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Re: Re: Θέμα: [EMHL] the problem

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  • Francois Rideau
    Dear Alexey Thank you for your remarks. I will try to draw all the envelope with Cabri and maybe SketchPad if possible. Friendly Francois ... [Non-text
    Message 1 of 4 , Mar 1, 2008
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      Dear Alexey
      Thank you for your remarks.
      I will try to draw all the envelope with Cabri and maybe SketchPad if
      possible.
      Friendly
      Francois
      2008/2/29 Alexey.A.Zaslavsky <zasl@...>:

      >
      >


      [Non-text portions of this message have been removed]
    • Nikolaos Dergiades
      Dear Alexey, and Francois This is very interesting. If the points P, Q are on the sides of angle A in order to have (AP+AQ)/(APQ) = (AB+BC+CD+DA)/(ABCD) =
      Message 2 of 4 , Mar 3, 2008
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        Dear Alexey, and Francois
        This is very interesting.
        If the points P, Q are on the sides of angle A
        in order to have
        (AP+AQ)/(APQ) = (AB+BC+CD+DA)/(ABCD) = constant
        the line PQ passes through a fix point A' on the
        A_bisector of ABCD.
        Similarly we have the points B', C', D' on the
        bisectors of angles B, C, D of ABCD.

        The construction of these points is as follows:
        On the semiline BA we take the point T such that
        AT = BC+CD+DA. The parallel from D to AC meets
        BA at S. The parallel from S to CT meets BC at U
        and the parallel from U to BA meets the A bisector
        at the required point A'.
        The parallel from A' to AD gives on D_bisector
        the point D'. The parallel from D' to DC gives
        of C_bisector the point C'. The parallel from C'
        to BC gives on B_bisector the point B' and the
        quadrilateral A'B'C'D' has parallel sides to
        the corresponding sides of ABCD.

        Let S1 be the intersection of lines AD, BC and
        let S2 be the intersection of lines AB, CD.
        Let A be between S1 and D and
        let A be between S2 and B.
        The line A'D' is the reflection of B'C' in the
        bisector of angle AS1B and the line
        A'B' is the reflection of line C'D' in the
        bisector of angle AS2D.

        If the points P, Q are on sides BC, AD then
        the envelope of PQ is a conic that is tangent
        to the lines AD, BC, A'D', B'C', CD', C'D and the
        center O1 of this conic is the center of the rhombus
        formed by the lines AD, BC, A'D', B'C'.
        We can find the tangency points with these lines
        and then construct the conic.
        for example if line CD' meets AD at C" then the
        perspector P1 of this inconic in triangle C"S1C
        is found as the isotomic conjugate of the
        anticomplement of O1. Then the lines
        S1P1, C"P1 and C'P1 give the points of tangency
        with the sides of triangle C"S1C.
        The true envelope is the arc of this conic from
        the tangency point of CD' to the tangency
        point of line C'D.
        Similarly if the points P, Q are on sides AB, CD
        then the true envelope of PQ is an arc of another
        conic defined as previously
        with center the center of the rhombus formed
        by the lines AB, CD, A'B', C'D', from the
        tangency point of this conic with the line BC'
        to the tangency point with the line B'C.

        Best regards
        Nikos Dergiades



        > Dear Nikos and Francois!
        > If the points P, Q aren't on the opposite sides then
        > the line PQ pass
        > through fixed point. In fact let P be on AB, Q on
        > AD. Then we have
        > (AP+AQ)/S(APQ)=
        > P(ABCD)/S(ABCD)=const. So the common point of PQ and
        > bissector of A is
        > fixed.
        > This follows that the envelop of all lines PQ is
        > formed by the arcs of
        > conics and the segments of lines. So it isn't
        > smooth. Is it convex?
        >
        > Sincerely
        > Alexey
        >
        >
        >
        >
        > Antivirus scanning: Symantec Mail Security for SMTP.
        >
        >
        > Yahoo! Groups Links
        >
        >
        >
        >
        >




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