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

[peterdirichlet@xxxxxxxxxxxxxx]

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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