1978 AHSME Problems/Problem 20

Problem 20

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

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

Solution

From the equation $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a},$ we add $2$ to each fraction to get $\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}.$ We perform casework on $a+b+c:$

If $a+b+c\neq0,$ then $a=b=c,$ from which $x=\frac{(2a)(2a)(2a)}{a^3}=8.$ However, this contradicts the precondition $x<0.$

If $a+b+c=0,$ then $x=\frac{(-c)(-a)(-b)}{abc}=\boxed{\textbf{(A) }{-}1}.$

~MRENTHUSIASM

1978 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1978 AHSME Problems/Problem 20

Problem 20

Solution

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