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Reflections

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  • Floor van Lamoen
    Dear all, Let us start with our friend ABC. Let us consider the lines a_c through C and a_b through B such that the reflection of a_c through AB and the
    Message 1 of 30 , Jan 11, 2000
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      Dear all,

      Let us start with our friend ABC.

      Let us consider the lines a_c through C and a_b through B such that the
      reflection of a_c through AB and the reflection of a_b through AC are
      the same line a_a. So a_b is: a_c reflected in AB and the result
      reflected in AC.

      In the same way we can make triples b_i and c_i for i = a,b,c.

      The triangle formed by a_a, b_b and c_c is of course the reflection
      triangle (A reflected in BC, B in AC and C in AB).

      "Problem" for you:
      The lines a_b and a_c intersect in a point A_1. Idem B_1 and C_1.
      Triangle A_1B_1C_1 is perspective to ABC. Identify the perspector.

      The point A_1 can sometimes be seen as the point where a laser beam
      meets itself when sent from C to B via reflection in AB and AC.

      The easiest way to solve this problem is of course to use
      P-perpendicularity.

      Kind regards,
      Floor van Lamoen <f.v.lamoen@...>.

      PS. Barry, does adding the address after my name help you to read the
      address?.
    • Floor van Lamoen
      Dear all, ... I will have to add my excuses on bad English to the ones that our French and Greek friends gave. For example, Reflection in ... is a Dutchism
      Message 2 of 30 , Jan 12, 2000
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        Dear all,

        I posted this problem:

        > Let us start with our friend ABC.
        >
        > Let us consider the lines a_c through C and a_b through B such that the
        > reflection of a_c through AB and the reflection of a_b through AC are
        > the same line a_a. So a_b is: a_c reflected in AB and the result
        > reflected in AC.
        >
        > In the same way we can make triples b_i and c_i for i = a,b,c.
        >
        > The triangle formed by a_a, b_b and c_c is of course the reflection
        > triangle (A reflected in BC, B in AC and C in AB).
        >
        > "Problem" for you:
        > The lines a_b and a_c intersect in a point A_1. Idem B_1 and C_1.
        > Triangle A_1B_1C_1 is perspective to ABC. Identify the perspector.
        >
        > The point A_1 can sometimes be seen as the point where a laser beam
        > meets itself when sent from C to B via reflection in AB and AC.
        >
        > The easiest way to solve this problem is of course to use
        > P-perpendicularity.

        I will have to add my excuses on bad English to the ones that our French
        and Greek friends gave. For example, "Reflection in ..." is a Dutchism
        for "Reflection through ...".

        The solution:

        Consider P=(f:g:h) as base for perpendicularity.

        The point CP/\AB = (f:g:0) so the reflection of C through AB is the
        reflection of C through (f:g:0) is (2f:2g:-f-g). This point I call C',
        the C-vertex of the P-reflection triangle.

        Similarly A' is (-g-h-:2g:2h).

        The point X=A'C'/\AB is found as ( 3fh-g : 2g(1+h) : 0 ).
        The point Y=A'C'/\BC is found as ( 0 : 2g(1+f) : 3fh-g ).

        Line CX and BY are of course [2g(1+h) : g-3fh : 0] and [0 : g-3fh :
        2g(1+f)] respectively.

        2g(1+f)(1+h)
        These lines intersect in B_1 = ( 1+f : ------------ : 1+h ).
        3fh-g

        Finding similar results for A_1 and C_1 we see that the perspector of
        ABC and A_1B_1C_1 must be (1+f:1+g:1+h), which is the point dividing
        segment GP in ratio 1:3. It is the P-nine point center.

        Best regards,
        Floor van Lamoen <f.v.lamoen AT wxs.nl>
      • Antreas P. Hatzipolakis
        Let ABC be a triangle and Pa,Pb,Pc points on the altitudes AHa, BHb, CHc, such that: APa = t*AHa, BPb = t*BHb, CPc = t*CHc A /| / | / | / Pa / |
        Message 3 of 30 , Apr 1, 2002
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          Let ABC be a triangle and Pa,Pb,Pc points on the
          altitudes AHa, BHb, CHc, such that:

          APa = t*AHa, BPb = t*BHb, CPc = t*CHc

          A
          /|\
          / | \
          / | \
          / Pa \
          / | \
          / | \
          B------Ha-----C


          1. A1 := (reflection of BC in BPa) /\ (reflection of BC in CPa)
          B1, C1 similarly.

          The triangles ABC, A1B1C1 are perspective.

          Perspector in Normals:

          cosA
          (--------------- ::), where T(t) = t^2 / (1 - 2t)
          1 + T(t)sin^2A

          [The locus of the perspectors is Jerabek c/h.]

          2. A2 := (reflection of BA in BPa) /\ (reflection of CA in CPa)
          B2, C2 similarly.

          The triangles ABC, A2B2C2 are perspective.

          Perspector in Normals:

          1 1 - T(t)sin^2A
          (------ * ------------------ ::)
          cosA 1 - T'(t)sin^2A

          1 - 2t + (1 - t)^2
          where T(t) = -------------------
          1 - 2t

          t^2 - 2
          T'(t) = ----------
          t - 2


          3. A3 := (reflection of BPa in BA) /\ (reflection of CPa in CA)
          B3, C3 similarly.

          The triangles ABC, A3B3C3 are perspective.

