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Re: [EMHL] Construction problem

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  • Gilles Boutte
    An other solution : A1B1C1is the given first Brocard triangle of the unknown triangle ABC. Construct first the the circumcircle (O1) and the centroid G of
    Message 1 of 5 , Aug 31, 2002
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      An other solution :


      A1B1C1is the given first Brocard triangle of the unknown triangle ABC.

      Construct first the the circumcircle (O1) and the centroid G of A1B1C1,
      it is the centroid of ABC.

      The line GA1 intersects the circle (O1) at A2. Construct similarly B2
      and C2.
      So A2B2C2 is the second Brocard triangle of ABC.

      Let Bc be the reflection of B2 wrt the tangent to (O1) at C1.
      Let Cb be the reflection of C2 wrt the tangent to (O1) at B1.

      The circumcircles of A2BcC1 and A2B1Cb are intersecting at A2 and at the
      vertex A of the triangle, we have to construct.

      Best regards,

      Gilles Boutte

      >[DK]
      >
      >Johnson, page 280:
      >
      >===
      >The solution depends onthe fact that any triangle is inversively similar to
      >its first Brocard triangle.
      >We locate the Brocard points of the given triangle, also its first Brocard
      >triangle; then by similarity we locate on the circumcircle of the given
      >triangle the Brocard points of the required triangle.
      >The vertices of the latter can then be found at once.
      >===
      >
      >Is this what you did?
      >
      >Regards,
      >Dick Klingens
      >
      >[AD]
      >
      >Dear all
      >Given the first Brocard triangle of triangle ABC,how to construct
      >the triangle ABC ?
      >
      >My solution has become unnecessarily long.Will anybody tell me a precise
      >solution of it?
      >
      >Yours faithfully
      >Atul.A.Dixit
      >
    • Atul Dixit
      Dear Dick and Gilles Thank you.Actually my solution is somewhat similar(though not efficient) as that given by Boutte Gilles.Actually,I didn t thought of using
      Message 2 of 5 , Sep 2, 2002
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        Dear Dick and Gilles

        Thank you.Actually my solution is somewhat similar(though not efficient) as
        that given by Boutte Gilles.Actually,I didn't thought of using the second
        Brocard triangle.

        Yours faithfully
        Atul.A.Dixit

        >From: "Dick Klingens" <dklingens@...>
        >Reply-To: Hyacinthos@yahoogroups.com
        >To: <Hyacinthos@yahoogroups.com>,<atul_dixie@...>
        >Subject: RE: [EMHL] Construction problem
        >Date: Sat, 31 Aug 2002 19:00:02 +0200
        >
        >Johnson, page 280:
        >
        >===
        >The solution depends onthe fact that any triangle is inversively similar to
        >its first Brocard triangle.
        >We locate the Brocard points of the given triangle, also its first Brocard
        >triangle; then by similarity we locate on the circumcircle of the given
        >triangle the Brocard points of the required triangle.
        >The vertices of the latter can then be found at once.
        >===
        >
        >Is this what you did?
        >
        >Regards,
        >Dick Klingens
        >
        >::-----Original Message-----
        >::From: Atul Dixit [mailto:atul_dixie@...]
        >::Sent: Saturday, August 31, 2002 6:42 PM
        >::To: Hyacinthos@yahoogroups.com
        >::Subject: [EMHL] Construction problem
        >::
        >::
        >::Dear all
        >:: Given the first Brocard triangle of triangle ABC,how to construct
        >::the triangle ABC ?
        >::
        >::My solution has become unnecessarily long.Will anybody tell me a precise
        >::solution of it?
        >::
        >::Yours faithfully
        >::Atul.A.Dixit
        >::
        >::
        >::
        >::_________________________________________________________________
        >::Send and receive Hotmail on your mobile device: http://mobile.msn.com
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        >::
        >




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