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

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=== Problem 20 ===
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== Problem 20 ==
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If <math>a,b,c</math> are non-zero real numbers such that <cmath>\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a},</cmath> and <cmath>x=\frac{(a+b)(b+c)(c+a)}{abc},</cmath> and <math>x<0,</math> then <math>x</math> equals
  
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>,
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<math>\textbf{(A) }{-}1\qquad
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{(B) }{-}2\qquad
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\textbf{(C) }{-}4\qquad
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\textbf{(D) }{-}6\qquad
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\textbf{(E) }{-}8    </math>  
  
<math>\textbf{(A) }-1\qquad
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==Solution==
\textbf{(B) }-2\qquad
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From the equation <cmath>\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a},</cmath> we add <math>2</math> to each fraction to get <cmath>\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}.</cmath>
\textbf{(C) }-4\qquad
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We perform casework on <math>a+b+c:</math>
\textbf{(D) }-6\qquad
 
\textbf{(E) }-8    </math>  
 
  
===Solution===
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* If <math>a+b+c\neq0,</math> then <math>a=b=c,</math> from which <math>x=\frac{(2a)(2a)(2a)}{a^3}=8.</math> However, this contradicts the precondition <math>x<0.</math>
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* If <math>a+b+c=0,</math> then <math>x=\frac{(-c)(-a)(-b)}{abc}=\boxed{\textbf{(A) }{-}1}.</math>
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~MRENTHUSIASM
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== See also ==
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{{AHSME box|year=1978|n=I|num-b=19|num-a=21}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 11:24, 24 March 2024

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

See also

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