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Contrieved Euler line construction

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  • shokoshu2
    This should be self-explaining... http://pic-hoster.net/view/8336/threesq.gif ...but if not: 0. Given a triangle ABC. 1. Choose some rectangle shape r (3:1 in
    Message 1 of 5 , Mar 10 3:22 AM
      This should be self-explaining...
      http://pic-hoster.net/view/8336/threesq.gif
      ...but if not:
      0. Given a triangle ABC.
      1. Choose some rectangle shape r (3:1 in the example).
      2. Inscribe a rectangle WXYZ with WX/XY=r into ABC:
      W goes on CA, X on AB and YZ lies on BC.
      3. Join the midpoint of YZ with A. Call this line g1.
      4. Repeat construction for the other sides to get
      g2 and g3.
      5. The g_i concur in a point P.
      (Proof: Apply Pythagoras, ray and Ceva theorems.
      Juggle a bit, I could do it in one minute on piece
      and paper, which I haven't done since years. Thus
      a purely geometric proof shouldn't be hard either.)
      7. The Ceva ratios are linear in r. Obviously,
      r=0 (YZ=BC) gives #2 and r=oo (Y=Z) #4. I thus
      strongly suspect P lies on the Euler line for any r.

      An additional question would be whether you could
      take the midpoint of WX [yes, guess why :-)]
      or the midpoint of the rectangle for the construction
      instead. Again I suspect yes (and probably any ratio
      R of the height of P with respect to the rectangle
      will do - again by ray theorem).

      Hauke
    • chris.vantienhoven
      Dear Hauke, You really surprised me with this nice problem! I investigated it and got some results. 1. The 3:1-rectangle has a length/width-ratio of 3. 2. The
      Message 2 of 5 , Mar 10 8:33 AM
        Dear Hauke,

        You really surprised me with this nice problem!
        I investigated it and got some results.
        1. The 3:1-rectangle has a length/width-ratio of 3.
        2. The points in your drawing lies on these lines;
        X20.X1327, X140.X3316, X590.X1131, X1132.X3068, X1328.X3091, X1656.X3317.
        3. This point is not on the Eulerline.
        But special is that it is on the Kiepert Hyperbola.
        4. The barycentrics of this point are: (SB + 3K) (SC + 3K),
        where SB = (a^2 ¨C b^2 + c^2) / 2, (a^2 + b^2 - c^2) / 2, K = Area of Triangle.
        5. When we take a 3:1-rectangle outside the Reference Triangle we get a corresponding point on these lines:
        X20.X1328, X140.X3317, X615.X1132, X1131.X3069, X1327.X3091, X1656.X3316.
        Again a point on the Kiepert Hyperbola.
        The barycentrics are: (SB - 3K) (SC - 3K).
        I think it is not without meaning that in both cases there is adding or subtracting of 3 times the TriangleArea in the barycentrics!
        6. When changing the length/width-ratio there developes a locus.
        The locus of these points is surprisingly (or not) the Kiepert Hyperbola!
        7. Let¡¯s name the length/width-ratio ¡°n¡±.
        When n=1 the perspector=X1131.
        When n=-1 the perspector=X1132.
        When n=2 the perspector=X485.
        When n=-2 the perspector=X486.
        When n=4 the perspector=X3316.
        When n=-4 the perspector=X3317.
        When n=2/3 ¡Ì3 the perspector=X13.
        When n=-2/3 ¡Ì3the perspector=X14.
        When n=2¡Ì3 the perspector=X17.
        When n=-2¡Ì3 the perspector=X18.
        Sofar my results.
        There must be more!
        Best Regards,

