[obm-l] Re: [obm-l] Uma soluçao bonitinha da IMO da India

[peterdirichlet@xxxxxxxxxxxxxx]

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)
>
>
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>
>
>
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TRANSIRE SVVM PECTVS MVNDOQUE POTIRE
CONGREGATI EX TOTO ORBE MATHEMATICI OB SCRIPTA INSIGNIA TRIBVERE
Medalha Fields(John Charles Fields)


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