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Fwd: [EMHL] Neuberg line?

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  • Francois Rideau
    ... From: Francois Rideau Date: Sat, Mar 29, 2008 at 2:02 PM Subject: Re: [EMHL] Neuberg line? To: Nikolaos Dergiades
    Message 1 of 1 , Mar 29, 2008
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      ---------- Forwarded message ----------
      From: Francois Rideau <francois.rideau@...>
      Date: Sat, Mar 29, 2008 at 2:02 PM
      Subject: Re: [EMHL] Neuberg line?
      To: Nikolaos Dergiades <ndergiades@...>


      Dear friends
      At last, I found again a book talking about the " Neuberg line" with the
      name " droite des divisions proportionnelles".
      Of course that's the Lalesco book: Geometrie du triangle, written around
      1910.
      By the way, Trajan describes very well what I told you in my last post about
      the extended similitudes of course not in my barbaric bourbakist language.
      As Neuberg was a little older, I am nearly sure he knew this line.

      The problem was: given 2 lines L = AB and L' = A'B' meeting in O to find a
      construction of the point M' = f(M) where <f> is the affine map from L onto
      L' such that:
      f(A) = A' and f(B) = B'
      So one chooses 2 new lines D and D' not parallel.
      The parallel to D through M and the parallel to D' though M' meet in a point
      P and it is easy to see that the locus of P when M moves on L is another
      line that we can also name "line of the proportional divisions wrt D and
      D'".
      When D is perpendicular to L and D' perpendicular to D', then we find again
      the Neuberg line.
      Another choice valid in any affine (not necessarily euclidian) plane is to
      take D = L' and D' = L.
      In this case if we look at <f> as a homographic correspondence between
      lines L and L', then the line so obtained is just the homographic axis.

      In the case of 3 points moving on the sides of ABC with uniform speeds, we
      get 3 Neuberg lines that is to say lines of proportional divisions wrt the
      directions perpendicular to the ABC sides and in general these 3 lines meet
      in a point the Neuberg point of which the vertices of the pedal triangle are
      homologous in the motions.

      Of course we can choose other directions to get 3 other lines of
      proportional divisions also meeting in general in one point and so on...

      So given a triangle A'B'C' there is in general a unique time t such that the
      triangles a(t)b(t)c(t) and A'B'C' are parallelogic and of course many
      special cases for which no such triangles exist or in the contrary
      infinitely many such triangles exist, this last case beeing of course the
      most interesting to construct.

      Friendly
      Francois

      2008/3/29 Francois Rideau <francois.rideau@...>:

      Dear Nikos
      > Can you tell me the number and the year of Crux where I can find your
      > problem n�2458?
      > Friendly
      > Francois
      >
      > 2008/3/27 Nikolaos Dergiades <ndergiades@...>:
      >
      > Dear Francois,
      > > I didn't know that Neuberg had used this line
      > > and I don't know if there is a name for this.
      > >
      > > I had also used this line when I proposed
      > > problem 2458 for Crux Mathematicorum.
      > > I had used for this line the name "anatomic line"
      > > of the pair (AB, A'B') since the orthogonal
      > > projections M, M' of a point of this line divide
      > > AB and A'B' in the same ratio.
      > >
      > > If the points A, B, C, D are arbitrary concyclic
      > > points then the Neuberg (anatomic) line of the pair
      > > (AB, CD) is the line passing through the center
      > > O of the circle and the intersection of lines AD, BC.
      > > Best regards
      > > Nikos Dergiades
      > >
      > >
      > > > Dear friends
      > > >
      > > > I ask you here a question of terminology.
      > > >
      > > > In the euclidian plane, we have 2 pairs of points
      > > > (A, B) and (A', B').
      > > >
      > > > For the sake of simplicity, I suppose that the lines
      > > > L = (AB) and L' =
      > > > (A'B') meet on a point O.
      > > > I look at the affine map f : L --> L' such that f(A)
      > > > = A' and f(B) = B'
      > > > If M is on the line L and M' = f(M), then we have
      > > > MA/MB = M'A'/M'B' as the
      > > > affine map f preserves ratios.
      > > > One says that the pair (M, M') is a pair of
      > > > homologous points of the map
      > > > <f>.
      > > > The perpendiculars at M to line L and at M' to line
      > > > L' meet in a point P and
      > > > it is easy to see that the locus of P is a line D.
      > > > I think that Neuberg was the first to use
      > > > frequently this line D, so I
      > > > would like to name this line, the Neuberg line of
      > > > the map <f> but maybe you
      > > > have already met it and I would be happy if you have
      > > > another idea to name
      > > > it.
      > > >
      > > > Of course this Neuberg line is useful to prove
      > > > theorems.
      > > > For example, if you take 3 pairs of points (A, B),
      > > > (A', B'), (A", B"), you
      > > > get 3 Neuberg lines and these 3 lines belongs to a
      > > > linear pencil of lines,
      > > > that is to say generally meet on a same point.
      > > > It is nearly obvious but you still have to write
      > > > down the proof!
      > > >
      > > > Friendly
      > > > Francois
      > > >
      > > >
      > > > [Non-text portions of this message have been
      > > > removed]
      > > >
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