$x^2+xy+y^2=c$ $;$ $y^2+yz+z^2=a$ $;$ $z^2+xz+x^2=b$ $;$ $xy+yz+zx=0$ $.$
Find relation between a,b,c(eliminate x,y,z)
I came across this question while i was solving an olympiad book
My approach was to first substitute $x$ from the 4th equation and further eliminate $y,z$ but it was very tough to get the result with this approach
Secondly i got up to these three equations:-
$x^2+y^2+z^2=(a+b+c)/2;$ $(x+y+z)^2=(a+b+c)/2;$ $(x+y)^2+(x+z)^2+(y+z)^2=(a+b+c);$
But i cannot figure out how could we eliminate $x,y,z$ using this approach .hence i was stuck at this point and also tried some more methods like subtracting two equations etc...
It is not a homework problem.
Adding up the first $3$ equations,
$2(x^2+y^2+z^2)+xy+yz+zx=a+b+c$
So, $2[x^2+y^2+z^2-(xy+yz+zx)]=a+b+c$
Therefore, $(x-y)^2+(y-z)^2+(z-x)^2=a+b+c$.
and $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=\dfrac12(a+b+c)$
Note that $(x-z)(x+y+z)=x^2-z^2+xy-yz=c-a$
$(x-z)^2=\dfrac{(c-a)^2}{(x+y+z)^2}=\dfrac{2(c-a)^2}{a+b+c}$.
Similarly, $(y-x)^2=\dfrac{2(a-b)^2}{a+b+c}$ and $(z-y)^2=\dfrac{2(b-c)^2}{a+b+c}$.
Therefore, $\displaystyle \dfrac{2(c-a)^2}{a+b+c}+\dfrac{2(a-b)^2}{a+b+c}+\dfrac{2(b-c)^2}{a+b+c}=a+b+c$.
$2[(c-a)^2+(a-b)^2+(b-c)^2]=(a+b+c)^2$
$4(a^2+b^2+c^2-ab-bc-ca)=(a^2+b^2+c^2+2ab+2bc+2ca)$
$3(a^2+b^2+c^2)-6(ab+bc+ca)=0$
$a^2+b^2+c^2-2(ab+bc+ca)=0$