AMC 1834 - If $a, b $ and $c$ are nonzero numbers such that (a+b-c)/c=(a-b+c)/b=(-a+b+c)/a and x=((a+b)(b+c)(c+a))/abc and x<0, then x=? I am getting stuck on this question:
If a, b, and c, are nonzero numbers such that $\frac{a+b-c}{c}$=$\frac{a-b+c}{b}$=$\frac{-a+b+c}{a}$ and x=$\frac{(a+b)(b+c)(c+a)}{abc}$ and x<0, then x=?
What I have done so far: Consider, if $\frac{a}{b}$=$\frac{c}{d}$, and (b+d) is non-zero, then $\frac{a+c}{b+d}$=$\frac{a}{b}$=$\frac{c}{d}$. So with that fact, I did... $\frac{a+b-c}{c}$=$\frac{a-b+c}{b}$ $\implies$
$\frac{2a}{c+b}$ and $\frac{2a}{c+b}$=$\frac{-a+b+c}{a}$$\implies$ $\frac{(a+b+c)}{a+b+c}$ = 1. Then I took the original equations and did $\frac{a+b-c}{c}$=1, so $a+b=2c$ , and similarly, $(a+c)=2b$ and $b+c=2a$. So plugging that into x=$\frac{(a+b)(b+c)(c+a)}{abc}$ we have $x=\frac{(2a)(2b)(2c)}{abc}$, which is $x=\frac{8(abc)}{abc}=8$. This is wrong though, because it says in the problem x<0. So how do I do this?
 A: $$\frac{a+b-c}{c}=\frac{a-b+c}{b}$$
$$ab+b^2-bc=ac-bc+c^2\implies a(b-c)=(b+c)(c-b)$$
$$\implies b=c \quad  OR\quad a+b+c=0$$
If we take $b=c$, $$\frac{a}{c}=\frac{-a+2c}{a}\quad using (1)and (3).$$
$$\implies ({\frac{a}{c}})^2=1\quad OR\quad -2$$
Obviosuly $-2$ isn't possible, $a=c$ will give $x=1$ and $a=-c$ will give $x=0$.
So, $a+b+c=0\implies a=-(b+c), b=-(a+c), c=-(a+b)\implies x=-1.$
A: Hint: $$a=-\frac{b}{2},b=b,c=-\frac{b}{2}$$ solves your problem and $$x=-1$$
A: You correctly derived:
$$\frac{2a}{c+b}=\frac{-a+b+c}{a}$$
Your mistake is here:

Then I took the original equations and did $\frac{a+b-c}{c}=1$

Instead, denote: $\frac{b+c}{a}=t$, then the above equation becomes:
$$\frac2{t}=-1+t \Rightarrow t^2-t-2=0 \Rightarrow t_1=-1 \ \ \text{and} \ \ t_2=2.$$ 
Also note that:
$$\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\iff \\
\frac{a+b}{c}-1=\frac{a+c}{b}-1=\frac{b+c}{a}-1 \iff \\
\frac{a+b}{c}=\frac{a+c}{b}=\frac{b+c}{a}=t.$$
Hence:
$$x=\frac{(a+b)(b+c)(c+a)}{abc}=t^3=(-1)^3=-1;\\
x=\frac{(a+b)(b+c)(c+a)}{abc}=t^3=2^3\not< 0.$$
