Para aplicar Erd�s-Mordell neste problema,e bom que ponhamos tudo para dentro!Tra�ando os paralelogramos DEFM,NFAB,PBCD,e o triangulo XYZ tal que ZY e perpendicular a BP,XZ a DM e XY a FN,da para reescrever a desigualdade.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 portanto vamos provar que XM+YN+ZP>=BN+BP+DP+DM+FM+FN.
Vamos quebrar isso em dois casos: 1)M,N e P coincidem.E ai e so usar Erd�s-Mordell em XYZ direto 2)MNP e um triangulo.Ai fica mais complicado...Mas a ideia e a mesma. Vamos colocar um espelho na bissetriz do angulo ZXY e pegar as imagens de X e de Z(X' e Z' nesta ordem).Considere os pes das perpendiculares de X e M(H e G nesta ordem).Sejam tambem x=YZ,y=ZX e z =XY as medidas dos lados de XYZ. Entao [XZ'M]+[Z'Y'M]+[Y'XM]=[XY'Z'],ou x*XH=x*MG+y*FM+z*MD.Agora vamos usar a desigualdade triangular em XMG e a desigualdade hipotenusa>cateto em XHG(o angulo em H e reto):XM+MG>=XG>=XH,logo XM>=(z/x)*DM+(y/x)*FM. E nao e dificil concluir que XM+YN+ZP>=z/x*DM+y/x*FM+x/y*FN+z/y*BN+y/z*BP+x/z*DP.Daqui sai o fim do problema. Depois eu continuo. ATEEEEEEE!!!!!!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 solu�ao e a oficial.A mais bonita e a de Ciprian Manolescu,o unico >>> Perfect Score da prova.Ele usou a famosa Desigualdade de Erd�s-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 ========================================================================= Instru��es 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 � <[EMAIL PROTECTED]> =========================================================================
