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Contribuciones en otras web de Geometría del Triángulo

Angel Montesdeoca

(Última actualización: )

  |   Laboratorio Virtual de Triángulos con Cabri   |   Cubics in the Triangle Plane   |   Encyclopedia of Quadri Figures   |   Bicentric Pairs   |   Encyclopedia Triangle Centers   |   Otras   |  

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


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

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


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


  • 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)

  • 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).
    (Q003 de Higher Degree Related Curves. Bernard Gibert)

  • Other locus properties of Q066 quartic

    1. Let P be a point and PaPbPc its cevian triangle, HaHbHcthe orthic triangle (cevian triangle of H). The parallel through the point Pb to the sideline HaHb and the parallel through the point Pc to the sideline HaHc intersect at A'. The points B' and C' defined likewise. The triangles ABC and A'B'C' are perspective if and nly if P lies on the Stammler quartic together with the circumcircle (Angel Motesdeoca, Hyacinthos #21817).
    (Q066 de Higher Degree Related Curves. Bernard Gibert)

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

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

  • 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)

  • 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)

  • 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)

  • 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 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)

  • 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 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)

  • 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)

  • 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)

  • 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)

  • 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

  • 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).

  • 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)

  • Locus properties of Q018

    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 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)


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


  • QA-P38: Montesdeoca-Hutson Point

    QA-P38 is the Perspector of the Reference Quadrangle with the Cyclocevian Quadrangle. The Cyclocevian Quadrangle CC1.CC2.CC3.CC4 is defined by: CCi = Cyclocevian Conjugate of Pi wrt Pj.Pk.Pl for all permutations of (i,j,k,l) ∈ (1,2,3,4). The definition of Cyclocevian Conjugate can be found at Ref-13.
    This point was separately found by Angel Montesdeoca (Hyacinthos message 21075, see[11]) and Randy Hutson (private mail to author EQF) in the same week (June, 2012).
    (
    Encyclopedia of Quadri-Figures )

  • QA-Co/1: QA-DT-Conic Perspector

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

  • QA-P16: QA-Harmonic Center

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

    ( Mostrar/Ocultar figura )

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


  • P(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.


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


  • 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(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(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(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(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(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(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(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(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(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(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(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)
    + a^12(14b10 + 29b8c2 + 36b6c4 + 36b4c6 + 29b2c8 + 14c10)
    - a10(84b^12 + 67b10c2 + 56b8c4 + 48b6c6 + 56b4c8 + 67b2c10 + 84c12)
    + a8(82b14 - 23b12c2 - 31b10c4 - 19b8c6 - 19b6c8 - 31b4c10 - 23b2c^12+ 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(5501).png


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

    Barycentrics    a^2(a^6(b^2+c^2)+2a^4(b^2+c^2)^2-a^2(4b^6+7b^4c^2+7b^2c^4+4c^6)+b^8-3b^6c^2-2b^4c^4-3b^2c^6+c^8 + 4S^3(2a^2-b^2-c^2) csc ω) : :
    X(8160) = 3 X[3] - X[1671] = 3 X[1670] + X[1671] = 2 X[1671] - 3 X[8161] = 2 X[1670] + X[8161]


    Let ABC be a triangle, and let A'B'C' be the cevian triangle of X(6). Let U be the line through A' parallel to AB, and let V be the line through A' parallel to AC. Let U = U∩BC and V' = V∩AB. The 4 points B, C, U', V' lie on a circle, (O)A. Define (O)B and (O)C cyclically. Let M be the circle tangent to (O)A, (O)B, (O)C that encompasses them; call M the outer Montesdeoca-Lemoine circle. Let M' be the circle tangent to (O)A, (O)B, (O)C that is encompassed by each of them; call M' the inner Montesdeoca-Lemoine circle. Then X(8160) is the center of M, and X(8161) is the center of M'. The contact points of M with (O)A, (O)B, (O)C are the vertices of a triangle perspective to ABC, with perspector X(1343). Likewise, the contact points of M' with (O)A, (O)B, (O)C are the vertices of a triangle perspective to ABC, with perspector X(1342). (Based on notes from Angel Montesdeoca, October 2, 2015)

    The points X(8160) and X(8161) are labeled Z1 and Z2, respectively, in this sketch: Hechos Geométricos 02/10/2015.

