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 <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 <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 | ||
| − | <math>\textbf{(A) }-1\qquad | + | <math>\textbf{(A) }{-}1\qquad |
| − | \textbf{(B) }-2\qquad | + | \textbf{(B) }{-}2\qquad |
| − | \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== | ||
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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> | * 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> | ||
| − | * If <math>a+b+c=0,</math> then <math>x=\frac{(-c)(-a)(-b)}{abc}=\boxed{\textbf{(A) }-1}.</math> | + | * If <math>a+b+c=0,</math> then <math>x=\frac{(-c)(-a)(-b)}{abc}=\boxed{\textbf{(A) }{-}1}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
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{{AHSME box|year=1978|n=I|num-b=19|num-a=21}} | {{AHSME box|year=1978|n=I|num-b=19|num-a=21}} | ||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | |||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 11:24, 24 March 2024
Problem 20
If 
are non-zero real numbers such that ![\[\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a},\]](http://latex.artofproblemsolving.com/9/7/3/9733280bec67e5196a2784002ad5f93db82b8d47.png)
and ![\[x=\frac{(a+b)(b+c)(c+a)}{abc},\]](http://latex.artofproblemsolving.com/1/a/8/1a811c35549cfac436bf409b437791e9f6cbab38.png)
and 
then 
equals
Solution
From the equation we add 
to each fraction to get ![\[\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}.\]](http://latex.artofproblemsolving.com/5/9/0/59023322e09a64cfb215ac0c0cdc98d676ea8cfc.png)
We perform casework on 
- If
then 
from which 
However, this contradicts the precondition 
then 
from which 
However, this contradicts the precondition 
- If
then 
then 
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
