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[obm-l] [obm-l] Uma soluçao bonitinha da IMO da India(reformulados)
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 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 Erdös-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 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)
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