## Re: A Nagel-Thebault configuration

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Message 1 of 20 , Jan 7, 2008
Dear friends

are related to the current problem and show a
connection of this problem to the equal inradii
in partial triangles set-up: Nagel cevian extended
is the equal inradii cevian in the triangle on the
same base with vertex in the intersection of Nagel
cevian and circumcircle. Another interesting fact:
incenter is one of the limiting points in the
coaxal system of two Thebault circles.

Cosmin, the problem with JSTOR search is that
the fresh stuff is not in the database yet,
there is a delay of up to several years depending
on a publication.

Regards,
Armpist.
• Dear Armpist and Cosmin ... Using this remark, it becomes obvious to find the second circle (O ) touching the 6 Thebault circles : The three radical axis of
Message 2 of 20 , Jan 8, 2008
Dear Armpist and Cosmin

> Another interesting fact:
> incenter is one of the limiting points in the
> coaxal system of two Thebault circles.

Using this remark, it becomes obvious to find the second circle (O')
touching the 6 Thebault circles :
The three radical axis of the three pairs of Thebault circle concur at
the Cosmin point T; so, the circle (T) centered at T and going through
I is orthogonal to the six Thebault circles which are invariant in the
inversion wrt (T). Hence, the inverse of (O) in (T) touches the 6
circles too.
This is much more clear than my previous explanation.
It remains to find a synthetic proof of the fac that the inverse of P
in (T) lies on (O)
Thank you again
Jean-Pierre
• ... touching ... (note ... point ... 2r) - ... Dear Jean-Pierre, Cosmin and all, Your first result above was mentioned and proved by Nguyen Minh Ha in a
Message 3 of 20 , Jan 8, 2008
--- In Hyacinthos@yahoogroups.com, "jpehrmfr" <jean-
pierre.ehrmann@...> wrote:
>
> Dear Hyacinthists
> A'B'C' is the cevian triangle of the Nagel point.
> Consider the six corresponding Thébault circles (the circles
touching
> A'A, A'B and the circumcircle internally, ....)
> Then :
> 1) The six Thébault circles are congruent to the incircle
> 2) Their 6 centers lie on the circle with center O,radius R-r
(note
> that this circle goes through X_1145, reflection of the Feuerbach
point
> in the Spieker center)
> 3) The six circles touch externally the circle with center (O,R-
2r) -
> this circle goes through the Nagel point -
> I suspect that this is not new. Have you head of this?
> I'll upload soon a sketch in Hyacinthos file.
> Friendly. Jean-Pierre
>
> PS : I apologize for a possible duplication
>
Dear Jean-Pierre, Cosmin and all,

Your first result above was mentioned and proved by Nguyen Minh Ha
in a Vietnamese Mathematics magazine, which was pblished in 2006.
This paper also contains a nice synthetic proof of Sawayama-Thebault
theorem. You can see it at:
http://toantuoitho.nxbgd.com.vn/toantuoitho2/report.asp?
soph=42&id=11&true=true ,
http://toantuoitho.nxbgd.com.vn/toantuoitho2/report.asp?
soph=43&id=11&true=true . But unfortunately, it's in Vietnamese.

Fiendly, Son Hong Ta.
• Dear Son and Armpist, Thank you both for the references to the particular cases. Son, does Nguyen Min Ha mentiones anything about the general configuration in
Message 4 of 20 , Jan 9, 2008
Dear Son and Armpist,

Thank you both for the references to the particular cases.
Son, does Nguyen Min Ha mentiones anything about the general
configuration in his paper or by our first result you meant only the
congruence of the Thebault circles in the Nagel case?

