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Re: [EMHL] LOCUS

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  • Francois Rideau
    Dear Antreas What do you mean by circle (Pb, PbP)? Friendly Francois ... [Non-text portions of this message have been removed]
    Message 1 of 22 , Apr 13, 2010
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      Dear Antreas
      What do you mean by circle (Pb, PbP)?
      Friendly
      Francois

      On Wed, Apr 7, 2010 at 1:11 PM, Antreas <anopolis72@...> wrote:

      >
      >
      > Let ABC be a triangle, P a point and
      > PaPbPc the pedal triangle of P.
      >
      > Let A' be the other than P intersection of
      > the circles (Pb, PbP) and (Pc, PcP)
      > and similarly B', C'.
      >
      > Which is the locus of P such that:
      >
      > 1. ABC, A'B'C' are perspective
      >
      > 2. PaPbPc, A'B'C' are perspective
      >
      > 3. ABC, A'B'C' are orthologic.
      >
      > The triangles PaPbPc, A'B'C' are orthologic.
      > One orthologic center is P. The other one?
      >
      > Antreas
      >
      >
      >


      [Non-text portions of this message have been removed]
    • Antreas Hatzipolakis
      Dear Francois Circle with center Pb and radius PbP APH On Tue, Apr 13, 2010 at 6:17 PM, Francois Rideau ... -- http://anopolis72000.blogspot.com/
      Message 2 of 22 , Apr 13, 2010
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        Dear Francois

        Circle with center Pb and radius PbP

        APH

        On Tue, Apr 13, 2010 at 6:17 PM, Francois Rideau
        <francois.rideau@...> wrote:
        > Dear Antreas
        > What do you mean by circle (Pb, PbP)?
        > Friendly
        > Francois
        >
        > On Wed, Apr 7, 2010 at 1:11 PM, Antreas <anopolis72@...> wrote:
        >
        >>
        >>
        >> Let ABC be a triangle, P a point and
        >> PaPbPc the pedal triangle of P.
        >>
        >> Let A' be the other than P intersection of
        >> the circles (Pb, PbP) and (Pc, PcP)
        >> and similarly B', C'.
        >>
        >> Which is the locus of P such that:
        >>
        >> 1. ABC, A'B'C' are perspective
        >>
        >> 2. PaPbPc, A'B'C' are perspective
        >>
        >> 3. ABC, A'B'C' are orthologic.
        >>
        >> The triangles PaPbPc, A'B'C' are orthologic.
        >> One orthologic center is P. The other one?
        >>
        >> Antreas
        >>
        >>
        >>
        >
        >
        > [Non-text portions of this message have been removed]
        >
        >
        >
        > ------------------------------------
        >
        > Yahoo! Groups Links
        >
        >
        >
        >



        --
        http://anopolis72000.blogspot.com/
      • Antreas
        Let ABC be a triangle, P a point, A B C the pedal triangle of P and L the Euler line of ABC. Let La,Lb,Lc be the reflections of L in the PA ,PB ,PC , resp.
        Message 3 of 22 , Aug 3, 2010
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          Let ABC be a triangle, P a point, A'B'C'
          the pedal triangle of P and L the Euler
          line of ABC.

          Let La,Lb,Lc be the reflections of L in the
          PA',PB',PC', resp.

          Which is the locus of P such that
          ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

          Antiedal triangle version:

          Let ABC be a triangle, P a point, A'B'C'
          the antipedal triangle of P and L the Euler
          line of A'B'C'. Let La,Lb,Lc be the reflections
          of L in the PA,PB,PC, resp.

          Which is the locus of P such that
          A'B'C', Triangle bounded by (La,Lb,Lc) are parallelogic?

          Are the simple points O,H,I lying on the locus?

