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O.VI.29

 
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Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Sun Apr 19, 2009 6:16 pm    Post subject: O.VI.29 Reply with quote

Se considera triunghiul isoscel ABC(AC=AB) si punctele E, F \in BC astfel incat B\in [FC], C \in [BE] si FB=CE=AB. Fie O punctul de intersectie a bisectoarelor unghiurilor ABF si ACE. Sa se arate ca:
a) OF=OA=OE;
b) A, O si mijlocul lui BC sunt coliniare.

Probleme date la olimpiade, RMT 1/1998
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Mateescu Constantin
Newton


Joined: 21 Apr 2009
Posts: 398
Location: Londra/Pitesti

PostPosted: Sun Apr 26, 2009 11:19 am    Post subject: Reply with quote

a)  m(\angle FBO)=180^{\circ}-\frac{m(\angle FBA)}{2}

 m(\angle ABO)=m(\angle CBO)+m(\angle ABC)=\frac{m(\angle ABF)}{2}+180^{\circ}-m(\angle ABF)=180^{\circ}-\frac{m(\angle FBA)}{2}

\Longrightarrow \angle FBO\equiv\angle ABO\ (1)
Deoarece AB=BF, [OB] latura comuna a \triangle FBO si \triangle ABO si cu (1)

\Longrightarrow \triangle FBO\equiv\triangle ABO \Longrightarrow OA=OF

Analog se arata ca OA=OE \Longrightarrow OA=OE=OF


b) Fie M mijlocul [BC]. Atunci AM este mediatoarea segmentului [BC]\ (2)

FM=FB+BM,\ EM=CE+MC, FB=CE si BM=CE
\Longrightarrow FM=EM, deci M este mijlocul [EF]

\triangle OEF este isoscel, deci M se afla pe mediatoarea [BC]\ (3)
Din (2) si (3) \Longrightarrow A, M si O sunt coliniare.
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