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

Re: [EMHL] Fuhrmann

Expand Messages
  • Ignacio Larrosa Canestro
    Saturday, January 15, 2005 9:53 PM [GMT+1=CET], ... In Episodes in Nineteenth and Twentieth Cantury Euclidean Geometry (chapter six: The Fuhrmann Circle ),
    Message 1 of 13 , Jan 16, 2005
    • 0 Attachment
      Saturday, January 15, 2005 9:53 PM [GMT+1=CET],
      Nikolaos Dergiades <ndergiades@...> escribio:

      > Sorry the following I said
      >> The following site
      >>
      > http://felix.unife.it/Root/d-Mathematics/d-The-mathematician/t-Mathematicians-A-Z#F
      >>
      >> gives that he was a Danish mathematician
      >
      > that is a Danish mathematician is not correct.
      >
      > Nikolaos Dergiades

      In "Episodes in Nineteenth and Twentieth Cantury Euclidean Geometry"
      (chapter six: "The Fuhrmann Circle"), Ross Honsberger saids that Wilhelm
      Fuhrmann was a 19th-century German geometer (1833-1904).

      Saludos,

      Ignacio Larrosa Canestro
      A Coruna (Espana)
      ilarrosa@...
    • ndergiades
      Dear friends, If ABC is a triangle and P = f(n) is an interior point with barycnetris (AP^n : BP^n : CP^n) then it is obvious that f(0) = G the centroid of
      Message 2 of 13 , Jan 16, 2005
      • 0 Attachment
        Dear friends,

        If ABC is a triangle and P = f(n) is an interior point
        with barycnetris (AP^n : BP^n : CP^n)
        then it is obvious that f(0) = G the centroid of ABC.
        Which is the point f(-1)?
        Can we find the points f(1), f(2), f(-2)?

        Best regards
        Nikolaos Dergiades
      • jpehrmfr
        Dear Nikolaos ... Some of them have been discussed in Haycinthos (note that sometimes there are many solutions - or no solution - for a given n) n = -1
        Message 3 of 13 , Jan 16, 2005
        • 0 Attachment
          Dear Nikolaos
          > If ABC is a triangle and P = f(n) is an interior point
          > with barycnetris (AP^n : BP^n : CP^n)
          > then it is obvious that f(0) = G the centroid of ABC.
          > Which is the point f(-1)?
          > Can we find the points f(1), f(2), f(-2)?

          Some of them have been discussed in Haycinthos (note that sometimes
          there are many solutions - or no solution - for a given n)
          n = -1
          Vect(AM)/AM +... = 0 gives the first Fermat point if the angles of
          A,B,C are <120, nothing elsewhere.
          n= -2
          If z,za,zb,zc are the complex affixes of P,A,B,C, this means that
          1/(z-a)+1/(z-b)+1/(z-c)=0. Hence two solutions : the focii of the
          Steiner inellipse
          n = 2
          The solutions are lying on the Kiepert hyperbola of the medial
          triangle.
          More precisely : if we consider the circular circumcubic through X
          (13),X(14),X(15),X(16),X(39),X(76),X(538),X(755) (Bernard???) the
          solutions are exactly the common points - apart X(39) - of the cubic
          and the hyperbola who are inside ABC. Hence, there exists at most 5
          solutions. Of course, a proof neads a very awful computation and, in
          any case, I don't know the number of solutions.
          n = 1 seems very complicated
          Friendly. Jean-Pierre
        • Bernard Gibert
          Dear Nikolaos and Jean-Pierre, ... What a coincidence ! I ve been working on this subject these latter days for n = +/-2. The cubic is K290 = O(X39)
          Message 4 of 13 , Jan 16, 2005
          • 0 Attachment
            Dear Nikolaos and Jean-Pierre,

            > [ND] If ABC is a triangle and P = f(n) is an interior point
            > > with  barycnetris (AP^n : BP^n : CP^n)
            > > then it is obvious that f(0) = G the centroid of ABC.
            > > Which is the point f(-1)?
            > > Can we find the points f(1), f(2), f(-2)?
            >
            > [snip]
            > n = 2
            > The solutions are lying on the Kiepert hyperbola of the medial
            > triangle.
            > More precisely : if we consider the circular circumcubic through X
            > (13),X(14),X(15),X(16),X(39),X(76),X(538),X(755) (Bernard???) the
            > solutions are exactly the common points - apart X(39) - of the cubic
            > and the hyperbola who are inside ABC. Hence, there exists at most 5
            > solutions. Of course, a proof neads a very awful computation and, in
            > any case, I don't know the number of solutions.

