let A' = L_b∩L_c, B' = L_c∩L_a, C' = L_a∩L_b. The lines AA', BB', CC' concur in X(14491).

A barycentric equation for the ellipse ℰ_a:

2 a⁴ x²+a⁴ x y+a⁴ x z+2 a⁴ y z-3 a² b² x²-a² b² x y+a² b² x z-3 a² c² x²+a² c² x y-a² c² x z+b⁴ x²-2 b² c² x²+c⁴ x²=0.

X(14633) = MIDPOINT OF X(2029) AND X(2039)

Barycentrics 2*a^4 + b^4 + c^4 + (2*a^2 + b^2 + c^2)*Sqrt(a^4 - a^2*b^2 + b^4 - a^2*c^2 - b^2*c^2 + c^4) : :

In the plane of a triangle ABC, let
F1 and F2 be the real foci of the Steiner inellipse
A1 = AF1∩BC
A2 = AF2∩BC
Ga = conic tangent to AF1 at A1, to AF2 at A2, and to F1F2
Ta = Ga∩F1F2
Ao = center of Ga, and define Bo and Co cyclically
The lines AAo, BBo, CCo concur in X(14633), and the lines ATa, BTb, CTc concur in X(6177). See X(6177) and X(14633). (Angel Montesdeoca, July 21, 2023)

X(14683) = SINGULAR FOCUS OF THE CUBIC K753

Barycentrics 3*a^6 - 3*a^4*b^2 + a^2*b^4 - b^6 - 3*a^4*c^2 + a^2*b^2*c^2 + b^4*c^2 + a^2*c^4 + b^2*c^4 - c^6 : :

Let T the Steiner triangle, E the Euler line of ABC, and P the parabola inscribed in T with directrix E. The focus of P is X(14683). (Angel Montesdeoca, February 14, 2020) (HG100220)

X(15109) = X(3)X(6)∩X(115)X(2963)

Barycentrics a^8-3 a^6 (b^2+c^2+3 a^4 (b^4+b^2 c^2+c^4))-a^2 (b^2-c^2)^2 (b^2+c^2) : :

In the plane of a triangle ABC, let
Lab = reflection of AB in the perpendicular bisector of AC
Lab = reflection of AC in the perpendicular bisector of AB
A' = Lab ∩ Lac
A'' = the point, other than A', of intersection of the circle having diameter AA' and the circle {{A',B,C}}; define B'' and C'' cyclically.
Then X(15109) is the finite fixed point of the affine transformation that takes ABC onto A''B''C''. Moreover, the lines AA'',BB'',CC'' concur in the Kosnita point, X(54). (Angel Montesdeoca, April 21, 2023)
See X(15109) punto fijo de una transformación afín

X(15316) = ISOGONAL CONJUGATE OF X(3542)

Barycentrics a^2 (a^2 - b^2 - c^2)/(a^6 - 3 a^4 (b^2 + c^2) - (b^2 - c^2)^2 (b^2 + c^2) + a^2 (3 b^4 - 2 b^2 c^2 + 3 c^4)) : :

Let DEF be the anticevian triangle of X(3), and let h be the affine transformation that carries ABC onto DEF. For any line ℒ through X(3), let ℒ' be the image of ℒ under h, and let U = ℒ∩ℒ'. Then as ℒ rotates around X(3), the line h(U)h^-1(U) pass through X(15316). See X(15316). (Angel Montesdeoca, March 7, 2021)

X(15692) = X(2) + 4 X(3)

Barycentrics 13*a^4 - 14*a^2*b^2 + b^4 - 14*a^2*c^2 - 2*b^2*c^2 + c^4 : :

Given a triangle ABC and a point P, let c(P) be the circumconic having perspector P, and let A' be the center of the conic that passes through P and is tangent to c(P) at B and C. Define points B' and C' cyclically. The triangles ABC and A'B'C' are orthologic if and only P is on the sextic (passing through X(2), X(6)) having this barycentric equation:

Cyclic sum [ y z(17(b^2-c^2) x^4- ((15a^2+31b^2-19c^2) y-(15a^2-19b^2+31c ^2) z)x^3+ a^2(y+z)(y-z)^3) ]= 0.

If P is the centroid then the orthologic center of A'B'C' with respect to ABC is X(15692). The reciprocal orthologic center is X(4). If P is the symmedian then the orthologic center of ABC with respect to A'B'C' is X(18535). The reciprocal orthologic center is X(3). (Angel Montesdeoca, December 12, 2022)

See X(15692) y X(18535) centros ortológicos

X(15805) = MIDPOINT OF X(3) AND X(3527)

Barycentrics a^2 (a^8-4 a^6 b^2+6 a^4 b^4-4 a^2 b^6+b^8-4 a^6 c^2-4 a^4 b^2 c^2+16 a^2 b^4 c^2-8 b^6 c^2+6 a^4 c^4+16 a^2 b^2 c^4+14 b^4 c^4-4 a^2 c^6-8 b^2 c^6+c^8) : :

In the plane of a triangle ABC, let
DEF = orthic triangle of ABC
(O) = circle with center D and pass-thorugh point A
A2 = (O)∩AB
A3 = (O)∩AC
(Oa) = circle through B, C, A2, A3
A' = center of (Oa), and define B' and C' cyclically
Then X(15805) is the finite fixed point of the affine transformation that carries ABC onto A'B'C'. (Angel Montesdeoca, December 3, 2022)

See Angel Montesdeoca, X(15805) punto fijo de una transformación afín.

X(16186) = X(3)X(49)∩X(122)X(125)

Barycentrics a^2(b^2-c^2)^2SA(3SA^2-S^2) : :

Sean T_aT_bT_c el triángulo tangencial, A' la intersección de la reflexión de AB en T_aT_c con a la reflexión de AC en T_aT_b, B' la intersección de la reflexión de BC en T_bT_a con a la reflexión de BA en T_bT_c, y C' la intersección de la reflexión de CA en T_cT_b con a la reflexión de CB en T_cT_a.
El punto fijo (propio) de la transformación afín que aplica ABC en A'B'C' es X(16186)

See Angel Montesdeoca, HG010218.

X(16187) = X(2)X(98)∩X(3)X(5646)

Barycentrics a^2(a^4-a^2(b^2+c^2)+10b^2c^2) : :

Dado un triángulo ABC con simediano K, la antiparalela a BC, respecto a AB y a AC, corta en B_a y C_a a AC y AB, respectivamente; la antiparalela a CA, respecto a BC y a BA, corta en C_b y A_b a BA y BC, respectivamente; la antiparalela a AB, respecto a CA y a CB, corta en A_c y B_c a CB y CA, respectivamente. El baricentro P_G de A_bB_cC_a y el baricentro U_G de A_bB_cC_a, forman un par bicéntrico.
La 'crosssum' de P_G y U_G es X(16187).

See Angel Montesdeoca, HG050218.

X(16188) = COMPLEMENT OF X(842)

Barycentrics 2a^12(b^2+c^2)- 4a^10(b^2+c^2)^2+ a^8(b^6+11b^4c^2+11b^2c^4+c^6)+ 2a^6(b^8-5b^6c^2-5b^2c^6+c^8)- a^4(2b^10-7b^8c^2+3b^6c^4+3b^4c^6-7b^2c^8+ 2c^10)+ 2a^2(b^12-3b^10c^2+2b^8c^4+2b^4c^8-3b^2c^10+c^12)-(b^2-c^2)^4(b^6+c^6) : :

See Angel Montesdeoca, HG120218.

X(16236) = X(1)X(631)∩X(7)X(519)

Barycentrics (a-2b-2c)(5a-b-c)/(b+c-a) : :

See Angel Montesdeoca, HG200218.

Dado un triángulo ABC, sean DEF el triángulo de contacto interior, D', E' y F', los puntos donde las rectas AD, BE y CF vuelven a cortar a la circunferencia inscrita a ABC.
Las rectas p_a, p_b y p_c son las perpendiculares por D', E' y F' a AD, BE y CF, respetivamente. Sea A₂B₂C₂ el triángulo delimitado por las rectas p_a, p_b y p_c.
El centro de perspectividad de DEF y A₂B₂C₂ es X(16236).

X(17749) = MOSES (0, 1, 1, -1, 0, -2, 0, 0, 0) POINT

Barycentrics a*(a^2*b + a*b^2 + a^2*c - a*b*c - 2*b^2*c + a*c^2 - 2*b*c^2) : :

Let M_aM_bM_c, DEF, and A'B'C' be the medial, intouch and 1st circumperp triangles, respectively. Let D', E' and F' be the orthogonal projections of M_a, M_b and M_c on EF, FD and DE, respectively. Then A'B'C' and D'E'F' are bilogic and the orthology center of A'B'C' with respect to D'E'F' is X(17749). The reciprocal orthologic center is X(10). (Angel Montesdeoca, October 29, 2019). X(17749) como centro ortológico

X(18221) = PERSPECTOR OF THESE TRIANGLES: 1st ZANIAH AND 5th MIXTILINEAR

Barycentrics 3*a^4-8*(b+c)*a^3-2*(b^2+4*b*c+c^2)*a^2+8*(b^2-c^2)*(b-c)*a-(b^2-c^2)^2 : :

Let A'B'C' be the intouch triangle and let A"B"C" be the Euler triangle of A'B'C'. The finite fixed point of the affine transformation that carries ABC onto A''B''C'' is X(18221), i.e. X(18221) = ATFF(ABC, A"B"C"); see the preamble just before X(10129)). (Angel Montesdeoca, March 25, 2022).
See El centro X(18221) como punto fijo de una transformación afín

X(18279) = X(4)X(74)∩?X(5)X(523)

Barycentrics a^14 (b^2+c^2)-3 a^12 (b^4+c^4)-(b^2-c^2)^6 (b^4+3 b^2 c^2+c^4)+a^10 (b^6+2 b^4 c^2+2 b^2 c^4+c^6)+a^2 (b^2-c^2)^4 (3 b^6-b^4 c^2-b^2 c^4+3 c^6)-a^6 (b^2-c^2)^2 (5 b^6+4 b^4 c^2+4 b^2 c^4+5 c^6)-a^8 (-5 b^8+12 b^6 c^2-12 b^4 c^4+12 b^2 c^6-5 c^8)-a^4 (b^2-c^2)^2 (b^8-11 b^6 c^2+10 b^4 c^4-11 b^2 c^6+c^8) : :

See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 27613. Hyacinthos 27613 .

Let ABC be a triangle, N is the NPC center, A'B'C' is the pedal triangle of N.
The Hatzipolakis axes of the triangles AB'C', BC'A' and CA'B' bound a triangle A''B''C''. Then the circumcenter of the triangle A''B''C'' is X(18279), lies on the Euler line of ABC.

X(18280) = (name pending))

Barycentrics a^2 (a^11+a^10 (b+c) + a^9 (-5 b^2+b c-5 c^2) - a^8 (5 b^3+4 b^2 c+4 b c^2+5 c^3) +a^7 (10 b^4-4 b^3 c+11 b^2 c^2-4 b c^3+10 c^4) + a^6 (10 b^5+6 b^4 c+9 b^3 c^2+9 b^2 c^3+6 b c^4+10 c^5) + a^5 (-10 b^6+6 b^5 c-4 b^4 c^2+9 b^3 c^3-4 b^2 c^4+6 b c^5-10 c^6) - a^4 (10 b^7+4 b^6 c+2 b^5 c^2+5 b^4 c^3+5 b^3 c^4+2 b^2 c^5+4 b c^6+10 c^7) + a^3 (5 b^8-4 b^7 c-5 b^6 c^2-b^5 c^3+2 b^4 c^4-b^3 c^5-5 b^2 c^6-4 b c^7+5 c^8) + a^2 (b-c)^2 (5 b^7+11 b^6 c+10 b^5 c^2+11 b^4 c^3+11 b^3 c^4+10 b^2 c^5+11 b c^6+5 c^7) - a (b^2-c^2)^4 (b^2-b c+c^2) - (b-c)^4 (b+c)^5 (b^2-b c+c^2)) : :

See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 27614.

Let ABC be a triangle with incenter I. The reflection of Hatzipolakis axis of triangle IBC in line BC is da. Similarly, we have db and dc. The lines da,db,dc bound triangle A'B'C'. Circumcenter of triangle A'B'C' lies on OI line of ABC.
The circumcenter of triangle A'B'C' lies on OI line of ABC and is:
X(18280) = R (3 R^2-4 s^2)*X(1) + 2 (r R^2+2 R^3+4 r s^2)*X(3)

X(18535) = HOMOTHETIC CENTER OF THESE TRIANGLES: EHRMANN MID AND 3rd ANTIPEDAL OF X(3)

Barycentrics a^2((a^2 - b^2 - c^2)^2 - 16b^2c^2)/(b^2 + c^2 - a^2) : :

Given a triangle ABC and a point P, let c(P) be the circumconic having perspector P, and let A' be the center of the conic that passes through P and is tangent to c(P) at B and C. Define points B' and C' cyclically. The triangles ABC and A'B'C' are orthologic if and only P is on the sextic (passing through X(2), X(6)) having this barycentric equation:

Cyclic sum [ y z(17(b^2-c^2) x^4- ((15a^2+31b^2-19c^2) y-(15a^2-19b^2+31c ^2) z)x^3+ a^2(y+z)(y-z)^3) ]= 0.

If P is the centroid then the orthologic center of A'B'C' with respect to ABC is X(15692). The reciprocal orthologic center is X(4). If P is the symmedian then the orthologic center of ABC with respect to A'B'C' is X(18535). The reciprocal orthologic center is X(3). (Angel Montesdeoca, December 12, 2022)

X(15692) y X(18535) centros ortológicos (10/12/2022)

X(18866) = (name pending)

Barycentrics a^16-2 a^14 (b^2+c^2)-3 a^12 (b^4-4 b^2 c^2+c^4)-(b^2-c^2)^4 (b^2+c^2)^2 (2 b^4-3 b^2 c^2+2 c^4)+2 a^10 (4 b^6-5 b^4 c^2-5 b^2 c^4+4 c^6)-2 a^6 (b^2-c^2)^2 (5 b^6-2 b^4 c^2-2 b^2 c^4+5 c^6)+a^8 (b^8-15 b^6 c^2+29 b^4 c^4-15 b^2 c^6+c^8)+a^4 (b^2-c^2)^2 (3 b^8+2 b^6 c^2-15 b^4 c^4+2 b^2 c^6+3 c^8)+4 a^2 (b^2-c^2)^2 (b^10-b^8 c^2+2 b^6 c^4+2 b^4 c^6-b^2 c^8+c^10) : :

ADGEOM #4604 (Tran Quang Hung, May 27, 2018)

Let ABC be a triangle with circumcenter O.
P is on its Euler line.
Oa, Ob, Oc are isogonal conjugate of O wrt triangles PBC, PCA, PAB.
Then centroid of OaObOc lies on Euler line of ABC.

ADGEOM #4608 (Angel Montesdeoca, May 28, 2018)

If P is on its Euler line such that OP:PH=t , then centroid Q of OaObOc ( lies on Euler line of ABC) is:
Q = (a^2-b^2) (a^2-c^2) (a^4-(b^2-c^2)^2) t (a^16 t^2-(b^2-c^2)^6 (b^2+c^2)^2 t^2-a^14 (b^2+c^2) t (1+3 t)+a^2 (b^2-c^2)^4 (b^2+c^2) t (-2 b^2 c^2 (-2+t)+b^4 (1+3 t)+c^4 (1+3 t))-a^6 (b^2-c^2)^2 (b^2+c^2) (b^4 (-5+t) t+c^4 (-5+t) t+b^2 c^2 (-5+7 t-4 t^2))-a^8 (b^4 c^4 (-7+14 t-12 t^2)+b^6 c^2 (3-9 t+4 t^2)+b^2 c^6 (3-9 t+4 t^2))-a^4 (b^2-c^2)^2 (2 b^8 t (2+t)+2 c^8 t (2+t)+b^6 c^2 (2+5 t-3 t^2)+b^2 c^6 (2+5 t-3 t^2)+b^4 c^4 (5-6 t+6 t^2))+a^10 (b^2+c^2) (b^4 (-5+t) t+c^4 (-5+t) t-b^2 c^2 (1-5 t+8 t^2))+a^12 (2 b^4 t (2+t)+2 c^4 t (2+t)+b^2 c^2 (1+9 t^2)))-(-a^2+b^2+c^2) (a^8 t-2 a^6 (b^2+c^2) t+a^4 b^2 c^2 (-2+3 t)-(b^2-c^2)^2 (b^2 c^2 (-1+t)+b^4 t+c^4 t)+a^2 (b^2+c^2) (b^2 c^2 (1-4 t)+2 b^4 t+2 c^4 t)) ((b^2-c^2) (-a^2 c^2-a^4 t+(b^2-c^2)^2 t) (a^4 (-2 b^4+b^2 c^2 (1-2 t))-a^8 t-(b^2-c^2)^3 (b^2+c^2) t+a^6 (2 c^2 t+b^2 (1+t))+a^2 (b^2-c^2) (-b^2 c^2 (-2+t)+2 c^4 t+b^4 (1+t)))+(-b^2+c^2) (-a^2 b^2-a^4 t+(b^2-c^2)^2 t) (a^4 (-2 c^4+b^2 c^2 (1-2 t))-a^8 t+(b^2-c^2)^3 (b^2+c^2) t+a^6 (2 b^2 t+c^2 (1+t))-a^2 (b^2-c^2) (-b^2 c^2 (-2+t)+2 b^4 t+c^4 (1+t)))) : ... : ...

If P=X(2)
[X(18866) = ] Q = a^16-2 a^14 (b^2+c^2)-3 a^12 (b^4-4 b^2 c^2+c^4)-(b^2-c^2)^4 (b^2+c^2)^2 (2 b^4-3 b^2 c^2+2 c^4)+2 a^10 (4 b^6-5 b^4 c^2-5 b^2 c^4+4 c^6)-2 a^6 (b^2-c^2)^2 (5 b^6-2 b^4 c^2-2 b^2 c^4+5 c^6)+a^8 (b^8-15 b^6 c^2+29 b^4 c^4-15 b^2 c^6+c^8)+a^4 (b^2-c^2)^2 (3 b^8+2 b^6 c^2-15 b^4 c^4+2 b^2 c^6+3 c^8)+4 a^2 (b^2-c^2)^2 (b^10-b^8 c^2+2 b^6 c^4+2 b^4 c^6-b^2 c^8+c^10) : ... : ...

X(19210) = X(3) OF 1st ANTI-SHARYGIN TRIANGLE

Barycentrics a^4 (-a^2+b^2+c^2)^2 (-a^4-b^4+b^2 c^2+a^2 (2 b^2+c^2)) (a^4-b^2 c^2+c^4-a^2 (b^2+2 c^2)) : :

Let L be the line through X(3) parallel to BC. Let A_b = L∩AB, and A_c = L∩AC, and let D = midpoint of AX(4). Let A' be the point, other than A, in which the circles {{B,D,A_b}} and {{C, D,A_c}} meet. Define B' and C' cyclically. The finite fixed point of the affine transformation that carries ABC onto A'B'C' is X(19210). (Angel Montesdeoca, April 13, 2021)
See El centro X(19210) como punto fijo de una transformación afín

X(19657) = (name pending)

Barycentrics a^2 (a-b) (a-c) (a^10+2 a^9 b-3 a^8 b^2-8 a^7 b^3+2 a^6 b^4+12 a^5 b^5+2 a^4 b^6-8 a^3 b^7-3 a^2 b^8+2 a b^9+b^10+a^9 c+4 a^8 b c+5 a^7 b^2 c-a^6 b^3 c-9 a^5 b^4 c-9 a^4 b^5 c-a^3 b^6 c+5 a^2 b^7 c+4 a b^8 c+b^9 c-4 a^8 c^2-6 a^7 b c^2+3 a^6 b^2 c^2+6 a^5 b^3 c^2+2 a^4 b^4 c^2+6 a^3 b^5 c^2+3 a^2 b^6 c^2-6 a b^7 c^2-4 b^8 c^2-4 a^7 c^3-11 a^6 b c^3-6 a^5 b^2 c^3+6 a^4 b^3 c^3+6 a^3 b^4 c^3-6 a^2 b^5 c^3-11 a b^6 c^3-4 b^7 c^3+6 a^6 c^4+4 a^5 b c^4-5 a^4 b^2 c^4-10 a^3 b^3 c^4-5 a^2 b^4 c^4+4 a b^5 c^4+6 b^6 c^4+6 a^5 c^5+11 a^4 b c^5+6 a^3 b^2 c^5+6 a^2 b^3 c^5+11 a b^4 c^5+6 b^5 c^5-4 a^4 c^6+2 a^3 b c^6+4 a^2 b^2 c^6+2 a b^3 c^6-4 b^4 c^6-4 a^3 c^7-5 a^2 b c^7-5 a b^2 c^7-4 b^3 c^7+a^2 c^8-2 a b c^8+b^2 c^8+a c^9+b c^9) (a^10+a^9 b-4 a^8 b^2-4 a^7 b^3+6 a^6 b^4+6 a^5 b^5-4 a^4 b^6-4 a^3 b^7+a^2 b^8+a b^9+2 a^9 c+4 a^8 b c-6 a^7 b^2 c-11 a^6 b^3 c+4 a^5 b^4 c+11 a^4 b^5 c+2 a^3 b^6 c-5 a^2 b^7 c-2 a b^8 c+b^9 c-3 a^8 c^2+5 a^7 b c^2+3 a^6 b^2 c^2-6 a^5 b^3 c^2-5 a^4 b^4 c^2+6 a^3 b^5 c^2+4 a^2 b^6 c^2-5 a b^7 c^2+b^8 c^2-8 a^7 c^3-a^6 b c^3+6 a^5 b^2 c^3+6 a^4 b^3 c^3-10 a^3 b^4 c^3+6 a^2 b^5 c^3+2 a b^6 c^3-4 b^7 c^3+2 a^6 c^4-9 a^5 b c^4+2 a^4 b^2 c^4+6 a^3 b^3 c^4-5 a^2 b^4 c^4+11 a b^5 c^4-4 b^6 c^4+12 a^5 c^5-9 a^4 b c^5+6 a^3 b^2 c^5-6 a^2 b^3 c^5+4 a b^4 c^5+6 b^5 c^5+2 a^4 c^6-a^3 b c^6+3 a^2 b^2 c^6-11 a b^3 c^6+6 b^4 c^6-8 a^3 c^7+5 a^2 b c^7-6 a b^2 c^7-4 b^3 c^7-3 a^2 c^8+4 a b c^8-4 b^2 c^8+2 a c^9+b c^9+c^10) : :

See Alexandr Skutin, Peter Moses, and Angel Montesdeoca, Hyacinthos 27780 and Hyacinthos 27781.