        Chris van Tienhoven




        --- In Hyacinthos@yahoogroups.com, "shokoshu2" <fc3a501@...> wrote:
        >
        > This should be self-explaining...
        > http://pic-hoster.net/view/8336/threesq.gif
        > ...but if not:
        > 0. Given a triangle ABC.
        > 1. Choose some rectangle shape r (3:1 in the example).
        > 2. Inscribe a rectangle WXYZ with WX/XY=r into ABC:
        > W goes on CA, X on AB and YZ lies on BC.
        > 3. Join the midpoint of YZ with A. Call this line g1.
        > 4. Repeat construction for the other sides to get
        > g2 and g3.
        > 5. The g_i concur in a point P.
        > (Proof: Apply Pythagoras, ray and Ceva theorems.
        > Juggle a bit, I could do it in one minute on piece
        > and paper, which I haven't done since years. Thus
        > a purely geometric proof shouldn't be hard either.)
        > 7. The Ceva ratios are linear in r. Obviously,
        > r=0 (YZ=BC) gives #2 and r=oo (Y=Z) #4. I thus
        > strongly suspect P lies on the Euler line for any r.
        >
        > An additional question would be whether you could
        > take the midpoint of WX [yes, guess why :-)]
        > or the midpoint of the rectangle for the construction
        > instead. Again I suspect yes (and probably any ratio
        > R of the height of P with respect to the rectangle
        > will do - again by ray theorem).
        >
        > Hauke
        >
      • Angel
        ... Dear Hauke P lies on the Kiepert hyperbola for any r Angel
        Message 3 of 5 , Mar 10 9:05 AM
          --- In Hyacinthos@yahoogroups.com, "shokoshu2" <fc3a501@...> wrote:
          >
          > This should be self-explaining...
          > http://pic-hoster.net/view/8336/threesq.gif
          > ...but if not:
          > 0. Given a triangle ABC.
          > 1. Choose some rectangle shape r (3:1 in the example).
          > 2. Inscribe a rectangle WXYZ with WX/XY=r into ABC:
          > W goes on CA, X on AB and YZ lies on BC.
          > 3. Join the midpoint of YZ with A. Call this line g1.
          > 4. Repeat construction for the other sides to get
          > g2 and g3.
          > 5. The g_i concur in a point P.
          > (Proof: Apply Pythagoras, ray and Ceva theorems.
          > Juggle a bit, I could do it in one minute on piece
          > and paper, which I haven't done since years. Thus
          > a purely geometric proof shouldn't be hard either.)
          > 7. The Ceva ratios are linear in r. Obviously,
          > r=0 (YZ=BC) gives #2 and r=oo (Y=Z) #4. I thus
          > strongly suspect P lies on the Euler line for any r.
          >
          > An additional question would be whether you could
          > take the midpoint of WX [yes, guess why :-)]
          > or the midpoint of the rectangle for the construction
          > instead. Again I suspect yes (and probably any ratio
          > R of the height of P with respect to the rectangle
          > will do - again by ray theorem).
          >
          > Hauke
          >

          Dear Hauke

          P lies on the Kiepert hyperbola for any r

          Angel
        • shokoshu2
          ... Well, that s a nice result too (one shouldn t try to guess a locus from 2 points - I m the Bob Beamon of Jumping To Conclusions :-) Hauke P.S. Cf.
          Message 4 of 5 , Mar 12 1:20 AM
            --- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
            >
            > Dear Hauke
            >
            > P lies on the Kiepert hyperbola for any r
            >
            Well, that's a nice result too (one shouldn't try
            to guess a locus from 2 points - I'm the
            Bob Beamon of Jumping To Conclusions :-)

            Hauke

            P.S. Cf.
            http://mathworld.wolfram.com/KiepertHyperbola.html
            - are the ratios r directly related to the angle
            omega given in the list? Is my construction just
            a rephrasing of Lemoines problem?
          • chris.vantienhoven
            Dear Hauke, dear friends, In connection of the construction of Hauke I made another observation. Kiepert (1869) showed that the lines connecting the vertices
            Message 5 of 5 , Mar 12 10:07 AM
              Dear Hauke, dear friends,

              In connection of the construction of Hauke I made another observation.

              Kiepert (1869) showed that the lines connecting the vertices of the given triangle and the corresponding peaks of the isosceles triangles concur. The locus of this point is the Kiepert Hyperbola.
              Let ha= length of perpendicular isosceles triangle on side a.
              And let n = (a/ha)/2.
              This ratio is because of the construction identical for all sides.

              The special thing is that the coordinates for the constructed point on the Kiepert Hyperbola have a general barycentric coordinate-formula: (SB - n.K)(SC - n.K). Only depending on ratio n.
              SB=(a^2 - b^2 + c^2)/2, SC=(a^2 + b^2 - c^2)/2, K = Area.

              This ratio n is the same n as used in the Hauke-constrution.
              So both constructions generate the same points and have the same locus, being the Kiepert Hyperbola. And have the same type of barycentric coordinate-formula.

