1978 AHSME Problems/Problem 20

Revision as of 21:35, 4 September 2021 by MRENTHUSIASM (talk | contribs) (The solution has some flaws. If b=c, then it is possible for (a,b,c)=(-2t,t,t), from which we obtain the correct answer -1. I will rewrite a solution with complete symmetry.)

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

EDITING IN PROGRESS

~MRENTHUSIASM

See also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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