Fie \( a,b,c\in\mathbb{C} \) distincte doua cate doua astfel incat
\( (a-b)^5+(b-c)^5+(c-a)^5=0 \). Aratati ca \( a,b,c \) sunt afixele unui triunghi echilateral.
Afixele unui triunghi echilateral
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- DrAGos Calinescu
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Demonstratie : Notam \( x=\frac{a-b}{a-c} \) si \( y=\frac{b-c}{a-c} \) . Atunci \( x+y=1 \) si, din enunt obtinem : \( x^5+y^5=1 \)DrAGos Calinescu wrote:Fie \( a\ ,\ b\ ,\ c\ \in\ \mathbb{C}\ ,\ a\ \ne\ b\ \ne\ c\ \ne\ a \) astfel incat \( (a-b)^5+(b-c)^5+(c-a)^5=0 \).
Aratati ca \( a\ ,\ b\ ,\ c \) sunt afixele unui triunghi echilateral .
sau \( x^5+(1-x)^5=1\ \Longleftrightarrow\ 5x(x-1)(x^2-x+1)=0 \) . Cum \( x\ne 0 \) si \( x\ne 1 \) (in caz contrar \( a=b \) sau \( b=c \) ) ,
obtinem : \( x^2-x+1=0 \) , deci \( x^3=-1 \) si \( |x|=1 \) . Analog \( |y|=1 \) , deci \( |a-b|=|b-c|=|c-a| \) de unde rezulta concluzia .
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Claudiu Mindrila
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Solutie. Daca \( x=a-b \), \( y=b-c \), \( z=c-a \) atunci \( x+y+z=x^{5}+y^{5}+z^{5}=0 \). Fie polinomul \( P\left(t\right)=t^{3}-mt^{2}+nt-p \) care are radacinile \( x,\ y,\ z \). Din relatiile lui Viete avem \( m=0 \) , deci \( P\left(t\right)=t^{3}+nt-p \). Avem:
\( \left\{ \begin{array}{c}
P\left(x\right)=0\\
P\left(y\right)=0\\
P\left(z\right)=0\end{array}\Longrightarrow\left\{ \begin{array}{c}
x^{3}=p-nx\\
y^{3}=p-ny\\
z^{3}=p-nz\end{array}\Longrightarrow\left\{ \begin{array}{c}
x^{5}=px^{2}-nx^{3}\\
y^{5}=py^{2}-ny^{3}\\
z^{5}=pz^{2}-nz^{3}\end{array}\right.\right.\right. \)
Sumand aceste relatii si tinand cont de ipoteza si de faptul ca \( a^{3}+b^{3}+c^{3}-3abc=\left(a+b+c\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right),\ \forall a,\ b,\ c\in\mathbb{C} \) avem ca
\( p\left(x^{2}+y^{2}+z^{2}\right)=n\left(x^{3}+y^{3}+z^{3}\right)=3np\Longrightarrow p\left(0^{2}-2n\right)=3np\Longrightarrow5np=0 \), si cum \( p\neq0 \) avem ca \( n=0\Longleftrightarrow a^{2}+b^{2}+c^{2}=ab+bc+ca \), ceea ce incheie problema.
Observatie. A se vedea si urmatoarea problema:
\( \left\{ \begin{array}{c}
P\left(x\right)=0\\
P\left(y\right)=0\\
P\left(z\right)=0\end{array}\Longrightarrow\left\{ \begin{array}{c}
x^{3}=p-nx\\
y^{3}=p-ny\\
z^{3}=p-nz\end{array}\Longrightarrow\left\{ \begin{array}{c}
x^{5}=px^{2}-nx^{3}\\
y^{5}=py^{2}-ny^{3}\\
z^{5}=pz^{2}-nz^{3}\end{array}\right.\right.\right. \)
Sumand aceste relatii si tinand cont de ipoteza si de faptul ca \( a^{3}+b^{3}+c^{3}-3abc=\left(a+b+c\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right),\ \forall a,\ b,\ c\in\mathbb{C} \) avem ca
\( p\left(x^{2}+y^{2}+z^{2}\right)=n\left(x^{3}+y^{3}+z^{3}\right)=3np\Longrightarrow p\left(0^{2}-2n\right)=3np\Longrightarrow5np=0 \), si cum \( p\neq0 \) avem ca \( n=0\Longleftrightarrow a^{2}+b^{2}+c^{2}=ab+bc+ca \), ceea ce incheie problema.
Observatie. A se vedea si urmatoarea problema:
Daca \( a,\ b,\ c\in\mathbb{C} \) a. i. \( \left(a-b\right)^{7}+\left(b-c\right)^{7}+\left(c-a\right)^{7}=0 \) atunci \( a,\ b,\ c \) sunt afixele varfurilor unui triunghi echilateral.
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste