If a, b, c are nonzero real 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 equals

1978 AMC 12 Problem 20

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20.

a,

b,

c

are nonzero real numbers such that

\begin{aligned} \frac{a+b-c}{c} &=\frac{a-b+c}{b}\\ &=\frac{-a+b+c}{a}, \end{aligned}

and

x=\frac{(a+b)(b+c)(c+a)}{abc},

and

x\lt0,

then

x

equals

-1

-2

-4

-6

-8

Set the three equal fractions to

t

and compare pairs of the resulting equations

The comparison forces either

a=b=c

a+b+c=0

; use the sign of

x

Let the common value be

t.

The first two resulting equations imply

(b-c)(t+2)=0,

and cyclic comparisons give the analogous relations. Thus either

a=b=c,

which gives

x=8,

t=-2.

In the latter case

a+b=-c,

b+c=-a,

and

c+a=-b,

x=\frac{(-c)(-a)(-b)}{abc}=-1.

The condition

x\lt0

selects the latter value.

Therefore, the correct answer is A.