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Paracevian perspector ?

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  • Eric Danneels
    Dear Hyacinthians, I wonder if the following general perspectivity is known: Consider a triangle ABC and 2 different points P and Q Let A B C be the cevian
    Message 1 of 4 , Jul 23, 2004
      Dear Hyacinthians,

      I wonder if the following general perspectivity is known:

      Consider a triangle ABC and 2 different points P and Q
      Let A'B'C' be the cevian triangle of P wrt triangle ABC
      The parallel to BC through Q intersects AP at A*
      The parallel to AP through Q intersects BC at A#
      Define B*, B#, C* and C# similarly

      ==> The triangles A*B*C* and A#B#C# are perspective

      Let P = (p : q : r), Q = (u : v : w) (in barycentrics)
      then the perspector S has coordinates
      ( p.(q + r).u.v.w + p.r.r.u.u.v + p.q.u.u.w -q.r.u.u.u : cyclic...)

      I found more than 280 combinations of points P and Q in the ETC
      which lead to a perspector also in the ETC

      What intrigues me most about this conjugation is the following
      property:

      With a given point P there seem to be 2 points Q1 and Q2 that lead
      to the same perspector S and moreover it seems that S is the
      midpoint of Q1 and Q2.

      However I can't see why, using the formula above

      Greetings from Bruges

      Eric Danneels
    • Bernard Gibert
      Dear Eric ... Your S is the midpoint of Q and P-Ceva conj. of Q. Best regards Bernard [Non-text portions of this message have been removed]
      Message 2 of 4 , Jul 23, 2004
        Dear Eric

        > [ED] I wonder if the following general perspectivity is known:
        >
        > Consider a triangle ABC and 2 different points P and Q
        > Let A'B'C' be the cevian triangle of P wrt triangle ABC
        > The parallel to BC through Q intersects AP at A*
        > The parallel to AP through Q intersects BC at A#
        > Define B*, B#, C* and C# similarly
        >
        > ==> The triangles A*B*C* and A#B#C# are perspective
        >
        > Let P = (p : q : r), Q = (u : v : w) (in barycentrics)
        > then the perspector S has coordinates
        > ( p.(q + r).u.v.w + p.r.r.u.u.v + p.q.u.u.w -q.r.u.u.u : cyclic...)

        ^^^^^^^^^^^^^^^^^^^^^^^+ p r u u v (typo) + etc

        > I found more than 280 combinations of points P and Q in the ETC
        > which lead to a perspector also in the ETC
        >
        > What intrigues me most about this conjugation is the following
        > property:
        >
        > With a given point P there seem to be 2 points Q1 and Q2 that lead
        > to the same perspector S and moreover it seems that S is the
        > midpoint of Q1 and Q2.
        >
        > However I can't see why, using the formula above


        Your S is the midpoint of Q and P-Ceva conj. of Q.

        Best regards

        Bernard

        [Non-text portions of this message have been removed]
      • jpehrmfr
        Dear Eric and Bernard ... cyclic...) ... lead ... [BG] ... This leads to a construction of Q if we know P and S but I don t understand why we cannot get 4
        Message 3 of 4 , Jul 23, 2004
          Dear Eric and Bernard
          > > [ED] I wonder if the following general perspectivity is known:
          > >
          > > Consider a triangle ABC and 2 different points P and Q
          > > Let A'B'C' be the cevian triangle of P wrt triangle ABC
          > > The parallel to BC through Q intersects AP at A*
          > > The parallel to AP through Q intersects BC at A#
          > > Define B*, B#, C* and C# similarly
          > >
          > > ==> The triangles A*B*C* and A#B#C# are perspective
          > >
          > > Let P = (p : q : r), Q = (u : v : w) (in barycentrics)
          > > then the perspector S has coordinates
          > > ( p.(q + r).u.v.w + p.r.r.u.u.v + p.q.u.u.w -q.r.u.u.u :
          cyclic...)
          >
          > ^^^^^^^^^^^^^^^^^^^^^^^+ p r u u v (typo) + etc
          >
          > > I found more than 280 combinations of points P and Q in the ETC
          > > which lead to a perspector also in the ETC
          > >
          > > What intrigues me most about this conjugation is the following
          > > property:
          > >
          > > With a given point P there seem to be 2 points Q1 and Q2 that
          lead
          > > to the same perspector S and moreover it seems that S is the
          > > midpoint of Q1 and Q2.
          > >
          > > However I can't see why, using the formula above

          [BG]
          > Your S is the midpoint of Q and P-Ceva conj. of Q.

          This leads to a construction of Q if we know P and S but I don't
          understand why we cannot get 4 solutions both by both reflected of
          each other wrt S.
          Note that when M moves on a line L, the Cevian quotient P/M moves on
          the conic C(P,L) going through the feet of the cevians of P and the
          feet of the cevians of the trilinear pole of L.
          Now draw through S the lines L,L' going through the infinite points
          of the circumconic going through P and S (if the conic is an
          ellipse, Q cannot exist)
          The solutions are the common points of L and C(P,L) and the common
          points of L' and C(P,L').
          Once more, I don't understand why it should be impossible to get 4
          real solutions.
          Friendly. Jean-Pierre
        • Darij Grinberg
          ... Cheers, Eric, for finding such a beautiful theorem! It was indeed too beautiful for me to leave it without a proper synthetic proof, so I have been looking
          Message 4 of 4 , Sep 1, 2004
            In Hyacinthos message #10135, Eric Danneels wrote:

            >> Consider a triangle ABC and 2 different
            >> points P and Q
            >> [...]
            >> The parallel to BC through Q intersects
            >> AP at A*
            >> The parallel to AP through Q intersects
            >> BC at A#
            >> Define B*, B#, C* and C# similarly
            >>
            >> ==> The triangles A*B*C* and A#B#C# are
            >> perspective

            Cheers, Eric, for finding such a beautiful
            theorem! It was indeed too beautiful for me to
            leave it without a proper synthetic proof, so
            I have been looking for one, and finally I have
            found a rather long synthetic proof with Ceva
            and Menelaos. You can find it in the note
            "On the paracevian perspector" on my website

            http://de.geocities.com/darij_grinberg

            Let me tell in brief how the proof goes: At
            first, I construct the cevian triangle A'B'C'
            of the point P with respect to the triangle
            ABC. Then, I show that the lines through the
            points A', B', C' parallel to the lines A*A#,
            B*B#, C*C# concur at one point R. Finally, I
            prove (and this is very easy) that the
            midpoint R' of the segment QR lies on the
            lines A*A#, B*B#, C*C#, and thus these lines
            concur, i. e. the triangles A*B*C* and A#B#C#
            are perspective. Actually, the point R is the
            P-Ceva conjugate of the point Q; this fact
            (which is due to Bernard Gibert, Hyacinthos
            message #10136) I didn't prove in my note,
            but it can be easily obtained using the
            auxiliary points I constructed there.

            Sincerely,
            Darij Grinberg
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