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16179Re:[EMHL] the problem

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  • Nikolaos Dergiades
    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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