Nagel (=extouch) triangle

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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
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

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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