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Re: [EMHL] Re: a geometry problem regarding equilater triangle!

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  • Floor en Lyanne van Lamoen
    Dear Georg and Jean-Pierre, ... Let A*B*C* be the the orthic triangle. The midpoints of BC, B C and B*C* are collinear. On the other hand A, the midpoint of
    Message 1 of 2 , Sep 30, 2003
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      Dear Georg and Jean-Pierre,

      > > 1)In the triangle ABC we consider A',B' and C' the simetric points
      > of
      > > A,B,C regarding the opposite sides. Prove that the triangle ABC
      > and
      > > A'B'C' have the same G point(weight point) if and only if the
      > triangle
      > > ABC is equilateral!
      >
      > Roughly, I would say that the fixed point of the affine mapping
      > ABC->A'B'C' is the symedian point and that symedian point = centroid
      > <=> the triangle is equilateral
      > Surely, there exists something more elementary and, may be, shorter.
      > Friendly. Jean-Pierre
      >

      Let A*B*C* be the the orthic triangle. The midpoints of BC, B'C' and
      B*C* are collinear. On the other hand A, the midpoint of B*C* and the
      symmedian point are collinear. So the symmedian point and G must be both
      on the same Cevian from A. Since these Cevians are normally symmetric
      through the angle bisector, this means that ABC must be isosceles with
      top angle A. In the same way ABC isosceles with top angle B. Hence ABC
      is equilateral.

      Kind regards,
      Sincerely,
      Floor.
    • Floor en Lyanne van Lamoen
      Dear Georg and Jean-Pierre, ... The solution I wrote this morning assumed that A AG shouls be collinear. That is a wrong assumption, so basicly all I wrote was
      Message 2 of 2 , Oct 1, 2003
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        Dear Georg and Jean-Pierre,

        > > > 1)In the triangle ABC we consider A',B' and C' the simetric points
        > > of
        > > > A,B,C regarding the opposite sides. Prove that the triangle ABC
        > > and
        > > > A'B'C' have the same G point(weight point) if and only if the
        > > triangle
        > > > ABC is equilateral!
        > >
        > > Roughly, I would say that the fixed point of the affine mapping
        > > ABC->A'B'C' is the symedian point and that symedian point = centroid
        > > <=> the triangle is equilateral
        > > Surely, there exists something more elementary and, may be, shorter.
        > > Friendly. Jean-Pierre
        > >

        The solution I wrote this morning assumed that A'AG shouls be collinear.
        That is a wrong assumption, so basicly all I wrote was useless.

        Another try.

        Let us assume that ABC and A'B'C' share centroid G. Then, since the
        centroid of A'B'C' is the reflection of the centroid of ABC through the
        centroid of the orthic triangel A*B*C*, we have that A*B*C* must have
        the same centroid G.

        This means that the Euler lines of A*B*C* and ABC must coincide: apart
        from the shared G also the ninepointcenter N lies on both Euler lines.

        This means that the orthocenter H of ABC lies on the Euler line of
        A*B*C*. The orthocenter is an incenter or an excenter of A*B*C*. On the
        other hand GN:NH = 1:3, or G*O*:O*Ix* = 1:3. This latter relation should
        easily lead analyticly to A*B*C* is equilateral. So G*=O*, hence G=N,
        hence G=H and ABC is equilateral as well.

        Kind regards,
        Sincerely,
        Floor.
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