PRIMEIRA PARTE:Para aplicar Erdös-Mordell neste problema,e bom que ponhamos tudo para dentro!Traçando os paralelogramos DEFM,NFAB,PBCD,e o triangulo XYZ tal que ZYe perpendicular a BP,XZ a DM e XY a FN,da para reescrever a desigualdade.Prove que XM+YN+ZP>=BN+BP+DP+DM+FM+FN. Essa e a proxima missao.ATEEEEEEEEEEE!!!!!Ploft!Peterdirichlet
-- Mensagem original -- >Esse problema foi considerado "O Imortal"(o menos respondido de toda a historia >da IMO).apenas 2 romenos e 4 armenios resolveram-no.TODA A EQUIPE CHINESA >ZEROU ESSE.IMO 1996 > > > >Problem 5 > >Let ABCDEF be a convex hexagon such that AB is parallel to DE, BC is parallel >to EF, and CD is parallel to FA. Let RA, RC, RE denote the circumradii of >triangles FAB, BCD, DEF respectively, and let p denote the perimeter of >the hexagon. Prove that: > > RA + RC + RE >= p/2. > > > >Solution > > >The starting point is the formula for the circumradius R of a triangle ABC: >2R = a/sin A = b/sin B = c/sin C. [Proof: the side a subtends an angle 2A >at the center, so a = 2R sin A.] This gives that 2RA = BF/sin A, 2RC = BD/sin >C, 2RE = FD/sin E. It is clearly not true in general that BF/sin A > BA >+ AF, although it is true if angle FAB >= 120, so we need some argument >that involves the hexagon as a whole. > >Extend sides BC and FE and take lines perpendicular to them through A and >D, thus forming a rectangle. Then BF is greater than or equal to the side >through A and the side through D. We may find the length of the side through >A by taking the projections of BA and AF giving AB sin B + AF sin F. Similarly >the side through D is CD sin C + DE sin E. Hence: > > 2BF >= AB sin B + AF sin F + CD sin C + DE sin E. Similarly: > > 2BD >= BC sin B + CD sin D + AF sin A + EF sin E, and > > 2FD >= AB sin A + BC sin C + DE sin D + EF sin F. > >Hence 2BF/sin A + 2BD/sin C + 2FD/sin E >= AB(sin A/sin E + sin B/sin A) >+ BC(sin B/sin C + sin C/sin E) + CD(sin C/sin A + sin D/sin C) + DE(sin >E/sin A + sin D/sin E) + EF(sin E/sin C + sin F/sin E) + AF(sin F/sin A >+ sin A/sin C). > >We now use the fact that opposite sides are parallel, which implies that >opposite angles are equal: A = E, B = E, C = F. Each of the factors multiplying >the sides in the last expression now has the form x + 1/x which has minimum >value 2 when x = 1. Hence 2(BF/sin A + BD/sin C + FD/sin E) >= 2p and the >result is proved. > >Essa soluçao e a oficial.A mais bonita e a de Ciprian Manolescu,o unico > Perfect Score da prova.Ele usou a famosa Desigualdade de Erdös-Mordell.Depois >eu envio a resposta dele.ATEEEEEEEEE.Peterdirichlet > > > > > > >TRANSIRE SVVM PECTVS MVNDOQUE POTIRE >CONGREGATI EX TOTO ORBE MATHEMATICI OB SCRIPTA INSIGNIA TRIBVERE >Medalha Fields(John Charles Fields) > > >------------------------------------------ >Use o melhor sistema de busca da Internet >Radar UOL - http://www.radaruol.com.br > > > >========================================================================= >Instruções para entrar na lista, sair da lista e usar a lista em >http://www.mat.puc-rio.br/~nicolau/olimp/obm-l.html >O administrador desta lista é <[EMAIL PROTECTED]> >========================================================================= > TRANSIRE SVVM PECTVS MVNDOQUE POTIRE CONGREGATI EX TOTO ORBE MATHEMATICI OB SCRIPTA INSIGNIA TRIBVERE Medalha Fields(John Charles Fields) ------------------------------------------ Use o melhor sistema de busca da Internet Radar UOL - http://www.radaruol.com.br ========================================================================= Instruções para entrar na lista, sair da lista e usar a lista em http://www.mat.puc-rio.br/~nicolau/olimp/obm-l.html O administrador desta lista é <[EMAIL PROTECTED]> =========================================================================