X(19658) = ANTIGONAL IMAGE OF X(484)

Barycentrics (a^2+a b+b^2-c^2) (a^2-b^2+a c+c^2) (a^3-a^2 b-a b^2+b^3+a^2 c-a b c+b^2 c-a c^2-b c^2-c^3) (a^3+a^2 b-a b^2-b^3+a^2 c-a b c+b^2 c-a c^2+b c^2-c^3) (a^3+a^2 b-a b^2-b^3-a^2 c-a b c-b^2 c-a c^2+b c^2+c^3) : :

See Alexandr Skutin, Peter Moses, and Angel Montesdeoca, Hyacinthos 27780 and Hyacinthos 27781.

X(19659) = X(2)X(3)∩X(32)X(6178)

Barycentrics (-a^2 - b^2 - c^2) (a^2 b^2 - b^4 + a^2 c^2 + 2 b^2 c^2 - c^4) - 2 a^2 (a^2 - b^2 - c^2) sqrt(a^4 + b^4 + c^4- b^2 c^2 - a^2 b^2 - a^2c^2) : :

See Angel Montesdeoca, HG180618.

X(19660) = X(2)X(3)∩X(32)X(6177)

Barycentrics (-a^2 - b^2 - c^2) (a^2 b^2 - b^4 + a^2 c^2 + 2 b^2 c^2 - c^4) + 2 a^2 (a^2 - b^2 - c^2) sqrt(a^4 + b^4 + c^4- b^2 c^2 - a^2 b^2 - a^2c^2) : :

See Angel Montesdeoca, HG180618.

X(19661) = X(2)X(1285)∩X(5)X(598)

Barycentrics 16 a^4 - a^2 (b^2 + c^2) + (b^2 + c^2)^2 : :

See Angel Montesdeoca, HG180618.

X(19662) = MIDPOINT OF X(115) AND X(599)

Barycentrics 2a^6-6a^4(b^2+c^2)+ 3a^2(b^4+4b^2c^2+c^4)-7b^6+3b^4c^2+3b^2c^4-7c^6 : :

See Angel Montesdeoca, HG180618.

X(19663) = X(39)X(597)∩X(384)X(11638)

Barycentrics a^8(b^2+c^2)+ a^6(6b^4-20b^2c^2+6c^4)+ 3a^4(b^6+2b^4c^2+2b^2c^4+c^6)- a^2(b^2+c^2)^2(2b^4+b^2c^2+2c^4)+b^2c^2(b^2+c^2)^3 : :

See Angel Montesdeoca, HG180618.

X(20031) = X(98)X(6530)∩X(107)X(685)

Barycentrics (a^2-c^2)(a^4-(b^2-c^2)^2)^2 (a^10 -a^8(2b^2+c^2) +a^6(2b^4+c^4) -a^4(2b^6-2b^4c^2+c^6) +a^2b^4(b^2-c^2)^2 +b^4c^2(b^2-c^2)^2 ) : :

See Angel Montesdeoca, HG260618.

X(20115) = (name pending)

Barycentrics a^2 (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4)/(2 a^8-6 a^6 b^2+7 a^4 b^4-4 a^2 b^6+b^8-6 a^6 c^2+4 a^2 b^4 c^2-4 b^6 c^2+7 a^4 c^4+4 a^2 b^2 c^4+6 b^4 c^4-4 a^2 c^6-4 b^2 c^6+c^8) : :

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 27820.

X(20116) = X(1)X(1170)∩X(7)X(79)

Barycentrics a (-a^4 (b+c)+5 a^2 b c (b+c)+2 a^3 (b^2+c^2)-2 a (b-c)^2 (b^2+3 b c+c^2)+(b-c)^2 (b^3+c^3)) : :

See Kadir Altintas and Angel Montesdeoca, ADGEOM 4766.

X(20117) = X(1)X(6920)∩X(2)X(5693)

Barycentrics a (a^5 (b+c)-a^4 (b+c)^2-(b^2-c^2)^2 (b^2+b c+c^2)+a^3 (-2 b^3+b^2 c+b c^2-2 c^3)+a (b-c)^2 (b^3+c^3)+a^2 (2 b^4+3 b^3 c-2 b^2 c^2+3 b c^3+2 c^4)) : :

See Kadir Altintas and Angel Montesdeoca, ADGEOM 4766.

X(20118) = X(1)X(12619)∩X(2)X(12739)

Barycentrics (a+b-c) (a-b+c) (a^4 (b+c)+a^2 b c (b+c)+2 a (b^2-c^2)^2-2 a^3 (b^2+c^2)-(b-c)^2 (b+c)^3) : :

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 27840.

X(20121) = X(1)X(7)∩X(948)X(4031)

Barycentrics (a+b-c) (a-b+c) (a^2-5 a (b+c)+4 (b-c)^2) : :

See Kadir Altintas and Angel Montesdeoca, ADGEOM 4791.

X(20123) = X(30)X(146)∩X(74)X(18317)

Barycentrics (a^2-b^2-c^2) (2 a^4-a^2 (b^2+c^2)-(b^2-c^2)^2)/(a^8-4 a^6 (b^2+c^2)+a^4 (6 b^4+b^2 c^2+6 c^4)+a^2 (-4 b^6+b^4 c^2+b^2 c^4-4 c^6)+(b^2-c^2)^2 (b^4+4 b^2 c^2+c^4)) : :

See Angel Montesdeoca, ADGEOM 4801.

X(20124) = EULER LINE INTERCEPT OF X(3258)X(18285)

Barycentrics 4 a^16 - 17 a^14 (b^2 + c^2) - (b^2 - c^2)^6 (11 b^4 + 14 b^2 c^2 + 11 c^4) + a^12 (43 b^4 + 10 b^2 c^2 + 43 c^4) + a^10 (-89 b^6 + 27 b^4 c^2 + 27 b^2 c^4 - 89 c^6) + a^2 (b^2 - c^2)^4 (37 b^6 + 17 b^4 c^2 + 17 b^2 c^4 + 37 c^6) - a^4 (b^2 - c^2)^2 (23 b^8 - 80 b^6 c^2 - 75 b^4 c^4 - 80 b^2 c^6 + 23 c^8) + a^8 (115 b^8 + 4 b^6 c^2 - 162 b^4 c^4 + 4 b^2 c^6 + 115 c^8) - a^6 (59 b^10 + 71 b^8 c^2 - 121 b^6 c^4 - 121 b^4 c^6 + 71 b^2 c^8 + 59 c^10) : :

See Antreas Hatzipolakis, César Lozada and Angel Montesdeoca Hyacinthos 27871 and Hyacinthos 27872.

X(20183) = X(9)X(362)∩X(10)X(164)

Barycentrics a (a^5 - a^4 (b + c) - 2 a^3 (b^2 + 10 b c + c^2) + 2 a^2 (b^3 + 7 b^2 c + 7 b c^2 + c^3) + a (b - c)^2 (b^2 + 6 b c + c^2) - (b - c)^4 (b + c) + 2 Sqrt[ b c (a + b - c) (a - b + c)] (-5 a^2 (b + c) + (b - c)^2 (b + c) + 4 a (b^2 + b c + c^2)) - 2 Sqrt[-a c (a - b - c) (a + b - c) ] (a^3 + a^2 (4 b + c) + c (b^2 - c^2) - a (5 b^2 + 4 b c + c^2)) - 2 Sqrt[ a b (a - b + c) (-a + b + c)] (a^3 - b^3 + b c^2 + a^2 (b + 4 c) - a (b^2 + 4 b c + 5 c^2))) : :

See Angel Montesdeoca, HG110718.

X(20477) = X(2)X(53)∩X(3)X(95)

Barycentrics a^8-3*(b^2+c^2)*a^6+3*(b^4+c^4)*a^4-(b^4-c^4)*(b^2-c^2)*a^2+2*b^2*c^2*(b^2-c^2)^2 : :

Let A'B'C' be the tangential triangle. Let La be the reflection of line B'C' in the perpedicular bisector of BC, and define Lb and Lc cyclically. Let A" = Lb∩Lc, and define B" and C" cyclically. Triangle A"B"C" is also the tangential triangle of the dual-of-orthic triangle, and X(20477) = X(7)-of-A"B"C". (Randy Hutson, August 29, 2018)

See Angel Montesdeoca, HG220718.

X(20478) = EULER LINE INTERCEPT OF X(578)X(14374)

Barycentrics a*(2*a*(a^8+8*a^4*b^2*c^2-2*(b^2+c^2)*a^6+2*(b^4-3*b^2*c^2+c^4)*(b^2+c^2)*a^2-(b^4+4*b^2*c^2+c^4)*(b^2-c^2)^2)*S*OH-(3*a^10-5*(b^2+c^2)*a^8-2*(b^4-8*b^2*c^2+c^4)*a^6+2*(b^2+c^2)*(3*b^4-7*b^2*c^2+3*c^4)*a^4-(b^2-c^2)^2*(b^4+8*b^2*c^2+c^4)*a^2-(b^4-c^4)*(b^2-c^2)^3)*b*c) : :

See Angel Montesdeoca, HG220718.

X(20479) = EULER LINE INTERCEPT OF X(578)X(14375)

Barycentrics a*(-2*a*(a^8+8*a^4*b^2*c^2-2*(b^2+c^2)*a^6+2*(b^4-3*b^2*c^2+c^4)*(b^2+c^2)*a^2-(b^4+4*b^2*c^2+c^4)*(b^2-c^2)^2)*S*OH-(3*a^10-5*(b^2+c^2)*a^8-2*(b^4-8*b^2*c^2+c^4)*a^6+2*(b^2+c^2)*(3*b^4-7*b^2*c^2+3*c^4)*a^4-(b^2-c^2)^2*(b^4+8*b^2*c^2+c^4)*a^2-(b^4-c^4)*(b^2-c^2)^3)*b*c) : :

See Angel Montesdeoca, HG220718.

X(20480) = X(110)X(382)∩X(476)X(15646)

Barycentrics 4 a^16 - 10 a^14 (b^2 + c^2) + a^12 (-6 b^4 + 49 b^2 c^2 - 6 c^4) + 5 a^10 (8 b^6 - 11 b^4 c^2 - 11 b^2 c^4 + 8 c^6) - a^8 (40 b^8 + 23 b^6 c^2 - 135 b^4 c^4 + 23 b^2 c^6 + 40 c^8) + 3 a^6 (b^2 - c^2)^2 (2 b^6 + 25 b^4 c^2 + 25 b^2 c^4 + 2 c^6) + a^4 (b^2 - c^2)^2 (10 b^8 - 6 b^6 c^2 - 57 b^4 c^4 - 6 b^2 c^6 + 10 c^8) - 2 a^2 (b^2 - c^2)^4 (2 b^6 + 5 b^4 c^2 + 5 b^2 c^4 + 2 c^6) - 4 b^2 c^2 (b^2 - c^2)^6 : :

See Angel Montesdeoca, HG220718 and Hyacinthos 27999.

X(20481) = X(2)X(6)∩X(3)X(111)

Barycentrics a^2(a^4+2a^2(b^2+c^2)+b^4-16b^2c^2+c^4) : :

See Angel Montesdeoca, HG030818.

X(20581) = X(1741)X(8758)∩X(2331)X(7649)

Barycentrics a (b^5 - b^4 c - b c^4 + c^5 + (b^4 - 2 b^2 c^2 + c^4) a + (-2 b^3 + 4 b^2 c + 4 b c^2 - 2 c^3) a^2 + (-2 b^2 - 4 b c - 2 c^2) a^3 + (b + c) a^4 + a^5) (b^5 + b^4 c - 2 b^3 c^2 - 2 b^2 c^3 + b c^4 + c^5 + (b^4 - 4 b^3 c + 6 b^2 c^2 - 4 b c^3 + c^4) a + (-2 b^3 + 2 b^2 c + 2 b c^2 - 2 c^3) a^2 + (-2 b^2 - 2 c^2) a^3 + (b + c) a^4 + a^5) : :

See Angel Montesdeoca, HG040818.

X(20587) = (name pending)

Barycentrics (2 a^4 - 2 a^2 b^2 + a^2 b c - b^3 c - 2 a^2 c^2 + 2 b^2 c^2 - b c^3) (2 a^4 - 2 a^2 b^2 - a^2 b c + b^3 c - 2 a^2 c^2 + 2 b^2 c^2 + b c^3) (a^6 + a^2 b^4 - 2 b^6 - 2 a^2 b^2 c^2 + 2 b^4 c^2 + a^2 c^4 + 2 b^2 c^4 - 2 c^6) : :

See Angel Montesdeoca, HG040818.

X(20588) = X(1)X(1167)∩X(2)X(15298)

Barycentrics a (a - b - c) (a^4 - 2 a^2 b^2 + b^4 + 2 a b^2 c - 2 b^3 c - 2 a^2 c^2 + 2 a b c^2 + 2 b^2 c^2 - 2 b c^3 + c^4) : :

See Angel Montesdeoca, HG040818.

X(20794) = X(3)X(69)∩X(6)X(694)

Barycentrics (a^2 b^2 + a^2 c^2 - b^2 c^2) sin 2A : :

Let T = anticevian triangle of X(3). Then X(20794) is the perspector, with respect to T, of the pivotal conic of the conico-pivotal cubic cK(#3,X394). (Angel Montesdeoca, May 4, 2019) HG040519

X(21230) = COMPLEMENT OF X(195)

Barycentrics (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) (a^6-3 a^4 b^2+3 a^2 b^4-b^6-3 a^4 c^2+3 a^2 b^2 c^2+b^4 c^2+3 a^2 c^4+b^2 c^4-c^6) : :

Let U be the circle with center X(5) and pass-through point A, and let meets the perpendicular bisector of segment BC in two points: one of them is the U-antipode of A and the other we denote by A'. Define B' and C' cyclically. Then X(21230) is the circumcenter of triangle A'B'C'. (Angel Montesdeoca, March 27, 2019) Hyacinthos #28933

X(21265) = (name pending)

Barycentrics a^20 (b^2+c^2)-(b^2-c^2)^10 (b^2+c^2)-a^18 (7 b^4+10 b^2 c^2+7 c^4)+10 a^16 (2 b^6+3 b^4 c^2+3 b^2 c^4+2 c^6)+11 a^6 (b^2-c^2)^4 (5 b^8+10 b^6 c^2+12 b^4 c^4+10 b^2 c^6+5 c^8)+a^2 (b^2-c^2)^6 (8 b^8-3 b^6 c^2-11 b^4 c^4-3 b^2 c^6+8 c^8)-a^14 (27 b^8+28 b^6 c^2+26 b^4 c^4+28 b^2 c^6+27 c^8)+a^12 (7 b^10-24 b^8 c^2-34 b^6 c^4-34 b^4 c^6-24 b^2 c^8+7 c^10)-a^4 (b^2-c^2)^4 (28 b^10-b^8 c^2-28 b^6 c^4-28 b^4 c^6-b^2 c^8+28 c^10)-a^8 (b^2-c^2)^2 (63 b^10+111 b^8 c^2+128 b^6 c^4+128 b^4 c^6+111 b^2 c^8+63 c^10)+a^10 (35 b^12+55 b^10 c^2+36 b^8 c^4+36 b^6 c^6+36 b^4 c^8+55 b^2 c^10+35 c^12) : :

See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 28059.

X(21267) = COMPLEMENT OF X(15519)

Barycentrics a^3-2 a^2 (b+c)+a (9 b^2-14 b c+9 c^2)-4 (b-c)^2 (b+c) : :

Recalling that triangle centers are functions, at (a,b,c) = (6,9,13), the values of X(21267) and X(4691) are equal.

See Tran Quang Hung, Angel Montesdeoca, and César Lozada, ADGEOM 4851 and ADGEOM 4855.

X(21268) = MIDPOINT ON X(4) AND X(5962)

Barycentrics 2 a^16 -7 a^14 (b^2+c^2) +4 a^12 (2 b^4+5 b^2 c^2+2 c^4) -a^10 (3 b^6+17 b^4 c^2+17 b^2 c^4+3 c^6) -2 a^8 b^2 c^2 (b^4-10 b^2 c^2+c^4) +3 a^6 (b^2-c^2)^2 (b^2+c^2)^3 -4 a^4 (b^2-c^2)^2 (2 b^8-b^6 c^2+2 b^4 c^4-b^2 c^6+2 c^8) +a^2 (b^2-c^2)^4 (7 b^6+b^4 c^2+b^2 c^4+7 c^6) -2 (b^2-c^2)^6 (b^4+b^2 c^2+c^4) : :

See Tran Quang Hung and Angel Montesdeoca, ADGEOM 4859.

X(21269) = MIDPOINT ON X(265) AND X(14989)

Barycentrics 4 a^16 -9 a^14 (b^2+c^2) +a^12 (-6 b^4+44 b^2 c^2-6 c^4) +14 a^10 (2 b^6-3 b^4 c^2-3 b^2 c^4+2 c^6) -3 a^8 (5 b^8+15 b^6 c^2-44 b^4 c^4+15 b^2 c^6+5 c^8) -a^6 (9 b^10-71 b^8 c^2+63 b^6 c^4+63 b^4 c^6-71 b^2 c^8+9 c^10) +a^4 (b^2-c^2)^2 (4 b^8+14 b^6 c^2-63 b^4 c^4+14 b^2 c^6+4 c^8) +6 a^2 (b^2-c^2)^4 (b^6-2 b^4 c^2-2 b^2 c^4+c^6) -(b^2-c^2)^6 (3 b^4+7 b^2 c^2+3 c^4) : :

See Tran Quang Hung and Angel Montesdeoca, ADGEOM 4867.

X(21664) = X(118)X(133)∩X(119)X(2804)

Barycentrics (2 a b c - a^2 (b + c) + (b - c)^2 (b + c))^2 (a^4 - (b^2 - c^2)^2) : :

See Angel Montesdeoca, HG210618.

X(21665) = X(4)X(150)∩X(25)X(917)

Barycentrics b^2c^2(-2a^3+a^2(b+c)+(b-c)^2(b+c))^2(a^4-(b^2-c^2)^2) : :

See Angel Montesdeoca, HG210618.

X(21666) = X(4)X(151)∩X(25)X(1311)

Barycentrics b^2c^2(b-c)^2(-a+b+c)^2(a^2+b^2-c^2)(a^2-b^2+c^2) : :

See Angel Montesdeoca, HG210618.

X(21667) = X(3)X(8754)∩X(5)X(131)

Barycentrics 2 a^14 (b^2 + c^2) - a^12 (11 b^4 + 2 b^2 c^2 + 11 c^4) + 2 a^10 (12 b^6 + b^4 c^2 + b^2 c^4 + 12 c^6) - a^8 (25 b^8 + 8 b^6 c^2 - 18 b^4 c^4 + 8 b^2 c^6 + 25 c^8) + 2 a^6 (5 b^10 + 5 b^8 c^2 - 6 b^6 c^4 - 6 b^4 c^6 + 5 b^2 c^8 + 5 c^10) + a^4 (b^2 - c^2)^2 (3 b^8 - 8 b^6 c^2 - 14 b^4 c^4 - 8 b^2 c^6 + 3 c^8) - 2 a^2 (b^2 - c^2)^4 (2 b^6 - b^4 c^2 - b^2 c^4 + 2 c^6) + (b^2 - c^2)^8 : :

See Angel Montesdeoca, HG250618.

