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Re: LOCUS

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  • Pierre
    Dear Antreas, ... Here, examples are : any, except from circumcircle. 0. OaObOc, HaHbHc : circumcircle and K001, Neuberg cubic. Examples : 1, 3, 4, 13, 14, 15,
    Message 1 of 22 , Jan 24, 2010
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      Dear Antreas,

      --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
      > Let ABC be a triangle, P a point, Oa,Ob,Oc and Ha,Hb,Hc the
      > circumcenters, orthocenters of PBC,PCA,PAB, resp.
      > Let A'B'C', A"B"C" be the cevian, pedal triangles of P, resp.
      > Which is the locus of P such that:
      > 0. OaObOc, HaHbHc ... WNR
      > 1. cevian, OaObOc
      > 2. cevian, HaHbHc
      > 3. pedal , OaObOc
      > 4. pedal , HaHbHc
      > are perspective?

      Here, examples are : any, except from circumcircle.

      0. OaObOc, HaHbHc : circumcircle and K001, Neuberg cubic. Examples : 1, 3, 4, 13, 14, 15, 16, 30, 74, 370, 399, 484, 616, 617, 1138, 1157, 1263, 1276, 1277, 1337, 1338, 2132, 2133, 3065, 3440, 3441, 3464, 3465, 3466, 3479, 3480, 3481, 3482, 3483, 3484

      1. cevian, OaObOc : circumcircle and 5th degree curve
      (q^2*c^2-r^2*b^2)*(b^2+c^2-a^2)*p^3+(r^2*a^2-p^2*c^2)*(c^2+a^2-b^2)*q^3+(p^2*b^2-q^2*a^2)*(a^2+b^2-c^2)*r^3=0. Examples : 1, 2, 4, 13, 14, 357, 1113, 1114, 1134, 1136, 1156

      2. cevian, HaHbHc : circumcircle and 7th degree curve. Examples : 4.

      3. pedal , OaObOc : circumcircle and K006 ORTHOCUBIC, pK(X6, X4). Examples : 1, 3, 4, 46, 90, 155, 254, 371, 372, 485, 486, 487, 488.

      4. pedal , HaHbHc : ever. Perspector : P.

      5. anticev , OaObOc : circumcircle and 7th degree. Examples : 1, 6, 13, 14, 399 (1->1, 6->3)

      6. anticev, HaHbHc : 8th degree. Examples : 1,4 (1->9, 4->4)

      7. circumcev , OaObOc : ever. Examples : (P, persp) : [2, 1296], [3, 110], [4, 110], [5, 1291], [6, 1296], [13, 110], [14, 110], [15, 110], [16, 110], [20, 112]...
      In fact, persp= X(110) when P belongs to K001

      Formula : a^2/((q^2*a^2-p^2*b^2)*S[b]*r+(-r^2*a^2+p^2*c^2)*S[c]*q+(q^2*c^2-r^2*b^2)*a^2*p), etc with S[a]=Conway symbol.

      8. circumcev, HaHbHc : circumcircle and 9th degree. Examples : 1.

      Best regards,
      Pierre.
    • Antreas Hatzipolakis
      Dear Pierre Thanks! ... This quintic looks interesting. Is it already listed by Bernard? APH [Non-text portions of this message have been removed]
      Message 2 of 22 , Jan 25, 2010
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        Dear Pierre

        Thanks!



        > --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>,
        > "Antreas" <anopolis72@...> wrote:
        > > Let ABC be a triangle, P a point, Oa,Ob,Oc and Ha,Hb,Hc the
        > > circumcenters, orthocenters of PBC,PCA,PAB, resp.
        > > Let A'B'C', A"B"C" be the cevian, pedal triangles of P, resp.
        > > Which is the locus of P such that:
        > > 0. OaObOc, HaHbHc ... WNR
        > > 1. cevian, OaObOc
        > > are perspective?
        >
        > Here, examples are : any, except from circumcircle.
        >
        > 0. OaObOc, HaHbHc : circumcircle and K001, Neuberg cubic. Examples : 1, 3,
        > 4, 13, 14, 15, 16, 30, 74, 370, 399, 484, 616, 617, 1138, 1157, 1263, 1276,
        > 1277, 1337, 1338, 2132, 2133, 3065, 3440, 3441, 3464, 3465, 3466, 3479,
        > 3480, 3481, 3482, 3483, 3484
        >
        > 1. cevian, OaObOc : circumcircle and 5th degree curve
        > (q^2*c^2-r^2*b^2)*(b^2+c^2-a^2)*p^3+(r^2*a^2-p^2*c^2)*(c^2+a^2-b^2)*q^3+(p^2*b^2-q^2*a^2)*(a^2+b^2-c^2)*r^3=0.
        > Examples : 1, 2, 4, 13, 14, 357, 1113, 1114, 1134, 1136, 1156
        >
        >
        This quintic looks interesting. Is it already listed by Bernard?

        APH


        [Non-text portions of this message have been removed]
      • Pierre
        Dear Antreas, ... Yes, you are right ! Q003, the Euler-Morley Quintic. Since there are 95 identified points on Q003, it remains only to obtain 95
        Message 3 of 22 , Jan 25, 2010
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          Dear Antreas,
          --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:

          > > 1. cevian, OaObOc : circumcircle and 5th degree curve
          > > (q^2*c^2-r^2*b^2)*(b^2+c^2-a^2)*p^3+(r^2*a^2-p^2*c^2)*(c^2+a^2-b^2)*q^3+(p^2*b^2-q^2*a^2)*(a^2+b^2-c^2)*r^3=0.
          > > Examples : 1, 2, 4, 13, 14, 357, 1113, 1114, 1134, 1136, 1156
          > >
          > This quintic looks interesting. Is it already listed by Bernard?