          APH
        • Angel
          ... ABC and Triangle bounded by (La,Lb,Lc) are parallelogic = the parallels through A,B,C to La,Lb,Lc, resp. concur on a point U U=X(265) for every point P of
          Message 4 of 22 , Aug 4, 2010
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            --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
            >
            > Let ABC be a triangle, P a point, A'B'C'
            > the pedal triangle of P and L the Euler
            > line of ABC.
            >
            > Let La,Lb,Lc be the reflections of L in the
            > PA',PB',PC', resp.
            >
            > Which is the locus of P such that
            > ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

            > APH
            >


            ABC and Triangle bounded by (La,Lb,Lc) are parallelogic =
            the parallels through A,B,C to La,Lb,Lc, resp. concur on a point U

            U=X(265) for every point P of the plane


            Angel
          • Antreas
            Let ABC be a triangle, P a point, PaPbPc the pedal (or cevian) triangle of P, and (Ia),(Ib),(Ic) the excircles. Let (Oa) be the circle passing through Pb, Pc
            Message 5 of 22 , Sep 7, 2010
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              Let ABC be a triangle, P a point, PaPbPc the pedal (or cevian)
              triangle of P, and (Ia),(Ib),(Ic) the excircles.

              Let (Oa) be the circle passing through Pb, Pc and touching
              externally (or internally) the circle (Ia) at Qa.

              Similarly (Ob),(Oc) and Qb,Qc.

              Which is the locus of P such that

              1. ABC, OaObOc

              2. ABC, QaQbQc

              are perspective?

              (There are four variations:
              pedal/cevian - externally/internally)

              APH
            • Angel
              Dear Antreas TWO PARTICULAR CASES ... - PaPbPc the cevian triangle of G (PaPbPc the medial triangle). Let (Oa) be the circle passing through Pb, Pc and
              Message 6 of 22 , Sep 9, 2010
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                Dear Antreas

                TWO PARTICULAR CASES
                ----------------------------------------

                - PaPbPc the cevian triangle of G (PaPbPc the medial triangle).

                Let (Oa) be the circle passing through Pb, Pc and touching
                internally the circle (Ia) at Qa. Similarly (Ob),(Oc) and Qb,Qc.

                Perspector of ABC and QaQbQc:

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

                Radical center of (Oa), (Ob), (Oc):

                (a (a^2(b + c) + 2 a (b^2 + 3 b c + c^2) + (b + c)^3): ... : ...)

                -----------------------

                - PaPbPc the cevian triangle of H (PaPbPc the orthic triangle)
                Let (Oa) be the circle passing through Pb, Pc and touching
                internally the circle (Ia) at Qa. Similarly (Ob),(Oc) and Qb,Qc.

                Perspector of ABC and QaQbQc:

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

                Radical center of (Oa), (Ob), (Oc): X(1829) = ZOSMA TRANSFORM OF X(10)

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

                Best regards,

                Angel

                --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
                >
                > Let ABC be a triangle, P a point, PaPbPc the pedal (or cevian)
                > triangle of P, and (Ia),(Ib),(Ic) the excircles.
                >
                > Let (Oa) be the circle passing through Pb, Pc and touching
                > externally (or internally) the circle (Ia) at Qa.
                >
                > Similarly (Ob),(Oc) and Qb,Qc.
                >
                > Which is the locus of P such that
                >
                > 1. ABC, OaObOc
                >
                > 2. ABC, QaQbQc
                >
                > are perspective?
                >
                > (There are four variations:
                > pedal/cevian - externally/internally)
                >
                > APH
                >
              • Antreas
                Let ABC be a triangle, P a point and Ab,Ac points on AB,AC, resp. (on the same size of BC) such that: area(PBC) = area(PBAb) = area(PCAc). Similarly the points
                Message 7 of 22 , May 5, 2011
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                  Let ABC be a triangle, P a point and Ab,Ac points on AB,AC, resp.
                  (on the same size of BC) such that:
                  area(PBC) = area(PBAb) = area(PCAc).

                  Similarly the points Bc,Ba and Ca,Cb.

                  Which is the locus of P such that the triangles: ABC, Triangle
                  bounded by AbAc,BcBa,CaCb are perspective?
                  (Special case: AbAc,BcBa,CaCb are concurrent)

                  APH
                • Francisco Javier
                  Dear Antreas: The lines AbAc,BcBa,CaCb are always parallel to the trilinear polar of P. Best regards, Francisco Javier.
                  Message 8 of 22 , May 5, 2011
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                    Dear Antreas:

                    The lines AbAc,BcBa,CaCb are always parallel to the trilinear polar of P.

                    Best regards,

                    Francisco Javier.

