Difference between revisions of "1978 AHSME Problems/Problem 20"
MRENTHUSIASM (talk | contribs) |
(→See also) |
||
(7 intermediate revisions by one other user not shown) | |||
Line 2: | Line 2: | ||
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== |
− | + | 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> | |
− | + | We perform casework on <math>a+b+c:</math> | |
− | <cmath> | ||
− | \frac{a+b}{c} | ||
− | |||
− | |||
− | a | ||
− | |||
− | |||
− | |||
− | * If <math>a+b+c= | + | * 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>b= | + | * If <math>a+b+c=0,</math> then <math>x=\frac{(-c)(-a)(-b)}{abc}=\boxed{\textbf{(A) }{-}1}.</math> |
− | + | ~MRENTHUSIASM | |
− | |||
− | ~MRENTHUSIASM | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
== See also == | == See also == | ||
{{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 12:24, 24 March 2024
Problem 20
If are non-zero real numbers such that
and
and
then
equals
Solution
From the equation we add
to each fraction to get
We perform casework on
- If
then
from which
However, this contradicts the precondition
- If
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.