X(21668) = X(6)X(1511)∩X(1297)X(13398)

Barycentrics a^2 (a^14 b^2-5 a^12 b^4+9 a^10 b^6-5 a^8 b^8-5 a^6 b^10+9 a^4 b^12-5 a^2 b^14+b^16+a^14 c^2-6 a^12 b^2 c^2+20 a^10 b^4 c^2-29 a^8 b^6 c^2+17 a^6 b^8 c^2-8 a^4 b^10 c^2+10 a^2 b^12 c^2-5 b^14 c^2-5 a^12 c^4+20 a^10 b^2 c^4-40 a^8 b^4 c^4+36 a^6 b^6 c^4-13 a^4 b^8 c^4-8 a^2 b^10 c^4+10 b^12 c^4+9 a^10 c^6-29 a^8 b^2 c^6+36 a^6 b^4 c^6-8 a^4 b^6 c^6+3 a^2 b^8 c^6-11 b^10 c^6-5 a^8 c^8+17 a^6 b^2 c^8-13 a^4 b^4 c^8+3 a^2 b^6 c^8+10 b^8 c^8-5 a^6 c^10-8 a^4 b^2 c^10-8 a^2 b^4 c^10-11 b^6 c^10+9 a^4 c^12+10 a^2 b^2 c^12+10 b^4 c^12-5 a^2 c^14-5 b^2 c^14+c^16) : :

See Angel Montesdeoca, HG250618.

X(21975) = COMPLEMENT OF X(3459)

Barycentrics (a^4-a^2 b^2+b^4-2 a^2 c^2-2 b^2 c^2+c^4) (a^4-2 a^2 b^2+b^4-a^2 c^2-2 b^2 c^2+c^4) (a^8-4 a^6 b^2+6 a^4 b^4-4 a^2 b^6+b^8-4 a^6 c^2+5 a^4 b^2 c^2+a^2 b^4 c^2-2 b^6 c^2+6 a^4 c^4+a^2 b^2 c^4+2 b^4 c^4-4 a^2 c^6-2 b^2 c^6+c^8) : :

Let Q be the pedal curve (a limaçon of Pascal) of the circle with center X(7728) and radius |OH|, with respect to X(265). Let A'B'C' be the triangle formed by the tangents at A,B,C to Q. Then the triangles ABC and A'B'C' are perspective, and their perspector is X(2963). Moreover, X(21975) is the only finite fixed point of the affine transformation that maps a triangle ABC onto A'B'C'. See X(21975). (Angel Montesdeoca, September 10, 2019). See HG080919.

X(22100) = X(5)X(524)∩X(7812)X(9487)

Barycentrics 5 a^10-21 a^8 (b^2+c^2) +a^6 (34 b^4+28 b^2 c^2+34 c^4)+a^4 (-31 b^6+15 b^4 c^2+15 b^2 c^4-31 c^6) +15 a^2 (b^2-c^2)^2 (b^4-b^2 c^2+c^4) - (b^2-c^2)^2 (2 b^6-3 b^4 c^2-3 b^2 c^4+2 c^6) : :

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 28146.

X(22101) = X(2)X(195)∩X(6)X(3459)

Barycentrics a^2 (a^20-10 a^18 (b^2+c^2)+(b^2-c^2)^8 (b^4-b^2 c^2+c^4)+a^16 (45 b^4+69 b^2 c^2+45 c^4)-2 a^2 (b^2-c^2)^6 (5 b^6-b^4 c^2-b^2 c^4+5 c^6)-4 a^14 (30 b^6+49 b^4 c^2+49 b^2 c^4+30 c^6)+a^12 (210 b^8+278 b^6 c^2+303 b^4 c^4+278 b^2 c^6+210 c^8)-6 a^10 (42 b^10+26 b^8 c^2+27 b^6 c^4+27 b^4 c^6+26 b^2 c^8+42 c^10)-2 a^6 (b^2-c^2)^2 (60 b^10+2 b^8 c^2-3 b^6 c^4-3 b^4 c^6+2 b^2 c^8+60 c^10)+a^4 (b^2-c^2)^2 (45 b^12-84 b^10 c^2+24 b^8 c^4+29 b^6 c^6+24 b^4 c^8-84 b^2 c^10+45 c^12)+a^8 (210 b^12-100 b^10 c^2+2 b^8 c^4+b^6 c^6+2 b^4 c^8-100 b^2 c^10+210 c^12)) : :

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 28146.

X(22270) = X(3)X(6748)∩X(97)X(631)

Barycentrics 1/(a^8-4a^6(b^2+c^2)+6a^4(b^4+b^2c^2+c^4) -4a^2(b^2-c^2)^2(b^2+c^2)+(b^2-c^2)^2(b^4-4b^2c^2+c^4)) : :

See Antreas Hatzipolakis and Angel Montesdeoca Hyacinthos Hyacinthos 28174 and HG060918.

X(22277) = (A,B,C,X(2); A',B',C',X(10)) COLLINEATION IMAGE OF X(8), WHERE A'B'C' = ANTICEVIAN TRIANGLE OF X(37)

Barycentrics a^2 (b + c) (a b - b^2 + a c + b c - c^2) : :

In the plane of a triangle ABC, let
I = incenter;
DEF = intouch triangle;
(O) = circumcircle;
P = perpendicular bisector of segment ID;
A_i = points of intersection of P and (O), and define B_i and C_i cyclically, for i = 1, 2;
D_i = reflection of D in IA_i, for i = 1, 2;
A' = E₁E₂∩F₁F₂, and define B' and C' cyclically;
The lines DA', EB', FC' concur in X(22277). See X(22277). (Angel Montesdeoca, August 6, 2022) and HG060822.

X(23050) = X(1)X(475)∩X(9)X(8750)

Barycentrics a (a^2+b^2-c^2) (a^2-b^2+c^2) (a^3-a^2 b+a b^2-b^3-a^2 c+2 a b c-b^2 c+a c^2-b c^2-c^3) : :

See Angel Montesdeoca, HG300818.

X(23051) = X(10)X(400)∩X(19)X(38)

Barycentrics a/(3a^2+b^2+c^2) : :

See Angel Montesdeoca, HG300818.

X(23052) = X(1)X(19)∩X(4)X(3663)

Barycentrics a(a^4-(b^2-c^2)^2)(a^4+ 2a^2(b^2+c^2)-3b^4-2b^2c^2-3c^4) : :

See Angel Montesdeoca, HG300818.

X(23053) = X(2)X(6)∩X(671)X(3524)

Barycentrics 17a^4-20a^2(b^2+c^2)+11b^4-26b^2c^2+11c^4 : :

See Angel Montesdeoca, HG300818.

X(23054) = X(1992)X(16509)∩X(4232)X(8860)

Barycentrics 1/(19a^4-40a^2b^2+13b^4-40a^2c^2-10b^2c^2+13c^4) : :

See Angel Montesdeoca, HG300818.

X(23055) = X(2)X(6)∩X(98)X(11172)

Barycentrics 11a^4-8a^2(b^2+c^2)+5b^4-14b^2c^2+5c^4 : :

See Angel Montesdeoca, HG300818.

X(23056) = = X(926)X(2170)∩X(2246)X(4845)

Barycentrics a(b-c)^2(b+c-a)^2(5a^2-4a(b+c)-(b-c)^2) : :

Let ABC be a triangle of incenter I, the perpendicular by I to AI intersects AC and AB in Ab and Ac, respectively.
The points Bc, Ba, Ca and Cb are defined cyclically.

The circles (AbBaBc), (BcCbCa) and (CaAcAb) concur in P.
The circles (AbBaAc), (BcCbBa) and (CaAcCb) cuncur in U.

Points P and U form a bicentric pair,
P = f(a,b,c) : f(b,c,a) : f(c,a,b) and Q = f(a,c,b) : f(b,a,c) : f(c,b,a)
with f(a,b,c) = a (a-b-c) (b-c) (2 a^2-2 a b-a c+b c-c^2) (a^2+a b-2 b^2-2 a c+b c+c^2).

The bicentric difference of P and U is X(23056).

See Emmanuel José García and Angel Montesdeoca, AdGeom4943 and HG100918.

X(23057) = X(1)X(22254)∩X(145)X(3716)

Barycentrics a(b-c)(4a^2-5a(b+c)+b^2+4bc+c^2) : :

Let ABC be a triangle of incenter I, the perpendicular by I to AI intersects AC and AB in Ab and Ac, respectively.
The points Bc, Ba, Ca and Cb are defined cyclically.

The circles (AbBaBc), (BcCbCa) and (CaAcAb) concur in P.
The circles (AbBaAc), (BcCbBa) and (CaAcCb) cuncur in U.

Points P and U form a bicentric pair,
P = f(a,b,c) : f(b,c,a) : f(c,a,b) and Q = f(a,c,b) : f(b,a,c) : f(c,b,a)
with f(a,b,c) = a (a-b-c) (b-c) (2 a^2-2 a b-a c+b c-c^2) (a^2+a b-2 b^2-2 a c+b c+c^2).

The bicentric sum of P and U is X(23057).

See Emmanuel José García and Angel Montesdeoca, AdGeom4943 and HG100918.

X(23058) = X(1)X(1146)∩X(4)X(9)

Barycentrics (b+c-a)(a^3+a(b-c)^2-2(b-c)(b^2-c^2)) : :

Let DEF be the medial triangle of ABC. Let Ha be the hyperbola with foci E and F that passes through A, and define Hb and Hc cyclically. The three hyperbolas have in common two points, and their midpoint is X(23058).

See Kadir Altintas and Angel Montesdeoca, HG110918.

X(23244) = ALTINTAS-MONTESDEOCA MIDPOINT

Barycentrics a(a-b-c)^3(a^6-3a^4(b-c)^2+3a^2(b-c)^2(b^2+6b c+c^2)-16a b(b-c)^2c(b+c)-(b-c)^4(b^2+6b c+c^2)) : :

Let DEF be the exdtouch triangle of ABC. Let Ha be the hyperbola with foci E and F that passes through A, and define Hb and Hc cyclically. The three hyperbolas have in common two points, and their midpoint is X(23244).

See Kadir Altintas and Angel Montesdeoca, HG150918 (Sep 17, 2018).

X(23286) = X(125)-CROSS CONJUGATE OF X(3)

Barycentrics a^2 (b^2-c^2) (a^2-b^2-c^2) (a^4-2 a^2 b^2+b^4-a^2 c^2-b^2 c^2) (a^4-a^2 b^2-2 a^2 c^2-b^2 c^2+c^4) : :

Let L be the line through X(3) parallel to BC, and let A' = L∩AX(100). Define B' and C' cyclically. The triangle A'B'C' is parallelogic to ABC (by construction), at X(3) and X(14380), and X(23286) is the unique fixed point of the affine transformation that maps ABC onto A'B'C'. (Angel Monesdeoca, June 27, 2020)
HG270620

X(23358) = MIDPOINT OF X(3) AND X(2917)

Barycentrics a^2 (a^14-3 a^12 b^2+a^10 b^4+5 a^8 b^6-5 a^6 b^8-a^4 b^10+3 a^2 b^12-b^14-3 a^12 c^2+5 a^10 b^2 c^2+a^8 b^4 c^2-5 a^6 b^6 c^2+4 a^4 b^8 c^2-4 a^2 b^10 c^2+2 b^12 c^2+a^10 c^4+a^8 b^2 c^4+2 a^6 b^4 c^4-3 a^4 b^6 c^4-a^2 b^8 c^4+5 a^8 c^6-5 a^6 b^2 c^6-3 a^4 b^4 c^6+4 a^2 b^6 c^6-b^8 c^6-5 a^6 c^8+4 a^4 b^2 c^8-a^2 b^4 c^8-b^6 c^8-a^4 c^10-4 a^2 b^2 c^10+3 a^2 c^12+2 b^2 c^12-c^14) : :

See Antreas Hatzipolakis, Peter Moses and Angel Montesdeoca, Hyacinthos 28295 and Hyacinthos 28296.

X(23408) = (name pending)

Barycentrics 2 a^16-7 a^14 (b^2+c^2)-(b^2-c^2)^6 (b^4-3 b^2 c^2+c^4)+4 a^12 (b^4+3 b^2 c^2+c^4)+2 a^2 (b^2-c^2)^4 (3 b^6-b^4 c^2-b^2 c^4+3 c^6)+2 a^10 (8 b^6-b^4 c^2-b^2 c^4+8 c^6)+a^8 (-35 b^8+3 b^6 c^2-6 b^4 c^4+3 b^2 c^6-35 c^8)-a^4 (b^2-c^2)^2 (18 b^8+4 b^6 c^2-b^4 c^4+4 b^2 c^6+18 c^8)+a^6 (33 b^10-21 b^8 c^2+5 b^6 c^4+5 b^4 c^6-21 b^2 c^8+33 c^10) : :

See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 28297.

X(23409) = MIDPOINT OF X(5) AND X(13163)

Barycentrics 2 a^10+15 a^2 b^2 c^2 (b^2-c^2)^2-3 a^8 (b^2+c^2)-(b^2-c^2)^4 (b^2+c^2)-2 a^6 (b^4-b^2 c^2+c^4)-a^4 (-4 b^6+17 b^4 c^2+17 b^2 c^4-4 c^6) : :

See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 28297.

X(23410) = MIDPOINT OF X(2) AND X(13490)

Barycentrics 2 a^10-2 a^6 (b^2-c^2)^2+10 a^2 b^2 c^2 (b^2-c^2)^2-3 a^8 (b^2+c^2)-(b^2-c^2)^4 (b^2+c^2)+2 a^4 (2 b^6-7 b^4 c^2-7 b^2 c^4+2 c^6) : :

See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 28297.

X(23411) = MIDPOINT OF X(546) AND X(9825)

Barycentrics 2 a^10+16 a^2 b^2 c^2 (b^2-c^2)^2-3 a^8 (b^2+c^2)-(b^2-c^2)^4 (b^2+c^2)-2 a^6 (b^4+c^4)+4 a^4 (b^6-4 b^4 c^2-4 b^2 c^4+c^6) : :

See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 28297.

X(23617) = CEVAPOINT OF X(6) AND X(9)

Barycentrics a/((b-c)^2+a(b+c)) : :

Let P = X(23617). Let Pa be the isotomic conjugate of P with respect to AX(6)X(9), and define Pb and Pc cyclically. Then ABC and PaPbPc are perspective in X(13622). (Angel Montesdeoca, September 23, 2018)

X(23618) = CEVAPOINT OF X(7) AND X(9)

Barycentrics 1/((b+c-a)(a^2(b+c)-2a(b-c)^2+(b-c)^2(b+c))) : :

Let P = X(23618). Let Pa be the isotomic conjugate of P with respect to AX(7)X(9), and define Pb and Pc cyclically. Then ABC and PaPbPc are perspective in X(3255). (Angel Montesdeoca, September 23, 2018)

X(23709) = X(3)X(95)∩X(1154)X(10606)

Barycentrics a^2 (-(b^2-c^2)^2+a^2 (b^2+c^2)) (a^16-5 a^14 (b^2+c^2)+b^2 c^2 (b^2-c^2)^4 (b^4+4 b^2 c^2+c^4)+3 a^12 (3 b^4+7 b^2 c^2+3 c^4)-a^10 (5 b^6+33 b^4 c^2+33 b^2 c^4+5 c^6)+a^8 (-5 b^8+23 b^6 c^2+38 b^4 c^4+23 b^2 c^6-5 c^8)-a^4 (b^2-c^2)^2 (5 b^8+7 b^6 c^2+8 b^4 c^4+7 b^2 c^6+5 c^8)+a^2 (b^2-c^2)^2 (b^10-b^8 c^2-4 b^6 c^4-4 b^4 c^6-b^2 c^8+c^10)+a^6 (9 b^10-7 b^8 c^2-14 b^6 c^4-14 b^4 c^6-7 b^2 c^8+9 c^10)) : :

See Tran Quang Hung, Angel Montesdeoca, and Ercole Suppa, Hyacinthos 28317 and Hyacinthos 28318.

X(23869) = MIDPOINT OF X(1) AND X(6788)

Barycentrics a^3 (b+c)-a^2 (-3 b^2+10 b c-3 c^2)+a (b^3+c^3)-(b^2-c^2)^2 : :

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 28355 and HG270918.

Let ABC be a triangle.

Denote:

Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.
A1, B1, C1 = points on IA, IB, IC such that:

AA1/AI = BB1/BI = CC1/CI = t

The perpendicular to AI at A1 intersects NbNc at A*
The perpendicular to BI at B1 intersects NcNa at B*
The perpendicular to CI at C1 intersects NaNb at C*

The points A*, B*, C* are collinear.

The envelope of the lines A*B*C* as t varies is the parabola of focus X(23869).

X(25466) = (A,B,C,X(2); A',B',C',X(2)) COLLINEATION IMAGE OF X(12), WHERE A'B'C' = GEMINI TRIANGLE 21

Barycentrics -a^2 b^2 + b^4 - 4 a^2 b c - 2 a b^2 c - a^2 c^2 - 2 a b c^2 - 2 b^2 c^2 + c^4 : :

Let DEF be the medial triangle. The circle that touches all three incircles of the triangles AEF, BFD, CDE and encompasses them touches in points that we denote by them in A', B', C', respectively. The lines DA', EB' , FC' concur in X(25466). See X(25466). (Angel Montesdeoca, November 29, 2019)
(HG291119)

X(28443) = EULER LINE INTERCEPT OF X(36)X(4870)

Barycentrics a (3 a^6-3 a^5 (b+c)+a^4 (-6 b^2+b c-6 c^2)-2 b c (b^2-c^2)^2+6 a^3 (b^3+c^3)+a^2 (3 b^4+b^3 c+6 b^2 c^2+b c^3+3 c^4)-3 a (b^5-b^4 c-b c^4+c^5)) : :

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 28681.

X(30332) = ENDO-HOMOTHETIC CENTER OF THESE TRIANGLES: ANTI-HUTSON INTOUCH AND 2nd EHRMANN

Barycentrics 7*a^3-5*(b+c)*a^2+(b-c)^2*a-3*(b^2-c^2)*(b-c) : :

Let Γ_a be the circle passing through X(7) tangent to AB and AC such that the touchpoints do not lie on the Adams circle. Define Γ_b and Γ_c cyclically. Then X(30332) is the center of the circle tangent to the circles Γ_a, Γ_b Γ_c .and encompassing (Angel Montesdeoca, June 2, 2023)
HG010623

X(30443) = X(30)X(21651)∩X(51)X(5895)

Barycentrics a^2 ( a^12 (b^2+c^2)-4 a^10 (b^2-c^2)^2+5 a^8 (b^2-c^2)^2 (b^2+c^2) -40 a^6 b^2 c^2 (b^2-c^2)^2-5 a^4 (b^2-c^2)^2 (b^6-9 b^4 c^2-9 b^2 c^4+c^6) +4 a^2 (b^2-c^2)^2 (b^8-2 b^6 c^2-14 b^4 c^4-2 b^2 c^6+c^8) -(b^2-c^2)^4 (b^6+7 b^4 c^2+7 b^2 c^4+c^6)) : :

Let ABC be a triangle with de Longchamps point L. A'B'C' is pedal triangle of L. L'=X(30443) is de Longchamps point of triangle A'B'C'.

See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 28754.

X(31383) = MARTA POINT

Barycentrics 3a^6-a^4(b^2+c^2)-a^2(b^2-c^2)^2-(b^2-c^2)^2(b^2+c^2) : :

Let ABC be a triangle with centroid G and orthocenter H.
The line parallel to BC for the point Pa, such that APa:PaH= t, intersects the sides AC and AB respectively at Ab and Ac.
Let Da be the point of intersection of the lines BAb and CAc.
Let La be the line joining the orthocenters of the triangles BDaAc and CDaAb.
Lb and Lb are defined similarly.
Let G' be the centroid of A'B'C', the triangle bound by La,Lb,Lc.
Let F be the fixed point (not on the line at infinity) of the affine transformation that applies ABC in A'B'C'.

Then, as t varies, the line G'F passes through a fixed point Q=X(31383). See Angel Montesdeoca, Hyacinthos 28832 and Centro ortológico y punto fijo de una afinidad.

X(31392) = REFLECTION OF X(5) IN X(14051)

Barycentrics (a^4+(b^2-c^2)^2-a^2 (b^2+2 c^2)) (a^12-(b^2-c^2)^6-a^10 (4 b^2+3 c^2)+a^8 (5 b^4+5 b^2 c^2+3 c^4)-a^6 (3 b^4 c^2+2 b^2 c^4)+a^2 (b^2-c^2)^2 (4 b^6-6 b^4 c^2-3 b^2 c^4+3 c^6)+a^4 (-5 b^8+9 b^6 c^2+b^4 c^4+4 b^2 c^6-3 c^8)) : :

Let ABC be a triangle.
Denote:
(O1) ,(O2), (O3) = the circumcircles of NBC, NCA, NAB, resp.
(Oa) ,(Ob), (Ob) = the reflections of (O1) ,(O2), (O3) in BC, CA, AB, resp.
They concurr at D = antigonal conjugate of N = X(1263).

The circumcircles of DAOa, DBOb, DCOc are coaxial.
2nd (other than D=X(1263)) intersection is X(31392) = REFLECTION OF X(5) IN X(14051).

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 28850.