            What a coincidence ! I've been working on this subject these latter
            days for n = +/-2.

            The cubic is K290 = O(X39) (orthopivotal cubic).
            I have proven that the solutions are inside the in-ellipse with
            perspector X(76).

            I conjecture that there is one real solution and 4 imaginary.

            Best regards

            Bernard

            [Non-text portions of this message have been removed]
          • Paris Pamfilos
            Dear friends, here is a property of X(37), for which a would appreciate any references. Construct the minimal in perimeter/area equilateral hexagon, inscribed
            Message 5 of 13 , Jan 17, 2005
            • 0 Attachment
              Dear friends,
              here is a property of X(37), for which a would appreciate any references.

              Construct the minimal in perimeter/area equilateral hexagon, inscribed in a
              triangle ABC and having two vertices on each side. The minimal hexagon is
              symmetric. Its sides are either parts of, or parallel to, the sides of the
              triangle and its center of symmetry coincides with the triangle-center
              X(37).

              Look at
              http://server.math.uoc.gr/~pamfilos/eGallery/problems/HexadivisionSymmetric.
              html
              for a related picture

              Another related question is about the inscribed conic (with center at x(37))
              is it well known?
              Do we know some other properties of this conic?
            • jpehrmfr
              Dear Nikolaos and Bernard [ND] If ABC is a triangle and P = f(n) is an interior point ... [JPE] ... through X ... the ... cubic ... most 5 ... and, in ... [BG]
              Message 6 of 13 , Jan 17, 2005
              • 0 Attachment
                Dear Nikolaos and Bernard
                [ND] If ABC is a triangle and P = f(n) is an interior point
                > > > with  barycnetris (AP^n : BP^n : CP^n)
                > > > then it is obvious that f(0) = G the centroid of ABC.
                > > > Which is the point f(-1)?
                > > > Can we find the points f(1), f(2), f(-2)?
                [JPE]
                > > n = 2
                > > The solutions are lying on the Kiepert hyperbola of the medial
                > > triangle.
                > > More precisely : if we consider the circular circumcubic
                through X
                > > (13),X(14),X(15),X(16),X(39),X(76),X(538),X(755) (Bernard???)
                the
                > > solutions are exactly the common points - apart X(39) - of the
                cubic
                > > and the hyperbola who are inside ABC. Hence, there exists at
                most 5
                > > solutions. Of course, a proof neads a very awful computation
                and, in
                > > any case, I don't know the number of solutions.
                [BG]
                > The cubic is K290 = O(X39) (orthopivotal cubic).
                > I have proven that the solutions are inside the in-ellipse with
                > perspector X(76).
                >
                > I conjecture that there is one real solution and 4 imaginary.

                If P = x:y:z, let Fa(P) = zBP^2-yCP^2; Fb(P) = xCP^2-zAP^2;
                Fc(P)= yAP^2-xBP^2
                Then Fa(P)=0, Fb(P)=0; Fc(P)=0 are three circular cubics and they
                have 5 other common points apart the circular points at infinity.
                These points are the required points.
                Now Fa(P)+Fb(P)+Fc(P) = 0 is the Kiepert hyperbola of the medial
                triangle and
                Fa(P)/a^2+Fb(P)/b^2+Fc(P)/c^2=0 is K290 (thanks Bernard)
                The cubic and the hyperbola intersect at X(39); their 5 other common
                points are the required points
                Using the parametrization x=1/(SB+t)+1/(SC+t), ...cyclic of the
                hyperbola, we get a ugly equation of degree 5 for t. The problem is
                to find the number of real roots (at least one, of course) of the
                equation.
                Note that, as the Galois group of the equation is S(5), our 5 points
                are not even conic constructible thus there is no easier way to
                construct these points than the way above.
                Friendly. Jean-Pierre
              • Nikolaos Dergiades
                Dear friends, ... Hence we have the following construction of X(37). We extend the sides AB, AC to BC , CB such that BC =BC=CB = a. From C we draw C B ||=
                Message 7 of 13 , Jan 17, 2005
                • 0 Attachment
                  Dear friends,

                  --- Paris Pamfilos <pamfilos@...> wrote:
                  >
                  > Dear friends,
                  > here is a property of X(37), for which a would
                  > appreciate any references.
                  >
                  > Construct the minimal in perimeter/area equilateral
                  > hexagon, inscribed in a
                  > triangle ABC and having two vertices on each side.
                  > The minimal hexagon is
                  > symmetric. Its sides are either parts of, or
                  > parallel to, the sides of the
                  > triangle and its center of symmetry coincides with
                  > the triangle-center
                  > X(37).

