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Re: tangents

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  • ben_goss_ro
    ... 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
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      --- 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
      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.
    • Jean-Louis Ayme
      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
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        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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      • Francois Rideau
        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
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          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]
          >
          >
          >


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