- --- 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

D.

> circumcircle. The tangent at A of circle (K) meets the line BC at

> 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

This was used in a Romanian TST in 2002. Only one student solved it

> only if AM is bisector of angle DAB.

(some of you might have heard of him: Valentin Vornicu, the

administrator of Mathlinks). The problem got me really depressed and

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 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

_____________________________________________________________________________

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[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 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]