                  Hence we have the following construction of X(37).
                  We extend the sides AB, AC to BC', CB' such that
                  BC'=BC=CB'= a.
                  From C' we draw C'B" ||= CB'= a and
                  from B' we draw B'C" ||= BC'= a
                  The segments BC", CB", B'C' have the same midpoint A*.
                  Similarly define the points B*, C*.
                  The distance of B from AC is asinC.
                  The distance of C" from AC is asinA
                  Hence the distance of A* from AC is a(sinA+sinC)/2.
                  Similarly
                  the distance of A* from AB is a(sinA+sinB)/2.
                  This means that the line AA* passes through the point
                  with trilinears (b+c : c+a : a+b) and this is X(37).
                  Hence the triangles ABC, A*B*C* are perspective
                  and the perspector is X(37).

                  Best regards
                  Nikolaos Dergiades




                  ____________________________________________________________
                  Do You Yahoo!?
                  Αποκτήστε τη δωρεάν @... διεύθυνση σας στο http://www.otenet.gr
                • Francois Rideau
                  We can also notice that the point of barycentrics (x(y+z),y(z+x),x(y+z)) is always the complement of the isotomic conjugate of the point of barycentrics
                  Message 8 of 13 , Jan 17, 2005
                  • 0 Attachment
                    We can also notice that the point
                    of barycentrics (x(y+z),y(z+x),x(y+z))
                    is always the complement of the isotomic conjugate
                    of the point of barycentrics (x,y,z), hence there is an affine
                    construction of the first point from the second,
                    here an affine construction of X(37) from the incenter X(1)
                    Best Regards
                    François Rideau
                  • Eric Danneels
                    ... Nikolaos s construction leads to many other perspectors. Using the following slightly changed notation: Let Ca be the point on AB such that BCa = BC (with
                    Message 9 of 13 , Jan 18, 2005
                    • 0 Attachment
                      --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades
                      <ndergiades@y...> wrote:
                      > Hence we have the following construction of X(37).
                      > We extend the sides AB, AC to BC', CB' such that
                      > BC'=BC=CB'= a.
                      > From C' we draw C'B" ||= CB'= a and
                      > from B' we draw B'C" ||= BC'= a
                      > The segments BC", CB", B'C' have the same midpoint A*.
                      > Similarly define the points B*, C*.
                      > The distance of B from AC is asinC.
                      > The distance of C" from AC is asinA
                      > Hence the distance of A* from AC is a(sinA+sinC)/2.
                      > Similarly
                      > the distance of A* from AB is a(sinA+sinB)/2.
                      > This means that the line AA* passes through the point
                      > with trilinears (b+c : c+a : a+b) and this is X(37).
                      > Hence the triangles ABC, A*B*C* are perspective
                      > and the perspector is X(37).

                      Nikolaos's construction leads to many other perspectors.

                      Using the following slightly changed notation:

                      Let Ca be the point on AB such that BCa = BC
                      (with B between A and Ca)
                      Define Cb, Ab, Ac, Bc and Ba similarly
                      A*, B* and C* are the midpoints of CaBa, AbCb and BcAc
                      A1 is the intersection of AcBa and CaAb
                      Define B1 and C1 similarly
                      CaBa, AbCb and BcAc form a triangle A2B2C2
                      BcCb, AcCa and BaAb form a triangle A3B3C3
                      A4 is the intersection of CbAc and BcCa
                      Define B4 and C4 similarly

                      we have the following perspectivities

                      1) Triangle ABC and A*B*C* ==> Perspector X37
                      2) Triangle ABC and A1B1C1 ==> Perspector X8
                      3) Triangle ABC and A2B2C2 ==> Perspector X65
                      4) Triangle ABC and A3B3C3 ==> Perspector X65
                      5) Triangle A2B2C2 and A*B*C* ==> Perspector X210
                      6) Triangle A1B1C1 and A2B2C2 ==> Perspector not in ETC
                      7) Triangle A1B1C1 and A3B3C3 ==> Perspector not in ETC
                      8) Triangle ABC and A4B4C4 ==> Perspector X8
                      9) Triangle A2B2C2 and A4B4C4 ==> Perspector not in ETC
                      10) Triangle A3B3C3 and A4B4C4 ==> Perspector not in ETC
                      ...

                      Best wishes for 2005

                      Eric



                      u(
                    Your message has been successfully submitted and would be delivered to recipients shortly.