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Nagel (=extouch) triangle

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  • Darij Grinberg
    I am interested in some questions on the Nagel (=extouch) triangle AaBbCc and the analogy to the Gergonne (=intouch) triangle A B C , where the points are
    Message 1 of 1 , Mar 5, 2003
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      I am interested in some questions on the Nagel (=extouch) triangle
      AaBbCc and the analogy to the Gergonne (=intouch) triangle A'B'C',
      where the points are defined as follows:

      Let ABC be a triangle. The point Aa is the point of tangency of the a-
      excircle with BC; analogously define Bb, Cc. The point Ab is the
      point of tangency of b-excircle with BC; analogously define Bc, Ca,
      Ba, Cb, Ac. The point A' is the point of tangency of the incircle
      with BC; analogously define B', C'.

      Here is a comparison of the Gergonne and Nagel triangles' properties.

      1. The Gergonne triangle is the cevian triangle of the Gergonne point
      and the pedal triangle of the incenter. - The Nagel triangle is the
      cevian triangle of the Nagel point and the pedal triangle of X(40)
      [this point is the reflection of the incenter in the circumcenter,
      was discussed here and should be called Bevan point, although I
      habitually call it Hadamard point].

      GENERALIZATION: The Gergonne and Nagel triangles are examples of
      isotomic triangles. These are two triangles A1B1C1 and A2B2C2, where
      A1 and A2 lie on BC and are the reflections of each other in the
      midpoint of BC, i. e. also BA1 = A2C, etc.. They need not be cevian
      triangles!
      However, if one of two isotomic triangles is the cevian triangle of
      some point P, the second one is the cevian triangle of the isotomic
      conjugate of P; and if one of two isotomic triangles is the pedal
      triangle of a point Q, the second one is the pedal triangle of the
      reflection of Q in the circumcenter of ABC.

      2. The Gergonne and Nagel triangles have equal areas.

      GENERALIZATION: Two isotomic triangles have always equal areas! (This
      looks great, but it is an easy corollary of the Routh theorem.)

      3. The medial triangle of the Gergonne triangle is perspective with
      ABC, the perspector being the incenter. - The medial triangle of the
      Nagel triangle is perspective with ABC, the perspector being the
      Mitten point X(9).

      GENERALIZATION: The medial triangle of the cevian triangle of any
      point P is perspective with ABC, the perspector being the
      isotomcomplement of P. (See Hyacinthos message #6423 for the
      definition of the isotomcomplement.) If two points P and P' are
      isotomic conjugates, their cevian triangles are isotomic triangles,
      and the isotomcomplements Q and Q' of P and P' are "isotomic-twin"
      points, i. e. two points whose anticomplements are isotomic
      conjugates.

      4. A triangle can be easily constructed from his Gergonne triangle. -
      But the construction of a triangle from his Nagel triangle is, as far
      as I know, still an unresolved question.

      What I have found are formulas for the sides of the Nagel triangle:

      Let ABC be a triangle with the "fourth side"

      d = D / R,

      where D = area of triangle ABC. (We easily get d = 2R sin A sin B sin
      C = a sin B sin C = b sin C sin A = c sin A sin B = abc / (4R²),
      which are just a few of the many equations involving d.)

      Then the squared sides of the Nagel triangle AaBbCc are

      BbCc ² = a(a-d); (1)
      CcAa ² = b(b-d); (2)
      AaBb ² = c(c-d). (3)

      This can be proven by the cosine law (in triangle ABbCc, we have ABb
      = s-c and ACc = s-b).

      The question is if these nice formulas give a construction of
      triangle ABC from the Nagel triangle AaBbCc. At first, it is easy to
      realize that if the angles A, B, C of triangle ABC are known, we can
      construct the triangle (because we can construct a triangle similar
      to ABC and its Nagel triangle and then undertake a similarity
      transformation). So let's rewrite (1), (2) and (3) as

      BbCc ² = 4R² sin² A (1 - sin B sin C);
      CcAa ² = 4R² sin² B (1 - sin C sin A);
      AaBb ² = 4R² sin² C (1 - sin A sin B).

      Now, if triangle AaBbCc is given, we must compute the angles A, B, C
      from the mutual ratios of sin² A (1 - sin B sin C), sin² B (1 - sin C
      sin A) and sin² C (1 - sin A sin B), and from the relation

      4 sin² A sin² B sin² C
      = 2 sin² B sin² C + 2 sin² C sin² A + 2 sin² A sin² B
      - sin^4 A sin^4 B sin^4 C

      which connects the angles of an arbitrary triangle ABC.

      If this computation is possible in terms of 2nd degree radicals, then
      ABC is constructible. Unfortunately, I have never studied Galois
      Theory, so I hope that one of the experts will help.

      Darij Grinberg
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