                    --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
                    >
                    > Let ABC be a triangle, P a point and Ab,Ac points on AB,AC, resp.
                    > (on the same size of BC) such that:
                    > area(PBC) = area(PBAb) = area(PCAc).
                    >
                    > Similarly the points Bc,Ba and Ca,Cb.
                    >
                    > Which is the locus of P such that the triangles: ABC, Triangle
                    > bounded by AbAc,BcBa,CaCb are perspective?
                    > (Special case: AbAc,BcBa,CaCb are concurrent)
                    >
                    > APH
                    >
                  • Francisco Javier
                    And, if lines AbAc, BcBa, CaCb intersect the lines BC, CA, AB at A , B , C , then lines AA , BB , CC are parallel to the polar trilineal of isotomic conjugate
                    Message 9 of 22 , May 6, 2011
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                      And, if lines AbAc, BcBa, CaCb intersect the lines BC, CA, AB at A', B', C', then lines AA', BB', CC' are parallel to the polar trilineal of isotomic conjugate of P.

                      --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
                      >
                      > Dear Antreas:
                      >
                      > The lines AbAc,BcBa,CaCb are always parallel to the trilinear polar of P.
                      >
                      > Best regards,
                      >
                      > Francisco Javier.
                      >
                      > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
                      > >
                      > > Let ABC be a triangle, P a point and Ab,Ac points on AB,AC, resp.
                      > > (on the same size of BC) such that:
                      > > area(PBC) = area(PBAb) = area(PCAc).
                      > >
                      > > Similarly the points Bc,Ba and Ca,Cb.
                      > >
                      > > Which is the locus of P such that the triangles: ABC, Triangle
                      > > bounded by AbAc,BcBa,CaCb are perspective?
                      > > (Special case: AbAc,BcBa,CaCb are concurrent)
                      > >
                      > > APH
                      > >
                      >
                    • Antreas
                      Let ABC be a triangle and A B C , A B C the cevian, pedal triangles of P, resp. Denote: Ab, Ac = the reflections of A in BB , CC Bc, Ba = the reflections of
                      Message 10 of 22 , Feb 25 3:37 PM
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                        Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
                        triangles of P, resp.

                        Denote:

                        Ab, Ac = the reflections of A' in BB', CC'

                        Bc, Ba = the reflections of B' in CC', AA'

                        Ca, Cb = the reflections of C' in AA', BB'

                        Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent at P = common circumcenter of the triangles)

                        La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.

                        For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
                        (antipode of Feuerbach point)

                        For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
                        on the pedal circle of H (=NPC).

                        Which is the point of concurrence?

                        In general:
                        Which is the locus of P such that the lines La,Lb,Lc
                        are concurrent?

                        Antreas
                      • Angel
                        Dear Antreas, If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of Feuerbach point on the incircle. If P=H the lines La,Lb,Lc concur in X(1986)=
                        Message 11 of 22 , Feb 26 4:57 AM
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                          Dear Antreas,

                          If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of Feuerbach point on the incircle.

                          If P=H the lines La,Lb,Lc concur in X(1986)= HATZIPOLAKIS REFLECTION POINT (Antreas Hatzipolakis,Hyacinthos 7868,9/12/03;coordinates by Barry Wolk,Hyacinthos 7876,9/13/03)

                          X(1986)=(a^2(4SA^2-b^2c^2)(a^2(SA^2-SB SC)-SA(c^2-b^2)^2)/SA: ... : ...)


                          In general:
                          The lines La,Lb,Lc are concurrent if P is on the algebraic curve (Gamma) of degree nine (SA, SB, SC usual Conway notation):


                          (SA+SB)^3(SA+SC)(SA*SB-2SA*SC+SB*SC)x^6y^3-

                          (SA+SB)^3(SA+SC)(5SA*SB+5SA*SC-4SB*SC)x^5y^4+

                          (SA+SB)^3(SB+SC)(5SA*SB-4SA*SC+5SB*SC)x^4y^5-

                          (SA+SB)^3(SB+SC)(SA*SB+SA*SC-2SB*SC)x^3y^6+

                          (SA+SB)^2x^2y^2z((-SA-SC)(-4SA^2SB+SA(5SA-3SB)SC+(SA+SB)SC^2)x^4-
                          2(SA+SC)(SA(SB-SC)^2+5SA^2(SB+SC)+SB*SC(SB+SC))x^3y-
                          4(SA-SB)SC(SB*SC+SA(SB+SC))x^2y^2+
                          2(SB+SC)(SA^2(SB+SC)+SB*SC(5SB+SC)+SA(5SB^2-2SB*SC+SC^2))x*y^3+
                          (SB+SC)(-SA(4SB-SC)(SB+SC)+SB*SC(5SB+SC))y^4)+

