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# Contribuciones en otras web de Geometría del Triángulo

#### Angel Montesdeoca

(Última actualización: )

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

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

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

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

• 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 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 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 K010

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

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

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)

• 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 property of K088

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)

• Locus properties of K128

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

• 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).

• 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).

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

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

( 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.

• 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).

• 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 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).

• 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)

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).

• 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).

• 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.

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

• QA-P38: Montesdeoca-Hutson Point
• 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.

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

See Hechos Geometricos 21/08/13 and Anopolis 885

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

• 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 NAB be the nine-point circle of AB'X(1), and define NBC and NCA cyclically. Let NAC be the nine-point circle of AC'X(1), and define NBA and NCB cyclically. Then P(111) is the radical center of NAB, NBC, NCA, and U(111) is defined symmetrically; i.e., as the radical center of NAC, NBA, NCB.    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

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

• X(57) ISOGONAL CONJUGATE OF X(9)

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

Let Va, Vb, Vc be the antipodes of V=X(40) in the circles (VBC), (VCA), (VAB), respectively. The lines AVa, BVb, CVc 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 IaIbIc be the excentral triangle. The Brocard axes of BCIa, CAIb, ABIc concur in X(101). (Randy Hutson, February 10, 2016)

X(101) = intersection of Lemoine axes of 1st & 2nd Montesdeoca bisector triangles.

• 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(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 Va, Vb, Vc be the antipodes of V=X(40) in the circles (VBC), (VCA), (VAB), respectively. The triangle VaVbVc 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 VaVbVc. (Angel Montesdeoca, October 15, 2019). See Antipodales del punto de Bevan y el centro X(57).

• 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(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 Lb be the perpendicular to B at I, and let Lc be the perpendicular to CI at I. Let Ab = Lb∩AH and Ac = Lc∩AH. Let U be the circle (I, Ab, Ac). Let Mb be the point, other than I, where BI meets U, and let Mc be the point, other than I, where CI meets U. Let ℓa be the line MbMc, and define ℓb and ℓc cyclically. The lines ℓa, ℓb, ℓc concur in X(962). (Angel Montesdeoca, March 12, 2020). See Otra construcción del punto de Longuet-Higgins

• 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 AB = A'∩AB and AC = A'∩AC. Define BC and CA cyclically, and define BA and CB cyclically. The six points AB, BC, CA, AB, BC, CA 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(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 Ha be the orthocenter of IBC. Let A1 be the point, other than A, where AI meets the circumcircle. Let ta be the tangent to the circle (DHaA1) at Ha, and define tb and tc cyclically. The lines ta, tb , tc concur in X(1320). (Angel Montesdeoca, September 3, 2020)

• 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(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(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(2217) = X(2)-ISOCONJUGATE OF X(573)

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

• 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 = SA, v = SB, w = SC, 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(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)

• 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 TATBTC the tangential triangle of ABC. Let DA = E∩TBTC, and define DB and DC cyclically. Let A' = DBTB∩DCTC, 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(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(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(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(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 MA be the midpoint of the (possibly nonreal) points in which the circumcircle of ABC meets the line B'C', and define MB and MC cyclically. The triangle MAMBMC 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 ωA of ABC is tangent to BC at A'. Similarly define ωB, ωC, B', C'. Consider two tangent lines ℓa, ℓa' from A' to ωB, ωC which are different from BC. Similarly define ℓb, ℓb', ℓc, ℓc'. Then ℓa, ℓa', ℓb, ℓb', ℓc, ℓc' 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(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 NA be the nine-point circle of the triangle IBC, and define NB and NC cyclically. Let RA be the reflection of NA in the line AI, and define RB and RC cyclically. Then X(5048) is the radical center of the circles RA, RB, RC.      (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 AaBaCa be the intouch triangle, so that the lines BI and CI intersect BbCc at Ba and Ca, respectively. Let A1A2A3 be the intouch triangle of AaBaCa. Define B1B2B3 and C1C2C3 cyclically. Let da be the radical axis of (A1B3C2) and (A1A2A3). Let ea be the radical axis of (A1B3C2) and (AaBbCc). Define db, dc and eb, ec cyclically. Then the triangles A'B'C' and A"B"C" formed by the lines da, db, dc and the lines ea, eb, ec, respectively, are homothetic and the homothetic center is X(5173). (Angel Montesdeoca, April 1, 2020)

See X(1000) como centro de perspectividad.

