[obm-l] 2006 USAMO, problem 6 [era: USA math olympiad]

[Luís Lopes <qed_texte@xxxxxxxxxxx>]

Sauda,c~oes,

Último do dia.

[]'s
Luís



>From: Steve Sigur
>Reply-To: Subject: Re: [EMHL] 2006 USAMO, problem 6
>Date: Thu, 20 Apr 2006 16:48:31 -0400
>
>Dear François and Quang Tuan,
>
>You both found the two ways that I found, one by angle chasing and
>Miquel and one by symmetry. The position of the point is independent
>of E and F.
>
>François, I bet the folds who made the problem did not consider E and
>F as distracting elements. None of the Georgia students (8 of them)
>who took this test got this problem.
>
>Steve
>
>
>On Apr 19, 2006, at 10:53 PM, Quang Tuan Bui wrote:
>
> > Dear Steve and All,
> >   (One typo correction of Steve: E and F be points on AD and BC
> > respectively, such that AE/ED = BF/FC)
> >   It is very nice configuration. I try to proof as follow:
> >   We denote intersection of BC and AD as U.
> >   Use Miquel theorem for quadrilateral we have:
> >   - Circle(SBF), circle(SAE) concur with circle(ABU), circle(EFU)
> >   - Circle(TCF), circle(TDE) concur with circle(CDU), circle(EFU)
> >   So now problem is:
> >   Let's ABCD is quadrilateral, E, F is midpoint of AD, BC and AD,
> > BC intersect at U then three following circles concur
> >               circle(ABU), circle(CDU), circle(EFU)
> >   Let circle(ABU), circle(CDU) concur at V, we should proof UVEF is
> > concyclic.
> >   Draw perpendicular bisectors of EF and UV. They intersection at
> > W. Note that center of circle(ABU) as Oa, and center of circle(CDU)
> > as Oc then Oa, Oc are on perpendicular bisector of UV. From Oa, Ob,
> > W draw perpendiculars to FU at Ka, Kb, M. Easy to show
> >   M is midpoint of FU. That means W is on the perpendicular
> > bisector of FU. Analogously we have W is on the perpendicular
> > bisector of EV. So W is center of circle pass through UVEF. Please
> > note that W is midpoint of OaOb.
> >
> >   Remark:
> >   1. Miquel theorem about four circles concurrency can be easy and
> > elementary proof by inscribed angles.
> >   2. There are many other circles pass through concurrent point of
> > the problem (by Miquel theorem).
> >   3. You can see Steiner proof for Miquel theorem in FG (Ten
> > theorems of complete quadrilateral proof by Steiner)
> >   Best regards,
> >   Bui Quang Tuan
> >
> >
> > Steve Sigur wrote:  This question was on
> > today's USA math olympiad. Enjoy.
> >
> >
> > Let ABCD be a quadrilateral, and let E and F be points on AC and BC,
> > respectively, such that AE/ED = B/FC. Ray FE meets rays BA and CD at
> > S and T, respectively. Prove that the circumcircles of SAE, SBF, TCF,
> > and TDE pass through a common point.
> >
> > A neat schema.
> >
> > Steve Sigur


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