                          (SA+SB)x*y*z^2((SA+SC)^2(SA*SB(5SA+SB)-(4SA-SB)(SA+SB)SC)x^5-
                          (SA+SC)(SA^2(19SB-14SC)(SB+SC)+SB^2SC(7SB+11*SC)+
                          SA*SB(7SB^2-8SB*SC-3SC^2))x^3y^2+(SB+SC)*
                          (SA^2SB(7SA+19SB)+SA(7SA^2-8SA*SB+5SB^2)SC+
                          (11SA-14SB)(SA+SB)SC^2)x^2y^3-(SB+SC)^2(SA*SB(SA+5SB)+
                          (SA-4SB)(SA+SB)SC)y^5)+

                          (SA+SB)z^3((-(SA+SC)^3)(-2SA*SB+
                          (SA+SB)SC)x^6+2(SA+SC)^2(SA(SB-SC)^2+5SA^2(SB+SC)+
                          SB*SC(SB+SC))x^5y-(SA+SC)(SA^2(14SB-19SC)*

                          (SB+SC)-SB*SC^2(11SB+7SC)+SA*SC(3SB^2+8SB*SC-7SC^2))*
                          x^4y^2+(SB+SC)(SA^2(14SB-11SC)(SB+SC)-
                          SB*SC^2(19SB+7SC)+SA*SC(-5SB^2+8SB*SC-7SC^2))*
                          x^2y^4-2(SB+SC)^2(SA^2(SB+SC)+SB*SC(5SB+SC)+
                          SA(5SB^2-2SB*SC+SC^2))x*y^5+(SB+SC)^3(-2SA*SB+(SA+SB)SC)*y^6)+

                          ((SA+SB)(SA+SC)^3(-4SB*SC+5SA(SB+SC))x^5+

                          4SB(SA-SC)(SA+SC)^2(SB*SC+SA(SB+SC))x^4y-

                          (SA+SC)(SB+SC)(SA^2SB(7SA+11SB)+
                          SA(7SA^2-8SA*SB-3SB^2)SC+(19SA-14SB)(SA+SB)SC^2)x^3y^2+

                          (SA+SC)(SB+SC)(SA^2(11SB-14SC)(SB+SC)+
                          SB^2SC(7SB+19SC)+SA*SB(7SB^2-8SB*SC+5SC^2))x^2y^3-

                          4SA(SB-SC)(SB+SC)^2(SB*SC+SA(SB+SC))x*y^4-

                          (SA+SB)(SB+SC)^3(5SA*SB-4SA*SC+5SB*SC)y^5)z^4-

                          (SA+SC)(SB+SC)((SA+SC)x^2-(SB+SC)y^2)((SA+SC)(-4SA*SB+5(SA+SB)SC)x^2+
                          2(SA*SB(SA+SB)+(SA-SB)^2SC+5(SA+SB)SC^2)x*y+
                          (SB+SC)(-4SA*SB+5(SA+SB)SC)y^2)z^5+

                          (SA+SC)(SB+SC)((SA+SC)(-2SB*SC+SA(SB+SC))x-(SB+SC)(SA(SB-2SC)+SB*SC)y)*(SA*x^2+SB*y^2+SC(x+y)^2)z^6 =0


                          (There must be a better simplification of this equation!)


                          This curve (Gamma) contains points isodynamic (X13, X14) but the corresponding triangles AAbAc, BBcBa and CCaCb are equilateral.

                          The triangle center X(74) -isogonal conjugate of Euler infinity point- is on the curve (Gamma) and the Euler Lines of A'AbAc, B'BcBa, C'CaCb
                          passing through A'', B'', C'' resp. (concurrent at X74 = common circumcenter of the triangles).


