[EMHL] Re: semiperimeter problem
- Dear Alexey and Jean-Pierre
> >> Let ABCD be convex quadrilateral, and X, Y - points on segmentsBC
> > and AD,not on
> >> 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
> sides, but on lines, passing trough corresponding sides, themeasures in the
> sum are oriented. And now minimum always exsists - and isconstructing
> exactly in that way as you have pointed out.[JPE]
> It should true if we suppose that X lies on the ray OBC and Y onthe
> ray OAD such as BX+AB+AY = XC+CD+YD where BX, AY, XC, YD aresigned.
> Suppose that X,Y move respectively on two rays from O,say R, R'such
> as OX + OY = k (constant)when
> 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 whichmeans
> that they are the projections of F upon the two rays (oranother approach I think is the following:
> equivalently that OX = OY = k/2)
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.