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
========================================================================= Instruções para entrar na lista, sair da lista e usar a lista em http://www.mat.puc-rio.br/~nicolau/olimp/obm-l.html =========================================================================