                          Best regards
                          Angel Montesdeoca

                          --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
                          >
                          > Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
                          > triangles of P, resp.
                          >
                          > Denote:
                          >
                          > Ab, Ac = the reflections of A' in BB', CC'
                          >
                          > Bc, Ba = the reflections of B' in CC', AA'
                          >
                          > Ca, Cb = the reflections of C' in AA', BB'
                          >
                          > Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent at P = common circumcenter of the triangles)
                          >
                          > La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.
                          >
                          > For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
                          > (antipode of Feuerbach point)
                          >
                          > For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
                          > on the pedal circle of H (=NPC).
                          >
                          > Which is the point of concurrence?
                          >
                          > In general:
                          > Which is the locus of P such that the lines La,Lb,Lc
                          > are concurrent?
                          >
                          > Antreas
                          >
                        • Angel
                          More information on the algebraic curve of degree nine (Gamma): - Passes through the vertices of the triangle ABC. - The vertices are multiple points of order
                          Message 12 of 22 , Feb 26 11:43 AM
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                            More information on the algebraic curve of degree nine (Gamma):

                            - Passes through the vertices of the triangle ABC.

                            - The vertices are multiple points of order 3.

                            - The real tangents at the vertices of ABC intersect at the point X(74).


                            Angel M.

                            --- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
                            >
                            >
                            > Dear Antreas,
                            >
                            > If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of Feuerbach point on the incircle.
                            >
                            > If P=H the lines La,Lb,Lc concur in X(1986)= HATZIPOLAKIS REFLECTION POINT (Antreas Hatzipolakis,Hyacinthos 7868,9/12/03;coordinates by Barry Wolk,Hyacinthos 7876,9/13/03)
                            >
                            > X(1986)=(a^2(4SA^2-b^2c^2)(a^2(SA^2-SB SC)-SA(c^2-b^2)^2)/SA: ... : ...)
                            >
                            >
                            > In general:
                            > The lines La,Lb,Lc are concurrent if P is on the algebraic curve (Gamma) of degree nine (SA, SB, SC usual Conway notation):
                            >
                            >
                            > (SA+SB)^3(SA+SC)(SA*SB-2SA*SC+SB*SC)x^6y^3-
                            >
                            > (SA+SB)^3(SA+SC)(5SA*SB+5SA*SC-4SB*SC)x^5y^4+
                            >
                            > (SA+SB)^3(SB+SC)(5SA*SB-4SA*SC+5SB*SC)x^4y^5-
                            >
                            > (SA+SB)^3(SB+SC)(SA*SB+SA*SC-2SB*SC)x^3y^6+
                            >
                            > (SA+SB)^2x^2y^2z((-SA-SC)(-4SA^2SB+SA(5SA-3SB)SC+(SA+SB)SC^2)x^4-
                            > 2(SA+SC)(SA(SB-SC)^2+5SA^2(SB+SC)+SB*SC(SB+SC))x^3y-
                            > 4(SA-SB)SC(SB*SC+SA(SB+SC))x^2y^2+
                            > 2(SB+SC)(SA^2(SB+SC)+SB*SC(5SB+SC)+SA(5SB^2-2SB*SC+SC^2))x*y^3+
                            > (SB+SC)(-SA(4SB-SC)(SB+SC)+SB*SC(5SB+SC))y^4)+
                            >
                            > (SA+SB)x*y*z^2((SA+SC)^2(SA*SB(5SA+SB)-(4SA-SB)(SA+SB)SC)x^5-
                            > (SA+SC)(SA^2(19SB-14SC)(SB+SC)+SB^2SC(7SB+11*SC)+
                            > SA*SB(7SB^2-8SB*SC-3SC^2))x^3y^2+(SB+SC)*
                            > (SA^2SB(7SA+19SB)+SA(7SA^2-8SA*SB+5SB^2)SC+
                            > (11SA-14SB)(SA+SB)SC^2)x^2y^3-(SB+SC)^2(SA*SB(SA+5SB)+
                            > (SA-4SB)(SA+SB)SC)y^5)+
                            >
                            > (SA+SB)z^3((-(SA+SC)^3)(-2SA*SB+
                            > (SA+SB)SC)x^6+2(SA+SC)^2(SA(SB-SC)^2+5SA^2(SB+SC)+
                            > SB*SC(SB+SC))x^5y-(SA+SC)(SA^2(14SB-19SC)*