• 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(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 AB be the reflection of A' in line BB', and define BC and CA cyclically. Let AC be the reflection of A' in line CC', and define BA and CB cyclically. Let L be the Euler line of ABC, let LA be the Euler line of AABAC, and define LB and LC cyclically. Let MA be the reflection of LA in AA', and define MB and MC cyclically. The lines MA, MB, MC concur in X(5494). Moreover, the four Euler lines L, LA, LB, LC 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 LA be the line through A' perpendicular to line AA', and define LB add LC cyclically. Let
UA = reflection of LA in AA'
UB = reflection of LA in BB'
UC = reflection of LA in CC'

VA = reflection of LB in AA'
VB = reflection of LB in BB'
VC = reflection of LB in CC'

WA = reflection of LC in AA'
WB = reflection of LC in BB'
WC = reflection of LC in CC'

TA = triangle formed by the lines in UA, UB, UC
TB = triangle formed by the lines in VA, VB, VC
TC = triangle formed by the lines in WA, WB, WC

OA = circumcenter of TA, OB = circumcenter of TA, OC = circumcenter of TA, O = X(3) = circumcenter of ABC. The points O, OA, OB, OC 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 LA be the line through A' perpendicular to line AA', and define LB add LC cyclically. Using the notation at X(5495), let MA be the line parallel to UA through B', and define MB and MC cyclically. Let A'' = MB∩MC, and define B'' and C'' cyclically. Let OA = circumcenter of A''B'C', and define OB and OC cyclically. Then the points X(1), OA, OB, OC 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 OA, OB, OC 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 NA be the nine-point center of the triangle BCO, where O = X(3), and define NB and NC cyclically. The nine-point center of the triangle NANBNC 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 IA be the A-excenter of a triangle ABC and let NA be the nine-point center of IABC. Define NB and NC cyclically. The circumcenter of NANBNC 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 NA be the nine-point center of NB'C', and define NB and NC cyclically. The nine-point center of NANBNC 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 NA be the nine-point center of NBC, and define NB and NC cyclically. The circumcenter of NANBNC 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 LA be the reflection of L in line BC, and define LB and LC 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 DA be the reflection of D in line BC, and define DB and DC cyclically. Let HA = LB∩DC, and define HB and HC cyclically. Let MA = LC∩DB, and define MB and MC cyclically. The triangles HAHBHC and MAMBMC 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(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)(271/2b2c2SA + S(S2 + 9SASA)

Let AP, BP, CP be the cevians of a point P in the plane of a triangle ABC. Let AB be a point on BP and AC a point on CP such that the triangle AABAC is equilateral. The lines ABAC, BCBA, CBCA 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)(271/2b2c2SA - S(S2 + 9SASA)

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 NA be the nine-point center of A'BC, and let OA be the circumcircle of NABC. Define OB and OC cyclically. The circles OA, OB, OC 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/[(a2SA - 2SBSC)(S2 - 3SASA)]

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 + 7S2A + 2SBSC)    (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 IA be the reflection of A'' in line AI. Define IB and IC cyclically. Then ABC and IAIB IC are orthologic triangles, and X(5691) is the ABC-orthology center of IAIB IC.     (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) = SBSC(5a2SA - SBSC)

Let ABC be a triangle, let PA be the polar of A with respect to the circle with diameter BC, and define PB and PC cyclically. Let AB = PA∩AB and AC = PA∩AC, and define BC, CA, BA, and CB cyclically. The six points AB, AC, BC, BA, CA, CB lie on, and define, the Montesdeoca conic. (Angel Montesdeoca, June 23, 2014)

A barycentric equation for the Montesdeoca conic is found from AB = SC : 0 : 2SA and AC = SB : 2SA : 0 to be as follows:

2(S2Ax2 + S2By2 + S2Cz2) - 5(SBSCyz + SCSAzx + SASBxy) = 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 (KA), (KB), (KC) be the circles described at X(5973) in association with the Hung-Feuerbach circle at X(5974). Let LA be the radical axis of (Ap) and (KA), and define LB and LC cyclically. The lines LA, LB, (LC 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 OAB be the circumcenter of ABA', and define OBC and OCA cyclically; let OAC be the circumcenter of ACA', and define OBA and OCB cyclically. Let OA be the circumcenter of triangle AOABOAC, and define OB and OC cyclically. Hatzipolakis proposed, and Montesdeoca proved, that the Euler lines concur, in X(186), and that the orthocenter of triangle OAOBOC, 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 OA be the circumcenter of A'BC, and define OB and OC 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 OA be the circumcenter of A'BC, and define OB and OC cyclically. The circumcircles of the four triangles ABC, AB''C'', A''BC'', A"B''C concur in X(6099). The lines OAA'', OBB'', OCC'' 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 OA be the circumcenter of A'BC, and define OB and OC 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 AB be the reflection of A in line OB, where O = circumcenter of ABC, and define BC and CA cyclically. Let AC be the reflection of A in line OC, and define BA and CB cyclically. Let A' be the nine-point center of triangle AABAC, 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 AB be the reflection of A in line OB, where O = circumcenter of ABC, and define BC and CA cyclically. Let AC be the reflection of A in line OC, and define BA and CB cyclically. Let A' be the nine-point center of triangle AABAC, 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 NA be the reflection of X(5) in the A-altitude, and define NB and NC cyclically. Then X(6102) is the orthocenter of NANBNC. 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(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 rA be a positive valued function of a,b,c, and define rB and rC cyclically. Let (A) denote the circle with center A and radius rA, and define (B) and (C) cyclically. Let P be the radical center of (A), (B), (C). Let rP be a positive valued function of a,b,c, let LA be the radical axis of (A), (B), (P), and define LB and LC cyclically. Let A' = LB∩LC, 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 (rA,rB,rC,rP) hyperbola. The triangle A'B'C' is the Dao (rA,rB,rC,rP) triangle, and the perspector D of ABC and A'B'C' is the Dao (rA,rB,rC,rP) perspector.

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

• 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 LA be the Euler line of triangle IBC, and define LB and LC 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 LB and the circle (C, |CS|). Let B3 be the point, other than S, of intersection of the line LC 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(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 LB be the line through I perpendiculat to AC, and let AB = LB∩BC and BA = LB∩BA. Define BC and CA cyclically, and define CB and AC cyclically. Let A' be the circumcenter of IABAC, 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 OA = reflection of X(3) in line AX(4), let HA = reflection of X(4) in OA, let MA = midpoint of segment OAHA, and define MB and MC cyclically. Then MAMBMC 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(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)

• 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 - X = 3 X + X = 2 X - 3 X = 2 X + X

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 - X = X + 3 X = 2 X - 3 X = 2 X + X

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 - a1/3b1/3c1/3 : b - a1/3b1/3c1/3 : c - a1/3b1/3c1/3

(Contributed by Angel Montesceoca, Ocftober 9, 2015) Hechos Geométricos 09/10/2015.
Let ABC be a triangle and k a real number. Let BA on line AB and CA on line AC be points such that BACA is parallel to BC at distance |kr(A)|, where r(A) is the inradius of triangle ABACA. Points CB, AB, AC, BC are defined cyclically. The six points BA, CA, CB, AB, AC, BC 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(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 CA = L∩AB, and define AB and BC cyclically. Let L' be the line through D parallel to AB, and let BA = L'∩CA, and define CB and AC cyclically. Let A'B'C' be the triangle having sidelines BACA, CBAB, ACBC. 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(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 - 4 X = 5 X - 8 X = 7 X - 10 X = X + 5 X = X - 10 X = X + 2 X

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(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 - 2 X

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 - 3 X

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    a2/3b2/3c2/3SBSC - a2SA(SASBSC)1/3 : :
X(10221) = -2(SASBSC)1/3*X(3) + a2/3b2/3c2/3*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 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 - (r^2+5rR+6R^2-s^2)*X

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

Barycentrics 7 a^2-3 (b^2+c^2) : :
X(11008) = 9 X - 10 X, 6 X - 5 X, 4 X - 3 X, 7 X - 8 X, 7 X - 6 X, 4 X - 7 X, 3 X - 5 X, 2 X - 3 X, 19 X - 18 X, 19 X - 12 X, 11 X - 12 X, 11 X - 10 X, 11 X - 9 X, 11 X - 6 X

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 - R X

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 : :