          Perspector in Normals:

          1 1 - T(t)sin^2A
          (------ * ------------------ ::)
          cosA 1 - T'(t)sin^2A


          Where T(t) = 2t / (1 + t), T'(t) = 2 / t


          Antreas

          A mantinada (= Cretan distich) in modern style I learned
          in the days (last Friday to Monday) I was in Crete:

          MODEM qa balw sto xwrio, COMPUTER sto mhtato,
          sto INTERNET qa to poulw to gala twn probatw.

          aph
        • ndergiades
          Dear Antreas, you must add in your mantinada ... kai me EMAIL s olh th gh qa steilw to mantato, me mpalwqies ton tolmhth HACKER qa balw katw. Nikos
          Message 4 of 30 , Apr 2, 2002
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            Dear Antreas,

            you must add in your mantinada

            > A mantinada (= Cretan distich) in modern style I learned
            > in the days (last Friday to Monday) I was in Crete:
            >
            > MODEM qa balw sto xwrio, COMPUTER sto mhtato,
            > sto INTERNET qa to poulw to gala twn probatw.

            kai me EMAIL s'olh th gh qa steilw to mantato,
            me mpalwqies ton tolmhth HACKER qa balw katw.

            Nikos



            --- In Hyacinthos@y..., "Antreas P. Hatzipolakis" <xpolakis@m...>
            wrote:
            > Let ABC be a triangle and Pa,Pb,Pc points on the
            > altitudes AHa, BHb, CHc, such that:
            >
            > APa = t*AHa, BPb = t*BHb, CPc = t*CHc
            >
            > A
            > /|\
            > / | \
            > / | \
            > / Pa \
            > / | \
            > / | \
            > B------Ha-----C
            >
            >
            > 1. A1 := (reflection of BC in BPa) /\ (reflection of BC in CPa)
            > B1, C1 similarly.
            >
            > The triangles ABC, A1B1C1 are perspective.
            >
            > Perspector in Normals:
            >
            > cosA
            > (--------------- ::), where T(t) = t^2 / (1 - 2t)
            > 1 + T(t)sin^2A
            >
            > [The locus of the perspectors is Jerabek c/h.]
            >
            > 2. A2 := (reflection of BA in BPa) /\ (reflection of CA in CPa)
            > B2, C2 similarly.
            >
            > The triangles ABC, A2B2C2 are perspective.
            >
            > Perspector in Normals:
            >
            > 1 1 - T(t)sin^2A
            > (------ * ------------------ ::)
            > cosA 1 - T'(t)sin^2A
            >
            > 1 - 2t + (1 - t)^2
            > where T(t) = -------------------
            > 1 - 2t
            >
            > t^2 - 2
            > T'(t) = ----------
            > t - 2
            >
            >
            > 3. A3 := (reflection of BPa in BA) /\ (reflection of CPa in CA)
            > B3, C3 similarly.
            >
            > The triangles ABC, A3B3C3 are perspective.
            >
            > Perspector in Normals:
            >
            > 1 1 - T(t)sin^2A
            > (------ * ------------------ ::)
            > cosA 1 - T'(t)sin^2A
            >
            >
            > Where T(t) = 2t / (1 + t), T'(t) = 2 / t
            >
            >
            > Antreas
            >
            > A mantinada (= Cretan distich) in modern style I learned
            > in the days (last Friday to Monday) I was in Crete:
            >
            > MODEM qa balw sto xwrio, COMPUTER sto mhtato,
            > sto INTERNET qa to poulw to gala twn probatw.
            >
            > aph
          • Michael Lambrou
            Dear Antreas and Nikos, ... (free translation) In my village I ll set a MODEM, in my ranch a COMPUTER through INTERNET I ll sell, my cow s milk and butter
            Message 5 of 30 , Apr 2, 2002
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              Dear Antreas and Nikos,

              >-----Original Message-----
              >From: ndergiades [mailto:ndergiades@...]
              >Sent: Tuesday, April 02, 2002 11:51 AM
              >To: Hyacinthos@yahoogroups.com
              >Subject: [EMHL] Re: Reflections

              >
              >> A mantinada (= Cretan distich) in modern style I learned
              >> in the days (last Friday to Monday) I was in Crete:
              >>
              >> MODEM qa balw sto xwrio, COMPUTER sto mhtato,
              >> sto INTERNET qa to poulw to gala twn probatw.


