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OLM Neamt 1998

 
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alex2008
Leibniz


Joined: 19 Oct 2008
Posts: 489
Location: Tulcea

PostPosted: Sat Nov 08, 2008 11:35 pm    Post subject: OLM Neamt 1998 Reply with quote

Daca a,b,c \in\mathbb{Q}_+^* si \frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}, calculati \frac{a^2+b^2+c^2}{ab+ac+bc}.
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Andi Brojbeanu
Newton


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

PostPosted: Mon Apr 20, 2009 6:52 pm    Post subject: Reply with quote

\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{a+b+b+c+a+c}=\frac{a+b+c}{2(a+b+c)}=\frac{1}{2}\Rightarrow 
a+b=2c; a+c=2b; b+c=2a.
De unde c=2a-b si c=2a-b\rightarrow2a-b=2b-a\Rightarrow 3a=3b\Rightarrow a=b, adica ab=a\cdot a=a^2; bc=b\cdot b=b^2; ac=c\cdot c=c^2.
Atunci \frac{a^2+b^2+c^2}{ab+bc+ac}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1.
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Virgil Nicula
Euler


Joined: 28 Sep 2007
Posts: 672

PostPosted: Mon Apr 20, 2009 10:46 pm    Post subject: Reply with quote

\frac {a}{b+c}=\frac {b}{c+a}=\frac {c}{a+b}=\frac {a+b+c}{2(a+b+c)}=\frac 12=\frac {a^2}{a(b+c)}= \frac {b^2}{b(c+a)}=\frac {c^2}{c(a+b)}=\frac {a^2+b^2+c^2}{2(ab+bc+ca)}\ \Longrightarrow\ \frac {a^2+b^2+c^2}{ab+bc+ca}=1\ .
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