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 2

We equate the first two expressions (More generally, we can equate any two expressions): $\frac{a+b-c}{c}=\frac{a-b+c}{b}.$ We add $1$ to both sides, then rearrange: $\begin{align*} \frac{a+b}{c} &= \frac{a+c}{b} \\ ab+b^2 &= ac+c^2 \\ \bigl(ab-ac\bigr)+\bigl(b^2-c^2\bigr) &= 0 \\ a(b-c)+(b+c)(b-c) &= 0 \\ (a+b+c)(b-c) &= 0, \end{align*}$ from which $a+b+c=0$ or $b=c.$

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

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

~Pega969 (Solution)

~MRENTHUSIASM (Revision)

The second solution gives us $a=b=c$ , and $x=\frac{8a^3}{a^3}=8$ , which is not negative, so this solution doesn't work.

Therefore, $x=-1\Rightarrow\boxed{A}$ .

EDITING IN PROGRESS

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 2

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