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) } | + | <math>\textbf{(A) }-1\qquad |
− | \textbf{(B) } | + | \textbf{(B) }-2\qquad |
− | \textbf{(C) } | + | \textbf{(C) }-4\qquad |
− | \textbf{(D) } | + | \textbf{(D) }-6\qquad |
− | \textbf{(E) } | + | \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) } | + | * If <math>a+b+c=0,</math> then <math>x=\frac{(-c)(-a)(-b)}{abc}=\boxed{\textbf{(A) }-1}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 03:38, 26 November 2021
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 |
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