              (free translation)

              In my village I' ll set a MODEM, in my ranch a COMPUTER
              through INTERNET I'll sell, my cow's milk and butter


              >
              > kai me EMAIL s'olh th gh qa steilw to mantato,
              > me mpalwqies ton tolmhth HACKER qa balw katw.
              >

              (free translation)

              via E-MAIL to all I will let it be known
              if I see a HACKER, I will gun him down


              Michael
            • Antreas P. Hatzipolakis
              Dear Michael, ... It is sheeps milk. Cretans are not cow-boys ! (Although they have much in common with Texans ! :-) A nice book about Cretan shepherds: David
              Message 6 of 30 , Apr 2, 2002
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                Dear Michael,

                [APH]:

                >>> A mantinada (= Cretan distich) in modern style I learned
                >>> in the days (last Friday to Monday) I was in Crete:
                >>>
                >>> MODEM qa balw sto xwrio, COMPUTER sto mhtato,
                >>> sto INTERNET qa to poulw to gala twn probatw.
                >
                >

                [ML]:

                > (free translation)
                >
                > In my village I' ll set a MODEM, in my ranch a COMPUTER
                > through INTERNET I'll sell, my cow's milk and butter
                >

                It is sheeps' milk. Cretans are not cow-boys !
                (Although they have much in common with Texans ! :-)

                A nice book about Cretan shepherds:

                David Outerbridge - Julie Thayer: The Last Shepherds.
                New York, N.Y.: The Viking Press, 1979
                (In pp. 108 - 130: Miltiades Xsilouris "voskos" Anoyia)

                [ND]:

                >
                >>
                >> kai me EMAIL s'olh th gh qa steilw to mantato,
                >> me mpalwqies ton tolmhth HACKER qa balw katw.
                >>
                >

                [ML]:

                > (free translation)
                >
                > via E-MAIL to all I will let it be known
                > if I see a HACKER, I will gun him down
                >
                >

                And two very good books about modern Cretan social life:

                Michael Herzfeld: Poetics of Manhood.
                Contest and Identity in a Cretan Mountain Village.
                Princeton, N.J.: Princeton University Press, 1985.

                Michael Herzfeld: A Place in History. Social and Monumental
                Time in a Cretan Town.
                Princeton, N.J.: Princeton University Press, 1991.


                Antreas
              • Antreas P. Hatzipolakis
                Let ABC be a triangle, P a point and p a line passing through P. Let A , B , C be the reflections of A,B,C in p. Which is the locus of p such that ABC, A B C
                Message 7 of 30 , Jun 17, 2002
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                  Let ABC be a triangle, P a point and p a line passing through P.

                  Let A', B', C' be the reflections of A,B,C in p.

                  Which is the locus of p such that ABC, A'B'C' are rthologic?
                  (the whole plane?)

                  APH
                • ndergiades
                  Dear Antreas, yes the locus is the whole plane. If S is the orthopole of p and H is the reflection of H = Orthocenter of ABC in S then the perpendicular
                  Message 8 of 30 , Jun 17, 2002
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                    Dear Antreas, yes the locus is the whole plane.
                    If S is the orthopole of p and H' is the reflection of
                    H = Orthocenter of ABC in S
                    then the perpendicular from A' to BC passes through H'
                    and the same for the perpendiculars from B', C' to CA, AB.
                    Hence ABC, A'B'C' are orthologic.

                    Best regards
                    Nikos Dergiades

                    --- In Hyacinthos@y..., Antreas P. Hatzipolakis <xpolakis@m...> wrote:
                    > Let ABC be a triangle, P a point and p a line passing through P.
                    >
                    > Let A', B', C' be the reflections of A,B,C in p.
                    >
                    > Which is the locus of p such that ABC, A'B'C' are rthologic?
                    > (the whole plane?)
                    >
                    > APH
                  • jpehrmfr
                    Dear Antreas, ... Two indirectly similar triangles are allways orthologic. Friendly. Jean-Pierre
                    Message 9 of 30 , Jun 17, 2002
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                      Dear Antreas,

                      > Let ABC be a triangle, P a point and p a line passing through P.
                      >
                      > Let A', B', C' be the reflections of A,B,C in p.
                      >
                      > Which is the locus of p such that ABC, A'B'C' are rthologic?
                      > (the whole plane?)

                      Two indirectly similar triangles are allways orthologic.
                      Friendly. Jean-Pierre
                    • xpolakis@netscape.net
                      Let ABC be a triangle P a point and PaPbPc the pedal triangle of P. Which is the locus of P such that the orthocenter of PaPbPc is lying on the Euler line of
                      Message 10 of 30 , Aug 12, 2003
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                        Let ABC be a triangle P a point and PaPbPc the pedal triangle
                        of P.

                        Which is the locus of P such that the orthocenter of PaPbPc
                        is lying on the Euler line of ABC?

                        I think that it is equivalent to:

                        Which is the locus of P such that the reflections
                        of the Euler line of ABC in the sides of PaPbPc
                        are concurrent.

                        Antreas


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                      • Darij Grinberg
                        Dear Antreas, ... Yes, it is equivalent. Thank you for the interesting problem. The locus is a cubic with equation a² SA (b²-c²) (b²z+c²y)
                        Message 11 of 30 , Aug 12, 2003
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                          Dear Antreas,

                          In Hyacinthos message #7495, you wrote:

                          >> Let ABC be a triangle P a point and PaPbPc the pedal
                          >> triangle of P.
                          >>
                          >> Which is the locus of P such that the orthocenter of
                          >> PaPbPc is lying on the Euler line of ABC?
                          >>
                          >> I think that it is equivalent to:
                          >>
                          >> Which is the locus of P such that the reflections
                          >> of the Euler line of ABC in the sides of PaPbPc
                          >> are concurrent.

