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Difference between revisions of "1978 AHSME Problems/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
Line 6: Line 7:
 
\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 18:31, 4 September 2021

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

Take the first two expressions (you can actually take any two expressions): $\frac{a+b-c}{c}=\frac{a-b+c}{b}$.

$\frac{a+b}{c}=\frac{a+c}{b}$

$ab+b^2=ac+c^2$

$a(b-c)+b^2-c^2=0$

$(a+b+c)(b-c)=0$

$\Rightarrow a+b+c=0$ OR $b=c$

The first solution gives us $x=\frac{(-c)(-a)(-b)}{abc}=-1$.

The second solution gives us $a=b=c$, and $x=\frac{8a^3}{a^3}=8$, which is not negative, so this solution doesn't work.

Therefore, $x=-1\Rightarrow\boxed{A}$.

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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