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11448[EMHL] Re: semiperimeter problem

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  • ndergiades
    Aug 2, 2005
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      Dear Alexey and Jean-Pierre

      > >> Let ABCD be convex quadrilateral, and X, Y - points on segments
      > > and AD,
      > >> so that YA+AB+BX=YD+DC+CX. Can we construct XY of maximum or
      > > minimum length?
      > >> Can we construct it syntheticly?
      > Can we make some generalization of the problem? i.e. X and Y lay
      not on
      > sides, but on lines, passing trough corresponding sides, the
      measures in the
      > sum are oriented. And now minimum always exsists - and is
      > exactly in that way as you have pointed out.

      > It should true if we suppose that X lies on the ray OBC and Y on
      > ray OAD such as BX+AB+AY = XC+CD+YD where BX, AY, XC, YD are
      > Suppose that X,Y move respectively on two rays from O,say R, R'
      > as OX + OY = k (constant)
      > Then the line XY envelopes a parabola touching R and R', with axis
      > the bisector of (R,R') and focus F the second fixed point of the
      > circles OXY;so 2.OF.cos(u/2) = k where u = <R,R'.
      > In our particular case, we have
      > k = 1/2 (OA+OB-AB+OC+OD+CD) = (OI+OJ)cos(u/2) with I,J as above.
      > Hence F is the midpoint of IJ; XY has no maximum and is minimum
      > X and Y lie on the tangent at the vertex of the parabola which
      > that they are the projections of F upon the two rays (or
      > equivalently that OX = OY = k/2)

      another approach I think is the following:
      If PQ means signed and (PQ) not signed
      M1, M2 are the midpoints of BC, AD respectively
      then we want BX+(AB)+AY = XC+(CD)+YD or
      XB+XC+YA+YD = (AB)-(CD) = m = constant or 2XM1 + 2XM2 = m
      or if P is the intersection of the perpendicular to BC at X with the
      perpendicular to AD at Y
      [(PB)^2-(PC)^2]/(CB)+ [(PA)^2-(PD)^2]/(AD) = constant
      From the above we conclude that the locus of P is a line L (easily
      constructable from two points of the locus)
      and since the points X,P,Y,O are cyclic we have XY = OP.sin(u)
      and XY becomes minimum if P is the orthogonal projection of O on to

      If the lines R, R' are parallel then it is easy to prove that the
      minimum XY is the
      distance of the parallels.

      Best regards
      Nikolaos Dergiades
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