                          Yes, it is equivalent. Thank you for the interesting
                          problem.

                          The locus is a cubic with equation

                          a² SA (b²-c²) (b²z+c²y) (b²SBzx+c²SCxy-a²SAyz+b²c²x²)
                          + ... cyclic
                          = 0.

                          I don't know any remarkable points on this cubic except
                          of the circumcenter X(3).

                          In fact, the orthocenter of the pedal triangle of a
                          point with barycentrics ( x : y : z ) has barycentrics

                          ( a² (b²z+c²y) (b²SBzx+c²SCxy-a²SAyz+b²c²x²)
                          : ...
                          : ... ).

                          We can take some other lines instead of the Euler line.
                          The Brocard axis gives a cubic through the
                          circumcenter and the orthocenter, for instance.

                          The same question with cevian instead of pedal triangles
                          would be interesting too, but I have not tried yet.

                          Sincerely,
                          Darij Grinberg
                        • Bernard Gibert
                          Dear Antreas and Darij, [APH] ... [DG] ... Let L be the trilinear polar of point P. The locus of P such that the orthocenter of PaPbPc lies on L is a circular
                          Message 12 of 30 , Aug 12, 2003
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                            Dear Antreas and Darij,

                            [APH]
                            > >> Let ABC be a triangle P a point and PaPbPc the pedal
                            > >> triangle of P.
                            > >>
                            > >> Which is the locus of P such that the orthocenter of
                            > >> PaPbPc is lying on the Euler line of ABC?

                            [DG]
                            > We can take some other lines instead of the Euler line.
                            > The Brocard axis gives a cubic through the
                            > circumcenter and the orthocenter, for instance.


                            Let L be the trilinear polar of point P.
                            The locus of P such that the orthocenter of PaPbPc lies on L is a
                            circular circum-cubic.
                            All those cubics form a mesh generated by three decomposed members Ca,
                            Cb, Cc.

                            If KaKbKc is the tangential triangle, Ca is the union of the line KbKc
                            and the circle centered at Ka passing through B and C.

                            Those cubics are not very exciting. The "best choice" would be P=H and
                            L=orthic axis where we obtain a nK0 with pole X571, root X6.
                            The real point at infinity is X924 and the cubic contains also X110 and
                            the centers of the Apollonian circles.

                            Best regards

                            Bernard

                            [Non-text portions of this message have been removed]
                          • Darij Grinberg
                            Dear Antreas, ... Now I have. The orthocenter of the cevian triangle of a point P with barycentrics ( 1/x : 1/y : 1/z ) has barycentrics ( (a²x + SCy + SBz)
                            Message 13 of 30 , Aug 12, 2003
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                              Dear Antreas,

                              In Hyacinthos message #7498, I wrote:

                              >> The same question with cevian instead of pedal triangles
                              >> would be interesting too, but I have not tried yet.

                              Now I have. The orthocenter of the cevian triangle of a
                              point P with barycentrics ( 1/x : 1/y : 1/z ) has
                              barycentrics

                              ( (a²x + SCy + SBz) (2SAyz + b²y² + c²z² - a²x²)
                              : ...
                              : ... ).

                              The locus of the point P so that this orthocenter lies on
                              any special line must be a sextic. If the special line
                              passes through the circumcenter of triangle ABC, this
                              sextic decomposes into the union of the circumcircle [see
                              next message] and a quartic which is actually the
                              isogonal conjugate of a conic through the circumcenter.
                              If our special line is the Brocard axis, this quartic has
                              equation

                              a²(b²-c²)y²z² + b²(c²-a²)z²x² + c²(a²-b²)x²y² = 0.

                              This quartic passes through X(1), X(2) and X(4) and is the
                              isogonal conjugate of the conic

                              a²(b²-c²)x² + b²(c²-a²)y² + c²(a²-b²)z² = 0,

                              which known as the Stammler hyperbola.

                              Sincerely,
                              Darij Grinberg
                            • Darij Grinberg
                              Dear Antreas, ... ... where ( x : y : z ) are the barycentrics of the point; not ( 1/x : 1/y : 1/z ). Sorry for the ambiguous use of variables. Sincerely,
                              Message 14 of 30 , Aug 12, 2003
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                                Dear Antreas,

                                In Hyacinthos message #7505, I wrote:

                                >> If our special line is the Brocard axis, this quartic has
                                >> equation
                                >>
                                >> a²(b²-c²)y²z² + b²(c²-a²)z²x² + c²(a²-b²)x²y² = 0.

                                ... where ( x : y : z ) are the barycentrics of the point;
                                not ( 1/x : 1/y : 1/z ).

                                Sorry for the ambiguous use of variables.

                                Sincerely,
                                Darij Grinberg
                              • Antreas P. Hatzipolakis
                                Let ABC be a triangle, HaHbHc its orthic triangle and P a point. Which is the locus of P such that the reflections of PHa, PHb, PHc in HHa, HHb, HHc, resp are
                                Message 15 of 30 , Sep 12, 2003
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                                  Let ABC be a triangle, HaHbHc its orthic triangle
                                  and P a point.

