## [EMHL] Re: semiperimeter problem

Expand Messages
• 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, 2005
• 0 Attachment
Dear Alexey and Jean-Pierre

[AM]
> >> Let ABCD be convex quadrilateral, and X, Y - points on segments
BC
> >> 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
constructing
> 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 on
the
> ray OAD such as BX+AB+AY = XC+CD+YD where BX, AY, XC, YD are
signed.
> Suppose that X,Y move respectively on two rays from O,say R, R'
such
> 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
when
> X and Y lie on the tangent at the vertex of the parabola which
means
> 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