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

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Latest revision as of 12:24, 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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