                                  Which is the locus of P such that the reflections
                                  of PHa, PHb, PHc in HHa, HHb, HHc, resp
                                  are concurrent?

                                  If the locus is the whole plane,
                                  then which is the locus of the point of
                                  concurrence, if P moves on a notable line (OH, OK,...)?

                                  Now, let P, P* be two isogonal points
                                  and PaPbPc the pedal triangle of P.

                                  Which is the locus of P such that the reflections of
                                  P*Pa, P*Pb, P*Pc in PPa, PPb, PPc, resp.
                                  are concurrent?

                                  (Variations: isotomic instead of isogonal;
                                  cevian instead of pedal)


                                  Greetings from Athens

                                  Antreas


                                  --
                                • Darij Grinberg
                                  Dear Antreas, ... The whole plane: the reflections concur at the isogonal conjugate of P with respect to the orthic triangle HaHbHc. In fact, HHa, HHb, HHc are
                                  Message 16 of 30 , Sep 12, 2003
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                                    Dear Antreas,

                                    In Hyacinthos message #7868, you wrote:

                                    >> Let ABC be a triangle, HaHbHc its orthic triangle
                                    >> and P a point.
                                    >>
                                    >> Which is the locus of P such that the reflections
                                    >> of PHa, PHb, PHc in HHa, HHb, HHc, resp
                                    >> are concurrent?

                                    The whole plane: the reflections concur at the isogonal
                                    conjugate of P with respect to the orthic triangle
                                    HaHbHc. In fact, HHa, HHb, HHc are the angle bisectors
                                    of the orthic triangle.

                                    Sincerely,
                                    Darij Grinberg
                                  • xpolakis
                                    Let ABC be a triangle, P a point, and Oa;Ha, Ob;Hb, Oc;Hc, the circumcenters, Orthocenters of PBC, PCA, PAB. The locus of P such that the lines OaHa, ObHb,
                                    Message 17 of 30 , Jan 30, 2004
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                                      Let ABC be a triangle, P a point, and Oa;Ha, Ob;Hb, Oc;Hc,
                                      the circumcenters, Orthocenters of PBC, PCA, PAB.

                                      The locus of P such that the lines
                                      OaHa, ObHb, OcHc [Euler Lines] are concurrent is well-known.

                                      Now, let O'a, H'a be the reflections of Oa, Ha in BC, resp.
                                      Similarly O'b, H'b; O'c,H'c.

                                      Which is the locus of P such that the lines

                                      1. O'aHa, O'bHb, O'cHc

                                      2. OaH'a, ObH'b, OcH'c

                                      are concurrent?


                                      Greetings from Athens

                                      Antreas
                                    • Bernard Gibert
                                      Dear Antreas, ... the McCay cubic + Linf + C(O,R) ... its inverse in the circumcircle, a nice bicircular quintic with a triple point at O where the tangents
                                      Message 18 of 30 , Jan 30, 2004
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                                        Dear Antreas,

                                        > [APH] Let ABC be a triangle, P a point, and Oa;Ha, Ob;Hb, Oc;Hc,
                                        > the circumcenters, Orthocenters of PBC, PCA, PAB.
                                        >
                                        > The locus of P such that the lines
                                        > OaHa, ObHb, OcHc [Euler Lines] are concurrent is well-known.
                                        >
                                        > Now, let O'a, H'a be the reflections of Oa, Ha in BC, resp.
                                        > Similarly O'b, H'b; O'c,H'c.
                                        >
                                        > Which is the locus of P such that the lines
                                        >
                                        > 1. O'aHa, O'bHb, O'cHc

                                        the McCay cubic + Linf + C(O,R)

                                        > 2. OaH'a, ObH'b, OcH'c

                                        its inverse in the circumcircle, a nice bicircular quintic with a
                                        triple point at O where the tangents make 60° angles with one another.

                                        > are concurrent?


                                        Best regards

                                        Bernard

                                        [Non-text portions of this message have been removed]
                                      • Antreas P. Hatzipolakis
                                        Dear Bernard ... Naturally we can replace the points O, H with another ones, but probably the case: O, K [instead of H] is most interesting (since the locus of
                                        Message 19 of 30 , Jan 30, 2004
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                                          Dear Bernard

                                          [APH]:
                                          >> [APH] Let ABC be a triangle, P a point, and Oa;Ha, Ob;Hb, Oc;Hc,
                                          >> the circumcenters, Orthocenters of PBC, PCA, PAB.
                                          >>
                                          >> The locus of P such that the lines
                                          >> OaHa, ObHb, OcHc [Euler Lines] are concurrent is well-known.
                                          >>
                                          >> Now, let O'a, H'a be the reflections of Oa, Ha in BC, resp.
                                          >> Similarly O'b, H'b; O'c,H'c.
                                          >>
                                          >> Which is the locus of P such that the lines
                                          >>
                                          >> 1. O'aHa, O'bHb, O'cHc

                                          [BG]:
                                          >the McCay cubic + Linf + C(O,R)

                                          [APH]:
                                          >> 2. OaH'a, ObH'b, OcH'c

                                          [BG]:
                                          >its inverse in the circumcircle, a nice bicircular quintic with a
                                          >triple point at O where the tangents make 60 deg. angles with one another.

