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Re: [EMHL] Polar of H

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  • Alexey.A.Zaslavsky
    ... Dear Alex! This fact follows from next theorem. Let two triangles ABC and A B C are perspective (AA , BB and CC concur in point P) and orthologic (the
    Message 1 of 13 , Nov 3, 2003
      >Dear friends!
      >The polar of orthocenter is perpendicular to Euler line.
      >May be it is well known fact, but short and syntnetic proof of it? Is it
      >well known also?
      >
      Dear Alex!
      This fact follows from next theorem. Let two triangles ABC and A'B'C' are
      perspective (AA', BB' and CC' concur in point P) and orthologic (the
      perpendiculars from A', B', C' to respectively BC, CA, AB concur in point
      Q). Then PQ is perpendicular to perspective axis. IF A'B'C' is the
      orthotriangle of ABC we obtain your fact. There are many another interesting
      corollary from this theorem. Also it has a 3-dimensial analogous.

      Sincerely Alexey
    • jpehrmfr
      Dear Alex and Alexey ... it? Is it ... A B C are ... (the ... in point ... interesting ... Just a little remark : in the present case, we have P = Q = H.
      Message 2 of 13 , Nov 3, 2003
        Dear Alex and Alexey

        > >The polar of orthocenter is perpendicular to Euler line.
        > >May be it is well known fact, but short and syntnetic proof of
        it? Is it
        > >well known also?
        > >
        > Dear Alex!
        > This fact follows from next theorem. Let two triangles ABC and
        A'B'C' are
        > perspective (AA', BB' and CC' concur in point P) and orthologic
        (the
        > perpendiculars from A', B', C' to respectively BC, CA, AB concur
        in point
        > Q). Then PQ is perpendicular to perspective axis. IF A'B'C' is the
        > orthotriangle of ABC we obtain your fact. There are many another
        interesting
        > corollary from this theorem. Also it has a 3-dimensial analogous.

        Just a little remark : in the present case, we have P = Q = H. Thus,
        you have to add to your theorem the fact that the perpendicular from
        A,B,C to B'C', C'A',A'B' concur at a point Q' on the line PQ.
        In the present case, Q' = O and we are done.
        Of course, the proof of the theorem above is much more difficult
        than a direct proof of the fact that the orthic axis is
        perpendicular to the Euler line.
        Another remark : if a point moves on a circumconic going through G,
        his trilinear polar has a fixed direction. Thus, the trilinear polar
        of M is perpendicular to the Euler line iff M lies on Kiepert
        hyperbola.
        Friendly. Jean-Pierre
      • xpolakis
        Dear Alex and Darij ... In general, if Q = (u:v:w) is a fixed point and P = (x:y:z) is a variable point, then the locus of P such that the trilinear polar of P
        Message 3 of 13 , Nov 4, 2003
          Dear Alex and Darij

          [Alex]:
          > >> The polar of orthocenter is perpendicular to Euler line.
          >

          In general, if Q = (u:v:w) is a fixed point and
          P = (x:y:z) is a variable point, then the locus of P such
          that the trilinear polar of P is perpendicular to QP line,
          is the cubic with trilinear equation (if my calculations
          were correct):

          x(y^2(u - wcosB) - z^2(u - vcosC)) + cyclic = 0

          If Q = O, then the locus is Darboux cubic, which contains H.


          [Darij]:

          > Yes, if you mean the trilinear polar. In fact, the trilinear
          > polar of the orthocenter of a triangle ABC is the radical
          > axis of the circumcircle and the nine-point circle of ABC.

          In general, which is the locus of P such that the
          trilinear polar of P is the radical axis of the circumcircle
          and the pedal (or cevian) circle of P?


          Antreas
        • Antreas P. Hatzipolakis
          Let ABC be a triangle, P a pointt, and P* its isogonal conjugate. Which is the locus of P such that the trilinear polars of P and P* are (1) perpendicular (2)
          Message 4 of 13 , Nov 4, 2003
            Let ABC be a triangle, P a pointt, and P* its isogonal
            conjugate.

            Which is the locus of P such that the trilinear polars
            of P and P* are (1) perpendicular (2) parallel?

            Antreas
            --
          • jpehrmfr
            Dear antreas ... As the trilinear polar of P has a fixed direction when P moves on a circumconic going through G, the trilinear polar of P and P* are parallel
            Message 5 of 13 , Nov 4, 2003
              Dear antreas
              > Which is the locus of P such that the trilinear polars
              > of P and P* are (1) perpendicular (2) parallel?

              As the trilinear polar of P has a fixed direction when P moves on a
              circumconic going through G, the trilinear polar of P and P* are
              parallel iff the line PP* goes through K. Hence, the locus (2) is
              the Grebe cubic - self-isogonal with pivot K - = K102 in Bernard's
              site.
              The other one is K107.
              Friendly. Jean-Pierre
            • Bernard Gibert
              Dear Antreas, ... I m not sure your computation is correct. The barycentric equation I have is : [ (SB u + a^2w)y - (SC u + a^2v)z] yz + cyclic = 0 It is a
              Message 6 of 13 , Nov 4, 2003
                Dear Antreas,

                > [APH] In general, if Q = (u:v:w) is a fixed point and
                > P = (x:y:z) is a variable point, then the locus of P such
                > that the trilinear polar of P is perpendicular to QP line,
                > is the cubic with trilinear equation (if my calculations
                > were correct):
                >
                >      x(y^2(u - wcosB) - z^2(u - vcosC)) + cyclic = 0
                >
                > If Q = O, then the locus is Darboux cubic, which contains H.