Let A' be the reflection of A in BC. Let A_b be the intersection of AC and the perpendicular bisector of A'B, and let A_c be the intersection of AB and the perpendicular bisector of A'C. Let A'' be the orthogonal projection of A' on A_bA_c, and define B'' and C'' cyclically. The lines AA'', BB'', CC'' concur in X(31392). (Angel Montesdeoca, July 16, 2020)
(El centro del triángulo X(31392))

X(31510) = MIDPOINT OF X(107) AND X(1304)

Barycentrics (a^2-b^2) (a^2-c^2) (2 a^14-2 a^12 (b^2+c^2)+9 a^8 (b^2-c^2)^2 (b^2+c^2)-8 a^4 (b^2-c^2)^4 (b^2+c^2)+(b^2-c^2)^6 (b^2+c^2)+a^10 (-7 b^4+16 b^2 c^2-7 c^4)+4 a^6 (b^2-c^2)^2 (b^4-5 b^2 c^2+c^4)+a^2 (b^2-c^2)^4 (b^4+8 b^2 c^2+c^4)) : :

Let ABC be a triangle.
Euler line meets BC, CA, AB at A', B', C', resp.
Perpendiculars from A', B', C' to BC, CA, AB resp. bound a triangle A"B"C".
Then X(1552) of A"B"C" lies on the Euler line of ABC: X(31510) = MIDPOINT OF X(107) AND X(1304)

See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 28858.

X(31511) = X(1)X(3)∩X(109)X(13138)

Barycentrics a (a-b) (a^2-(b-c)^2) (a-c) (2 a^7-a^6 (b+c)+a^4 (b-c)^2 (b+c)-(b-c)^2 (b+c)^5-6 a^3 (b^2-c^2)^2+4 a (b^2-c^2)^2 (b^2+c^2)+a^2 (b-c)^2 (b^3+7 b^2 c+7 b c^2+c^3)) : :

Let ABC be a triangle.
X(1)X(4) line of ABC meets BC, CA, AB at A', B', C'.
Perpendicular lines from A', B', C' to BC, CA, AB bound triangle A"B"C".
Then X(84) of A"B"C" lies on OI line of ABC: X(31511) = X(1)X(3)∩X(109)X(13138)

See Tran Quang Hung and Angel Montesdeoca, ADGEOM 5142.

X(31512) = X(4)X(8)∩X(11)X(901)

Barycentrics a^6+6 a^4 b c-2 a^5 (b+c)-2 a^3 b c (b+c)-(b-c)^4 (b+c)^2+a^2 b c (3 b^2-5 b c+3 c^2)+a (b-c)^2 (2 b^3-3 b^2 c-3 b c^2+2 c^3) : :

See Tran Quang Hung and Angel Montesdeoca, ADGEOM 5144.

X(31513) = X(4)X(8)∩X(11)X(901)

Barycentrics a^2 (a^8 b^2 c^2-2 a^2 b^4 c^4 (b^2+c^2)-a^6 (b^6+b^4 c^2+b^2 c^4+c^6)+a^4 (b^8+4 b^4 c^4+c^8)-b^2 c^2 (b^8-3 b^6 c^2+3 b^4 c^4-3 b^2 c^6+c^8)) : :

See Tran Quang Hung and Angel Montesdeoca, ADGEOM 5144.

X(31668) = (name pending)

Barycentrics a (a+b) (a+c) ((b-c)^6 (b+c)^7+(b^2-c^2)^4 (b^4-b^3 c+8 b^2 c^2-b c^3+c^4) a-(b-c)^2 (b+c)^3 (4 b^6-2 b^5 c+11 b^4 c^2-14 b^3 c^3+11 b^2 c^4-2 b c^5+4 c^6) a^2-(b^2-c^2)^2 (4 b^6-7 b^5 c-3 b^4 c^2+7 b^3 c^3-3 b^2 c^4-7 b c^5+4 c^6) a^3+(5 b^9-3 b^8 c+15 b^7 c^2+31 b^6 c^3-18 b^5 c^4-18 b^4 c^5+31 b^3 c^6+15 b^2 c^7-3 b c^8+5 c^9) a^4+(5 b^8-18 b^7 c-43 b^6 c^2-16 b^5 c^3+14 b^4 c^4-16 b^3 c^5-43 b^2 c^6-18 b c^7+5 c^8) a^5+b c (12 b^5+3 b^4 c-19 b^3 c^2-19 b^2 c^3+3 b c^4+12 c^5) a^6+b c (22 b^4+41 b^3 c+47 b^2 c^2+41 b c^3+22 c^4) a^7+(-5 b^5-13 b^4 c-17 b^3 c^2-17 b^2 c^3-13 b c^4-5 c^5) a^8+(-5 b^4-13 b^3 c-17 b^2 c^2-13 b c^3-5 c^4) a^9+2 (2 b^3+3 b^2 c+3 b c^2+2 c^3) a^10+(4 b^2+3 b c+4 c^2) a^11+(-b-c) a^12-a^13) : :

I call Kosnita line the line conneting NPC center X(5) and Kosnita point X(54) of a triangle.
Let ABC be a triangle with excenters Ia, Ib, Ic.Then refections L'a, L'b, L'c of Kosnita lines La, Lb, Lc of triangles IaBC, IbCA, and IcAB in lines BC, CA, and AB respectively are concurrent at X(110) = focus of Kiepert parabola.

If Aa=Lb ∩ Lc, Ab=L'a ∩ Lc, Ac=L'a∩ Lb, then the line AAa, BAb, CAc concur in a point Wa, and define Wb and Wc cyclically. WaWbWc is perspective to ABC at X(1290) and WaWbWc is perspective to AaBbCc at X(1290).
Then the three perspectrix of ABC and WaWbWc, ABC and AaBaBc, AaBbBc and WaWbWc concur in X(31668).

See Tran Quang Hung and Angel Montesdeoca, Hyacinthos 28896.

X(31676) = X(3)X(125)∩X(25)X(6344)

Barycentrics a^2SA (SA^2 + 5S^2)/(4SA^2-b^2c^2) : :

Let ABC be a triangle of circumcenter O and a point D variable on the circumcircle. Let Ha, Hb, Hc be the orthocenters of the triangles DBC, DCA, DAB, respectively.
The circle of diameter OHa cuts again the lines BHa and CHa in Hab and Hac, respectively.
The perpendicular bisector of HabHac passes through a fixed point Fa, on the perpendicular bisector of BC, when D varies over the circumcircle.
Let p_a be the polar of Fa with respect to the circumcircle. The polar p_b and p_c are defined cyclically.
The triangle A'B'C' formed by the lines p_a, p_b and p_c is perspective with ABC, with perspector X(31676)

See Nguyên Danh Lân and Angel Montesdeoca, Hyacinthos 28906, HG080319, AoPS.

X(31846) = X(324)X(547)∩X(343)X(15699)

Barycentrics 1/(5 a^8-17 a^6 (b^2+c^2)+a^4 (21 b^4+17 b^2 c^2+21 c^4)-11 a^2 (b^2-c^2)^2 (b^2+c^2)+(b^2-c^2)^2 (2 b^4-7 b^2 c^2+2 c^4)) : :

See Ioannis Panakis, Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 28926.

X(31868) = X(5)X(27684)∩X(2883)X(3845)

Barycentrics 4 a^16-22 a^14 b^2+45 a^12 b^4-31 a^10 b^6-30 a^8 b^8+76 a^6 b^10-63 a^4 b^12+25 a^2 b^14-4 b^16-22 a^14 c^2+58 a^12 b^2 c^2-27 a^10 b^4 c^2-21 a^8 b^6 c^2-56 a^6 b^8 c^2+156 a^4 b^10 c^2-119 a^2 b^12 c^2+31 b^14 c^2+45 a^12 c^4-27 a^10 b^2 c^4-18 a^8 b^4 c^4-20 a^6 b^6 c^4-81 a^4 b^8 c^4+207 a^2 b^10 c^4-106 b^12 c^4-31 a^10 c^6-21 a^8 b^2 c^6-20 a^6 b^4 c^6-24 a^4 b^6 c^6-113 a^2 b^8 c^6+209 b^10 c^6-30 a^8 c^8-56 a^6 b^2 c^8-81 a^4 b^4 c^8-113 a^2 b^6 c^8-260 b^8 c^8+76 a^6 c^10+156 a^4 b^2 c^10+207 a^2 b^4 c^10+209 b^6 c^10-63 a^4 c^12-119 a^2 b^2 c^12-106 b^4 c^12+25 a^2 c^14+31 b^2 c^14-4 c^16 : :

Let DEF be the cevian triangle of X(5). Let ℓ_b be the perpendicular to BE at E, and let ℓ_c be the perpendicular to CF at F. Let A' = ℓ_b∩ℓ_c, and define B' and C' cyclically. The triangles ABC and A'B'C' are orthologic (by construction), and the orthologic center of A'B'C' respect to ABC is X(31868). (Angel Montesdeoca, March 13, 2021)

X(31873) = X(125)X(546)∩X(133)X(1594)

Barycentrics 2 a^14 (b^2+c^2) +a^12 (-6 b^4+8 b^2 c^2-6 c^4) +a^10 (b^6-3 b^4 c^2-3 b^2 c^4+c^6) +a^8 (15 b^8-44 b^6 c^2+62 b^4 c^4-44 b^2 c^6+15 c^8) -a^6 (b^2-c^2)^2 (20 b^6+b^4 c^2+b^2 c^4+20 c^6) +a^4 (b^2-c^2)^2 (8 b^8+31 b^6 c^2-42 b^4 c^4+31 b^2 c^6+8 c^8) +a^2 (b^2-c^2)^4 (b^6-14 b^4 c^2-14 b^2 c^4+c^6) -(b^2-c^2)^6 (b^4+5 b^2 c^2+c^4) : :

See Vu Thanh Tung and Angel Montesdeoca, Hyacinthos 28942.

Let DEF be the pedal triangle of a point P wrt a triangle ABC.
Oa and Ha are the circumcenter and the orthocenter of AEF, resp.
Let α be a real number and A1 be the point such that: OaA1 = α OaHa. Define B1 and C1 cyclically.

(1) The triangle A1B1C1 are orthologic to both triangles ABC and DEF, with orthologic centers V and V', resp.

(2) When P fixed and α varies the orthologic centers move in fixed lines, L and L'.

When the point P traverses the Euler line of ABC the line L envelope the parabola tangent to Euler line at X(5), tangent to line X(4)X(51) at X(185), and tangent to line X(389)X(546) at X(10095).
X(31873) is the focus.

X(32006) = POINT II-CAPH OF X(69)

Barycentrics 3*a^4 - 3*b^4 + 2*b^2*c^2 - 3*c^4 : :

Let A'B'C' be the anticevian triangle of X(69). Let A" be the orthogonal projection of A' on the line through X(69) parallel to line BC, and define B" and C" cyclically. The lines AA", BB", CC" concur in X(32006) = Point II-Caph of X(69). See X(33006).png (Angel Montesdeoca, December 6, 2022)
See X(32006) y triángulo anticeviano del retrocentro.

X(32319) = X(4)X(19209)∩X(20)X(2979)

Barycentrics a^2 (b^2 c^2 (b^2 - c^2)^6 - 10 a^10 (b^2 - c^2)^2 (b^2 + c^2) - 2 a^2 (b^2 - c^2)^4 (b^2 + c^2)^3 + a^12 (2 b^4 - 5 b^2 c^2 + 2 c^4) + 5 a^8 (b^2 - c^2)^2 (4 b^4 + 7 b^2 c^2 + 4 c^4) - 4 a^6 (b^2 - c^2)^2 (5 b^6 + 9 b^4 c^2 + 9 b^2 c^4 + 5 c^6) + a^4 (b^2 - c^2)^2 (10 b^8 + 13 b^6 c^2 + 18 b^4 c^4 + 13 b^2 c^6 + 10 c^8)) : :

See Angel Montesdeoca, Hyacinthos 29000 and HG020519.

X(32320) = REFLECTION OF X(17434) IN X(647)

Barycentrics a^4 (a^2 - b^2 - c^2)^3 (b^2 - c^2) : :

See Angel Montesdeoca, Hyacinthos 29000 and HG020519.

X(32446) = X(2)X(2124)∩X(10)X(971)

Barycentrics (b+c-a)(a^5 (b+c)-3 a^4 (b-c)^2+2 a^3 (b-c)^2 (b+c)+2 a^2 (b-c)^2 (b^2+6 b c+c^2)-a (b-c)^2 (3 b^3+13 b^2 c+13 b c^2+3 c^3)+(b-c)^6) : :

See Angel Montesdeoca, Hyacinthos 29006, Hyacinthos 29006 and HG070519.

X(32486) = MIDPOINT OF X(1) AND X(5400)

Barycentrics a(a^5(b+c)-4a^4b c+a^3(-2b^3+3b^2c+3b c^2-2c^3)+5a^2b c(b-c)^2+a(b-c)^2(b^3-2b^2c-2b c^2+c^3)- b c(b^2-c^2)^2) : :

See Angel Montesdeoca, Hyacinthos 29011 and HG130519.

X(32487) = X(2292)X(3754)∩X(4535)X(18697)

Barycentrics a(b+c)^2/(2a^2-(b+c)^2) : :

See Angel Montesdeoca, Hyacinthos 29011 and HG130519.

X(33568) = X(2)X(824)∩X(4809)X(14402)

Barycentrics (b^3 - c^3) (-2 a^3 + b^3 + c^3)^2 : :

See Angel Montesdeoca, Hyacinthos 29181 and HG100719.

X(33569) = X(2)X(647)∩X(684)X(2491)

Barycentrics a^4(b^2-c^2)(b^4+c^4-a^2(b^2+c^2))^2(-a^4+2b^2c^2+a^2(b^2+c^2)) : :

See Angel Montesdeoca, Hyacinthos 29181 and HG100719.

X(33570) = X(2)X(650)∩X(647)X(1962)

Barycentrics a^2 (b - c) (b^2 + c^2 - a (b + c))^2 (-a^2 + 2 b c + a (b + c)) : :

See Angel Montesdeoca, Hyacinthos 29181 and HG100719.

X(33571) = X(2)X(216)∩X(1636)X(2972)

Barycentrics a^4 (b^2 - c^2)^2 (-a^2 + b^2 + c^2)^4 (-2 b^2 c^2 (b^2 - c^2)^2 + a^6 (b^2 + c^2) + a^2 (b^2 - c^2)^2 (b^2 + c^2) - 2 a^4 (b^4 - b^2 c^2 + c^4)) : :

See Angel Montesdeoca, Hyacinthos 29181 and HG100719.

X(33572) = X(2)X(92)∩X(7004)X(7117)

Barycentrics a^2 (b - c)^2 (a^3 + b^3 + b^2 c + b c^2 + c^3 - a^2 (b + c) - a (b^2 + c^2))^2 (a^4 (b + c) - a^2 (b - c)^2 (b + c) - 2 b (b - c)^2 c (b + c) + a (b^2 - c^2)^2 - a^3 (b^2 + c^2)) : :

See Angel Montesdeoca, Hyacinthos 29181 and HG100719.

X(33573) = REFLECTION OF X(14477) IN X(2)

Barycentrics (b - c)^2 (-a + b + c)^2 (2 a^2 - (b - c)^2 - a (b + c)) : :

See Angel Montesdeoca, Hyacinthos 29181 and HG100719.

X(33591) = EULER LINE INTERCEPT OF X(524)X(10282)

Barycentrics -10 a^10+21 a^8 (b^2+c^2)-2 a^6 (b^4+4 b^2 c^2+c^4)-4 a^4 (5 b^6-b^4 c^2-b^2 c^4+5 c^6)+4 a^2 (b^2-c^2)^2 (3 b^4+b^2 c^2+3 c^4)-(b^2-c^2)^4 (b^2+c^2) : :

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 29196.

X(34199) = (name pending)

Barycentrics 2 a^10 - 9 a^8 (b^2 + c^2) + 2 a^6 (5 b^4 + 9 b^2 c^2 + 5 c^4) + a^4 (4 b^6 - 9 b^4 c^2 - 9 b^2 c^4 + 4 c^6) - 3 a^2 (b^2 - c^2)^2 (4 b^4 + 3 b^2 c^2 + 4 c^4) + 5 (b^2 - c^2)^4 (b^2 + c^2) : :

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 29455 and HG270819.

X(34285) = isogonal conjugate of X(33586)

Barycentrics 1/(a^4+2a^2(b^2+c^2)-3b^4+2b^2c^2-3c^4) : :

Let A'B'C' and A"B"C" be the cevian and circumcevian triangles of the orthocenter. Let L_a be the radical axis of circles with segments BC and A'A" as diameters, and define L_b and L_c cyclically. The triangle formed by lines L_a, L_b, L_c is perspective to ABC, and the perspector is X(34285). (Angel Montesdeoca, September 5, 2019) See HGT

X(34355) = X(1)X(12619)∩X(106)X(18976)

Barycentrics (a-b+c) (a+b-c) (a^7 (b+c)+(b^2-c^2)^4+a^6 (b^2-12 b c+c^2)-a^2 (b^2-c^2)^2 (b^2-b c+c^2)-a (b-c)^2 (b+c)^3 (3 b^2-5 b c+3 c^2)+a^5 (-5 b^3+13 b^2 c+13 b c^2-5 c^3)-a^4 (b^4-11 b^3 c+30 b^2 c^2-11 b c^3+c^4)+a^3 (7 b^5-16 b^4 c+10 b^3 c^2+10 b^2 c^3-16 b c^4+7 c^5)) : :

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 29575 and HG031019.

Let Na, Nb, Nc denote the nine-point centers of the triangles IBC, ICA, IAB, and let A"B"C" be the reflection of intouch triangle in the line X(1)X(3). Then NaNbNc is similar to A''B''C'', and the center of similitude is X(34355). (Angel Montesdeoca, October 7, 2019)

X(34578) = CYCLOLOGIC CENTER S(X(1),X(2))

Barycentrics (a^2 + a*b + b^2 - 2*a*c - 2*b*c + c^2)*(a^2 - 2*a*b + b^2 + a*c - 2*b*c + c^2) : :

Let DEF be the intouch triangle and I_aI_bI_c the excentral-triangle. Let D' be the trisector nearest D of segment DI_a. Let L_a be the radical axis of (DEF) and (BCD'), and define L_b and L_c cyclically. The triangle A'B'C' formed by the lines L_a, L_b, L_c is perspective to ABC, and the perspector is X(34578). (Angel Montesdeoca, December 31, 2020)
In general, if D' is a point such that DD' / D'I_a = t (a constant real number) then the perspector of ABC and A'B'C', given by barycentrics

a/(a^3-a^2 (b+c) (1+2 t)-a (b^2 (1-4 t)+c^2 (1-4 t)-2 b c (1-2 t+2 t^2))-(b-c)^2 (b+c) (-1+2 t)) : : ,

lies on the circumhyperbola that is the isogonal conjugate of line X(1)X(6). See also X(39980). (Angel Montesdeoca, December 31, 2020)

See HGT301220

X(34815) = CENTER OF PERSPECONIC OF ABC AND CIRCUMCEVIAN TRIANGLE OF X(64)

Barycentrics a^2*(a^2 - b^2 - c^2)*(3*a^8 - 2*a^4*b^4 - 8*a^2*b^6 + 7*b^8 + 4*a^4*b^2*c^2 + 8*a^2*b^4*c^2 - 12*b^6*c^2 - 2*a^4*c^4 + 8*a^2*b^2*c^4 + 10*b^4*c^4 - 8*a^2*c^6 - 12*b^2*c^6 + 7*c^8) : :

Let A'B'C' be the circumcevian triangle of X(64). Let A"=AO∩B'C', and define B" and C" cyclically. The points A'A", B'B", C'C" concur in X(34815). More generally, suppose that a point P lies on the Darboux cubic, and let A'B'C' be the circumcevian triangle of P. Define B'' and C'' cyclically. The points A'A", B'B", C'C" concur in a point, Q(P). The appearance of (i,j) in the following list means that X(i) is on the Darboux cubic and Q(X(i)) = X(j): (1,991), (,3), (4,5890), (64,34815). (Angel Montesdeoca, August 3, 2021).
La cúbica de Darboux y triángulos circuncevianos

X(35015) = X(1)X(4)∩X(11)X(244)

Barycentrics (a - b - c)*(b - c)^2*(a^2*b - b^3 + a^2*c - 2*a*b*c + b^2*c + b*c^2 - c^3) : :

In the plane of a triangle ABC, let
DEF = orthic triangle;
F' = X(11) = Feuerbach point;
D' = reflection of D in AF;
Da = EF∩D'F';
Oa = circumcenter of ADDa, and define Ob and Oc cyclically;
The finite fixed point of the affine transformation that maps ABC onto OaOObOc is X(35015). (Angel Montesdeoca, May 26, 2022)
See HG250522

X(35445) = X(1)X(3) ∩ (line at infinity)

Barycentrics a*(5*a^2 - 4*a*b - b^2 - 4*a*c + 2*b*c - c^2) : :

Let B_c be the reflection of B in the internal angle bisector of angle C, and let C_b be the reflection of C in the internal angle bisector of angle B. Let Γ_a be the circle that passes through A, B_c, C_b. Let A' be the pole of BC, with respect to Γ_a and define B 'and C' cyclically. Then X(1156) is the perspector of A'B'C' and ABC, and X(35445) is the perspector of A'B'C' and the excentral triangle. (Angel Montesdeoca, January 13, 2021)
See HG120121.

