Prova do lema:veja que os triangulos DEF e DMF sao semelhantes,logo tem o mesmo circunraio.Mas XM e o diametro do triangulo DMF.Logo XM=2*RA.E reformule a desigualdade. Vamos quebrar isso em dois casos: 1)M,N e P coincidem.E ai e so usar Erdös-Mordell em XYZ. 2)MNP e um triangulo.Depois eu continuo. ATEEEEEEE!!!!!!loft1Peterdirichlet. -- Mensagem original --
>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]> =========================================================================