                                          [APH]:
                                          >> are concurrent?

                                          Naturally we can replace the points O, H with another ones,
                                          but probably the case: O, K [instead of H] is most interesting
                                          (since the locus of the Euler lines mentioned above, is the same with
                                          the locus of the Brocard axes)

                                          So, let O'a, K'a be the reflections of Oa, Ka in BC, resp.
                                          Similarly O'b, K'b; O'c,K'c, where K's are the Lemoines
                                          of PBC, PCA, PAB.

                                          Which is the locus of P such that the lines

                                          1. O'aKa, O'bKb, O'cKc

                                          2. OaK'a, ObK'b, OcK'c

                                          are concurrent?

                                          Greetings from Athens

                                          Antreas

                                          --
                                        • Antreas P. Hatzipolakis
                                          Let ABC be a triangle, P a point, A ,B ,C the reflections of P in BC,CA,AB, resp., and A ,B ,C the reflections of P in AA ,BB ,CC resp. Which is the locus
                                          Message 20 of 30 , Mar 12, 2004
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                                            Let ABC be a triangle, P a point, A',B',C' the reflections
                                            of P in BC,CA,AB, resp., and A",B",C" the reflections of P
                                            in AA',BB',CC' resp.

                                            Which is the locus of p such that ABC, A"B"C" are
                                            (1) Perspective (2) Orthologic ?

                                            Also, if PA" /\ BC = A*, PB" /\ CA = B*, PC" /\ AB = C*,
                                            then which is the locus of P such that A*,B*,C*, are collinear?

                                            Greetings from Athens

                                            Antreas

                                            --
                                          • Antreas P. Hatzipolakis
                                            A well-known theorem is: Let L be a line passing through H. The reflections La,Lb,Lc of L in the sidelines BC,CA,AB, resp. of ABC, are concurrent. Now, denote
                                            Message 21 of 30 , Jul 13, 2006
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                                              A well-known theorem is:

                                              Let L be a line passing through H.
                                              The reflections La,Lb,Lc of L in the sidelines BC,CA,AB, resp.
                                              of ABC, are concurrent.

                                              Now, denote Ma,Mb,Mc the reflections of BC,CA,AB in La,Lb,Lc, resp.

                                              For which line(s) L [passing through H] these reflections
                                              Ma,Mb,Mc are concurrent?

                                              In general:

                                              Let L be a line, La,Lb,Lc its reflections in BC,CA,AB,resp.,
                                              and Ma,Mb,Mc the reflections of BC,CA,AB in La,Lb,Lc, resp.

                                              Which is the envelope of L such that Ma,Mb,Mc be concurrent?

                                              Antreas
                                              --
                                            • xpolakis
                                              Let ABC be a triangle P = (x:y:z) a point and L a line through P. The Reflections of AP, BP, CP in L intersect BC, CA, AB at A ,B ,C , resp. The points
                                              Message 22 of 30 , Feb 11 1:44 AM
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                                                Let ABC be a triangle P = (x:y:z) a point and L a line through P.

                                                The Reflections of AP, BP, CP in L intersect BC, CA, AB
                                                at A',B',C', resp.

                                                The points A',B',C' are collinear.
                                                Denote this line as L(P).

                                                If L = PP* (ie the line passing through two isogonal points
                                                P,P*) which is the intersection of L(P) and L(P*) ?
                                                (Special case: P = H, P* = O)

                                                APH
                                              • Francois Rideau
                                                For the moment, I see that L(P) and L(P*) are tangent to the in-conic with foci P and P* but that s all I can do! Friendly Francois [Non-text portions of this
                                                Message 23 of 30 , Feb 11 9:02 AM
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                                                  For the moment, I see that L(P) and L(P*) are tangent to the in-conic with
                                                  foci P and P* but that's all I can do!
                                                  Friendly
                                                  Francois


                                                  [Non-text portions of this message have been removed]
                                                • Bernard Gibert
                                                  Dear Antreas and François, ... The intersection Q of L(P) and L(P*) is a 15th degree point with respect to the coordinates p:q:r of P ! Very ugly ! When P =
                                                  Message 24 of 30 , Feb 11 9:20 AM
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                                                    Dear Antreas and François,

                                                    > [APH] Let ABC be a triangle P = (x:y:z) a point and L a line through
                                                    > P.
                                                    >
                                                    > The Reflections of AP, BP, CP in L intersect BC, CA, AB
                                                    > at A',B',C', resp.
                                                    >
                                                    > The points A',B',C' are collinear.
                                                    > Denote this line as L(P).
                                                    >
                                                    > If L = PP* (ie the line passing through two isogonal points
                                                    > P,P*) which is the intersection of L(P) and L(P*) ?
                                                    > (Special case: P = H, P* = O)

                                                    The intersection Q of L(P) and L(P*) is a 15th degree point with
                                                    respect to the coordinates p:q:r of P ! Very ugly !

