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A perspector question

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  • Hauke Reddmann
    Let ABC and A B C be perspective trangles. Can ABC and A B C be similar *without* AB||A B etc.? Hauke
    Message 1 of 3 , Dec 5, 2008
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      Let ABC and A'B'C' be perspective trangles.
      Can ABC and A'B'C' be similar *without*
      AB||A'B' etc.?

      Hauke
    • Kostas Vittas
      Dear Hauke, I think the answer is yes. If we consider an arbitrary convex quadrilateral ABCD, we can construct a point P inwardly to it, such that the
      Message 2 of 3 , Dec 5, 2008
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        Dear Hauke, I think the answer is yes.

        If we consider an arbitrary convex quadrilateral ABCD, we can construct
        a point P inwardly to it, such that the triangles PAB, PDC to be
        similar, with for example <PAB = <PCD and <PBA = <PDC, as well known.

        ( The point P is the intersection point of two Appolonian circles with
        respect to the diagonals AC, BD of ABCD and ratio r = AB / CD ).

        Let now S be, the intersection point of the sidelines AD, BC and we
        define three points P', C', D', on the rays SP, SC, SD respectively,
        such that P'C' // PC and P'D' // PD.

        The triangles PAB, P'D'C', are perspective and similar as the problem
        states.

        Best regards, Kostas Vittas.


        --- In Hyacinthos@yahoogroups.com, "Hauke Reddmann" <fc3a501@...> wrote:
        >
        > Let ABC and A'B'C' be perspective trangles.
        > Can ABC and A'B'C' be similar *without*
        > AB||A'B' etc.?
        >
        > Hauke
        >
      • Hauke Reddmann
        ... A bit of computing also affirms that - a triangle similar to ABC can be rotated (1), scaled (1) and shifted (2 free variables), and the perspector just
        Message 3 of 3 , Dec 7, 2008
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          --- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...> wrote:
          >
          > Dear Hauke, I think the answer is yes.
          >
          A bit of computing also affirms that -
          a triangle similar to ABC can be rotated
          (1), scaled (1) and shifted (2 free variables),
          and the perspector just demands one equation.

          By ray law (or what it's called in English)
          the line through the barycenters M and M' of
          ABC and A'B'C' also goes through O. In the
          "standard" case of course a line through *any*
          two centers will go through O but it still
          would be interesting if, say, H,H' and O are
          collinear (H orthocenter) and still the sides
          of ABC and A'B'C' not parallel. Remember, there
          are still 3 variables to use up :-)

          Hauke
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