Given $x,y,z\in \mathbb{R} ^+$ , that are the solutions to the system of equations :
$$x^2+xy+y^2=57$$$$y^2+yz+z^2=84$$$$z^2+zx+x^2=111$$What is the value of $xy+3yz+5zx$?
(maphybich)
$\begin{cases} x^2 + xy + y^2 = 57\\ y^2+yz+z^2=84\\z^2+xz+x^2=111 \end{cases}$
It is noted that $57,84,111$ are in arithmetic progression with $D=27$
It is known that if $x^2 + xy + y^2, y^2+yz+z^2, z^2+xz+x^2$ are in arithmetic progression, so are $x,y,z$.
This finding emerges upon making the substitutions (if $x<y<z$) $x=y-d, z=y+d$ in the above
system to yield:
$\Longrightarrow \begin{cases} d^2 + 3y^2-3dy = 57\\ d^2+3y^2+3dy=84\\d^2+3y^2~~~~~~~~~=111 \end{cases}$
In the so-obtained system, each equation indeed differs sequentially by a constant amount $~~3dy~$ (however in
this case, in a wrong order, owing to the choice $x<y<z$
Trying $~y<x<z~$ with $y=x-d, z=x+d~$ re-establishes the correct order:
$\Longrightarrow \begin{cases} d^2 + 3x^2-3dx = 57\\ d^2+3x^2~~~~~~~~~~=84\\d^2+3x^2+3dx~=111 \end{cases}$
It can be derived therefrom that $~~3dx=27\Longrightarrow dx=9$
$\Longrightarrow \begin{cases} dx = 9 \\ d^2+3x^2=84\end{cases}$
$\Longrightarrow 3x^4-84x^2+81=0$ $\Longrightarrow x \in \{-3 \sqrt{3},-1, 1, 3 \sqrt{3}\}$
$\Longrightarrow (x,y,z)= (1,-8,10), (-1,8,-10), (3 \sqrt{3},2\sqrt{3}, 4\sqrt{3}), (-3 \sqrt{3},-2\sqrt{3}, -4\sqrt{3})$
Only acceptable solution $(x,y,z)= (3 \sqrt{3},2\sqrt{3}, 4\sqrt{3}) \in R_{+}$
(AP can be noted)
$\Longrightarrow xy+3yz+5zx=270$