## Re: tangents

Expand Messages
• ... D. ... Denote ... This was used in a Romanian TST in 2002. Only one student solved it (some of you might have heard of him: Valentin Vornicu, the
Message 1 of 4 , Apr 11 2:33 PM
• 0 Attachment
--- In Hyacinthos@yahoogroups.com, "orl_ml" <orlando.doehring@h...>
wrote:
> Let ABC be a triangle with AB  BC, AB < AC and let (K) is its
> circumcircle. The tangent at A of circle (K) meets the line BC at
D.
> Let (K_1) be a circle which tangents (K) and the sides AD, BD.
Denote
> M the point of tangent of (K_1) with BD. Show that AC = MC if and
> only if AM is bisector of angle DAB.

This was used in a Romanian TST in 2002. Only one student solved it
(some of you might have heard of him: Valentin Vornicu, the
I haven't though about it since 2002.

Assume AM is the bisector of DAB. Let C' be a point on BC s.t.
C'M=C'A. Then C' is th center of the Apollonius circle of ADB
corresponding to A and it's well-known that C'M^2=C'B*C"D so it's
natural to consider the inversion of pole C' and power C'M^2. This
inversion turns the line AD into the circle passing through A, B, C'
and invariates the circle (K_1), so the circle (ABC') is tangent to
the circle (K_1). It's eay to see that this means that (ABC')=(K)
because they have 2 common points, A and B, and they're both tangent
to a circle (K_1).

Assume now that CA=CM. We consider the inversion of pole C and power
CA^2 (=CM^2). Take D' on BC s.t. CB*CD'=CM^2 and D' is on the same
side of C as B. It's easy to see that in this case AM is the bisector
of D'AB. On the other hand, the circle (K_1), which is inveriant, is
tangent to both (K) and the image of (K), which is line AD', so (K_1)
is tangent to AD and AD', and it's clear that D=D' because otherwise
D and D' would be on different sides of M, and they are not so
because CB<CM and CB*CD'=CM^2=>CD'>CM and it was clear that CD>CM in
the first place.

I think this finishes it.
• Dear Orlando, Ben_Grosso, Valentin and Hyacinthists, recently I read again some old message and particularly # 9541 and 9657 of 2004. Therefor, I propose a new
Message 2 of 4 , Feb 13, 2008
• 0 Attachment
Dear Orlando, Ben_Grosso, Valentin and Hyacinthists,
recently I read again some old message and particularly # 9541 and 9657 of 2004.
Therefor, I propose a new synthetic proof of this difficult problem of TST 2002 in Roumanian.
You will see this proof on my site
http://perso.orange.fr/jl.ayme
vol. (2008) : Un remarquable résultat de Vladimir protassov
Sincerely
Jean-Louis

_____________________________________________________________________________
Ne gardez plus qu'une seule adresse mail ! Copiez vos mails vers Yahoo! Mail http://mail.yahoo.fr

[Non-text portions of this message have been removed]
• Dear Jean Louis Very nice proof! I need some precision on the french text: What do you mean by I centre de ABC ? I guess you mean I centre du cercle
Message 3 of 4 , Feb 13, 2008
• 0 Attachment
Dear Jean Louis
Very nice proof!
I need some precision on the french text:
What do you mean by " I centre de ABC"?
I guess you mean " I centre du cercle inscrit de ABC", that is to say: "I
is the incenter of ABC"?

Moreover, there is in general 2 circles through A and C and tangent
to your "blue" circle in T with your notation and the other in T'.
Let I be the ABC-incenter and Ib be the excenter wrt B.
Then Protassov says that the internal bissector of angle <ATC is on
incenter I but I think that the internal bissector of angle <AT'C is on
excenter Ib.
Friendly
Francois

On Feb 13, 2008 12:19 PM, Jean-Louis Ayme <jeanlouisayme@...> wrote:

> Dear Orlando, Ben_Grosso, Valentin and Hyacinthists,
> recently I read again some old message and particularly # 9541 and 9657 of
> 2004.
> Therefor, I propose a new synthetic proof of this difficult problem of TST
> 2002 in Roumanian.
> You will see this proof on my site
> http://perso.orange.fr/jl.ayme
> vol. (2008) : Un remarquable résultat de Vladimir protassov
> Sincerely
> Jean-Louis
>
> __________________________________________________________
> Ne gardez plus qu'une seule adresse mail ! Copiez vos mails vers Yahoo!
> Mail http://mail.yahoo.fr
>
> [Non-text portions of this message have been removed]
>
>
>

[Non-text portions of this message have been removed]
Your message has been successfully submitted and would be delivered to recipients shortly.