Difference between revisions of "1978 AHSME Problems/Problem 20"

(Created page with "=== Problem 20 === 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)...")
 
(Problem 20)
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=== Problem 20 ===
 
 
 
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 11:50, 3 July 2016

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