                                                    When P = O, it is X3258, the complement of X476.

                                                    Note that L(P) and L(P*) are parallel when P and P* are two points on
                                                    Kjp = K024.

                                                    Best regards

                                                    Bernard

                                                    [Non-text portions of this message have been removed]
                                                  • xpolakis
                                                    Dear Bernard Given that X3258 lies on the NPC of ABC (pedal circle of O and H), I make this generalization (conjecture): Let ABC be a triangle, and P,P* two
                                                    Message 25 of 30 , Feb 14 1:35 PM
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                                                      Dear Bernard

                                                      Given that X3258 lies on the NPC of ABC (pedal circle of O and H),
                                                      I make this generalization (conjecture):

                                                      Let ABC be a triangle, and P,P* two isogonal
                                                      conjugate points.

                                                      The reflections of AP,BP,CP in the line PP* intersect
                                                      BC,CA,AB, at three collinear points A',B',C'.

                                                      The reflections of AP*,BP*,CP* in the line PP* intersect
                                                      BC,CA,AB, at three collinear points A*,B*,C*.

                                                      Conjecture:

                                                      The Lines A'B'C' and A*B*C* intersect at a point
                                                      lying on the common pedal circle of P and P*.

                                                      Is it true for all P's, or only for some ones? Locus??

                                                      Antreas


                                                      > > [APH] Let ABC be a triangle P = (x:y:z) a point and L a line through
                                                      > > P.
                                                      > >
                                                      > > The Reflections of AP, BP, CP in L intersect BC, CA, AB
                                                      > > at A',B',C', resp.
                                                      > >
                                                      > > The points A',B',C' are collinear.
                                                      > > Denote this line as L(P).
                                                      > >
                                                      > > If L = PP* (ie the line passing through two isogonal points
                                                      > > P,P*) which is the intersection of L(P) and L(P*) ?
                                                      > > (Special case: P = H, P* = O)

                                                      [BG]
                                                      > The intersection Q of L(P) and L(P*) is a 15th degree point with
                                                      > respect to the coordinates p:q:r of P ! Very ugly !
                                                      >
                                                      > When P = O, it is X3258, the complement of X476.
                                                      >
                                                      > Note that L(P) and L(P*) are parallel when P and P* are two points on
                                                      > Kjp = K024.
                                                    • Bernard Gibert
                                                      Dear Antreas, ... This is true for any point P on the McCay cubic and on a tricircular isogonal 12th degree curve with six real asymptotes parallel to those of
                                                      Message 26 of 30 , Feb 15 12:50 AM
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                                                        Dear Antreas,

                                                        > [APH] Given that X3258 lies on the NPC of ABC (pedal circle of O and
                                                        > H),
                                                        > I make this generalization (conjecture):
                                                        >
                                                        > Let ABC be a triangle, and P,P* two isogonal
                                                        > conjugate points.
                                                        >
                                                        > The reflections of AP,BP,CP in the line PP* intersect
                                                        > BC,CA,AB, at three collinear points A',B',C'.
                                                        >
                                                        > The reflections of AP*,BP*,CP* in the line PP* intersect
                                                        > BC,CA,AB, at three collinear points A*,B*,C*.
                                                        >
                                                        > Conjecture:
                                                        >
                                                        > The Lines A'B'C' and A*B*C* intersect at a point
                                                        > lying on the common pedal circle of P and P*.
                                                        >
                                                        > Is it true for all P's, or only for some ones? Locus??

                                                        This is true for any point P on the McCay cubic and on a tricircular
                                                        isogonal 12th degree curve with six real asymptotes parallel to those
                                                        of the Thomson cubic and the Kjp cubic.

                                                        Best regards

                                                        Bernard

                                                        [Non-text portions of this message have been removed]
                                                      • xpolakis
                                                        Dear Francois [APH] ... [FR] ... Probably is interesting the envelope of the lines L(P) as P moves on the line L. Special Case: L = Euler Line. By the way,
                                                        Message 27 of 30 , Feb 17 10:16 AM
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                                                          Dear Francois

                                                          [APH]
                                                          >>Let ABC be a triangle P = (x:y:z) a point and L a line through P.
                                                          >>
                                                          >> The Reflections of AP, BP, CP in L intersect BC, CA, AB
                                                          >> at A',B',C', resp.
                                                          >>
                                                          >> The points A',B',C' are collinear.
                                                          >> Denote this line as L(P).
                                                          >>
                                                          >> If L = PP* (ie the line passing through two isogonal points
                                                          >> P,P*) which is the intersection of L(P) and L(P*) ?
                                                          >> (Special case: P = H, P* = O)

                                                          [FR]
                                                          > For the moment, I see that L(P) and L(P*) are tangent to
                                                          > the in-conic with
                                                          > foci P and P* but that's all I can do!

                                                          Probably is interesting the envelope of the lines L(P)
                                                          as P moves on the line L.
                                                          Special Case: L = Euler Line.

                                                          By the way, which point is the intersection of the trilinear
                                                          polars of two isogonal conjugate points P,P*?