X(36809) = (name pending)

Barycentrics (S^2+SB SC)/(SA(3S^2+5 SB SC)) : :
Barycentrics (a^2 (b^2+c^2)-(b^2-c^2)^2)/((b^2+c^2-a^2) (a^4+3 a^2 (b^2+c^2)-4 (b^2-c^2)^2)) : :

Let H be the orthocenter and M be the midpoint of AH. Let B_a and C_a be the orthogonal projections of B and C on CM and BM, respectively. Define C_b and A_c cyclically, and define A_b and B_c cyclically. Let A'B'C' be the triangle having sidelines B_aC_a, C_bA_b, A_cB_c. Then A'B'C' is perspective to ABC, and the perspector is X(36809).

See Angel Montesdeoca, Euclid 641 .

X(36922) = X(1)X(5791)∩X(2)X(14563)

Barycentrics (a - 2*b - 2*c)*(3*a^3 - a^2*b - 3*a*b^2 + b^3 - a^2*c + 2*a*b*c - b^2*c - 3*a*c^2 - b*c^2 + c^3) : :

In the plane of a triangle ABC, let
Ia = A-excenter = -a : b : c, and define Ib and Ic cyclically
Ta = AaBaCa = pedal triangle of Ia, and define Tb and Tc cyclically
T'a = A'aB'aC'a = Ta-cevian triangle of Ia, and define T'b and T'c cyclically
A' = radical center of A-excircle and the circles having diameters BB'a and CC'a, and define B' and C' cyclically.
The lines A'Aa, B'Bb, C'Cc concur in X(36922). (Angel Montesdeoca, January 24, 2023)

X(37084) = MIDPOINT OF X(3) AND X(14152)

Barycentrics a^4 (b^2-c^2) (b^2+c^2-a^2)^2 (a^4-2 a^2 (b^2+c^2)+b^4-b^2 c^2+c^4) : :

Dados un triángulo ABC y un punto P, sean P_aP_bP_c y P^aP^bP^c los triángulos ceviano y pedal de P, respectivamente.
Sean Q_a, Q_b, Q_c los puntos que dividen a los segmentos P_aP^a, P_bP^b, P_cP^c, resp., en la misma razón t.
Las perpendiculares por Q_a, Q_b, Q_c a BC, CA, AB, respectivamente, forman un triángulo A'B'C', directamente semejante a ABC.
El lugar geométrico de los centros de semejanza, cuando t varía, es una circunferencia que pasa por P.
Cuando P es el circuncentro, el centro de esta circunferencia es X(37084).

See Tran Quang Hung and Angel Montesdeoca, Euclid 703. and HG120320

X(37085) = X(6)X(826)∩X(512)X(1692)

Barycentrics a^4 (b^6 - c^6 + a^4 (-b^2 + c^2)) : :

Dados un triángulo ABC y un punto P, sean P_aP_bP_c y P^aP^bP^c los triángulos ceviano y pedal de P, respectivamente.
Sean Q_a, Q_b, Q_c los puntos que dividen a los segmentos P_aP^a, P_bP^b, P_cP^c, resp., en la misma razón t.
Las perpendiculares por Q_a, Q_b, Q_c a BC, CA, AB, respectivamente, forman un triángulo A'B'C', directamente semejante a ABC.
El lugar geométrico de los centros de semejanza, cuando t varía, es una circunferencia que pasa por P.
Cuando P es el simediano, el centro de esta circunferencia es X(37085).

See Tran Quang Hung and Angel Montesdeoca, Euclid 703. and HG120320

X(37283) = MIDPOINT OF X(12039) AND X(14810)

Barycentrics 2 a^8-7 a^2 b^2 c^2 (b^2+c^2)-2 a^4 (b^4+b^2 c^2+c^4) : :

Let ABC and A'B'C' be two orthologic triangles such that their orthologic centers coincide at symmedian point, X(6). Then ABC and A'B'C' are perspective, and their perspector P lies on the Jerabek hyperbola. Let P* be the isogonal conjugate of P, and O' the circumcenter of A'B'C'. If ABC remains fixed while A'B'C' varies, then the line P*O' envelopes a hyperbola passing through X(6), X(23), with center X(37283), and asymptotes X(37283)X(5004) and X(37283)X(5005). (Angel Montesdeoca, March 17, 2020)

See Angel Montesdeoca, Euclid 714.
Triángulos con centro ortológico común. HG150320

X(37518) = X(3)X(1392)∩X(4)X(1388)

Barycentrics a (9 a^6 - 15 a^5 (b + c) - a^4 (12 b^2 - 49 b c + 12 c^2) + 30 a^3 (b - c)^2 (b + c) - 15 a (b - c)^4 (b + c) - 3 a^2 (b - c)^2 (b^2 + 14 b c + c^2) + (b^2 - c^2)^2 (6 b^2 - 13 b c + 6 c^2) ) : :

See Angel Montesdeoca, Euclid 721.
El centro de la circunferencia de Hatzipolakis-Lozada como centro ortológico.

X(37593) = X(1)X(3)∩X(37)X(42)

Barycentrics a*(b + c)*(3*a + b + c) : :

Let DEF be intouch triangle. Let (Ob) be the circle centered at B that passes through D and F. Let (Oc) be the circle centered at C that passes through D and E. Let E' be the point of intersection, other than E, of the circles (AEF) and (Oc). Let F' be the point of intersection, other than F, of the circles (AEF) and (Ob). Let A' be the center of circle DE'F', and define B' and C' cyclically. The finite fixed point of the affine transformation that carries ABC onto A'B'C' is X(37593). (Angel Montesdeoca, September 24, 2020).
Una propiedad del centro del triángulo X(37593)

X(37754) = X(1)X(1956)∩X(48)X(163)

Barycentrics a^3 (a^2-b^2-c^2)^3 (b^2-c^2)^2 : :

See Angel Montesdeoca, Euclid 778 and HG030420.

X(37755) = ISOGONAL CONJUGATE OF X(2326)

Barycentrics a (b+c)^2 (a^2-b^2-c^2)/(b+c-a)^2 : :

See Angel Montesdeoca, Euclid 778 and HG030420.

X(37836) = MIDPOINT OF X(73) AND X(23361)

Barycentrics a^2 (2 a^2+a (b+c)-(b-c)^2) (a^2 (b+c)-a b c-b^3-c^3) : :

Dado un triángulo ABC, sean DEF el triángulo medial y A'B'C' el segundo triángulo circumperpendicular. Se consideran los trapecios isósceles AA'A_bE y AA'A_cF con AA' paralelo a A_bE y a A_cF. Los puntos B_c, B_a y C_a, C_b se definen cíclicamente.
El triángulo UVW formado por las rectas A_bA_c, B_cB_a y C_aC_b es perspectivo al triángulo medial. El centro de perspectividad es X(37836).

See Angel Montesdeoca, Euclid 796 and HG070420.

X(37837) = X(1)X(227)∩X(3)X(960)

Barycentrics a( -(b^2-c^2)^2 (b^2+c^2)+(b-c)^2 (3 b^3+b^2 c+b c^2+3 c^3) a+4 b (b-c)^2 c a^2+(-6 b^3+2 b^2 c+2 b c^2-6 c^3)a^3+(3 b^2-4 b c+3 c^2) a^4+3 (b+c) a^5-2 a^6) : :

Dado un triángulo ABC, sean DEF el triángulo medial y A'B'C' el segundo triángulo circumperpendicular. Se consideran los trapecios isósceles AA'A_bE y AA'A_cF con AA' paralelo a A_bE y a A_cF. Los puntos B_c, B_a y C_a, C_b se definen cíclicamente.
El triángulo UVW formado por las rectas A_bA_c, B_cB_a y C_aC_b y ABC son ortológicos . El centro de ortología de ABC respecto a UVW es el circuncentro y el centro de ortología de UVW respecto a ABC es X(378379).

See Angel Montesdeoca, Euclid 799 and HG070420.

X(37865) = PERSPECTOR OF THESE TRIANGLES: ABC AND JENKINS TANGENTIAL

Barycentrics (b+c)/(a^3 (b+c)+a^2 b c-a (b-c)^2 (b+c)-b c (b+c)^2) : :

The triangle bounded by the tangents at contact points between each Jenkins circles and its internally tangent excircles is here named the Jenkins Tangential Triangle.
- See Angel Montesdeoca Euclid 808.

X(37874) = PERSPECTOR OF THE PARALLELS-TANGENTIAL-CONIC-OF-X(4)

Barycentrics b^2*c^2*(a^4+2*(3*b^2-c^2)*a^2+(b^2-c^2)^2)*(a^4-2*(b^2-3*c^2)*a^2+(b^2-c^2)^2) : :

There are two parabolas having focus A and axis AB and passing through C; these parabolas also meet AC in two more points, denoted by A_c1 and A_c2. Likewise there are two parabolas having focus A, axis AC, and passing through B. They also meet in two other points on AB, denoted by A_b1 and A_b2. The diagonal points of the complete quadrangle A_b1A_b2A_c1A_c2 are A, the infinity point of BC, and a third point, denoted by A'. Define B' and C' cyclically. The parallels through A',B',C' to BC, CA, AB, respectively, form a triangle A"B"C" that is homothetic to ABC. The homothetic center is X(37874). (Angel Montesdeoca, February 23, 2022)
HG220222.

X(37998) = INFINITY POINT OF X(11)X(3835)

Barycentrics a (b-c) (-b^2 c^2+a^3 (b+c)-a^2 (2 b^2+b c+2 c^2)+a (b^3+b^2 c+b c^2+c^3)) : :

Let σ be the similarity transformation between the 1st & 2nd Montesdeoca bisector triangles (see P(160))
When P traverses the line X(1)X(6) (radical axis of the circumcircles of the 1st & 2nd Montesdeoca bisector triangles), then the lines σ(P)σ^-1(P) are parallel, with point at infinity at X(37998).
See Angel Montesdeoca Euclid 844 and HGT2017.

X(39530) = CENTER OF ((POLAR CIRCLE, LEMOINE AXIS))

Barycentrics (a^2 + b^2 - c^2)*(a^2 - b^2 + c^2)*(a^4 - a^2*b^2 - a^2*c^2 - 2*b^2*c^2)*(a^2*b^2 - b^4 + a^2*c^2 + 2*b^2*c^2 - c^4) : :

In the plane of a triangle ABC, let N=X(5), DEF = orthic triangle, A'B'C' = Euler triangle, A"=ND∩B'C', and define B" and C" cyclically. The lines A'A", B'B", C'C" concur in X(39530). (Angel Montesdeoca, August 2, 2021)
El centro del triángulo X(39530)

X(39704) = ISOTOMIC CONJUGATE OF X(3679)

Barycentrics (2*a + 2*b - c)*(2*a - b + 2*c) : :

In the plane of a tiangle ABC, let (BC) be the half-plane determined by the line BC that does not include A. Let Ab be the point in (BC) that satisfies |BAb| = |AC|, and let Ac be the point in (BC) that satisfies |BAc| = |AB|. Let La be the line of the diagonal points, other than A, of the complete quadrangle BCAbAc. Define Lb and Lc cyclically. Let A' = Lb∩Lc, and define B' and C' cyclically. Then A'B'C' is perspective to ABC, and the perspector is X(39704). (Angerl Montesdeoca, March 20, 2023)
See Angel Montesdeoca, Una propiedad del centro del triángulo X(39704).

X(39980) = ISOGONAL CONJUGATE OF X(3731)

Barycentrics a*(3*a + 3*b - c)*(3*a - b + 3*c) : :

Let DEF be the intouch triangle and I_aI_bI_c the excentral-triangle. Let D' be the trisector nearest I_a of segment DI_a. Let L_a be the radical axis of (DEF) and (BCD'), and define L_b and L_c cyclically. The triangle A'B'C' formed by the lines L_a, L_b, L_c is perspective to ABC, and the perspector is X(39980). See also X(34578). (Angel Montesdeoca, December 31, 2020)
See HG301220

X(40257) = TETRAHEDRAL PROJECTION OF ABC TO 2nd CIRCUMPERP TRIANGLE

Barycentrics a*(a^6-2*(b+c)*a^5-(b^2-4*b*c+c^2)*a^4+(b+c)*(4*b^2-7*b*c+4*c^2)*a^3-(b^2+5*b*c+c^2)*(b-c)^2*a^2-(b^2-c^2)*(b-c)*(2*b^2-3*b*c+2*c^2)*a+(b^2-c^2)*(b-c)*(b^3+c^3)) : :

In the plane of a triangle ABC, let
O = circumcenter;
A' = reflection of A in O, and define B'; and C' cyclically;
H = orthocenter;
A''B''C'' = circumcevian triangle of H wrt A''B''C'';
A_b = AC∩B'A'', and define B_c and C_a cyclically;
A_c = AB∩C'A'', and define B_a and C_b cyclically;
A_o = circumcenter of AA_bA_c, and define B_o and C_ocyclically;
σ = the affine transformation that maps ABC onto A_oB_oC_ocyclically;
I = incenter;
Then σ(I) = X(40257). Moreover, if X = x : y : z, then
σ(X) = (-a+b+c)^2 (a^4-2 a^2 (b-c)^2+(b-c)^2 (b^2-b c+c^2)) x-a^2 (a-c) c (a-b+c)^2 y-a^2 (a-b) b (a+b-c)^2 z : : . See X(40257). (Angel Montesdeoca, June 28, 2022)
See Angel Montesdeoca, HG270622.

X(41005) = CROSSPOINT OF X(69) AND X(264)

Barycentrics (a^2 - b^2 - c^2)*(a^4*b^2 - 2*a^2*b^4 + b^6 + a^4*c^2 + 4*a^2*b^2*c^2 - b^4*c^2 - 2*a^2*c^4 - b^2*c^4 + c^6) : :

In the plane of a triangle ABC, let
DEF = anticevian triangle of X(3);
A_b = orthogonal projection of D on AB;
A_c = orthogonal projection of D on AC;
D_b = BD∩A_bA_c;
D_c = CD∩A_bA_c;
A'=BD_c∩CD_b, and define B' and C' cyclically.
The lines AA', BB', CC' concur in X(41005). See figure (Angel Montesdeoca, January 2, 2022) and general case in HG291221.

X(41300) = MONTESDEOCA HOMOTHETIC CENTER

Barycentrics (b^2-c^2)(3 a^4 - 3 a^2 (b^2 + c^2)- 2 b^2 c^2 ) : :

X(41300) is the homothetic center of the triangles A_bB_cC_a and A_cB_aC_b defined at PU(196) in Bicentric Pairs of Points.
(See Un par bicéntrico sobre la tripolar de X(290))

X(41679) = TRILINEAR POLE OF THE BPC-LINE OF X(26)

Barycentrics a^2*(a^4-2*(b^2+c^2)*a^2+b^4+c^4)*(a^2-c^2)*(a^2-b^2+c^2)*(a^2-b^2)*(a^2+b^2-c^2) : :

In the plane of a triangle ABC, let
MaMbMc = medial triangle
DEF = orthic triangle
H = X(4) = orthocenter
Hb = line through H perpendicular to BH
Hab = Hb∩BC
Hc = line through H perpendicular to CH
Hac = Hc∩BC
Pb = line through Hac parallel to DMb
Pc = line through Hab parallel to DMc
Ab = Pb∩AC; define Bc and Ca cyclically
Ac = Pc∩AB; define Ba and Cb cyclically
A'B'C' = triangle with edges AbAc, BcBa, CaCb
Then X(41679) is the unique finite fixed point of the affine transformation that maps ABC onto A'B'C'. The triangles ABC and A'B'C' are perspective, and their perspector is X(51776). The perspector of the perspeconic of ABC and A'B'C' is X(51777), and the center of that conic is X(51778). (Angel Montesdeoca, September 12, 2022)
See HG090922

X(42789) = 1ST MONTESDEOCA-EULER POINT

Barycentrics a*(b*c*SB*SC + Sqrt[2]*a*SA*Sqrt[SA*SB*SC]) : :

The mapping X(3) + t X(4) ↦ 8 t SA SB SC X(3) + a^2 b^2 c^2 X(4) is an involution of the Euler line. The mapping includes {X(3), X(4)} and {X(1113), X(1114)} as involutory pairs. The fixed points of the mapping are
(a (b c SBSC + Sqrt[2] a SA Sqrt[SA SB SC]) : : and (a (b c SB SC - Sqrt[2] a SA Sqrt[SA SB SC]) : : ,
These points are real if and only if the reference triangle ABC is acute. (Angel Montesdeoca, April 26, 2021.) The points are here named the 1st Montesdeoca-Euler point and 2nd Montesdeoca-Euler point, respectively.
X(42789) lies on the curves K114 and Q023 and this line: {2, 3}
For a figure showing X(42789) and X(42790) on the curves K114 and Q023, see Euler line, points, and curves. (Angel Montesdeoca, April 27, 2021).

Una involución sobre la recta de Euler

X(42790) = 2nd MONTESDEOCA-EULER POINT

Barycentrics a*(b*c*SB*SC - Sqrt[2]*a*SA*Sqrt[SA*SB*SC]) : :

Una involución sobre la recta de Euler

X(43448) = X(2)X(99)∩X(4)X(6)

Barycentrics a^4 - 2*a^2*b^2 - 3*b^4 - 2*a^2*c^2 + 6*b^2*c^2 - 3*c^4 : :

In the plane of a triangle ABC, let
G = X(2) = centroid
H = X(4) = orthocenter
Ab = point of intersection, other than B, of BH and circle (G,B,C), and define Bc and Ca cyclically
Ac point of intersection, other than C, of CH and circle (G,B,C), and define Ba and Cb cyclically
A' = BcBa∩CaCb, and define B' and C' cyclically
Then A'B'C' is homothetic to the orthic triangle of ABC, and X(43448) is the homothetic center. (Angel Montesdeoca, January 21, 2023)

X(43448) como centro de homotecia

X(43919) = REFLECTION OF X(3) IN X(1624)

Barycentrics a^2*(a^12*b^2 - 5*a^10*b^4 + 10*a^8*b^6 - 10*a^6*b^8 + 5*a^4*b^10 - a^2*b^12 + a^12*c^2 - 6*a^10*b^2*c^2 + 3*a^8*b^4*c^2 + 14*a^6*b^6*c^2 - 15*a^4*b^8*c^2 + 3*b^12*c^2 - 5*a^10*c^4 + 3*a^8*b^2*c^4 - 20*a^6*b^4*c^4 + 10*a^4*b^6*c^4 + 21*a^2*b^8*c^4 - 9*b^10*c^4 + 10*a^8*c^6 + 14*a^6*b^2*c^6 + 10*a^4*b^4*c^6 - 40*a^2*b^6*c^6 + 6*b^8*c^6 - 10*a^6*c^8 - 15*a^4*b^2*c^8 + 21*a^2*b^4*c^8 + 6*b^6*c^8 + 5*a^4*c^10 - 9*b^4*c^10 - a^2*c^12 + 3*b^2*c^12) : :

In the plane of a triangle ABC, let
T_aT_bT_c = tangential triangle;
O_a = line through X(3) parallel to BC, which cuts T_bT_c at U_a;
L_a = line through U_a perpendicular to BC, and define L_b and L_c cyclically;
A' = L_b∩L_c, and define B' and C' cyclically.
Then A'B'C' is perspective to T_aT_bT_c, and the perspector is X(43919). See X(43919). (Angel Montesedeoca, March 15, 2022)

X(44860) = (name pending)

Barycentrics 2 a^18 (b^2-c^2)^2+b^4 c^4 (b^2-c^2)^6 (b^2+c^2)+a^16 (-9 b^6+13 b^4 c^2+13 b^2 c^4-9 c^6)-a^2 b^2 c^2 (b^2-c^2)^4 (2 b^8+3 b^6 c^2-6 b^4 c^4+3 b^2 c^6+2 c^8)+a^14 (14 b^8-5 b^6 c^2-44 b^4 c^4-5 b^2 c^6+14 c^8)+a^4 (b^2-c^2)^4 (b^10+12 b^8 c^2+9 b^6 c^4+9 b^4 c^6+12 b^2 c^8+c^10)-a^12 (5 b^10+30 b^8 c^2-47 b^6 c^4-47 b^4 c^6+30 b^2 c^8+5 c^10)+a^8 (b^2-c^2)^2 (13 b^10+3 b^8 c^2-19 b^6 c^4-19 b^4 c^6+3 b^2 c^8+13 c^10)-a^6 (b^2-c^2)^2 (6 b^12+17 b^10 c^2-18 b^8 c^4+14 b^6 c^6-18 b^4 c^8+17 b^2 c^10+6 c^12)-a^10 (10 b^12-48 b^10 c^2+25 b^8 c^4+30 b^6 c^6+25 b^4 c^8-48 b^2 c^10+10 c^12) : :

In a triangle ABC, let
La = perpendicular bisector of side BC, and define Lb and LCc cyclically;
Ab = La∩AC, Ac = La∩AB, Bc = Lb∩BA, Ba = Lb∩BC, Ca = Lc∩CB, Cb = Lc∩CA,
Then the circumcircles of ABC, T1 = AbBcCa and T2 = AcBaCb are coaxial
The radical axis common to these circles is the trlinear polar of X(44828).