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(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,
HaHbHc = orthic triangle,
Γa = circumcircle of EFHa,
Γb = circumcircle of FDHb,
Γc = circumcircle of DEHb,
U = circle tangent to and encompassing Γa, Γb, Γc.
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(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 da. The envelope of da when P moves on circumcircle is a circle (Oa). Define (Ob), (Oc) cyclically. The radical center of the circles (Oa), (Ob), (Oc) 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 Db, Dc be the circumcenters of ABD, ACD. The lines through X(3) parallel to DDb, DDc intersects the lines through I perpendicular to IB, IC at Ab, Ac respectively. Let (A") be the circle with center centered at A" and tangent to line AbAc, 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(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(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 ℰa be the ellipse with major axis the segment BC and length of the minor axis a/Sqrt. Let La be the polar of A wrt ℰa, and define Lb and Lc cyclically. Let A' = Lb∩Lc, B' = Lc∩La, C' = La∩Lb. The lines AA', BB', CC' concur in X(14491).

A barycentric equation for the ellipse ℰa:

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(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(16186) = X(3)X(49)∩X(122)X(125)

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

Sean TaTbTc el triángulo tangencial, A' la intersección de la reflexión de AB en TaTc con a la reflexión de AC en TaTb, B' la intersección de la reflexión de BC en TbTa con a la reflexión de BA en TbTc, y C' la intersección de la reflexión de CA en TcTb con a la reflexión de CB en TcTa.
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 Ba y Ca a AC y AB, respectivamente; la antiparalela a CA, respecto a BC y a BA, corta en Cb y Ab a BA y BC, respectivamente; la antiparalela a AB, respecto a CA y a CB, corta en Ac y Bc a CB y CA, respectivamente. El baricentro PG de AbBcCa y el baricentro UG de AbBcCa, forman un par bicéntrico.
La 'crosssum' de PG y UG es X(16187).

See Angel Montesdeoca, HG050218.

• 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 pa, pb y pc son las perpendiculares por D', E' y F' a AD, BE y CF, respetivamente. Sea A2B2C2 el triángulo delimitado por las rectas pa, pb y pc.
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 MaMbMc, DEF, and A'B'C' be the medial, intouch and 1st circumperp triangles, respectively. Let D', E' and F' be the orthogonal projections of Ma, Mb and Mc 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(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)) : :

• 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)) : :

• 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) : :

• 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)) : :

• 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.

• 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(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(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 Ab be the intersection of AC and the perpendicular bisector of A'B, and let Ac be the intersection of AB and the perpendicular bisector of A'C. Let A'' be the orthogonal projection of A' on AbAc, 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 Nguyęn 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(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(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(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 La be the radical axis of circles with segments BC and A'A" as diameters, and define Lb and Lc cyclically. The triangle formed by lines La, Lb, Lc 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(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 Ba and Ca be the orthogonal projections of B and C on CM and BM, respectively. Define Cb and Ac cyclically, and define Ab and Bc cyclically. Let A'B'C' be the triangle having sidelines BaCa, CbAb, AcBc. Then A'B'C' is perspective to ABC, and the perspector is X(36809).

See Angel Montesdeoca, Euclid 641 .

• 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 PaPbPc y PaPbPc los triángulos ceviano y pedal de P, respectivamente.
Sean Qa, Qb, Qc los puntos que dividen a los segmentos PaPa, PbPb, PcPc, resp., en la misma razón t.
Las perpendiculares por Qa, Qb, Qc 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 PaPbPc y PaPbPc los triángulos ceviano y pedal de P, respectivamente.
Sean Qa, Qb, Qc los puntos que dividen a los segmentos PaPa, PbPb, PcPc, resp., en la misma razón t.
Las perpendiculares por Qa, Qb, Qc 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(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'AbE y AA'AcF con AA' paralelo a AbE y a AcF. Los puntos Bc, Ba y Ca, Cb se definen cíclicamente.
El triángulo UVW formado por las rectas AbAc, BcBa y CaCb 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'AbE y AA'AcF con AA' paralelo a AbE y a AcF. Los puntos Bc, Ba y Ca, Cb se definen cíclicamente.
El triángulo UVW formado por las rectas AbAc, BcBa y CaCb 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(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.

• ================= 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--)

• ================= O t r a s ================

• 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.)