          Yes, you are right ! Q003, the Euler-Morley Quintic.

          Since there are 95 "identified" points on Q003, it remains "only" to obtain 95 interpretations of the property persp(cevian, OaObOc)...

          Best regards.
        • Antreas Hatzipolakis
          Dear Pierre ... Probably are interesting the variations with Orthologic instead of Perspective triangles. Most likely the loci are high degree curves, but it
          Message 4 of 22 , Jan 25, 2010
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            Dear Pierre


            > --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>,
            > "Antreas" <anopolis72@...> wrote:
            > > Let ABC be a triangle, P a point, Oa,Ob,Oc and Ha,Hb,Hc the
            > > circumcenters, orthocenters of PBC,PCA,PAB, resp.
            > > Let A'B'C', A"B"C" be the cevian, pedal triangles of P, resp.
            > > Which is the locus of P such that:
            > > 0. OaObOc, HaHbHc ... WNR
            > > 1. cevian, OaObOc
            > > 2. cevian, HaHbHc
            > > 3. pedal , OaObOc
            > > 4. pedal , HaHbHc
            > > are perspective?
            >
            > Here, examples are : any, except from circumcircle.
            >
            > 0. OaObOc, HaHbHc : circumcircle and K001, Neuberg cubic. Examples : 1, 3,
            > 4, 13, 14, 15, 16, 30, 74, 370, 399, 484, 616, 617, 1138, 1157, 1263, 1276,
            > 1277, 1337, 1338, 2132, 2133, 3065, 3440, 3441, 3464, 3465, 3466, 3479,
            > 3480, 3481, 3482, 3483, 3484
            >
            > 1. cevian, OaObOc : circumcircle and 5th degree curve
            > (q^2*c^2-r^2*b^2)*(b^2+c^2-a^2)*p^3+(r^2*a^2-p^2*c^2)*(c^2+a^2-b^2)*q^3+(p^2*b^2-q^2*a^2)*(a^2+b^2-c^2)*r^3=0.
            > Examples : 1, 2, 4, 13, 14, 357, 1113, 1114, 1134, 1136, 1156
            >
            > 2. cevian, HaHbHc : circumcircle and 7th degree curve. Examples : 4.
            >
            > 3. pedal , OaObOc : circumcircle and K006 ORTHOCUBIC, pK(X6, X4). Examples
            > : 1, 3, 4, 46, 90, 155, 254, 371, 372, 485, 486, 487, 488.
            >
            > 4. pedal , HaHbHc : ever. Perspector : P.
            >
            > 5. anticev , OaObOc : circumcircle and 7th degree. Examples : 1, 6, 13, 14,
            > 399 (1->1, 6->3)
            >
            > 6. anticev, HaHbHc : 8th degree. Examples : 1,4 (1->9, 4->4)
            >
            > 7. circumcev , OaObOc : ever. Examples : (P, persp) : [2, 1296], [3, 110],
            > [4, 110], [5, 1291], [6, 1296], [13, 110], [14, 110], [15, 110], [16, 110],
            > [20, 112]...
            > In fact, persp= X(110) when P belongs to K001
            >
            > Formula :
            > a^2/((q^2*a^2-p^2*b^2)*S[b]*r+(-r^2*a^2+p^2*c^2)*S[c]*q+(q^2*c^2-r^2*b^2)*a^2*p),
            > etc with S[a]=Conway symbol.
            >
            > 8. circumcev, HaHbHc : circumcircle and 9th degree. Examples : 1.
            >
            >

            Probably are interesting the variations with Orthologic instead of
            Perspective triangles.

            Most likely the loci are high degree curves, but it will be interesting to
            find simple points
            lying on these loci.

            For example:
            I think that in the the above case 0 [ ie OaObOc, HaHbHc be orthologic] the
            point I = Incenter
            is on the locus. If so we have a line passing through the two orthologic
            centers and the Schiffler point.

            Greetings

            Antreas


            [Non-text portions of this message have been removed]
          • Antreas
            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
            Message 5 of 22 , Apr 7, 2010
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              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
            • Angel
              ... A B C is the reflection triangle of P(x:y:z) in its pedal triangle. 1. ABC, A B C are perspective if and only if P lies on the orthocubic cubic (Antreas
              Message 6 of 22 , Apr 7, 2010
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                --- In Hyacinthos@yahoogroups.com, "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
                >


                A'B'C' is the reflection triangle of P(x:y:z) in its pedal triangle.

                1. ABC, A'B'C' are perspective if and only if P lies on the orthocubic cubic (Antreas P Hatzipolakis and Paul Yiu, Reflections in triangle geometry, Forum Geometricorum, 9 (2009) 301--348. Proposition 26)

                3. ABC, A'B'C' are orthologic if and only if P lies (infinity lines, circumcircle, McCay Cubic).


                The triangles PaPbPc, A'B'C' are orthologic.
                One orthologic center is P(x:y:z). The other one:
                a^2(2b^2c^2x^2 + c^2(a^2+b^2-c^2)x y + b^2(a^2-b^2+c^2)x z + (a^2b^2-b^4+a^2c^2+2b^2c^2-c^4)y z): ... : ....

                Angel
              • Francois Rideau
                Dear Antreas What do you mean by circle (Pb, PbP)? Friendly Francois ... [Non-text portions of this message have been removed]
                Message 7 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 8 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 9 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 10 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 11 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 12 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 13 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 14 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 15 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 16 of 22 , Feb 25, 2013
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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 17 of 22 , Feb 26, 2013
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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 18 of 22 , Feb 26, 2013
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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 19 of 22 , Feb 26, 2013
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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 20 of 22 , Feb 26, 2013
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                                            [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


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