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Re: Fermats and co

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  • Steve Sigur
    Bernard, Congratulations on a nice piece of work. Below I will outline my approach and give coordinates of some of the points you asked about. on 12/25/99
    Message 1 of 3 , Dec 28, 1999
      Bernard,

      Congratulations on a nice piece of work.

      Below I will outline my approach and give coordinates of some of the
      points you asked about.


      on 12/25/99 12:18 PM Bernard Gibert wrote

      >I have found a quite different approach to the Fermats,etc based on
      >projective geometry :
      >
      >Let's call (F) the pencil (faisceau in French) of conics with basis the two
      >Fermats Fn & Fs and the two Napoleons Nn & Ns. Kiepert hyperbola is
      >obviously one of the conics from (F).

      Another interesting one that comes to mind is the Evans conic, which goes
      through these points and the two isodynamics.

      >
      >the triangle KON (K = Lemoine, O = circumcenter, N = 9pts center i.e.
      >midpoint of [HO] ) is the autopolar triangle of (F) : each vertex is the
      >polar point of the opposite side w.r.t. any conic of (F). In particular, the
      >polar lines of K, O ,N w.r.t Kiepert hyperbola are ON (Euler line), KN (name
      >?), KO (Brocard line) but this is true for any conic.

      Very interesting and insightful.
      >
      >We know that (F) leads to a Desarguesean involution on the Brocard line KO
      >(and more generally on any line) with fixed points O and K obtained with the
      >two degenerate conics consisting of the pairs of lines concurrent in O and
      >K.

      Could you explain or give me a reference to Desarguesean involution. I
      can read French and English sources.

      >
      >Then, (g) being a conic of (F) intersecting the line KO in m and m', those
      >two points are in involution and therefore are inverse w.r.t the circle with
      >diameter OK : one is inside, the other outside unless they are on the circle
      >itself i.e. m = O or m = K.
      >
      >Let's call now :
      > m* and m'* the isogonal conjugates of m and m' (they are on Kiepert
      >hyperbola),
      > n = Hm inter Om* and n' = Hm' inter Om'* (they are on Kiepert
      >hyperbola as well) (H = orthocenter),
      > n* and n'* the isogonal conjugates of n and n' (they are on KO and
      >inverse w.r.t the circle with diameter OK).
      >
      >We have 8 points [not always distincts, see below]: 4 on KO and 4 on Kiepert
      >which are all simply related to a whole bunch of properties :
      >
      >1). nn' and m*m'* concur in K
      > mm'*, m*m', nn'*, n*n' concur in G
      > mn*, n'm'*, m*n concur in O
      > mn, m'n', m*n*, m'*n'* concur in H
      > m*n' and m'*n concur in N
      >2). two points on KO (m & m' or n & n') and the 4 on Kiepert are on a same
      >conic. (Pascal's theorem reciprocal)
      >3). m & m' are harmonic conjugates w.r.t O & K and there are many other
      >harmonic conjugates (it is a consequence of the polarity above)
      >
      >So, any point M chosen on KO leads to a group of 8 points (including M) and
      >if M is one of the two isodynamics, the 8 points will be the Fx, Nx, Ix, Nx*
      >with all the properties above.

      Again very nice but see how I come to these conclusions below by using
      isosceles Napoleons.

      >
      >****************************************
      >
      >One particular case is very interesting :
      >
      >we know that any polar line of a point m on KO w.r.t. any conic of (F) goes
      >through N so, if we take m = O (circumcenter), this polar line will meet
      >Kiepert in two points V1 & V2 and the tangent lines to Kiepert are OV1 &
      >OV2. In a previous message, I have called "Vecten points" those two points
      >(centers of concurrence of lines going through one vertex and the centre of
      >the square built externally or internally on the opposite side)
      >
      >then if we call Y1 = GV1 inter HV2 and Y2 = GV2 inter HV1, those two points
      >are on KO and are inverse w.r.t the circle with diameter OK.
      >In this situation, the 8 points are in 4 pairs of 2 merged points : V1, V2,
      >Y1, Y2 vertices of a quadrangle the diagonal triangle of which is HGK, and
      >this implies a lot of harmonic conjugacy between all those points and N & O.
      >
      >QUESTION : are those points Y1 & Y2 simply related to ABC ? are they known
      >under a specific name ?

      I worked out the barycentric coordinates of these points. Vn,s = :SA SC
      +- bb 2k + 4kk: = :N +- bb 2k:, where the notation is given below. This
      shows that the Vecten points are on the NK line.

      From the Mathematica oracle I got Y1,2 = :+- bbSB -bb 2k:. These are
      not named points but are clearly on the OK line, as you stated.

      Notation: S = (aa+bb+cc)/2, SB = S - bb, k = area of triangle.
      Coordinates of O = :bbSB: . Coordinates of N = :2 SA SC + bbSB: . Also
      4kk :1: = : bbSB + SA SC :. When coordinate are symmetrical, I only give
      the y coordinate.

      >
      >****************************************
      >
      >Another thing to study is based on the fact that the Kiepert hyperbola is
      >the isotomic conjugate of the line KG (G = centroid) therefore it is
      >possible to adapt all that has been said for the involution on KO to another
      >involution on KG but the situation is less familiar : new points have to be
      >identified and a new work has to be done. Maybe, someone will be
      >interested...
      >
      This is very interesting. Since all the properties you listed are
      projective, they are preserved by a projective mapping. We know the
      mapping that takes OK into GK. It is

      (:y:) -> (:y/bb:).

