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.ATEEE!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.ATE.Peterdirichlet
TRANSIRE SVVM PECTVS MVNDOQUE POTIRE
CONGREGATI EX TOTO ORBE MATHEMATICI OB SCRIPTA INSIGNIA TRIBVERE
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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
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