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Fie \( a\ ,\ b\ , \ c \) numere complexe distincte cu \( |a|=|b|=|c|\ >\ 0 \) . Sa se arate ca punctele de afixe \( a \) , \( b \) si \( c \)

sunt varfurile unui triunghi echilateral daca si numai daca exista \( k\in\mathbb{R}-\{1\} \) astfel incat :

\( |ka+b+c|=|a+kb+c|=|a+b+kc| \) .

\( |ka+b+c|=|a+kb+c|=|a+b+kc| <=> |(a+b+c)+(k-1)a|^2=|(a+b+c)+(k-1)b|^2=|(a+b+c)+(k-1)c|^2 \) (1) Sa notam \( a+b+c=s \) Relatia (1) devine:
\( |s+(k-1)a|^2=|s+(k-1)b|^2=|s+(k-1)c|^2<=>|s|^2+|k-1|^2|a|^2+(k-1)\overline{s}a+(k-1)s\overline{a}=|s|^2+|k-1|^2|b|^2+(k-1)\overline{s}b+(k-1)s\overline{b}=|s|^2+|k-1|^2|c|^2+(k-1)\overline{s}c+(k-1)s\overline{c} \)
\( <=>(k-1)(\overline{s}a+s\overline{a})=(k-1)(\overline{s}b+s\overline{b})=(k-1)(\overline{s}c+s\overline{c})<=>\overline{s}a+s\overline{a}=\overline{s}b+s\overline{b}=\overline{s}c+s\overline{c} \) (2)
Notam \( |a|=|b|=|c|=r \).
Cazul 1) daca \( s=0 \) ,deci \( a+b+c=0=>|-a|^2=|b+c|^2=>r^2=2r^2+\overline{b}c+b\overline{c}=>\overline{b}c+b\overline{c}=-r^2 \).Analog se arata ca \( \overline{c}a+c\overline{a}=\overline{a}b+a\overline{b}=-r^2 \).Deci \( 3r^2=|c|^2+|b|^2-\overline{b}c-b\overline{c}=|a|^2+|c|^2-\overline{c}a-c\overline{a}=|a|^2+|b|^2-\overline{a}b-a\overline{b}=>|a-b|=|b-c|=|c-a|=>a,b,c \) sunt afixele varfurilor unui triunghi echilateral.
Cazul 2) daca \( s\neq 0=>r^2(\frac{a}{s}+\frac{s}{a})=r^2(\frac{b}{s}+\frac{s}{b})=r^2(\frac{c}{s}+\frac{s}{c})=>\frac{a-b}{s}+\frac{s(b-a)}{ab}=0=>(a-b)(\frac{1}{s}-\frac{s}{ab})=0 \).Cum \( a\neq b=>ab=s^2 \).Analog \( bc=s^2 \) si \( ca=s^2 \).Deci \( ab=bc=ca \).
\( ab=bc=>ab-bc=0=>b(a-c)=0=>a=c \) sau \( b=0 \),ambele situatii fiind imposibile.
S-a demonstrat implicatia:"Fie a,b,c numere complexe distincte,cu \( |a|=|b|=|c|>0 \).Daca exista \( k\in\mathbb{R}-\{1\} \) astfel incat \( |ka+b+c|=|a+kb+c|=|a+b+kc| \),atunci \( a,b,c \) sunt afixele varfurilor unui triunghi echilateral".

Reciproc,daca stim ca \( a,b,c \) sunt afixele varfuriloe unui triunghi echilateral,atunci putem considera \( a=r,b=r\epsilon,c=r\epsilon^2 \),unde \( \epsilon \) e radacina nereala de ordinul 3 a unitatii. \( a+b+c=r(1+\epsilon+\epsilon^2)=0 \).De aici,prin calcul,reciproca se demonstreaza usor.