              I don't know if this is known. For me this is new.
              Here is a list of points on the Kiepert Hyperbola with corresponding ratio n.
              When n = INF then P = X2.
              When n = 0 then P = X4.
              When n = 2/3 V3 then P = X13.
              When n = - 2/3 V3 then P = X14.
              When n = 2 V3 then P = X15.
              When n = - 2 V3 then P = X18.
              When n = 1 then P = X1131.
              When n = - 1 then P = X1132.
              When n = 2 then P = X485.
              When n = - 2 then P = X486.
              When n = 3 then P = Hauke Point.
              When n = - 3 then P = 2nd Hauke Point.
              When n = 4 then P = X3316.
              When n = - 4 then P = X3317.
              Again Barycentric coordinates are: (SB - n.K)(SC - n.K).

              One special note: not every point on the Kiepert Hyperbola can be "catched" with this formula. Only the ones constructable as mentioned above.

              Best regards,

              Chris van Tienhoven



              --- In Hyacinthos@yahoogroups.com, "chris.vantienhoven" <van10hoven@...> wrote:
              >
              > Dear Hauke,
              >
              > You really surprised me with this nice problem!
              > I investigated it and got some results.
              > 1. The 3:1-rectangle has a length/width-ratio of 3.
              > 2. The points in your drawing lies on these lines;
              > X20.X1327, X140.X3316, X590.X1131, X1132.X3068, X1328.X3091, X1656.X3317.
              > 3. This point is not on the Eulerline.
              > But special is that it is on the Kiepert Hyperbola.
              > 4. The barycentrics of this point are: (SB + 3K) (SC + 3K),
              > where SB = (a^2 ¨C b^2 + c^2) / 2, (a^2 + b^2 - c^2) / 2, K = Area of Triangle.
              > 5. When we take a 3:1-rectangle outside the Reference Triangle we get a corresponding point on these lines:
              > X20.X1328, X140.X3317, X615.X1132, X1131.X3069, X1327.X3091, X1656.X3316.
              > Again a point on the Kiepert Hyperbola.
              > The barycentrics are: (SB - 3K) (SC - 3K).
              > I think it is not without meaning that in both cases there is adding or subtracting of 3 times the TriangleArea in the barycentrics!
              > 6. When changing the length/width-ratio there developes a locus.
              > The locus of these points is surprisingly (or not) the Kiepert Hyperbola!
              > 7. Let¡¯s name the length/width-ratio ¡°n¡±.
              > When n=1 the perspector=X1131.
              > When n=-1 the perspector=X1132.
              > When n=2 the perspector=X485.
              > When n=-2 the perspector=X486.
              > When n=4 the perspector=X3316.
              > When n=-4 the perspector=X3317.
              > When n=2/3 ¡Ì3 the perspector=X13.
              > When n=-2/3 ¡Ì3the perspector=X14.
              > When n=2¡Ì3 the perspector=X17.
              > When n=-2¡Ì3 the perspector=X18.
              > Sofar my results.
              > There must be more!
              > Best Regards,
              >
              > Chris van Tienhoven
              >
              >
              >
              >
              > --- In Hyacinthos@yahoogroups.com, "shokoshu2" <fc3a501@> wrote:
              > >
              > > This should be self-explaining...
              > > http://pic-hoster.net/view/8336/threesq.gif
              > > ...but if not:
              > > 0. Given a triangle ABC.
              > > 1. Choose some rectangle shape r (3:1 in the example).
              > > 2. Inscribe a rectangle WXYZ with WX/XY=r into ABC:
              > > W goes on CA, X on AB and YZ lies on BC.
              > > 3. Join the midpoint of YZ with A. Call this line g1.
              > > 4. Repeat construction for the other sides to get
              > > g2 and g3.
              > > 5. The g_i concur in a point P.
              > > (Proof: Apply Pythagoras, ray and Ceva theorems.
              > > Juggle a bit, I could do it in one minute on piece
              > > and paper, which I haven't done since years. Thus
              > > a purely geometric proof shouldn't be hard either.)
              > > 7. The Ceva ratios are linear in r. Obviously,
              > > r=0 (YZ=BC) gives #2 and r=oo (Y=Z) #4. I thus
              > > strongly suspect P lies on the Euler line for any r.
              > >
              > > An additional question would be whether you could
              > > take the midpoint of WX [yes, guess why :-)]
              > > or the midpoint of the rectangle for the construction
              > > instead. Again I suspect yes (and probably any ratio
              > > R of the height of P with respect to the rectangle
              > > will do - again by ray theorem).
              > >
              > > Hauke
              > >
              >
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