Best regards,
Cosmin
• ... the ... Dear Cosmin, He only mentioned about the congurence of the Thebault circles in the Nagel case. I ve read this paper again, and in this paper, he
Message 5 of 20 , Jan 10, 2008
--- In Hyacinthos@yahoogroups.com, "Cosmin Pohoata"
<pohoata_cosmin2000@...> wrote:
>
> Dear Son and Armpist,
>
> Thank you both for the references to the particular cases.
> Son, does Nguyen Minh Ha mentiones anything about the general
> configuration in his paper or by our first result you meant only
the
> congruence of the Thebault circles in the Nagel case?
>
>
> Best regards,
> Cosmin
>

Dear Cosmin,

He only mentioned about the congurence of the Thebault circles in
the Nagel case. I've read this paper again, and in this paper, he
said the author of the congurence is Ha Vu Anh, a teacher from Vinh
Phuc, Vietnam. Anyway, your results and proofs are very nice. Thanks
for sharing.

Best regards, Son Hong Ta.
• Dear Hyacinthists, Both these problems have easy algebraic solutions/constructions. One can easily find and from b+c=s ; bc=u^2 and .
Message 6 of 20 , Jan 18, 2008
Dear Hyacinthists, Both these problems have easy algebraic
solutions/constructions. One can easily find <b> and <c> from >b+c=s ; bc=u^2> and <b-c=d ; bc=v^2>. But... I am looking for an elegant "pure" geometrical approach. Is there one? Best regards, Luis
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• Dear Luis, [LL]: Both these problems have easy algebraic solutions/constructions. One can easily find b and c from b+c=s ; bc=u^2 and b-c=d ; bc=v^2. But... I
Message 7 of 20 , Jan 19, 2008
Dear Luis,

[LL]: Both these problems have easy algebraic solutions/constructions.
One can easily find b and c from b+c=s ; bc=u^2 and b-c=d ; bc=v^2.
But... I am looking for an elegant "pure" geometrical approach. Is
there one?

*** Here is one. The relation (b+c)^2 + (b-c)^2 = 4m_a^2 + a^2
shows that the two problems can be solved simultaneously.

Given a segment \$BC\$ of length a, midpoint D, and a perpendicular
segment A_0D = m_a, construct
(i) a semicircle center D through A_0,
(ii) the circle with diameter A_0B (with perpendicular diameter MN, M
and C on opposite sides of A_0B),
(iii) the circle, center M, through A_0 and B.

Let P be a point on (ii) with BP = b + c for P on the semicircle A_0MB
and BP=b-c for P on A_0NB.

Extend BP to intersect the circle (iii) at Q.
Construct the circle through Q with center B to intersect the
semicircle (i) at A.

Then ABC is the required triangle.

Best regards
Sincerely
Paul
• Dear Luis, My apology for omission of a factor 1/2 in the length of BP in the construction given in the prior posting (Hyacinthos 16048). [LL]: Both these
Message 8 of 20 , Jan 19, 2008
Dear Luis,

My apology for omission of a factor 1/2 in the length of BP in the
construction given in the prior posting (Hyacinthos 16048).

[LL]: Both these problems have easy algebraic solutions/constructions.
One can easily find b and c from b+c=s ; bc=u^2 and b-c=d ; bc=v^2.
But... I am looking for an elegant "pure" geometrical approach. Is
there one?

*** Here is one. The relation (b+c)^2 + (b-c)^2 = 4m_a^2 + a^2
shows that the two problems can be solved simultaneously.

Given a segment \$BC\$ of length a, midpoint D, and a perpendicular
segment A_0D = m_a, construct
(i) a semicircle center D through A_0,
(ii) the circle with diameter A_0B (with perpendicular diameter MN, M
and C on opposite sides of A_0B),
(iii) the circle, center M, through A_0 and B.

Let P be a point on (ii) with BP = (b + c)/2 for P on the semicircle
A_0MB and BP = (b-c)/2 for P on A_0NB.

Extend BP to intersect the circle (iii) at Q.
Construct the circle through Q with center B to intersect the
semicircle (i) at A.

Then ABC is the required triangle.

Best regards
Sincerely
Paul
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