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Topic: Appollonian problem for excircles (was: A Tucker circle...)
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Darij Grinberg

Posts: 5
Registered: 12/6/04
Appollonian problem for excircles (was: A Tucker circle...)
Posted: Oct 14, 2002 6:36 AM
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This is a quite long message on the circles touching the excircles of
a triangle, so I have divided it into four parts. The most theorems I
can't prove.


Part 1. Jenkins' Theorem
Part 2. "Appollonian circles" by Aeppli
Part 3. Comparison with Reidt-Wolff-Athen
Part 4. The "Kh" circle and its center X(970)


Reidt-Wolff-Athen "Elemente der Mathematik" (1960s) is a very
outstanding German schoolbook. Especially, I refer to the "Mittelstufe
2" geometry volume. It handles not only with the Euler line and the
nine-point circle, but even a proof of Feuerbach's theorem is
presented there. This is one of the standard proofs using inversion
and some lemmas; on the next page, there are some exercises.

One of them is the following:

15. a) Invert a triangle with its in- and excircles at the nine-point
circle. You get a circle tangle (original: "Kreisknäuel") consisting
of 8 circles, each of them touches 4 others!
b) Prove Jenkins' theorem (1883): Every circle extrenally touching two
excircles and internally touching the third excircle, passes through
the nine-point center. All together, there are three such circles.

In part 3, we will see that this theorem is false. The three circles
pass not through the nine-point center, but through the Spieker


There is a more systematical article on the problem:

Alfred Aeppli: "Das Taktionsproblem von Appollonius, angewandt auf die
vier Berührungskreise eines Dreiecks", Elemente der Mathematik 2/1958
pages 25-30.

Aeppli regards the circles that touch the three excircles and the ones
touching two excircles and the incircle. We, however, will only
concern the first.

After the Appollonian problem, there are at the most 8 circles which
touch the three excircles of triangle ABC. Three of them are in fact
lines - the sidelines BC, CA, AB. The fourth is the nine-point circle
"n". Further three "Kha", "Khb" and "Khc" pass through the Spieker
center H' of ABC, and their tangents at this Spieker center are
parallel to BC, CA, resp. AB (these are the circles in Jenkins'
theorem). The eight circle "Kh" is not such easy to find; Aeppli
states that H' is the outer center of similarity of this circle "Kh"
and the nine-point circle "n".

The center of the circle "Kh" lies on the Tucker line (= Brocard axis)
of ABC.

(This is not in Aeppli's article, but he says that the distances dA,
dB, dC of the vertices A, B, C to the line joining the center of "Kh"
with the circumcenter of ABC have the proportion

bb - cc cc - aa aa - bb
dA : dB : dC = ------- : ------- : -------.
aa bb cc

But this means that this terms

bb - cc cc - aa aa - bb
------- : ------- : -------
aa bb cc

are the barycentrics of this line. It is not very difficult to prove
that they are the barycentric of the Tucker line (= Brocard axis).

The radius "rh" of "Kh" is

/ a+b b+c c+a \
"rh" = ( --- + --- + --- + 1 ) "rn",
\ c a b /

where "rn" is the radius of the nine-point circle "n", i. e.

"rn" = abc / 8S (S = area of ABC).

These are the theorems of Aeppli. He doesn't give any proofs, founding
it by the sentence:

"As all calculations are completely elementary, we will show only the

Perhaps somebody can prove some of this theorems; I have no idea.
(Probably, it is really a good idea to use inversions.)


The comparison of Jenkins' theorem and Aeppli's theorem on "Kh1",
"Kh2" and "Kh3" causes some doubts. After Jenkins, the three circles
pass through the nine-point center, i. e. the circumcenter of the
medial (=complementary) triangle; after Aeppli, they pass through the
Spieker center, i. e. the incenter of the medial triangle. A test
using the dynamic geometry program showed that in fact, they do pass
through the Spieker center and not through the nine-point center.


Now we get back to the problem posted former in this newsgroup. Aeppli
states that the center X of the "Kh" circle lies on the Tucker line (=
Brocard axis).

This was also proved analytically by Roland Stärk (using a method
which is similar to tripolar coordinates).

It seems more is true: The circle "Kh" is a Tucker circle itself. I
still haven't seen any proof.

Since the Spieker center is the outer center of similarity of "Kh" and
"n", it is collinear with the center X of "Kh" and the nine-point
center of ABC.

So, the center X is the intersection of the Tucker line and the line
containing the nine-point center and the Spieker center (i. e. the
OI-line of the medial triangle). That means that it is the point
X(970) in Kimberling's list!

After Kimberling, the trilinears are f(a,b,c) : f(b,c,a) : f(c,a,b),

f(a,b,c) = a[a^3(b + c)^2 + a(ab + ac - 2bc)(b^2 + c^2)
- bc(b^3 + c^3) - a(b^4 + c^4) - (b^5 + c^5)].


Darij Grinberg

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