Let Q be a point on the Euler line of ABC and Q1, Q2 same to Q points of T1, T2, resp.-
Let Mq be the midpoint of Q1Q2 (it is a center of ABC).
Which point is its intersection with the Euler line of ABC?

See Antreas Hatzipolakis and Angel Montesdeoca, Euclid 2416 .
Figure in: https://amontes.webs.ull.es/imagenes/euclid2401.png

X(44915) = X(2)X(6)∩X(67)X(42007)

Barycentrics (2 a^2-b^2-c^2)(2 a^6-2 a^4 (b^2+c^2)-a^2 (b^4-4 b^2 c^2+c^4)+3 (b^2-c^2)^2 (b^2+c^2)) : :

Un centro del triángulo sobre GK
Given a triangle ABC, with centroid G = X(2), circumcenter O = X(3) and symmedian K = X(6), let A'B'C 'and A"B"C" be the cevian and anticevian triangles of X(524) (on the infinity line), isogonal conjugate of Parry's point (A", B", C" are the midpoints of the segments AA ', BB', CC ', respectively). Let DEF be the pedal triangle of O with respect to A"B"C".
Let AaAbAc, BbBcBa, and CcCaCb be the cevian triangles of D, E, and F, respectively. AaBbCc is the trilinear polar of X(524).
The lines AbAc, BcBa, CaCb concur in X(8030) - 2 X(599) (on the line GK), reflection of the centroid of the triangle A'B'C', X(8030), in the isotomic conjugate, X(599), of the perspector, X(598), of the inscribed ellipse of Lemoine.

X(44935) = X(4)X(18853)∩X(30)X(568)

Barycentrics 2 a^10-7 a^8 (b^2+c^2)+2 a^6 (5 b^4+18 b^2 c^2+5 c^4)-8 a^4 (b^6+b^4 c^2+b^2 c^4+c^6)+4 a^2 (b^2-c^2)^2 (b^4-4 b^2 c^2+c^4)-(b^2-c^2)^4 (b^2+c^2) : :

Reflexión de X(7667) en X(16657) como centro ortológico
See Angel Montesdeoca, Euclid 2601 .
Let ABC be a triangle and DEF the orthic triangle of ABC. The circle through the midpoints of EF, EC, FB, cuts again AC and AB in Ab and Ac, respectively, and points Bc, Ca, Ba, Cb are defined similarly. ABC and the triangle A'B'C' bounded by AbAc, BcBa, CaCb are orthologic. The orthology center of ABC and A'B'C' is the centroid and the orthology center of A'B'C' and ABC is the reflection of X(7667) in X(16657) ( i.e. , X(44935)=X(7667) - 2 X(16657)).

X(45131) = X(1)X(318)∩X(72)X(519)

Barycentrics a^6 (b+c)-2 a^5 b c-a^4 (b+c) (2 b^2-3 b c+2 c^2)+a^2 (b-c)^2 (b+c) (b^2+c^2)+2 a b c (b^2-c^2)^2-b c(b-c)^2 (b+c)^3 : :

See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 2662.

La is the line joining the midpoints of AH and BC.
The reflections of La, Lb, Lc in AI, BI, CI, resp. bound a triangle A*B*C*.
ABC and A*B*C* are parallelogic.
Parallelogic center (ABC, A*B*C*) = H.
Parallelogic center (A*B*C*, ABC) = X(45131).

X(45801) = X(6)X(523)∩X(141)X(525)

Barycentrics (b^2 - c^2)*(-a^6 + a^4*b^2 - a^2*b^4 + b^6 + a^4*c^2 + a^2*b^2*c^2 - b^4*c^2 - a^2*c^4 - b^2*c^4 + c^6) : :

In the plane of a triangle ABC, let
M = X(115), the center of the Kiepert hyperbola;
DEF = tangential triangle of the Kiepert hyperbola;
D' = reflection of D in BC, and define B' and F' cyclically; A' = AM∩E'F', and define B' and C' cyclically. The lines A'D', B'E', C'F' concur in X(45801) . (Angel Montesdeoca, August 17, 2022)
See Angel Montesdeoca, HG170822.

X(45976) = EULER LINE INTERCEPT OF X(1)X(6797)

Barycentrics a (a^6-a^5 (b+c)+a^4 (-2 b^2+3 b c-2 c^2)+2 a^3 (b^3+c^3)+a^2 (b^4-5 b^3 c+2 b^2 c^2-5 b c^3+c^4)-a (b^5-b^4 c-b c^4+c^5)+2 b c (b^2-c^2)^2) : :

Consider a point Q on the Neuberg cubic and La, Lb, Lc the concurrent Euler lines of QBC, QCA, QAB, resp.

Is there a point P on the Euler line of ABC such that the perpendiculars to same to P points Pa, Pb, Pc of La, Lb, Lc, resp. are concurrent?

For examp;e Q = I
Is there a point P on the Euler line of ABC such that the perpendiculars to Pa, Pb, Pc = same to P points on the Euler lines of IBC, ICA, IAB, resp., are concurrent?

If Q = I, then there is a point P= EULER LINE INTERCEPT OF X(1)X(6797) on the Euler line of ABC such that the perpendiculars to Pa, Pb, Pc = same to P points on the Euler lines of IBC, ICA, IAB, resp., are concurrent at Z= X(1)X(1389)∩X(4)X(496)=X(45977).

See Antreas Hatzipolakis and Angel Montesdeoca, Euclid 3114.
Figure

X(45977) = X(1)X(1389)∩X(4)X(496)

Barycentrics a*(a^6-(b+c)*a^5-(2*b-c)*(b-2*c)*a^4+2*(b+c)*(b^2-3*b*c+c^2)*a^3+(b-c)^2*(b^2-4*b*c+c^2)*a^2-(b^2-c^2)*(b-c)*(b^2-4*b*c+c^2)*a+(b^2-c^2)^2*b*c) : :

Consider a point Q on the Neuberg cubic and La, Lb, Lc the concurrent Euler lines of QBC, QCA, QAB, resp.

Is there a point P on the Euler line of ABC such that the perpendiculars to same to P points Pa, Pb, Pc of La, Lb, Lc, resp. are concurrent?

For examp;e Q = I
Is there a point P on the Euler line of ABC such that the perpendiculars to Pa, Pb, Pc = same to P points on the Euler lines of IBC, ICA, IAB, resp., are concurrent?

If Q = I, then there is a point P= EULER LINE INTERCEPT OF X(1)X(6797)=X(45976) on the Euler line of ABC such that the perpendiculars to Pa, Pb, Pc = same to P points on the Euler lines of IBC, ICA, IAB, resp., are concurrent at Z= X(1)X(1389)∩X(4)X(496).

See Antreas Hatzipolakis and Angel Montesdeoca, Euclid 3114.
Figure

X(46034) = X(4)X(6)∩X(20)X(1078)

Barycentrics a^8+8 a^6 (b^2+c^2)-2 a^4 (b^2+c^2)^2-8 a^2 (b^2-c^2)^2 (b^2+c^2)+(b^2-c^2)^2 (b^4-10 b^2 c^2+c^4) : :

See Angel Montesdeoca, euclid 3223.

Let G_A be the antipode of X(2) in the A-McCay circle, and define G_B and G_C cyclically; then G_AG_BG_C is homothetic to the McCay triangle at X(2). Let L_A be the line tangent to the A-McCay circle at G_A, and define L_B and L_C cyclically. Let A" = L_B∩L_C, and define B" and C" cyclically. Then A"B"C" is the antipedal triangle of X(2) with respect to G_AG_BG_C, and A"B"C" is inversely similar to ABC and homothetic to the 1st Brocard triangle. Also, X(7620) = X(2)-of-A"B"C" = X(376)-of G_AG_BG_C. (Randy Hutson, May 27, 2015)

Then ABC and A"B"C" are orthologic (Theorem 26 (Mihai Miculiţa)).

The orthologic center of ABC and A"B"C" (on the circumcircle) is X(98).
The orthologic center of A"B"C" and ABC (on the circumcircle of A"B"C") is :

X(4)X(6)∩X(20)X(1078)

X(46354) = X(3)X(15474∩X(942)X(39267)

Barycentrics (a - b - c)*(a^3 + a^2*b + a*b^2 + b^3 - a^2*c - 2*a*b*c - b^2*c - a*c^2 - b*c^2 + c^3)^2*(a^3 - a^2*b - a*b^2 + b^3 + a^2*c - 2*a*b*c - b^2*c + a*c^2 - b*c^2 + c^3)^2 : :

See Angel Montesdeoca, HG121221.
Dado un triángulo ABC, sean DEF el triángulo de contacto interior y P un punto no situado sobre la circunferencia inscrita, Γ. Sean A', B', C' las proyecciones de D, E,F desde P sobre Γ.
Las rectas AA', BB', CC' son concurrentes.
Para P=X(3), el punto de concurrencia es X(46354).

X(46355) = X(1)X(280)∩X(4)X(189)

Barycentrics (-a+b+c)*(a^3+(b-c)*a^2-(b-c)^2*a-(b+c)*(b^2-c^2))^2*(a^3-(b-c)*a^2-(b-c)^2*a+(b+c)*(b^2-c^2))^2 : :

See Angel Montesdeoca, HG121221.
Dado un triángulo ABC, sean DEF el triángulo de contacto interior y P un punto no situado sobre la circunferencia inscrita, Γ. Sean A', B', C' las proyecciones de D, E,F desde P sobre Γ.
Las rectas AA', BB', CC' son concurrentes.
Para P=X(4), el punto de concurrencia es X(46355).

X(46356) = X(1)X(44301)∩X(8)X(8051)

Barycentrics (a + b - c)*(a - b + c)*(a^2 + 2*a*b + b^2 - 6*a*c + 2*b*c + c^2)^2*(a^2 - 6*a*b + b^2 + 2*a*c + 2*b*c + c^2)^2 : :

See Angel Montesdeoca, HG121221.
Dado un triángulo ABC, sean DEF el triángulo de contacto interior y P un punto no situado sobre la circunferencia inscrita, Γ. Sean A', B', C' las proyecciones de D, E,F desde P sobre Γ.
Las rectas AA', BB', CC' son concurrentes.
Para P=X(8), el punto de concurrencia es X(46356).

X(46362) = X(1)X(4014)∩X(40)X(550)

Barycentrics a (a ^ 5 (b + c) +5 a ^ 3 bc (b + c) +5 bc (b ^ 2-c ^ 2) ^ 2 + 2 a ^ 4 (b ^ 2-8 b c + c ^ 2) + a ^ 2 (-2 b ^ 4 + 11 b ^ 3 c-10 b ^ 2 c ^ 2 + 11 bc ^ 3-2 c ^ 4) -a (b ^ 5 + 6 b ^ 4 c-3 b ^ 3 c ^ 2-3 b ^ 2 c ^ 3 + 6 bc ^ 4 + c ^ 5)) : :

Given a triangle ABC with the length of its sides a = |BC|, b = |CA|, c = |AB|, let Γa, Γb, Γc be the circles centered in the incenter, X(1), of radii a, b, c, respectively.
A1 is the pole of BC with respect to Γa and A2 is the point of intersection of the polar of B with respect to Γc with the polar of C with respect to Γb. Points B1, B2 and C1, C2 are defined cyclically.

The six points A1, A2, B1, B2, C1, C2 lie on a same conic, which passes through the centers of the triangle X(1), X(3), X(20), X(10454) and whose tangent at X(1) is X(1)X(256).

The center of this conic is X(46362).

See Angel Montesdeoca, Euclid 3557.

X(46363) = X(3)X(6)∩X(51)X(3089)

Barycentrics a^2 (-a^12 (b^2+c^2)-2 a^2 (b^2-c^2)^4 (2 b^4+b^2 c^2+2 c^4)+a^10 (4 b^4+6 b^2 c^2+4 c^4)+(b^2-c^2)^4 (b^6+b^4 c^2+b^2 c^4+c^6)+a^4 (b^2-c^2)^2 (5 b^6-17 b^4 c^2-17 b^2 c^4+5 c^6)-a^8 (5 b^6+17 b^4 c^2+17 b^2 c^4+5 c^6)+4 a^6 (7 b^6 c^2+2 b^4 c^4+7 b^2 c^6)) : :

Let ABC be a triangle and A'B'C' the pedal triangle of H
Denote
Ab,Ac = the orthogonal projections of A' on AB, AC resp.
M1 = the midpoint of AbAc
Ma = the midpoint of AA'

Similarly Mb, Mc and M2, M3.
MaMbMc, M1M2M3 are perspective at the center of the Taylor circle X(389) of ABC
Which point is the perspector X(389) wrt triangles MaMbMc, M1M2M3?
The perspector X(389) wrt triangle M1M2M3 is X(46363).

See Antreas Hatzipolakis and Angel Montesdeoca, Euclid 3570.

X(46364) = (name pending)

Barycentrics 4 a^34-25 a^32 (b^2+c^2)-(b^2-c^2)^12 (b^2+c^2)^3 (2 b^4+5 b^2 c^2+2 c^4)+a^30 (50 b^4+204 b^2 c^2+50 c^4)-a^28 (b^6+508 b^4 c^2+508 b^2 c^4+c^6)+a^26 (-116 b^8+82 b^6 c^2+2092 b^4 c^4+82 b^2 c^6-116 c^8)+2 a^2 (b^2-c^2)^10 (b^2+c^2)^2 (6 b^8-19 b^6 c^2-70 b^4 c^4-19 b^2 c^6+6 c^8)-2 a^18 (b^2-c^2)^4 (44 b^8-2315 b^6 c^2-7550 b^4 c^4-2315 b^2 c^6+44 c^8)+a^24 (65 b^10+1410 b^8 c^2-2691 b^6 c^4-2691 b^4 c^6+1410 b^2 c^8+65 c^10)+2 a^22 (111 b^12-564 b^10 c^2-2075 b^8 c^4+5888 b^6 c^6-2075 b^4 c^8-564 b^2 c^10+111 c^12)-a^4 (b^2-c^2)^8 (27 b^14-168 b^12 c^2+42 b^10 c^4+1507 b^8 c^6+1507 b^6 c^8+42 b^4 c^10-168 b^2 c^12+27 c^14)-a^20 (297 b^14+2824 b^12 c^2-13546 b^10 c^4+10681 b^8 c^6+10681 b^6 c^8-13546 b^4 c^10+2824 b^2 c^12+297 c^14)+a^16 (b^2-c^2)^2 (407 b^14+226 b^12 c^2-21124 b^10 c^4+17291 b^8 c^6+17291 b^6 c^8-21124 b^4 c^10+226 b^2 c^12+407 c^14)+2 a^6 (b^2-c^2)^6 (17 b^16+30 b^14 c^2+1076 b^12 c^4-430 b^10 c^6-3434 b^8 c^8-430 b^6 c^10+1076 b^4 c^12+30 b^2 c^14+17 c^16)+2 a^10 (b^2-c^2)^4 (62 b^16+439 b^14 c^2-5030 b^12 c^4-1079 b^10 c^6+15312 b^8 c^8-1079 b^6 c^10-5030 b^4 c^12+439 b^2 c^14+62 c^16)-2 a^14 (b^2-c^2)^2 (121 b^16+2364 b^14 c^2-7492 b^12 c^4-9756 b^10 c^6+26454 b^8 c^8-9756 b^6 c^10-7492 b^4 c^12+2364 b^2 c^14+121 c^16)-a^12 (b^2-c^2)^2 (59 b^18-2894 b^16 c^2-5351 b^14 c^4+42427 b^12 c^6-33217 b^10 c^8-33217 b^8 c^10+42427 b^6 c^12-5351 b^4 c^14-2894 b^2 c^16+59 c^18)-a^8 (b^2-c^2)^4 (61 b^18+1118 b^16 c^2-1185 b^14 c^4-12083 b^12 c^6+13113 b^10 c^8+13113 b^8 c^10-12083 b^6 c^12-1185 b^4 c^14+1118 b^2 c^16+61 c^18) : :

Let ABC be a triangle and A'B'C' the pedal triangle of H
Denote
Ab,Ac = the orthogonal projections of A' on AB, AC resp.
M1 = the midpoint of AbAc
Ma = the midpoint of AA'

Similarly Mb, Mc and M2, M3.
MaMbMc, M1M2M3 are perspective at the center of the Taylor circle X(389) of ABC
Which point is the perspector X(389) wrt triangles MaMbMc, M1M2M3?
The perspector X(389) wrt triangle M1M2M3 is X(46364).

See Antreas Hatzipolakis and Angel Montesdeoca, Euclid 3560.

X(46365) = (name pending)

Barycentrics a/(3*a^10*b^2 - 2*a^9*b^3 - 14*a^8*b^4 + 6*a^7*b^5 + 24*a^6*b^6 - 6*a^5*b^7 - 18*a^4*b^8 + 2*a^3*b^9 + 5*a^2*b^10 + 10*a^10*b*c + 2*a^9*b^2*c - 10*a^8*b^3*c + 2*a^7*b^4*c - 10*a^6*b^5*c + 10*a^5*b^6*c + 26*a^4*b^7*c - 18*a^3*b^8*c - 16*a^2*b^9*c + 4*a*b^10*c + 3*a^10*c^2 + 2*a^9*b*c^2 + 16*a^8*b^2*c^2 - 8*a^7*b^3*c^2 - 21*a^6*b^4*c^2 + 20*a^5*b^5*c^2 + 5*a^4*b^6*c^2 - 2*a^2*b^8*c^2 - 14*a*b^9*c^2 - b^10*c^2 - 2*a^9*c^3 - 10*a^8*b*c^3 - 8*a^7*b^2*c^3 + 30*a^6*b^3*c^3 - 24*a^5*b^4*c^3 - 26*a^4*b^5*c^3 + 40*a^3*b^6*c^3 + 10*a^2*b^7*c^3 - 6*a*b^8*c^3 - 4*b^9*c^3 - 14*a^8*c^4 + 2*a^7*b*c^4 - 21*a^6*b^2*c^4 - 24*a^5*b^3*c^4 + 26*a^4*b^4*c^4 - 24*a^3*b^5*c^4 - 3*a^2*b^6*c^4 + 30*a*b^7*c^4 - 4*b^8*c^4 + 6*a^7*c^5 - 10*a^6*b*c^5 + 20*a^5*b^2*c^5 - 26*a^4*b^3*c^5 - 24*a^3*b^4*c^5 + 12*a^2*b^5*c^5 - 14*a*b^6*c^5 + 4*b^7*c^5 + 24*a^6*c^6 + 10*a^5*b*c^6 + 5*a^4*b^2*c^6 + 40*a^3*b^3*c^6 - 3*a^2*b^4*c^6 - 14*a*b^5*c^6 + 10*b^6*c^6 - 6*a^5*c^7 + 26*a^4*b*c^7 + 10*a^2*b^3*c^7 + 30*a*b^4*c^7 + 4*b^5*c^7 - 18*a^4*c^8 - 18*a^3*b*c^8 - 2*a^2*b^2*c^8 - 6*a*b^3*c^8 - 4*b^4*c^8 + 2*a^3*c^9 - 16*a^2*b*c^9 - 14*a*b^2*c^9 - 4*b^3*c^9 + 5*a^2*c^10 + 4*a*b*c^10 - b^2*c^10) : :

See Angel Montesdeoca and Peter Moses, Euclid 3563.

X(46432) = X(3)X(6)∩X(115)X(393)

Barycentrics a^2 (a^6-3 a^2 b^4+2 b^6+6 a^2 b^2 c^2-2 b^4 c^2-3 a^2 c^4-2 b^2 c^4+2 c^6) : :

In the plane of a triangle ABC, let
I = X(1) = incenter
Ia, Ib, Ic = excenters
DEF = anticomplementary triangle
A1B1C1 = 1st circumperp triangle of DEF
A2B2C2 = 2nd circumperp triangle of DEF
A' = point of intersection, other than X(147), of the circles {{A1,Ib, IC}} and {{A2, X1,Ia}}, and define B' and C' cyclically.
The finite fixed point of the affine transformation from ABC to A'B'C' is X(46432). (Angel Montesdeoca, October 15, 2023).
See: El centro del triángulo X(46432) punto fijo de una afinidad

X(46864) = (name pending)

Barycentrics 1/(3 a^8-9 a^6 (b^2+c^2)+a^4 (9 b^4+b^2 c^2+9 c^4)+a^2 (-3 b^6+17 b^4 c^2+17 b^2 c^4-3 c^6)-9 b^2 c^2 (b^2-c^2)^2) : :

See Angel Montesdeoca, Euclid 4208 and HG010222.

X(46921) = (name pending)

Barycentrics 1/(5 a^8-24 a^6 (b^2+c^2)+a^4 (42 b^4+34 b^2 c^2+42 c^4)-32 a^2 (b^2-c^2)^2 (b^2+c^2)+3 (b^2-c^2)^2 (3 b^4-8 b^2 c^2+3 c^4)) : :

See Antreas Hatzipolakis and Angel Montesdeoca, euclid 4304..