                            >
                            > (SB+SC)-SB*SC^2(11SB+7SC)+SA*SC(3SB^2+8SB*SC-7SC^2))*
                            > x^4y^2+(SB+SC)(SA^2(14SB-11SC)(SB+SC)-
                            > SB*SC^2(19SB+7SC)+SA*SC(-5SB^2+8SB*SC-7SC^2))*
                            > x^2y^4-2(SB+SC)^2(SA^2(SB+SC)+SB*SC(5SB+SC)+
                            > SA(5SB^2-2SB*SC+SC^2))x*y^5+(SB+SC)^3(-2SA*SB+(SA+SB)SC)*y^6)+
                            >
                            > ((SA+SB)(SA+SC)^3(-4SB*SC+5SA(SB+SC))x^5+
                            >
                            > 4SB(SA-SC)(SA+SC)^2(SB*SC+SA(SB+SC))x^4y-
                            >
                            > (SA+SC)(SB+SC)(SA^2SB(7SA+11SB)+
                            > SA(7SA^2-8SA*SB-3SB^2)SC+(19SA-14SB)(SA+SB)SC^2)x^3y^2+
                            >
                            > (SA+SC)(SB+SC)(SA^2(11SB-14SC)(SB+SC)+
                            > SB^2SC(7SB+19SC)+SA*SB(7SB^2-8SB*SC+5SC^2))x^2y^3-
                            >
                            > 4SA(SB-SC)(SB+SC)^2(SB*SC+SA(SB+SC))x*y^4-
                            >
                            > (SA+SB)(SB+SC)^3(5SA*SB-4SA*SC+5SB*SC)y^5)z^4-
                            >
                            > (SA+SC)(SB+SC)((SA+SC)x^2-(SB+SC)y^2)((SA+SC)(-4SA*SB+5(SA+SB)SC)x^2+
                            > 2(SA*SB(SA+SB)+(SA-SB)^2SC+5(SA+SB)SC^2)x*y+
                            > (SB+SC)(-4SA*SB+5(SA+SB)SC)y^2)z^5+
                            >
                            > (SA+SC)(SB+SC)((SA+SC)(-2SB*SC+SA(SB+SC))x-(SB+SC)(SA(SB-2SC)+SB*SC)y)*(SA*x^2+SB*y^2+SC(x+y)^2)z^6 =0
                            >
                            >
                            > (There must be a better simplification of this equation!)
                            >
                            >
                            > This curve (Gamma) contains points isodynamic (X13, X14) but the corresponding triangles AAbAc, BBcBa and CCaCb are equilateral.
                            >
                            > The triangle center X(74) -isogonal conjugate of Euler infinity point- is on the curve (Gamma) and the Euler Lines of A'AbAc, B'BcBa, C'CaCb
                            > passing through A'', B'', C'' resp. (concurrent at X74 = common circumcenter of the triangles).
                            >
                            >
                            > Best regards
                            > Angel Montesdeoca
                            >
                            > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
                            > >
                            > > Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
                            > > triangles of P, resp.
                            > >
                            > > Denote:
                            > >
                            > > Ab, Ac = the reflections of A' in BB', CC'
                            > >
                            > > Bc, Ba = the reflections of B' in CC', AA'
                            > >
                            > > Ca, Cb = the reflections of C' in AA', BB'
                            > >
                            > > Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent at P = common circumcenter of the triangles)
                            > >
                            > > La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.
                            > >
                            > > For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
                            > > (antipode of Feuerbach point)
                            > >
                            > > For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
                            > > on the pedal circle of H (=NPC).
                            > >
                            > > Which is the point of concurrence?
                            > >
                            > > In general:
                            > > Which is the locus of P such that the lines La,Lb,Lc
                            > > are concurrent?
                            > >
                            > > Antreas
                            > >
                            >
                          • Antreas Hatzipolakis
                            Dear Angel Thank you. I came to this configuration trying to find three homocentric (concentric) circles but not by construction (ie not by taking a point as
                            Message 13 of 22 , Feb 26 12:15 PM
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                              Dear Angel

                              Thank you.

                              I came to this configuration trying to find three homocentric (concentric)
                              circles
                              but not by construction (ie not by taking a point as center).

