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Re: [EMHL] A problem of I. Yaglom

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  • Nikolaos Dergiades
    Dear Ricardo, If ABC is the required triangle and A B C is the given triangle then AA = BB = CC and the three lines AA , BB , CC pass through the first
    Message 1 of 2 , Nov 19, 2002
      Dear Ricardo,

      If ABC is the required triangle and A'B'C' is the
      given triangle then AA' = BB' = CC'
      and the three lines AA', BB', CC' pass through
      the first Fermat point F of triangle ABC.
      We construct on the sides of triangle A'B'C'
      equilateral triangles and let A", B", C" be the vertices
      of these triangles outside of A'B'C'.
      Since B'FC' = 120 and B'A"C' = 60
      the points F, B', A", C' are concyclic. (1)
      The rotation about C' by angle 60 moves
      B to A
      B' to A".
      Hence A"A = BB' = AA' and
      angle C'A"A = angle C'B'B
      = angle C'B'F
      = angle C'A"F because of (1)
      This means that the line A"A passes through F,
      the points A", A, A' are collinear
      and the required point A is the mid point of
      the segment A'A". Similarly we construct the
      points B, C of triangle ABC.

      Best regards
      Nikos Dergiades


      >Dear Hyacinthos:
      >
      >
      >Construct a triangle, given the tree points in the plane
      that are the outer vertcies of equilateral triangles
      constructed outward on the sides of the desired triangle.
      >
      >
      >Yaglom, I (1973): Geometric transformations / translated
      from the Russian by Allen Shields, A. Shenitzer (p 12)
      >
      >
      >Best regards
      >
      >Ricardo Barroso (Universidad de Sevilla)
    • Floor en Lyanne van Lamoen
      Dear Ricardo ... I believe this question originates from EMHL (Lemoine) himself, I was given the reference: E.M.H. Lemoine, Question 864, Nouvelles Annales de
      Message 2 of 2 , Nov 20, 2002
        Dear Ricardo

        [RC]:
        > Construct a triangle, given the tree points in the plane that are the outer vertcies of equilateral triangles constructed outward on the sides of the desired triangle.
        >
        > Yaglom, I (1973): Geometric transformations / translated from the Russian by Allen Shields, A. Shenitzer (p 12)


        I believe this question originates from EMHL (Lemoine) himself, I was
        given the reference:

        E.M.H. Lemoine, Question 864, Nouvelles Annales de Mathematiques 2-7
        191 (1868).

        A very easy solution can be found by applying the given in section 4 sub
        4 of

        F.M. van Lamoen and P. Yiu, The Kiepert pencil of Kiepert hyperbolas,
        Forum Geom., 1 (2001) 125-132.

        >From this source we conclude that if we first build equilateral
        triangles outwardly on the sides of ABC, resulting in a new triangle
        A'B'C' of the new vertices, and then build equilateral triangles
        inwardly on the sides of A'B'C', resulting in new vertices forming a
        triangle A"B"C", then ABC is the medial triangle of A"B"C".

        This yields a very easy construction!

        Kind regards,
        Sincerely,
        Floor.
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