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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>,
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== Problem 20 ==
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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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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
  
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<math>\textbf{(A) }-1\qquad
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<math>\textbf{(A) }{-}1\qquad
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\textbf{(B) }-2\qquad
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\textbf{(B) }{-}2\qquad
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\textbf{(C) }-4\qquad
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\textbf{(C) }{-}4\qquad
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\textbf{(D) }-6\qquad  
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\textbf{(D) }{-}6\qquad  
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\textbf{(E) }-8    </math>
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\textbf{(E) }{-}8    </math>  
  
 
==Solution==
 
==Solution==
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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>
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We perform casework on <math>a+b+c:</math>
  
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Take the first two expressions (you can actually take any two expressions):  <math>\frac{a+b-c}{c}=\frac{a-b+c}{b}</math>.  
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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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<math>\frac{a+b}{c}=\frac{a+c}{b}</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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<math>ab+b^2=ac+c^2</math>
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~MRENTHUSIASM
  
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<math>a(b-c)+b^2-c^2=0</math>
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== See also ==
  
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<math>(a+b+c)(b-c)=0</math>
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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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<math>\Rightarrow a+b+c=0</math> OR <math>b=c</math>
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{{MAA Notice}}
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The first solution gives us <math>x=\frac{(-c)(-a)(-b)}{abc}=-1</math>.
 
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The second solution gives us <math>a=b=c</math>, and <math>x=\frac{8a^3}{a^3}=8</math>, which is not negative, so this solution doesn't work.
 
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Therefore, <math>x=1\Rightarrow\boxed{A}</math>.
 

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 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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