Let ABC be a triangle and A'B'C' the pedal triangle of O

Denote

Ha = the orthocenter of AB'C'
Hab, Hac = the reflections of Ha in A'B', A'C', resp.

Similarly
Hbc, Hba and Hca, Hcb

The six points lie on a conic centered at O

[Angel Montesdeoca]:
Addendum:

Denote:
H'ab = HabA' /\ HacC, H'ac = HacA' /\ HabB

Similarly
H'bc, H'ba and H'ca, H'cb.

The six points H'ab, H'ac, H'bc, H'ba , H'ca, H'cb lie on a conic centered at X(5054).

Barycentric equation:

(3 a^8-20 a^6 (b^2+c^2)-44 a^2 (b^2-c^2)^2 (b^2+c^2)+(b^2-c^2)^2 (15 b^4-26 b^2 c^2+15 c^4)+a^4 (46 b^4+32 b^2 c^2+46 c^4)) x^2+(-6 a^8+44 a^6 (b^2+c^2)+68 a^2 (b^2-c^2)^2 (b^2+c^2)-2 (b^2-c^2)^2 (9 b^4-22 b^2 c^2+9 c^4)-8 a^4 (11 b^4+7 b^2 c^2+11 c^4)) y z + ... = 0.

Perspector P = (name pending) =

= 1/(5 a^8-24 a^6 (b^2+c^2)+a^4 (42 b^4+34 b^2 c^2+42 c^4)-32 a^2 (b^2-c^2)^2 (b^2+c^2)+3 (b^2-c^2)^2 (3 b^4-8 b^2 c^2+3 c^4)) : :

= lies on this line: {5054, 6748}

(6 - 9 - 13) - search numbers [2.76798777984726, 3.51561751656797, -0.0707573894921845]

X(46929) = X(4)X(3917)∩X(53)X(39662)

Barycentrics ((a^2 - b^2 - c^2)^2 - 16 b^2 c^2) /((a^2 - b^2 - c^2) (a^4 - 2 a^2 b^2 + b^4 - 2 a^2 c^2 - 6 b^2 c^2 + c^4)) : :

See Angel Montesdeoca, euclid 4323 and HG120222 .

X(47032) = EULER LINE INTERCEPT OF X(79)X(517)

Barycentrics a^7-a^6 (b+c)-a^5 (b^2-5 b c+c^2) +a^4 (b-c)^2 (b+c)-a^3 (b^2-b c+c^2) (b^2+4 b c+c^2) +a^2 (b-c)^2 (b+c)^3+a (b-c)^4 (b+c)^2 -(b-c)^4 (b+c)^3 : :

Let ABC be a triangle. Take Ba and Ca on AB and AC, respectively, on the negative side of BC (the region that does not contain A) such that BBa=CCa=BC.
Let Cab and Bac be the points where the circle (ABaCA) cuts again BCa and CBa, respectively.
The points Abc, Cba and Bca, Acb are defined cyclically.
The lines CabBac, AbcCba and BcaAcb form a triangle DEF. ABC and DEF are bilogic.
This perspector is X(5559).

The orthologic center of ABC wrt DEF is X(79) and the orthologic center of DEF wrt ABC is X(47032).

See Angel Montesdeoca, euclid 4468 and HG250222.

X(47033) = X(8)X(79)∩X(10)X(21)

Barycentrics a^4-a^3 (b+c)-(b^2-c^2)^2+a^2 b c+a (b^3+b^2 c+b c^2+c^3) : :

See Angel Montesdeoca, euclid 4482 and HG250222.

X(47034) = X(65)X(79)∩X(119)X(191)

Barycentrics a^10-a^9 (b+c)+a^8 (-3 b^2+4 b c-3 c^2)+2 a^7 (b-c)^2 (b+c)+a^6 (4 b^4-5 b^3 c+7 b^2 c^2-5 b c^3+4 c^4)+a^5 b c (3 b^3-2 b^2 c-2 b c^2+3 c^3)-a^4 (4 b^6+b^4 c^2-6 b^3 c^3+b^2 c^4+4 c^6)-a^3 (b-c)^2 (2 b^5-5 b^3 c^2-5 b^2 c^3+2 c^5)+a^2 (b^2-c^2)^2 (3 b^4-b^3 c-b c^3+3 c^4)+a (b-c)^4 (b+c)^3 (b^2-3 b c+c^2)-(b-c)^6 (b+c)^4 : :

See Angel Montesdeoca, euclid 4482 and HG250222.

X(47074) = X(2)X(187)∩X(30)X(14866)

Barycentrics 20 a^6-3 a^4 (b^2+c^2)-3 a^2 (7 b^4+2 b^2 c^2+7 c^4)+2 (b^6-6 b^4 c^2-6 b^2 c^4+c^6) : :

See Kadir Altintas and Angel Montesdeoca, euclid 4510 and HGT070322

X(47075) = X(2)X(187)∩X(22)X(2930)

Barycentrics 16 a^6+3 a^4 (b^2+c^2)-3 a^2 (5 b^4+b^2 c^2+5 c^4)-2 (b^2+c^2)^3 : :

Let G=X(2)-Centroid of ABC. A1B1C1-Circumcevian triangle of G. Gamma_A:Circle with diameter GA1. Define Gamma_B, Gamma_C cyclically. Tangent to Gamma_A at G intersects the Gamma_B and Gamma_C at Ab, Ac resp. Define Ba, Bc, Ca, Cb cyclically.
Then Ab, Ac, Ba, Bc, Ca, Cb lie on same circle Gamma.
Barycentric equation of Gamma:
(-a^8+26 a^6 (b^2+c^2)-6 a^4 (7 b^4+20 b^2 c^2+7 c^4)+(b^2+c^2)^2 (11 b^4-14 b^2 c^2+11 c^4)+a^2 (-58 b^6+78 b^4 c^2+78 b^2 c^4-58 c^6)) x^2+2 (-61 a^8+83 a^6 (b^2+c^2)+(b^2+c^2)^2 (5 b^4-26 b^2 c^2+5 c^4)+12 a^4 (7 b^4+11 b^2 c^2+7 c^4)+a^2 (-55 b^6+87 b^4 c^2+87 b^2 c^4-55 c^6)) y z+ ....
The center of Gamma is the midpoint of X(2) and circumcevian-isogonal conjugate of X(2), X(47075).

See Kadir Altintas and Angel Montesdeoca, Euclid 4511.

X(47101) = X(2)X(187)∩X(3)X(754)

Barycentrics 5 a^4-3 a^2 (b^2+c^2) -b^4-c^4 : :

Let G=X(2)-Centroid of ABC. O_A, O_B, O_C-Circumcenters of GBC, GCA, GAB resp. Gamma_A:Circle through points G, O_B, O_C. Define Gamma_B, Gamma_C cyclically. Tangent to Gamma_A at G intersects the Gamma_B and Gamma_C at Ab, Ac. Define Ba, Bc, Ca, Cb cyclically.
* Ab, Ac, Ba, Bc, Ca, Cb lie on same circle Gamma.
Barycentric equation of Gamma:
(a^2-2 (b^2+c^2)) (a^4-2 b^4+7 b^2 c^2-2 c^4+a^2 (b^2+c^2)) x^2+(17 a^6-32 a^4 (b^2+c^2)+5 (b^2-c^2)^2 (b^2+c^2)-2 a^2 (b^4+6 b^2 c^2+c^4)) y z+ ....
The center of Gamma is X(47101).

See Kadir Altintas and Angel Montesdeoca, Euclid 4525.

X(47102) = X(6)X(8354)∩X(20)X(538)

Barycentrics 9 a^4-4 a^2 (b^2+c^2)-3 b^4+2 b^2 c^2-3 c^4 : :

Let G=X(2)-Centroid of ABC. O_A, O_B, O_C-Circumcenters of GBC, GCA, GAB resp. Gamma_A:Circle through points G, O_B, O_C. Define Gamma_B, Gamma_C cyclically. Tangent to Gamma_A at G intersects the Gamma_B and Gamma_C at Ab, Ac. Define Ba, Bc, Ca, Cb cyclically.
* Ab, Ac, Ba, Bc, Ca, Cb lie on same circle Gamma.
A2=CaAc ∩ AbBa, B2=AbBa ∩ BcCb, C2=BcCb ∩ CaAc.
A3=CbAb ∩ AcBc, B3=AcBc ∩ BaCa, C3=BaCa ∩ CbAb.
The triangles A2B2C2 and A3B3C3 are bilogic, with Sondat line the line that passes through the centroid and the Schoute center.
** The perspector of A2B2C2 and A3B3C3 is X(2)
** The orthologic center of A3B3C3 wrt A2B2C2 is the center of circle (Ab, Ac, Ba, Bc, Ca, Cb): X(47101)
** The orthologic center of A2B2C2 wrt A3B3C3 is X(47102).

See Kadir Altintas See Kadir Altintas and Angel Montesdeoca, Euclid 4525.

X(47441) = MIDPOINT OF X(4) AND X(3345)

Barycentrics a^9*b + 3*a^8*b^2 - 8*a^6*b^4 - 6*a^5*b^5 + 6*a^4*b^6 + 8*a^3*b^7 - 3*a*b^9 - b^10 + a^9*c - 6*a^8*b*c + 4*a^6*b^3*c - 6*a^5*b^4*c + 8*a^4*b^5*c + 8*a^3*b^6*c - 4*a^2*b^7*c - 3*a*b^8*c - 2*b^9*c + 3*a^8*c^2 + 8*a^6*b^2*c^2 + 12*a^5*b^3*c^2 - 6*a^4*b^4*c^2 - 8*a^3*b^5*c^2 - 8*a^2*b^6*c^2 - 4*a*b^7*c^2 + 3*b^8*c^2 + 4*a^6*b*c^3 + 12*a^5*b^2*c^3 - 16*a^4*b^3*c^3 - 8*a^3*b^4*c^3 + 4*a^2*b^5*c^3 - 4*a*b^6*c^3 + 8*b^7*c^3 - 8*a^6*c^4 - 6*a^5*b*c^4 - 6*a^4*b^2*c^4 - 8*a^3*b^3*c^4 + 16*a^2*b^4*c^4 + 14*a*b^5*c^4 - 2*b^6*c^4 - 6*a^5*c^5 + 8*a^4*b*c^5 - 8*a^3*b^2*c^5 + 4*a^2*b^3*c^5 + 14*a*b^4*c^5 - 12*b^5*c^5 + 6*a^4*c^6 + 8*a^3*b*c^6 - 8*a^2*b^2*c^6 - 4*a*b^3*c^6 - 2*b^4*c^6 + 8*a^3*c^7 - 4*a^2*b*c^7 - 4*a*b^2*c^7 + 8*b^3*c^7 - 3*a*b*c^8 + 3*b^2*c^8 - 3*a*c^9 - 2*b*c^9 - c^10 : :

Dado un triángulo ABC y un punto P, no situado sobre sus lados; sean DEF el triángulo ceviano de P y D', E', F' los puntos en los que la tripolar de P cortan a los lados BC, CA, AB, respectivamente.
Sea A' la imagen de un punto mediante la transformación afín que aplica ABC en D'E'F'. Los puntos B' y C' se definen cíclicamente.
El centro ortológico de A'B'C' respecto a ABC, cuando P=X(34404) es X(47441).

See Angel Montesdeoca, HG260322.

X(47442) = MIDPOINT OF X(23) AND X(647)

Barycentrics a^2*(b - c)*(b + c)*(3*a^8 - 4*a^6*b^2 - 2*a^4*b^4 + 4*a^2*b^6 - b^8 - 4*a^6*c^2 + 9*a^4*b^2*c^2 - 3*a^2*b^4*c^2 - 2*b^6*c^2 - 2*a^4*c^4 - 3*a^2*b^2*c^4 + 4*b^4*c^4 + 4*a^2*c^6 - 2*b^2*c^6 - c^8) : :

Dados un triángulo ABC y un punto P. Sea DEF el triángulo cevianoSi ABC es un triángulo y P un punto, las rectas AP, BP y CP cortan a los lados opuestos del triángulo en A', B', C'. Al triángulo A'B'C' se le denomina triángulo ceviano de P.
Si (u:v:w) son las coordenadas baricéntricas de P, A'=(0:v:w). de P. La circunferencia (ABD) vuelve a cortar a AC en D_b y la circunferencia (ACD) vuelve a cortar a AB en D_c.
Sea d_B la tangente en B a la circunferencia que pasa por B, el punto medio de AD y el punto medio de CD_c. Así mismo, sea d_C la tangente en C a la circunferencia que pasa por C, el punto medio de AD y el punto medio de BD_b.
Consideremos los puntos A_b=AC ∩ d_B y A_c= AB ∩ d_C. Se definen cíclicamente, B₁, C₁; B_c, B_a y C_a, C_b.
Si P está sobre la hipérbola de Jerabek, los seis puntos A_b, A_c, B_c, B_a, C_a, C_b están sobre una misma cónica, con centro P_o en el eje órtico.
La recta P*Po envuelve una cónica, tangente a la recta de Euler en X(7482) y al eje órtico en X(2489). Su centro es X(47442),

See Angel Montesdeoca, HG230322.

X(47443) = X(2)X(18020)∩X(25)X(250)

Barycentrics a^2*(a - b)^3*(a + b)^3*(a - c)^3*(a + c)^3*(a^2 + b^2 - c^2)*(a^2 - b^2 + c^2) : :

Dados un triángulo ABC y un punto P. Sea DEF el triángulo cevianoSi ABC es un triángulo y P un punto, las rectas AP, BP y CP cortan a los lados opuestos del triángulo en A', B', C'. Al triángulo A'B'C' se le denomina triángulo ceviano de P.
Si (u:v:w) son las coordenadas baricéntricas de P, A'=(0:v:w). de P. La circunferencia (ABD) vuelve a cortar a AC en D_b y la circunferencia (ACD) vuelve a cortar a AB en D_c.
Sea d_B la tangente en B a la circunferencia que pasa por B, el punto medio de AD y el punto medio de CD_c. Así mismo, sea d_C la tangente en C a la circunferencia que pasa por C, el punto medio de AD y el punto medio de BD_b.
Consideremos los puntos A_b=AC ∩ d_B y A_c= AB ∩ d_C. Se definen cíclicamente, B₁, C₁; B_c, B_a y C_a, C_b.
Si P está sobre la hipérbola de Jerabek, los seis puntos A_b, A_c, B_c, B_a, C_a, C_b están sobre una misma cónica, con centro P_o en el eje órtico.
La recta P*Po envuelve una cónica, tangente a la recta de Euler en X(7482) y al eje órtico en X(2489). Su perspector es X(47443),

See Angel Montesdeoca, HG230322.

X(49304) = X(109)X(42552)∩X(1411)X(10703)

Barycentrics a (a-b) (a-c) (a^13-3 a^12 (b+c)+a^11 (b^2+13 b c+c^2)+a^10 (7 b^3-17 b^2 c-17 b c^2+7 c^3) -a^9 (11 b^4+5 b^3 c-47 b^2 c^2+5 b c^3+11 c^4) +a^8 (b^5+39 b^4 c-43 b^3 c^2-43 b^2 c^3+39 b c^4+c^5) +a^7 (14 b^6-44 b^5 c-16 b^4 c^2+93 b^3 c^3-16 b^2 c^4-44 b c^5+14 c^6) -a^6 (b-c)^2 (14 b^5+23 b^4 c-45 b^3 c^2-45 b^2 c^3+23 b c^4+14 c^5) -a^5 (b-c)^2 (b^6-36 b^5 c+3 b^4 c^2+56 b^3 c^3+3 b^2 c^4-36 b c^5+c^6) +a^4 (b-c)^2 (11 b^7-18 b^6 c-27 b^5 c^2+32 b^4 c^3+32 b^3 c^4-27 b^2 c^5-18 b c^6+11 c^7) -a^3 (b-c)^4 (7 b^6+17 b^5 c+b^4 c^2-13 b^3 c^3+b^2 c^4+17 b c^5+7 c^6) -a^2 (b-c)^4 (b^7-8 b^6 c-6 b^5 c^2+3 b^4 c^3+3 b^3 c^4-6 b^2 c^5-8 b c^6+c^7) +a (b-c)^6 (b+c)^2 (3 b^4-b^3 c+3 b^2 c^2-b c^3+3 c^4) -(b-c)^6 (b+c)^3 (b^4-b^3 c+2 b^2 c^2-b c^3+c^4)) : :

Let I be the incenter of a triangle ABC and DEF the intouch triangle. The circle centered at I passing through A intersects the ray DB in Ab and the ray DC in Ac. The circumcircles of ABAc and ACAb intersect in A', other than A. Define B' and C' cyclically.
The triangles ABC and A'B'C' are cyclologic (i.e. the circles (AB'C'), (BC'A'), and (CA'B) are concurrent in a common point). The cyclologic center of A'B'C' with respect to ABC is X(80), reflection of incenter in feuerbach point, and the cyclologic center of ABC with respect to A'B'C' is X(49304).

See Angel Montesdeoca, Euclid 5089 and HG180522

X(50443) = X(1)X(5)∩X(40)X(499)

Barycentrics (a-b-c) (a^3-3 a (b-c)^2-2 (b-c)^2 (b+c)) : :

The tangent to the incircle at a variable point T cuts the Euler circle at points T1 and T2.
** The locus of the point of intersection of the tangents from T1 and T2 (other than the tangent at T) to the incircle is a conic (C) tangent at the Feuberbach point to the incircle. Its center is X(50443).

See Angel Montesdeoca, Euclid 5183 and HGT060622

X(50444) = X(1)X(5)∩X(165)X(499)

Barycentrics (a-b-c) (a^3-5 a (b-c)^2-4 (b-c)^2 (b+c)) ::

The tangent to the incircle at a variable point T cuts the Euler circle at points T1 and T2. ** The locus of the point of intersection of the tangents from T1 and T2 (other than the tangent at T) to the incircle is a conic (C) tangent at the Feuberbach point to the incircle.
** The center of the reciprocal polar of I(r) with respect to (C) is X(50444).

See Angel Montesdeoca, Euclid 5183 and HGT060622

X(50574) = X(1)X(523)∩X(11)X(7137)

Barycentrics (b^2-c^2) (a^5-a^2 ((a+b-c) c^2+b^2 (a-b+c))-b^5-c^5+b^2 c^2 (a+b+c)) : :

In the affine homology, with center X(523) (isogonal conjugate of the focus of the Kiepert parabola), axis the line that passes through the centers of the Kiepert and Jerabek hyperbolas y ratio -3, the homologue of the incenter is X(50574) = (2r+R) X(12)-2(r+R)X(9276).

See Angel Montesdeoca, euclid 5222 and HG310521

X(50729) = X(2)X(11166)∩X(51)X(17430)

Barycentrics (a^2-2 (b^2+c^2)) (20 a^4+a^2 (b^2+c^2)-(b^2+c^2)^2) : :

Let ABC be a triangle and G, K the centroid, symmedian point, resp.
Let Gb, Kb be the orthogonal projections of B to AG and AK respectively, and Gc, Kc be the orthogonal projections of C to AG and AK respectively. Let A1 (on BC) be the intersection of GbKc and GcKb, and define B1 and C1 cyclically. Let A2 (on AH) be the intersection of GbKb and GcKc, and define B2 and C2 cyclically. The triangles A1B1C1 and A2B2C2 are orthologic; the orthologic center of A1B1C1 wrt A2B2C2 is X(50729).

See Angel Montesdeoca, Euclid 5247 and HG120622

X(50730) = X(2)X(47859)∩X(3)X(47591)

Barycentrics 20 a^10-19 a^8 (b^2+c^2)+a^6 (-145 b^4+122 b^2 c^2-145 c^4) +a^4 (149 b^6-153 b^4 c^2-153 b^2 c^4+149 c^6)+a^2 (17 b^8-100 b^6 c^2+198 b^4 c^4-100 b^2 c^6+17 c^8) -2 (b^2-c^2)^2 (11 b^6-21 b^4 c^2-21 b^2 c^4+11 c^6) : :

Let ABC be a triangle and G, K the centroid, symmedian point, resp.
Let Gb, Kb be the orthogonal projections of B to AG and AK respectively, and Gc, Kc be the orthogonal projections of C to AG and AK respectively. Let A1 (on BC) be the intersection of GbKc and GcKb, and define B1 and C1 cyclically. Let A2 (on AH) be the intersection of GbKb and GcKc, and define B2 and C2 cyclically. The triangles A1B1C1 and A2B2C2 are orthologic; the orthologic center of A2B2C2 wrt A1B1C1 is X(50730) .

See Angel Montesdeoca, Euclid 5247 and HG120622

X(51121) = X(514)X(3158)∩X(519)X(42050)

Barycentrics 3 a^4-4 b (b-c)^2 c-a^3 (b+c)+a^2 (-7 b^2+2 b c-7 c^2)+a (5 b^3-b^2 c-b c^2+5 c^3) : :

Let ABC be a triangle and P, Q two points. The perpendicular through P to BC intersects the parallel through Q to BC at A', and similarly B' and C'.
ABC and A'B'C' have the same symmedian point if Q=f(P), where f is the affine transformation with fixed points the symmedian and the points at infinity of the Kiepert hyperbola, which maps the orthocenter to the centroid.
f[X(1)] = X(51121).