                I'm not sure your computation is correct. The barycentric equation I
                have is :

                [ (SB u + a^2w)y - (SC u + a^2v)z] yz + cyclic = 0

                It is a circum-cubic K passing through G and having its asymptotes
                parallel to those of the Thomson cubic which is obtained with Q = O.

                1. When Q lies at infinity, K decomposes into the line at infinity and
                a circum-conic through G.

                2. K is a pK iff Q lies on Darboux and then, the pole W lies on
                Thomson, the pivot R on Lucas.
                We find a family of pK having all their asymptotes parallel to the
                Thomson cubic.

                question : what are the mappings Q -> W, Q -> R ?

                3. K is a nK0 iff Q lies on the circumcircle and then, the pole is Q
                and the root is the tripole of the Simson line of Q (a point on the
                Simson cubic).

                Best regards

                Bernard

                [Non-text portions of this message have been removed]
              • fredlangch
                Dear Bernard, ... It seems that Q - R is the classical Pinkernell transfo. (Same algebraïc formulas). Regards. Fred
                Message 7 of 13 , Nov 4, 2003
                  Dear Bernard,
                  > 2. K is a pK iff Q lies on Darboux and then, the pole W lies on
                  > Thomson, the pivot R on Lucas.
                  > We find a family of pK having all their asymptotes parallel to the
                  > Thomson cubic.
                  >
                  > question : what are the mappings Q -> W, Q -> R ?

                  It seems that Q -> R is the classical Pinkernell transfo.
                  (Same algebraïc formulas).
                  Regards.
                  Fred
                • Alexey.A.Zaslavsky
                  Dear Jean-Pierre! ... Of course this fact must be used. It is known that if the perpendicular from A,B,C to B C , C A ,A B concur at a point Q then the
                  Message 8 of 13 , Nov 5, 2003
                    Dear Jean-Pierre!
                    >
                    >Just a little remark : in the present case, we have P = Q = H. Thus,
                    >you have to add to your theorem the fact that the perpendicular from
                    >A,B,C to B'C', C'A',A'B' concur at a point Q' on the line PQ.
                    >In the present case, Q' = O and we are done.
                    >
                    Of course this fact must be used. It is known that if the perpendicular from
                    A,B,C to B'C', C'A',A'B' concur at a point Q then the erpendicular from
                    A',B',C' to BC, CA, AB concur at a point Q'. So if two triangles are
                    perspective and orthologic the theorem prove that the perspective center and
                    two orthologic centers are collinear. Another interesting fact: if Q=Q' then
                    AA', BB' and CC' concur. Infortunately, my proof is very hard.

                    Sincerely Alexey
                  • Bernard Gibert
                    Dear Fred, ... you re right ! W is the perspector of the pedal triangle of Q and the anticevian triangle of Q, R is the perspector of the pedal triangle of Q
                    Message 9 of 13 , Nov 5, 2003
                      Dear Fred,

                      > [BG]
                      > > 2. K is a pK iff Q lies on Darboux and then, the pole W lies on
                      > > Thomson, the pivot R on Lucas.
                      > > We find a family of pK having all their asymptotes parallel to the
                      > > Thomson cubic.
                      > >
                      > > question : what are the mappings Q -> W, Q -> R ?
                      > [FL]
                      > It seems that Q -> R is the classical Pinkernell transfo.
                      > (Same algebraïc formulas).

                      you're right !

                      W is the perspector of the pedal triangle of Q and the anticevian
                      triangle of Q,

                      R is the perspector of the pedal triangle of Q and ABC.

                      Thank you.

                      Best regards

                      Bernard

                      [Non-text portions of this message have been removed]
                    • Antreas P. Hatzipolakis
                      ... Dear Jean-Pierre There are more similar problems on tripolars, and probably we have discussed some of them in the past. Two other ones: Which is the locus
                      Message 10 of 13 , Nov 5, 2003
                        [APH]:

                        >> Which is the locus of P such that the trilinear polars
                        >> of P and P* are (1) perpendicular (2) parallel?

                        [JPE]:

                        >As the trilinear polar of P has a fixed direction when P moves on a
                        >circumconic going through G, the trilinear polar of P and P* are
                        >parallel iff the line PP* goes through K. Hence, the locus (2) is
                        >the Grebe cubic - self-isogonal with pivot K - = K102 in Bernard's
                        >site.
                        >The other one is K107.


                        Dear Jean-Pierre

                        There are more similar problems on tripolars,
                        and probably we have discussed some of them
                        in the past.

                        Two other ones:

                        Which is the locus of P such that the tripolars of
                        P, P*, and the line PP* are concurrent?
                        [P* = isog. conj. of P]

                        Which is the locus of P such that the tripolars
                        of P, P*, and P# are concurrent?
                        [P# = isot. conj. of P]

                        Greetings

                        Antreas


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