      When we do this, we get the following points.

      Ix -> isoFx
      Nx -> isoNx*
      Fx -> isoIx
      Nx* -> isoNx

      Where iso = isotomic conjugate.

      Under this transformation O -> isoH and G -> isoK so that the Euler line
      becomes
      the isoH - H line, an important line.



      ---------

      Isosceles Napoleons method.

      Choose an angle g and construct "isosceles Napoleons" on the sides of
      triangle ABC. These are triangles whose base angles are both g. They can
      be constructed both facing out from ABC or facing in. Let the vertices
      not on ABC be A', B', C'. It is known that AA', BB', CC' concur in a
      point on the Kiepert hyperbola. As g varies from 0-pi/2, the Kiepert
      hyperbola is swept out.

      We can identify the following special cases.

      g = pi/3 -> AA'/\BB'/\CC' = Fn
      g = -pi/3 -> AA'/\BB'/\CC' = Fs
      g = pi/6 -> AA'/\BB'/\CC' = Nn
      g = -pi/6 -> AA'/\BB'/\CC' = Ns
      g = pi/4 -> AA'/\BB'/\CC' = Vn
      g = -pi/4 -> AA'/\BB'/\CC' = Vs

      Here Kn,s will stand for an arbitrary pair of points on the Kiepert
      hyperbola.

      Kn,s = : 1/(cot B +- cot g) =:= 1/(SB +- 1k cot g) =:= :SA SC +- bb u +
      uu), where I have set u = 2k cot g.

      The Vecten case is very nice because cot g = 1.

      We know that Kn Kn-Ks Ks Kn + Ks are harmonic conjugate points.

      I have shown in other postings that Kn-Ks = K and that Kn+Ks is on the
      Euler line.

      Hence as angle g varies, Kn,s sweep out the Kiepert hyperbola and Kn+Ks
      sweeps out the Euler line.

      From here the concurrences are worked out as I have done in earlier posts.

      So this was my locus method.

      Steve
    • wolkb@xx.xxxxxxxxx.xx
      on 12/25/99 12:18 PM Bernard Gibert wrote ... Almost all of those collinearities given in 1) can be summarized by: n * m* n m n* m * n m K
      Message 2 of 3 , Jan 4, 2000
        on 12/25/99 12:18 PM Bernard Gibert wrote

        >I have found a quite different approach to the Fermats,etc based on
        >projective geometry :
        >
        >Let's call (F) the pencil (faisceau in French) of conics with basis the two
        >Fermats Fn & Fs and the two Napoleons Nn & Ns. Kiepert hyperbola is
        >obviously one of the conics from (F).
        >
        >the triangle KON (K = Lemoine, O = circumcenter, N = 9pts center i.e.
        >midpoint of [HO] ) is the autopolar triangle of (F) : each vertex is the
        >polar point of the opposite side w.r.t. any conic of (F). In particular, the
        >polar lines of K, O ,N w.r.t Kiepert hyperbola are ON (Euler line), KN (name
        >?), KO (Brocard line) but this is true for any conic.
        >
        >We know that (F) leads to a Desarguesean involution on the Brocard line KO
        >(and more generally on any line) with fixed points O and K obtained with the
        >two degenerate conics consisting of the pairs of lines concurrent in O and
        >K.
        >
        >Then, (g) being a conic of (F) intersecting the line KO in m and m', those
        >two points are in involution and therefore are inverse w.r.t the circle with
        >diameter OK : one is inside, the other outside unless they are on the circle
        >itself i.e. m = O or m = K.
        >
        >Let's call now :
        > m* and m'* the isogonal conjugates of m and m' (they are on Kiepert
        >hyperbola),
        > n = Hm inter Om* and n' = Hm' inter Om'* (they are on Kiepert
        >hyperbola as well) (H = orthocenter),
        > n* and n'* the isogonal conjugates of n and n' (they are on KO and
        >inverse w.r.t the circle with diameter OK).
        >
        >We have 8 points [not always distincts, see below]: 4 on KO and 4 on Kiepert
        >which are all simply related to a whole bunch of properties :
        >
        >1). nn' and m*m'* concur in K
        > mm'*, m*m', nn'*, n*n' concur in G
        > mn*, n'm'*, m*n concur in O
        > mn, m'n', m*n*, m'*n'* concur in H
        > m*n' and m'*n concur in N
        >2). two points on KO (m & m' or n & n') and the 4 on Kiepert are on a same
        >conic. (Pascal's theorem reciprocal)
        >3). m & m' are harmonic conjugates w.r.t O & K and there are many other
        >harmonic conjugates (it is a consequence of the polarity above)
        >
        >So, any point M chosen on KO leads to a group of 8 points (including M) and
        >if M is one of the two isodynamics, the 8 points will be the Fx, Nx, Ix, Nx*
        >with all the properties above.

        Almost all of those collinearities given in 1) can be summarized by:

        n'* m* n' m
        n* m'* n m'
        K H G O

        form a desmic triple of quadrilaterals.

        Of course other desmic triples can be found, but they usually involve
        one or more unnamed points. The special case m=Ix gives:

        Ns* Fn Ns In
        Nn* Fs Nn Is
        K H G O

        where all points are familiar. Note that each of the three quadrilaterals
        here is slightly collapsed, since three of its four vertices are collinear.
        (Nn*, Nn, N, Fs are collinear, as are Ns*, Ns, N, Fn.)
        --
        Barry Wolk <wolkb@...>
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