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Contribuciones en otras web de Geometría del Triángulo

Angel Montesdeoca


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

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


  • 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 Stammler 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)

  • 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 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 )

  • 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 )


  • 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 240813 and Anopolis 885

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

  • 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 210813 and Anopolis 860


  • 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(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}, (more pending)
    (Encyclopedia of Triangle Centers )

  • 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(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(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)