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