                              Antreas



                              On Tue, Feb 26, 2013 at 2:57 PM, Angel <amontes1949@...> wrote:

                              > **
                              >
                              >
                              >
                              > Dear Antreas,
                              >
                              > If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of
                              > Feuerbach point on the incircle.
                              >
                              > If P=H the lines La,Lb,Lc concur in X(1986)= HATZIPOLAKIS REFLECTION POINT
                              > (Antreas Hatzipolakis,Hyacinthos 7868,9/12/03;coordinates by Barry
                              > Wolk,Hyacinthos 7876,9/13/03)
                              >
                              > X(1986)=(a^2(4SA^2-b^2c^2)(a^2(SA^2-SB SC)-SA(c^2-b^2)^2)/SA: ... : ...)
                              >
                              > In general:
                              > The lines La,Lb,Lc are concurrent if P is on the algebraic curve (Gamma)
                              > of degree nine (SA, SB, SC usual Conway notation):
                              >
                              > (SA+SB)^3(SA+SC)(SA*SB-2SA*SC+SB*SC)x^6y^3-
                              >
                              > (SA+SB)^3(SA+SC)(5SA*SB+5SA*SC-4SB*SC)x^5y^4+
                              >
                              > (SA+SB)^3(SB+SC)(5SA*SB-4SA*SC+5SB*SC)x^4y^5-
                              >
                              > (SA+SB)^3(SB+SC)(SA*SB+SA*SC-2SB*SC)x^3y^6+
                              >
                              > (SA+SB)^2x^2y^2z((-SA-SC)(-4SA^2SB+SA(5SA-3SB)SC+(SA+SB)SC^2)x^4-
                              > 2(SA+SC)(SA(SB-SC)^2+5SA^2(SB+SC)+SB*SC(SB+SC))x^3y-
                              > 4(SA-SB)SC(SB*SC+SA(SB+SC))x^2y^2+
                              > 2(SB+SC)(SA^2(SB+SC)+SB*SC(5SB+SC)+SA(5SB^2-2SB*SC+SC^2))x*y^3+
                              > (SB+SC)(-SA(4SB-SC)(SB+SC)+SB*SC(5SB+SC))y^4)+
                              >
                              > (SA+SB)x*y*z^2((SA+SC)^2(SA*SB(5SA+SB)-(4SA-SB)(SA+SB)SC)x^5-
                              > (SA+SC)(SA^2(19SB-14SC)(SB+SC)+SB^2SC(7SB+11*SC)+
                              > SA*SB(7SB^2-8SB*SC-3SC^2))x^3y^2+(SB+SC)*
                              > (SA^2SB(7SA+19SB)+SA(7SA^2-8SA*SB+5SB^2)SC+
                              > (11SA-14SB)(SA+SB)SC^2)x^2y^3-(SB+SC)^2(SA*SB(SA+5SB)+
                              > (SA-4SB)(SA+SB)SC)y^5)+
                              >
                              > (SA+SB)z^3((-(SA+SC)^3)(-2SA*SB+
                              > (SA+SB)SC)x^6+2(SA+SC)^2(SA(SB-SC)^2+5SA^2(SB+SC)+
                              > SB*SC(SB+SC))x^5y-(SA+SC)(SA^2(14SB-19SC)*
                              >
                              > (SB+SC)-SB*SC^2(11SB+7SC)+SA*SC(3SB^2+8SB*SC-7SC^2))*
                              > x^4y^2+(SB+SC)(SA^2(14SB-11SC)(SB+SC)-
                              > SB*SC^2(19SB+7SC)+SA*SC(-5SB^2+8SB*SC-7SC^2))*
                              > x^2y^4-2(SB+SC)^2(SA^2(SB+SC)+SB*SC(5SB+SC)+
                              > SA(5SB^2-2SB*SC+SC^2))x*y^5+(SB+SC)^3(-2SA*SB+(SA+SB)SC)*y^6)+
                              >
                              > ((SA+SB)(SA+SC)^3(-4SB*SC+5SA(SB+SC))x^5+
                              >
                              > 4SB(SA-SC)(SA+SC)^2(SB*SC+SA(SB+SC))x^4y-
                              >
                              > (SA+SC)(SB+SC)(SA^2SB(7SA+11SB)+
                              > SA(7SA^2-8SA*SB-3SB^2)SC+(19SA-14SB)(SA+SB)SC^2)x^3y^2+
                              >
                              > (SA+SC)(SB+SC)(SA^2(11SB-14SC)(SB+SC)+
                              > SB^2SC(7SB+19SC)+SA*SB(7SB^2-8SB*SC+5SC^2))x^2y^3-
                              >
                              > 4SA(SB-SC)(SB+SC)^2(SB*SC+SA(SB+SC))x*y^4-
                              >
                              > (SA+SB)(SB+SC)^3(5SA*SB-4SA*SC+5SB*SC)y^5)z^4-
                              >
                              > (SA+SC)(SB+SC)((SA+SC)x^2-(SB+SC)y^2)((SA+SC)(-4SA*SB+5(SA+SB)SC)x^2+
                              > 2(SA*SB(SA+SB)+(SA-SB)^2SC+5(SA+SB)SC^2)x*y+
                              > (SB+SC)(-4SA*SB+5(SA+SB)SC)y^2)z^5+
                              >
                              > (SA+SC)(SB+SC)((SA+SC)(-2SB*SC+SA(SB+SC))x-(SB+SC)(SA(SB-2SC)+SB*SC)y)*(SA*x^2+SB*y^2+SC(x+y)^2)z^6
                              > =0
                              >
                              > (There must be a better simplification of this equation!)
                              >
                              > This curve (Gamma) contains points isodynamic (X13, X14) but the
                              > corresponding triangles AAbAc, BBcBa and CCaCb are equilateral.
                              >
                              > The triangle center X(74) -isogonal conjugate of Euler infinity point- is
                              > on the curve (Gamma) and the Euler Lines of A'AbAc, B'BcBa, C'CaCb
                              > passing through A'', B'', C'' resp. (concurrent at X74 = common
                              > circumcenter of the triangles).
                              >
                              > Best regards
                              > Angel Montesdeoca
                              >
                              >
                              > --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
                              > >
                              > > Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
                              > > triangles of P, resp.
                              > >
                              > > Denote:
                              > >
                              > > Ab, Ac = the reflections of A' in BB', CC'
                              > >
                              > > Bc, Ba = the reflections of B' in CC', AA'
                              > >
                              > > Ca, Cb = the reflections of C' in AA', BB'
                              > >
                              > > Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp.
                              > (concurrent at P = common circumcenter of the triangles)
                              > >
                              > > La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.
                              > >
                              > > For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
                              > > (antipode of Feuerbach point)
                              > >
                              > > For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
                              > > on the pedal circle of H (=NPC).
                              > >
                              > > Which is the point of concurrence?
                              > >
                              > > In general:
                              > > Which is the locus of P such that the lines La,Lb,Lc
                              > > are concurrent?
                              > >
                              > > Antreas
                              > >
                              >
                              >
                              >



