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Appollonian problem for excircles (was: A Tucker circle...)
Posted:
Oct 14, 2002 6:36 AM


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.
 CONTENTS 
Part 1. Jenkins' Theorem Part 2. "Appollonian circles" by Aeppli Part 3. Comparison with ReidtWolffAthen Part 4. The "Kh" circle and its center X(970)
 PART 1. JENKINS' THEOREM 
ReidtWolffAthen "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 ninepoint 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 ninepoint circle. You get a circle tangle (original: "KreisknÃÂ¤uel") 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 ninepoint 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 ninepoint center, but through the Spieker center.
 PART 2. "APPOLLONIAN CIRCLES" BY AEPPLI 
There is a more systematical article on the problem:
Alfred Aeppli: "Das Taktionsproblem von Appollonius, angewandt auf die vier BerÃÂ¼hrungskreise eines Dreiecks", Elemente der Mathematik 2/1958 pages 2530.
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 ninepoint 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 ninepoint 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 ninepoint 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 results."
Perhaps somebody can prove some of this theorems; I have no idea. (Probably, it is really a good idea to use inversions.)
 PART 3. COMPARISON WITH REIDTWOLFFATHEN 
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 ninepoint 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 ninepoint center.
 PART 4. THE "Kh" CIRCLE AND ITS CENTER X(970) 
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 StÃÂ¤rk (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 ninepoint center of ABC.
So, the center X is the intersection of the Tucker line and the line containing the ninepoint center and the Spieker center (i. e. the OIline 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), where
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)].
 END 
Darij Grinberg



