Loading ...
Sorry, an error occurred while loading the content.

[EMHL] Re: semiperimeter problem

Expand Messages
  • ndergiades
    Dear Alexey and Jean-Pierre [AM] ... BC ... not on ... measures in the ... constructing ... [JPE] ... the ... signed. ... such ... when ... means ... another
    Message 1 of 3 , Aug 2 11:18 AM
    • 0 Attachment
      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
    Your message has been successfully submitted and would be delivered to recipients shortly.