                              --
                              http://anopolis72000.blogspot.com/


                              [Non-text portions of this message have been removed]
                            • Antreas Hatzipolakis
                              [APH] ... More for P = H: The NPCs of A AbAc, B BcBa, C CaCb are concurrent on the NPC of A B C (On the Poncelet point of H with respect A B C ie the point
                              Message 14 of 22 , Feb 26 10:24 PM
                              • 0 Attachment
                                [APH]


                                >
                                > Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
                                > triangles of P, resp.
                                >
                                > Denote:
                                >
                                > Ab, Ac = the reflections of A' in BB', CC'
                                >
                                > Bc, Ba = the reflections of B' in CC', AA'
                                >
                                > Ca, Cb = the reflections of C' in AA', BB'
                                >
                                > Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent
                                > at P = common circumcenter of the triangles)
                                >
                                > La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.
                                >
                                > For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
                                > (antipode of Feuerbach point)
                                >
                                > For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
                                > on the pedal circle of H (=NPC).
                                >
                                >
                                >
                                More for P = H:

                                The NPCs of A'AbAc, B'BcBa, C'CaCb are concurrent on
                                the NPC of A'B'C' (On the Poncelet point of H with respect
                                A'B'C' ie the point of concurrence of the NPCs of
                                A'B'C', HB'C', HC'A', HA'B'. So we have seven concurrent NPCs)

                                The parallels through A,B,C to the (concurrent at H)
                                Euler lines Ea, Eb, Ec of A'AbAc, B'BcBa, C'CaCb, resp. concur on the
                                circumcircle of ABC on the antipode of the Euler line
                                reflection point. And the perpendiculars, on the Euler line reflection
                                point.


                                APH


                                [Non-text portions of this message have been removed]
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