Difference between revisions of "1978 AHSME Problems/Problem 20"
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If <math>a,b,c</math> are non-zero real numbers such that <math>\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}</math>, | If <math>a,b,c</math> are non-zero real numbers such that <math>\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}</math>, | ||
and <math>x=\frac{(a+b)(b+c)(c+a)}{abc}</math>, and <math>x<0</math>, then <math>x</math> equals | and <math>x=\frac{(a+b)(b+c)(c+a)}{abc}</math>, and <math>x<0</math>, then <math>x</math> equals | ||
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\textbf{(C) }-4\qquad | \textbf{(C) }-4\qquad | ||
\textbf{(D) }-6\qquad | \textbf{(D) }-6\qquad | ||
| − | \textbf{(E) }-8 </math> | + | \textbf{(E) }-8 </math> |
===Solution=== | ===Solution=== | ||
Revision as of 10:50, 3 July 2016
If 
are non-zero real numbers such that 
,
and 
, and 
, then 
equals
