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IMAC Juniori I 15 mai 2010 Subiectul I

Posted: Tue Jun 15, 2010 8:56 pm
by Andi Brojbeanu
Numerele reale \( x, y, z \) distincte doua cate doua, satisfac egalitatile: \( x^3-3x^2=y^3-3y^2=z^3-3z^2 \). Aflati \( x+y+z \).
R. Moldova

Metoda mai simpla :)

Posted: Wed Jun 16, 2010 12:21 pm
by Mr. Ady
\( x^3-3x^2=y^3-3y^2 \)
\( x^3-y^3=3x^2-3y^2 \)
\( (x-y)(x^2+xy+y^2)=3(x+y)(x-y) \)
Cum \( x \neq y \Rightarrow x^2+xy+y^2=3(x+y) \)
Analog \( y^2+zy+z^2=3(z+y) \) si \( z^2+zx+x^2=3(z+x) \)
\( \Rightarrow y^2+zy+z^2-z^2-zx-x^2=3(y-x) \)
\( \Leftrightarrow (y+x)(y-x)+z(y-x)=3(y-x) \)
\( \Leftrightarrow y+x+z=3 \)

Posted: Wed Jun 16, 2010 12:33 pm
by Horia Nicolaescu
O alta solutie ar fi urmatoarea.
Notam t=x^3-3x^2
Deci x,y,z sunt radacinile polinomului de grad 3:
P(a)=a^3-3a^2-t si aplicand relatiile lui viete obtinem x+y+z=3