    X(8160) lies on these lines: {3,6}, {30,5404}, {35,3238}, {36,3237}, {140,5403}, {1676,6683}, {2546,7786}

    X(8160) = midpoint of X(3) and X(1670)
    X(8160) = reflection of X(8161) in X(3)
    X(8160) = {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (3,1343,5092), (39,5092,8161)

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

    Barycentrics    a^2(a^6(b^2+c^2)+2a^4(b^2+c^2)^2-a^2(4b^6+7b^4c^2+7b^2c^4+4c^6)+b^8-3b^6c^2-2b^4c^4-3b^2c^6-c^8 - 4S^3(2a^2-b^2-c^2) csc ω) : :
    X(8161) = 3 X[3] - X[1670] = X[1670] + 3 X[1671] = 2 X[1670] - 3 X[8160] = 2 X[1671] + X[8160]


    See X(8160).

    X(8161) lies on these lines: {3,6}, {30,5403}, {35,3237}, {36,3238}, {140,5404}, {1677,6683}, {2547,7786}

    X(8161) = midpoint of X(3) and X(1671)
    X(8161) = reflection of X(8160) in X(3)
    X(8161) = {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (3,1342,5092), (39,5092,8160)

  • X(8183) = MONTESDEOCA DEGENERATE CONICS POINT

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

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

  • X(10033) =  24th HATZIPOLAKIS-MONTESDEOCA POINT

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

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

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

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

  • X(10034) =  25th HATZIPOLAKIS-MONTESDEOCA POINT

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

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

  • X(10035) =  26th HATZIPOLAKIS-MONTESDEOCA POINT

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

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

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


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

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

  • X(10036) =  27th HATZIPOLAKIS-MONTESDEOCA POINT

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

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

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

  • X(10105) =  28th HATZIPOLAKIS-MONTESDEOCA POINT
  • Barycentrics    a (a^7 b^2-3 a^5 b^4+3 a^3 b^6-a b^8-2 a^7 b c-6 a^6 b^2 c+3 a^5 b^3 c+13 a^4 b^4 c-8 a^2 b^6 c-a b^7 c+b^8 c+a^7 c^2-6 a^6 b c^2-16 a^5 b^2 c^2+9 a^4 b^3 c^2+15 a^3 b^4 c^2-2 a^2 b^5 c^2-b^7 c^2+3 a^5 b c^3+9 a^4 b^2 c^3+12 a^3 b^3 c^3+10 a^2 b^4 c^3+a b^5 c^3-3 b^6 c^3-3 a^5 c^4+13 a^4 b c^4+15 a^3 b^2 c^4+10 a^2 b^3 c^4+2 a b^4 c^4+3 b^5 c^4-2 a^2 b^2 c^5+a b^3 c^5+3 b^4 c^5+3 a^3 c^6-8 a^2 b c^6-3 b^3 c^6-a b c^7-b^2 c^7-a c^8+b c^8) : :

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

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

  • X(10106) =  29th HATZIPOLAKIS-MONTESDEOCA POINT

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

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


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

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


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

  • X(10107) =  30th HATZIPOLAKIS-MONTESDEOCA POINT

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

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


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

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


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

  • X(10108) =  31st HATZIPOLAKIS-MONTESDEOCA POINT

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

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


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

  • X(10122) =  32nd HATZIPOLAKIS-MONTESDEOCA POINT

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


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

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

  • X(10123) =  33rd HATZIPOLAKIS-MONTESDEOCA POINT

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


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

  • X(10124) =  34th HATZIPOLAKIS-MONTESDEOCA POINT

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


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


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

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

    Barycentrics    a^2 (a^14-4 a^12 (b^2+c^2)-4 a^8 b^2 c^2 (b^2+c^2)-b^2 c^2 (b^2-c^2)^4 (b^2+c^2)+a^10 (5 b^4+8 b^2 c^2+5 c^4)-a^2 (b^2-c^2)^2 (b^8-5 b^6 c^2-7 b^4 c^4-5 b^2 c^6+c^8)-5 a^6 (b^8-b^6 c^2-b^4 c^4-b^2 c^6+c^8)+a^4 (4 b^10-11 b^8 c^2-5 b^6 c^4-5 b^4 c^6-11 b^2 c^8+4 c^10)) : :
    X(10203) = 4R²X(3) - |OH²|2X(54)

    X(10203) is the nine-point center of the antipedal triangle of the Kosnita point. (Tran Quang Hung and Angel Montesdeoca, August 31, 2016; see Hyacinthos #24193):

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

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

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

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

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

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

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

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

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

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

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

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

  • X(10205) = (Name pending)

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

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

  • X(10206) = HUNG-MONTESDEOCA RADICAL CENTER

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

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

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

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

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

  • X(10207) = HUNG-MONTESDEOCA PERSPECTOR

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

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

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

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

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

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

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

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

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

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

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

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

  • X(10212) = 35th HATZIPOLAKIS-MONTESDEOCA POINT

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

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

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

  • X(10213) = 36th HATZIPOLAKIS-MONTESDEOCA POINT

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

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

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

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

    Barycentrics    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(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. )


  • ================= 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.)