See Angel Montesdeoca, Euclid 5290 and HG140319.

X(51122) = X(2)X(2418)∩X(3)X(538)

Barycentrics 3 a^4-7 a^2 (b^2+c^2)+8 b^2 c^2 : :

Let ABC be a triangle and P, Q two points. The perpendicular through P to BC intersects the parallel through Q to BC at A', and similarly B' and C'.
ABC and A'B'C' have the same symmedian point if Q=f(P), where f is the affine transformation with fixed points the symmedian and the points at infinity of the Kiepert hyperbola, which maps the orthocenter to the centroid.
f[X(3)] = X(51122).

See Angel Montesdeoca, Euclid 5290 and HG140319.

X(51123) = X(2)X(2418)∩X(5)X(7781)

Barycentrics 4 a^4-9 a^2 (b^2+c^2)+b^4+6 b^2 c^2+c^4 : :

Let ABC be a triangle and P, Q two points. The perpendicular through P to BC intersects the parallel through Q to BC at A', and similarly B' and C'.
ABC and A'B'C' have the same symmedian point if Q=f(P), where f is the affine transformation with fixed points the symmedian and the points at infinity of the Kiepert hyperbola, which maps the orthocenter to the centroid.
f[X(5)] = X(51123).

See Angel Montesdeoca, Euclid 5290 and HG140319.

X(51316) = ISOGONAL CONJUGATE OF X(37672)

Barycentrics 1/(3 a^4 - 6 a^2 b^2 + 3 b^4 - 6 a^2 c^2 + 2 b^2 c^2 + 3 c^4) : :

Given a triangle with orthocenter H=X(4), let DEF and D'E'F' be the cevian and precevian triangles of H, and D" be the symmetry of D with respect to D'. Let La be the radical axis of the circumference of diameter DD" and circumference (BCD"). The radical axes Lb and Lc are defined cyclically. The lines La,Lb,Lc form a perspective triangle with ABC with perspector W (= X(31316) ), the isogonal conjugate of X(37672) and the isotomic conjugate of X(32831).

See Angel Montesdeoca, Euclid 5336 and HG020722.

X(51341) = X(1)X(1696)∩X(6)X(7091)

Barycentrics a (a - b - c)/(a^3+a^2 (b+c)-a (b^2-6 b c+c^2)-(b+c)^3) : :

Let A' be the point in which the A-excircle is tangent to the circle that passes through vertices B and C, and define B' and C' cyclically. The lines AA', BB', CC' concur in X(5423). (Peter Moses, December 10, 2015). The tangents at A', B', C' to the corresponding excircle form a perspective triangle with ABC and the perspector is Z= X(51341)

See Angel Montesdeoca, Euclid 5356 and HG030722.

X(51342) = X(3)X(133)∩X(4)X(64)

Barycentrics (a^4-(b^2-c^2)^2) (a^12+a^10 (b^2+c^2)+18 a^6 (b^2-c^2)^2 (b^2+c^2)+2 (b^2-c^2)^4 (b^4+4 b^2 c^2+c^4)-7 a^4 (b^2-c^2)^2 (b^4+6 b^2 c^2+c^4)-4 a^8 (3 b^4-5 b^2 c^2+3 c^4)-a^2 (b^2-c^2)^2 (3 b^6-19 b^4 c^2-19 b^2 c^4+3 c^6)) : :

Let DEF be the Euler triangle of ABC. Let La be the perpendicular to OD through D, and define Lb, Lc, cyclically. Let A'B'C' be the triangle formed by the lines La, Lb, Lc. The center of the elipse, imagen of circumcircle for the affine transformation that maps the reference triangle ABC onto A'B'C', is X(51342).

See Angel Montesdeoca, Euclid 5364 and HG040722.

X(51513) = X(378)X(1116)∩X(460)X(512)

Barycentrics (b^2-c^2) (a^6 (b^2+c^2)-a^4 (b^2-c^2)^2-a^2 (b^2-c ^2)^2(b^2+c^2)+(b^2-c^2)^4) : :

F₁ and F₂ are a bicentric pair given by f(a,b,c)= b^2 (b^2-c^2) (a^2+b^2-c^2) (a^2-b^2+c^2)^2.

Bicentric sum F₁⊕F₂ is X(115).
X(115) is on the line F₁F₂. Therefore, the harmonic conjugate W of X(115), with respect to F₁ and F₂ is X(51513) (C.Kimberling, Triangle Centers and Central Triangles, Congressus Numerantium, 129 (1998). Exercise no6, p.46):
See Angel Montesdeoca, Euclid 5464 and HG070822.

X(51684) = (name pending)

Barycentrics 6 a+Sqrt(6) Sqrt(a^2+b^2+c^2) : :

See Angel Montesdeoca, Euclid 5486. HG300822

X(51685) = X(3624)X(4657)∩X(3966)X(4034)

Barycentrics (a-b-c)/(a^2+2 a b-b^2+2 a c-b c-c^2) : :

See Angel Montesdeoca, Euclid 5486. HG300822

X(51686) = X(1)X(25)∩X(28)X(86)

Barycentrics a^2/((-a^2+b^2+c^2) (a^2+b^2+2 b c+c^2)) : :

See Angel Montesdeoca, Euclid 5386. HG300822

X(51776) = X(2)X(11610)∩X(69)X(248)

Barycentrics a^2*(a^2 - b^2 - c^2)*(a^4 + b^4 - a^2*c^2 - b^2*c^2)*(a^4 - 2*a^2*b^2 + b^4 - 2*a^2*c^2 + c^4)*(a^4 - a^2*b^2 - b^2*c^2 + c^4)::

X(51776) is the perspector of ABC and the triangle A'B'C' constructed at X(41679).
See HG090922.

X(51777) = (name pending)

Barycentrics a^2 (a^4+b^4+c^4-2 a^2 (b^2+c^2)) /(a^16-5 a^14 (b^2+c^2)-(b^2-c^2)^4 (b^4+c^4)^2-a^4 (b^4-c^4)^2 (2 b^4-b^2 c^2+2 c^4)+a^12 (10 b^4+13 b^2 c^2+10 c^4)-a^10 (11 b^6+13 b^4 c^2+13 b^2 c^4+11 c^6)+a^8 (8 b^8+6 b^6 c^2+4 b^4 c^4+6 b^2 c^6+8 c^8)+a^6 (-3 b^10+b^8 c^2-2 b^6 c^4-2 b^4 c^6+b^2 c^8-3 c^10)+a^2 (b^2-c^2)^2 (3 b^10-b^8 c^2+2 b^6 c^4+2 b^4 c^6-b^2 c^8+3 c^10))

X(51777) is the perspector of ABC and the triangle A'B'C' constructed at X(41679).
See HG090922.

X(51778) = (name pending)

Barycentrics a^2 (a^22-7 a^20 (b^2+c^2)+a^18 (23 b^4+38 b^2 c^2+23 c^4)-(b^2-c^2)^4 (b^4+c^4)^2 (b^6+c^6)-a^16 (49 b^6+92 b^4 c^2+92 b^2 c^4+49 c^6)+a^14 (78 b^8+136 b^6 c^2+155 b^4 c^4+136 b^2 c^6+78 c^8)-2 a^12 (49 b^10+69 b^8 c^2+76 b^6 c^4+76 b^4 c^6+69 b^2 c^8+49 c^10)+a^10 (98 b^12+92 b^10 c^2+99 b^8 c^4+94 b^6 c^6+99 b^4 c^8+92 b^2 c^10+98 c^12)-2 a^8 (39 b^14+12 b^12 c^2+22 b^10 c^4+19 b^8 c^6+19 b^6 c^8+22 b^4 c^10+12 b^2 c^12+39 c^14)+a^2 (b^2-c^2)^2 (7 b^16-4 b^14 c^2+7 b^12 c^4+7 b^4 c^12-4 b^2 c^14+7 c^16)+a^6 (49 b^16-24 b^14 c^2+21 b^12 c^4+4 b^10 c^6+16 b^8 c^8+4 b^6 c^10+21 b^4 c^12-24 b^2 c^14+49 c^16)-a^4 (23 b^18-33 b^16 c^2+24 b^14 c^4-12 b^12 c^6+6 b^10 c^8+6 b^8 c^10-12 b^6 c^12+24 b^4 c^14-33 b^2 c^16+23 c^18)) : :

X(51778) is the perspector of ABC and the triangle A'B'C' constructed at X(41679).
See HG090922.

X(52014) = X(3)X(51)∩X(4)X(1609)

Barycentrics a^2 (a^6-3 a^4 (b^2+c^2)+3 a^2 (b^2-c^2)^2-(b^2-c^2)^2 (b^2+c^2)) (a^8-4 a^6 (b^2+c^2)+a^4 (6 b^4+8 b^2 c^2+6 c^4)-4 a^2 (b^6+c^6)+(b^2-c^2)^4) : :

See Angel Montesdeoca, Euclid 5508 and X(1609) punto fijo de una transformación afín.

X(52457) = X(2)X(7)∩X(3)X(1633)

Barycentrics (b+c-a) (a^4+(b-c)^4-2 a^2 (b^2+c^2)) : :

See Angel Montesdeoca, Euclid 5554.
See HG1122

X(52470) = X(3)X(6524)∩X(25)X(6526)

Barycentrics a^2 (a^8-4 a^6 (b^2+c^2)+a^4 (6 b^4+11 b^2 c^2+6 c^4)-2 a^2 (2 b^6+5 b^4 c^2+5 b^2 c^4+2 c^6)+b^8+3 b^6 c^2+8 b^4 c^4+3 b^2 c^6+c^8)/(b^2+c^2-a^2)^3 : :

See Angel Montesdeoca, Euclid 5559.
See HG181122.

X(52495) = X(1)X(210)∩X(58)X(1100)

Barycentrics a (4 a^3+13 a^2 (b+c)+2 a (6 b^2+11 b c+6 c^2)+3 (b+c)^3) : :

See Angel Montesdeoca, Euclid 5560.
See HG211122.

X(52521) = X(191)X(1050)∩X(21061)X(21936)

Barycentrics a^3 (b+c) (a^6 b c+a^4 (b^4-2 b^3 c+4 b^2 c^2-2 b c^3+c^4)+a^3 (3 b^5-b^4 c-b c^4+3 c^5)+a^2 (3 b^6-b^5 c-5 b^4 c^2+7 b^3 c^3-5 b^2 c^4-b c^5+3 c^6)+a (b^7-b^6 c-b^5 c^2+3 b^4 c^3+3 b^3 c^4-b^2 c^5-b c^6+c^7)+b^3 c^3 (b+c)^2) : :

See Angel Montesdeoca, Euclid 5567.
See La elipse de Hutson y el ortocentro del triángulo incentral.

X(52725) = X(2)X(3)∩X(8182)X(33900)

Barycentrics a^2 (a^8-24 a^4 b^2 c^2-4 a^6 (b^2+c^2)+a^2 (4 b^6+21 b^4 c^2+21 b^2 c^4+4 c^6)-b^8+11 b^6 c^2-12 b^4 c^4+11 b^2 c^6-c^8) : :

[Kadir Altintas]:
Let G=X(2) centroid of ABC. Circle (GBC) intersects the AB, AC at Ac, Ab resp. Define Ba, Bc, Ca, Cb cyclically.
O'A: inverse of circumcenter of ABcCb respect to circumcircle of ABC.
Define O'B, O'C cyclically.
* Symmedian point of O'AO'BO'C lies on Euler line of ABC.

[Angel Montesdeoca]:
The symmedian point of O'AO'BO'C is the reflection of X(10204) in X(3).

See Kadir Altintas and Angel Montesdeoca, Euclid 5598.

X(52792) = ISOGONAL CONJUGATE OF X(52792)

Barycentrics a^2*(a^4+4*b^4-2*a*b^2*c-5*b^2*c^2+c^4-a^2*(5*b^2+2*c^2))*(a^4+b^4-2*a*b*c^2-5*b^2*c^2+4*c^4-a^2*(2*b^2+5*c^2)) : :

See Angel Montesdeoca and Ivan Pavlov, Euclid 5622.

X(52793) = X(3)X(12)∩X(11)X(35)

Barycentrics 4*a^4+(b^2-c^2)^2-a^2*(5*b^2+2*b*c+5*c^2) : :

See Angel Montesdeoca and Ivan Pavlov, Euclid 5622.

X(52794) = X(8)X(44149)∩X(319)X(1232)

Barycentrics b^2*c^2*(4*a^4+(b^2-c^2)^2-a^2*(5*b^2+2*b*c+5*c^2)) : :

See Angel Montesdeoca and Ivan Pavlov, Euclid 5622.

X(52795) = X(5)X(993)∩X(10)X(547)

Barycentrics 2*a^4-7*a^2*b^2+5*b^4-2*a*b^2*c-7*a^2*c^2-2*a*b*c^2-10*b^2*c^2+5*c^4 : :

See Angel Montesdeoca and Ivan Pavlov, Euclid 5622.

X(52796) = X(389)X(3634)∩X(515)X(3819)

Barycentrics a^2*(2*a^3*b^2*c^2*(b+c)-2*a*b^2*(b-c)^2*c^2*(b+c)-a^6*(b^2+c^2)+(b^2-c^2)^2*(b^4+6*b^2*c^2+c^4)+a^4*(3*b^4+4*b^2*c^2+3*c^4)-a^2*(3*b^6+7*b^4*c^2+4*b^3*c^3+7*b^2*c^4+3*c^6)) : :

See Angel Montesdeoca and Ivan Pavlov, Euclid 5622.

X(52942) = EULER LINE INTERCEPT OF X(148)X(1992)

Barycentrics 11 a^4-2 a^2 (b^2+c^2)+16 b^2 c^2-7 (c^4 +b^4) : :

See Abdilkadir Altintaşand Angel Montesdeoca, Euclid 5683.
Simedianos de triángulos círculo cevianos sobre la recta de Euler

X(52943) = EULER LINE INTERCEPT OF X(315)X(15300)

Barycentrics 25 a^4-16 a^2 (b^2+c^2)+20 b^2 c^2-11 (c^4+b^4) : :

See Abdilkadir Altintaşand Angel Montesdeoca, Euclid 5683.
Simedianos de triángulos círculo cevianos sobre la recta de Euler

X(52944) = EULER LINE INTERCEPT OF X(9830)X(35369)

Barycentrics 47 a^4-20 a^2 (b^2+c^2)+52 b^2 c^2-25 (c^4+b^4) : :

See Abdilkadir Altintaşand Angel Montesdeoca, Euclid 5683.
Simedianos de triángulos círculo cevianos sobre la recta de Euler

X(52998) = X(98)X(9381) ∩X(110)X(6328)

Barycentrics 1/((b^2-c^2)(a^2-b^2-c^2) (a^8+a^4 b^2 c^2-2 a^6 (b^2+c^2)-(b^2-c^2)^2 (b^4+c^4)+a^2 (2 b^6-b^4 c^2-b^2 c^4+2 c^6))) ::

See Abdilkadir Altintaşand Angel Montesdeoca, Euclid 5705.
Una parábola con foco la reflexión de X(933) en la recta de Euler

X(53033) = X(2)X(39)∩X(20)X(636)

Barycentrics 3 a^4-2 a^2 (b^2+c^2)+3 b^4+2 b^2 c^2+3 c^4 : :

Sea ABC un triángulo con incentro I=X₁, ortocentro H=X₄ y triangulo medial M_aM_bM_c, la bisectriz interior en A corta la altura por C en D y la bisectriz exterior en A corta a la altura por B en E.
Las circunferencias (BDH) y (CEH) se vuelven a cortar en A_bc.
Sea ℓ_bc la recta M_aA_bc. Permutando B y C, en la construcción anterior, se define la recta ℓ_cb.
Procediendo cíclicamente sobre los vértices de ABC, se definen las rectas ℓ_ca, ℓ_ac, ℓ_ab, ℓ_ba.
Las seis rectas ℓ_bc, ℓ_cb, ℓ_ca, ℓ_ac, ℓ_ab, ℓ_ba son tangentes a una misma cónica con centro W = (r^2+4 r R-s^2)^2 X₁₄₁- 5 r^2 s^2 X₆₃₁= X(53033).
See Angel Montesdeoca, Euclid 5727 and Seis rectas tangentes a una cónica.

X(53841) = (name pending)

Barycentrics (a*(b+c))/(a*b*c*(b+c)+a*Sqrt[-a^2+2*(b+c)^2]*S+2*S^2) : :

See Angel Montesdeoca, Euclid 5796, and HG120323

X(53842) = (name pending)

Barycentrics (a*(b+c))/(a*b*c*(b+c)-a*Sqrt[-a^2+2*(b+c)^2]*S+2*S^2) : :

See Angel Montesdeoca, Euclid 5796, and HG120323

al menos, de colaboraciones

================= Art of Problems Solving ================

Easy but cute geometry (Fedir Yudin. Jul 17, 2020)

Let ABC be a triangle and A' be the reflection of A about BC. Let P and Q be points on AB and AC, respectively, such that PA'=PC and QA'=QB. Prove that the perpendicular from A' to PQ passes through the circumcenter of triangle ABC.

Let A" be the orthogonal projection of A' on PQ, and define B'' and C'' cyclically. The lines AA'', BB'', CC'' concur in X(31392).
(See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 28850, and also El centro del triángulo X(31392) -- In Spanish--)

================= The American Mathematical Monthly ================

11586[2011, 653].Proposed by Takis Konstantopoulos, Uppsala University, Uppsala, Sweden.

A Line and a Triangle Generate Three Convergent Point Sequences

Let A₀, B₀, and C₀ be noncollinear points in the plane. Let p be a line that meets lines B₀C₀, C₀A₀, and A₀B₀ at A*, B*, and C* respectively. For n ≥ 1, let A_n be the intersection of B*B_{n - 1} with C*C_{n - 1}, and define B_n and C_n similarly. Show that all three sequences converge, and describe their respective limits.

Editorial comment. As pointed out by several readers, we must assume that p does not pass through any of the three initial points. Also, this must be viewed in the projective plane, since it may happen that a point in one of the sequences is defined by the intersection of two parallel lines.

Solution II by Angel Montesdeoca, Univ.de La Laguna, Spain, and Angel Plaza, Univ.de Las Palmas de Gran Canaria, Spain.

Corresponding sides of triangles A₀B₀C₀ and A₁ B₁C₁ meet at points on line p, so by Desargues' s Theorem the two triangles are perspective from a point, say P. By induction, all subsequent triangles A_nB_n C_n are also mutually perspective from this same point P. Thus all the points A_n are collinear, all B_n are collinear, and all C_n are collinear. We claim that the limits of the respective sequences are the intersection points with p of the lines through the sequences. We show this for sequence {A_n}.
We use homogeneous coordinates (x : y : z) with A₀B₀C₀ as reference triangle, so that A₀ = (1 : 0 : 0), B₀ = (0 : 1 : 0), and C₀ = (0 : 0 : 1). We write UV for the line through U and V. Let the equation of p be q x + r y + s z = 0 for some choice of coefficients q, r, s. Then p intersects B₀C₀ (x = 0) at A* = (0 : -s : r). It intersects C₀A₀ (y = 0) at B* = (s : 0 : -q). It intersects A₀B₀ (z = 0) at C* = (-r : q : 0). Since point P is the intersection of A₀A₁ (r y = s z) and C₀C₁ (q x = r y), we have P = (1/q : 1/r : 1/s). Therefore A₁ = (1/q : -1/r : -1/s) = (r s : -sq : -qr), B₁ = (-r s : sq : -qr), and C₁ = (-r s : -sq : qr). Since A₂ is the intersection of B*B₁ (q x + 2 r y + s z = 0) with C*C₁ (q x + r y + 2 s z = 0), it follows that A₂ = (3 r s : -sq : -qr), and similarly for B₂ and C₂. The next points are A₃ =(5r s :-3sq :-3qr ), A₄ =(11r s :-5sq :-5qr ), and in general A_n =(a_n r s :-a_n-1sq :-a_n-1qr ) =((a_n /a_n-1)r s :-sq :-qr ), where {a_n} is defined recursively by a₀ =1, a_n =2a_n-1 +(-1)ⁿ, the Jacobsthal sequence. Take the limit as n → ∞ to find (since a_n /a_n-1 →2) that A_n → A* =(2r s :-sq :-qr ), so A* is the intersection of p with the line A₀P (r y =s z, same as A₀A₁), since its coordinates satisfy both equations.

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El Aleph. Blogs EL PAÍS La circunferencia de Feuerbach o por qué me encantan los triángulos. Miguel Ángel Morales (16/09/2016)

(... Y en este pdf tenéis información sobre cómo demostrar este resultado, entre otras muchas cosas relacionadas con geometría.)

Inscribed Conics and the Darboux Cubic Todor Zaharinov. International Journal of Computer Discovered Mathematics (IJCDM). Volume 4 pp.18-26

This paper is generalization of the note of Angel Montesdeoca [4, 10/01/2019] "La cúbica de Darboux y la elipse inscrita de Steiner"