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Re: [EMHL] Some Points

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  • Fred Lang
    Dear Andreas, ... Your result is ok, you can do the same with interior points: The circle (A,AA ) intersects the int. of AB at Abi and the int. of AC at Aci
    Message 1 of 9 , Oct 30, 2000
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      Dear Andreas,

      xpolakis@... a écrit :

      > There is another simple point in that figure with the circles (A, AA"),
      > (B,BB"), (C,CC") [AA", BB", CC": altitudes].
      >
      > Let me use a new point notation.
      >
      > The circle (A,AA") intersects the ext. of AB at Ab and the ext. of AC at Ac
      > (B,BB") BC Bc BA Ba
      > (C,CC") CA Ca CB Cb
      >
      > Ac Ab
      > \ /
      > \ /
      > \ /
      > Z \/A Y
      > /\
      > / \
      > / \
      > / \
      > Bc---------B--------C--------Cb
      > / \
      > / X \
      > / \
      > / \
      > Ba Ca
      >
      > Let X = BCa /\ CBa; Y = CAb /\ ACb, Z = ABc /\ BAc
      >
      > Then the triangles XYZ,ABC are in perspective.
      > (I found that the perspector is the centroid G)
      >
      > And BTW, is the radical center of these circles a remarkable point?
      >
      > Antreas
      >

      Your result is ok, you can do the same with interior points:

      The circle (A,AA") intersects the int. of AB at Abi and the int. of AC at Aci
      (B,BB") BC Bci BA Bai

      (C,CC") CA Cai CB Cbi

      Let Xi = BCai /\ CBai; Yi = CAbi /\ ACbi, Zi = ABci /\ BAci


      Then the triangles XiYiZi, ABC are in perspective.
      (I found that the perspector is the centroid G)

      But an interesting result for you is that the radical center is the

      X(389) = CENTER OF THE TAYLOR CIRCLE

      Why?

      I have a nice french little book: "La géométrie du triangle" de Yvonne et rené
      Sortais. Collectio Hermann 1987.

      Page 38:

      Soient I, J, K les pieds des hauteurs du triangle ABC.
      Soient M, N les projetés orthogonaux de I sur AC et AB.
      Soient P,Q les projetés orthogonaux de J sur AB et BC.
      Soient R,S les projetés orthogonaux de K sur BC et CA.
      Alors
      1) PS//BC, NR//CA, QM//AB
      2) Les six points MNPQRS sont cocycliques, ils définissent le cercle de Taylor de
      ABC.

      Page 40:
      Soient I'J'K' les milieux des côtés du triangle IJK
      Alors
      le centre du cercle de Taylor est le point de concours des trois bissectrices
      intérieures du triangle I'J'K'.

      You can probably find the join with your points.

      Amicalement de Lausanne.

      Fred
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