                                                          Greetings from Greece

                                                          APH
                                                        • Nikolaos Dergiades
                                                          Dear Andreas, in case of the Euler line the envelope is the inconic with foci H, O and perspector the isotomic conjugate of O. If the line L passes through the
                                                          Message 28 of 30 , Feb 17 3:39 PM
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                                                            Dear Andreas,
                                                            in case of the Euler line the envelope is
                                                            the inconic with foci H, O and perspector the
                                                            isotomic conjugate of O.
                                                            If the line L passes through the incenter I
                                                            then the envelope of L(P) is always the incircle.

                                                            Best regards
                                                            Nikos

                                                            > [APH]
                                                            > >>Let ABC be a triangle P = (x:y:z) a point and L a
                                                            > line through P.
                                                            > >>
                                                            > >> The Reflections of AP, BP, CP in L intersect BC,
                                                            > CA, AB
                                                            > >> at A',B',C', resp.
                                                            > >>
                                                            > >> The points A',B',C' are collinear.
                                                            > >> Denote this line as L(P).
                                                            >
                                                            > [FR]
                                                            > > For the moment, I see that L(P) and L(P*) are tangent
                                                            > to
                                                            > > the in-conic with
                                                            > > foci P and P* but that's all I can do!
                                                            >
                                                            > Probably is interesting the envelope of the lines L(P)
                                                            > as P moves on the line L.
                                                            > Special Case: L = Euler Line.
                                                            >
                                                            > By the way, which point is the intersection of the
                                                            > trilinear
                                                            > polars of two isogonal conjugate points P,P*?
                                                            >
                                                            > Greetings from Greece
                                                            >
                                                            > APH
                                                            >




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                                                          • xpolakis
                                                            Let ABC be a triangle, P a point, A*B*C* the cevian triangle of P, and L a line through P. The reflections of AP,BP,CP in L intersect the BC,CA,AB at the
                                                            Message 29 of 30 , May 31, 2009
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                                                              Let ABC be a triangle, P a point, A*B*C*
                                                              the cevian triangle of P, and L a line through P.

                                                              The reflections of AP,BP,CP in L intersect
                                                              the BC,CA,AB at the collinear points A',B',C', resp.
                                                              and the B*C*,C*A*,A*B* at the collinear points A",B",C", resp.

                                                              Which is the locus of the intersection of the lines
                                                              A'B'C' and A"B"C" as P moves on the line L?

                                                              Special Case: L = Euler Line of ABC.

                                                              Antreas


                                                              [APH]
                                                              > Let ABC be a triangle P = (x:y:z) a point and L a line through P.
                                                              >
                                                              > The Reflections of AP, BP, CP in L intersect BC, CA, AB
                                                              > at A',B',C', resp.
                                                              >
                                                              > The points A',B',C' are collinear.
                                                              > Denote this line as L(P).
                                                              >
                                                              > If L = PP* (ie the line passing through two isogonal points
                                                              > P,P*) which is the intersection of L(P) and L(P*) ?
                                                              > (Special case: P = H, P* = O)
                                                              >
                                                              > APH
                                                              >
                                                            • xpolakis
                                                              Let ABC be a triangle, P a point and L a line passing through P. The reflections of AP, BP, CP in L intersect BC, CA, AB in three collinear points. Call this
                                                              Message 30 of 30 , Jun 20, 2009
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                                                                Let ABC be a triangle, P a point and L a line
                                                                passing through P.

                                                                The reflections of AP, BP, CP in L
                                                                intersect BC, CA, AB in three collinear points.
                                                                Call this new line as f(L).

                                                                Reversely:

                                                                For a pointP and a line L, there are two lines
                                                                L1,L2, perpendicular at P, such that:
                                                                f(L1) = f(L2) = L.

                                                                Now, let L be the line tangent to incircle
                                                                at the Feuerbach Point.

                                                                Which are the perpendicular at I lines L1, L2,
                                                                so that f(L1) = f(L2) = L ?

                                                                Antreas





                                                                [APH]
                                                                > > >>Let ABC be a triangle P = (x:y:z) a point and L a
                                                                > > line through P.
                                                                > > >>
                                                                > > >> The Reflections of AP, BP, CP in L intersect BC,
                                                                > > CA, AB
                                                                > > >> at A',B',C', resp.
                                                                > > >>
                                                                > > >> The points A',B',C' are collinear.
                                                                > > >> Denote this line as L(P).
                                                                > >
                                                                > > Probably is interesting the envelope of the lines L(P)
                                                                > > as P moves on the line L.
                                                                > > Special Case: L = Euler Line.
                                                                > >
                                                                > > By the way, which point is the intersection of the
                                                                > > trilinear
                                                                > > polars of two isogonal conjugate points P,P*?
                                                                > >

                                                                [FR]
                                                                > > > For the moment, I see that L(P) and L(P*) are tangent
                                                                > > > to the in-conic with
                                                                > > > foci P and P* but that's all I can do!


                                                                [ND]
                                                                > in case of the Euler line the envelope is
                                                                > the inconic with foci H, O and perspector the
                                                                > isotomic conjugate of O.
                                                                > If the line L passes through the incenter I
                                                                > then the envelope of L(P) is always the incircle.
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