0

How would you solve this problem:

If $a,b,c$ are non-zero real 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 $x$ equals

$\textbf{(A) }-1\qquad \textbf{(B) }-2\qquad \textbf{(C) }-4\qquad \textbf{(D) }-6\qquad \textbf{(E) }-8$

Ayush
  • 19

1 Answers1

2

Suppose that the three equal expressions are all equal to $y$. Then we have $$\frac{a+b}{c}=\frac{c+a}{b}=\frac{b+c}{a}=y+1$$ and so $$x=(y+1)^3\ .$$ Also, adding the three equations $$a+b=c(y+1)\ ,\quad c+a=b(y+1)\ ,\quad b+c=a(y+1)$$ gives $$2(a+b+c)=(y+1)(a+b+c)\ .$$ There are two options:

  • if $y+1=2$ then $x=8$, but this is impossible as we are told $x<0$;
  • if $a+b+c=0$ then $-c=c(y+1)$ so $y+1=-1$ so $x=-1$.
David
  • 82,662