Question

If $\frac{2y+2z-x}{a}=\frac{2z+2x-y}{b}=\frac{2x+2y-z}{c},$ show that $\frac{x}{2b+2c-a}=\frac{y}{2c+2a-b}=\frac{z}{2a+2b-c}.$

Answer 1

Let $\frac{2y+2z-x}{a}=\frac{2z+2x-y}{b}=\frac{2x+2y-z}{c}=k$

$\therefore 2y+2z-x=ak⟶\left(I\right)$

$2z+2x-y=bk⟶\left(II\right)$

$2x+2y-z=ck⟶\left(III\right)$

Now multiply all equations by $2:$

$4y+4z-2x=2ak⟶\left(IV\right)$

$4z+4x-2y=2bk⟶\left(V\right)$

$4x+4y-2z=2ck⟶\left(VI\right)$

Now using $\left(V\right)+\left(VI\right)-\left(I\right):$

$4z+4x-2y+4x+4y-2z-2y-2z+x=k(2c+2b-a)$

$⟹9x=k(2b+2c-a)$

$⟹\frac{x}{(2b+2c-a)}=\frac{k}{9}⟶\left(a\right)$

Now using $\left(IV\right)+\left(VI\right)-\left(II\right):$

$4y+4z-2x+4x+4y-2z-2z-2x+y=k(2a+2c-b)$

$⟹9y=k(2a+2c-b)$

$⟹\frac{y}{(2a+2c-b)}=\frac{k}{9}⟶\left(b\right)$

Now using $\left(IV\right)+\left(V\right)-\left(III\right):$

$4y+4z-2x+4z+4x-2y-2x-2y+z=k(2a+2b-c)$

$⟹9z=k(2a+2b-c)$

$⟹\frac{z}{(2a+2b-c)}=\frac{k}{9}⟶\left(c\right)$

$\because a=b=c$

$\therefore \frac{x}{(2b+2c-a)}=\frac{y}{(2a+2c-b)}=\frac{z}{(2a+2b-c)}$