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

Revision as of 11:01, 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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 27: / Line 27: @@
 Therefore, <math>x=1\Rightarrow\boxed{A}</math>.
+== See also ==
+{{AHSME box|year=1978|n=I|num-b=19|num-a=21}}
+{{MAA Notice}}

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

Revision as of 11:01, 3 July 2016

Solution

See also