Difference between revisions of "1978 AHSME Problems/Problem 20"
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== Problem 20 == | == Problem 20 == | ||
| − | 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> | + | 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 | ||
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\textbf{(E) }-8 </math> | \textbf{(E) }-8 </math> | ||
| − | ==Solution== | + | ==Solution 2== |
| + | We equate the first two expressions (More generally, we can equate any two expressions): <cmath>\frac{a+b-c}{c}=\frac{a-b+c}{b}.</cmath> | ||
| + | We add <math>1</math> to both sides, then rearrange: | ||
| + | <cmath>\begin{align*} | ||
| + | \frac{a+b}{c} &= \frac{a+c}{b} \\ | ||
| + | ab+b^2 &= ac+c^2 \\ | ||
| + | \bigl(ab-ac\bigr)+\bigl(b^2-c^2\bigr) &= 0 \\ | ||
| + | a(b-c)+(b+c)(b-c) &= 0 \\ | ||
| + | (a+b+c)(b-c) &= 0, | ||
| + | \end{align*}</cmath> | ||
| + | from which <math>a+b+c=0</math> or <math>b=c.</math> | ||
| − | + | * If <math>a+b+c=0,</math> then <math>x=\frac{(-c)(-a)(-b)}{abc}=-1.</math> | |
| − | <math> | + | * If <math>b=c,</math> then <math>x=\frac{(a+b)(b+c)(c+a)}{abc}</math> |
| − | + | ~Pega969 (Solution) | |
| − | + | ~MRENTHUSIASM (Revision) | |
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The second solution gives us <math>a=b=c</math>, and <math>x=\frac{8a^3}{a^3}=8</math>, which is not negative, so this solution doesn't work. | The second solution gives us <math>a=b=c</math>, and <math>x=\frac{8a^3}{a^3}=8</math>, which is not negative, so this solution doesn't work. | ||
Therefore, <math>x=-1\Rightarrow\boxed{A}</math>. | Therefore, <math>x=-1\Rightarrow\boxed{A}</math>. | ||
| + | |||
| + | <b>EDITING IN PROGRESS</b> | ||
== See also == | == See also == | ||
Revision as of 21:03, 4 September 2021
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 2
We equate the first two expressions (More generally, we can equate any two expressions): ![\[\frac{a+b-c}{c}=\frac{a-b+c}{b}.\]](http://latex.artofproblemsolving.com/b/2/0/b208ea84869d4de36baca11391a6d0d6a7d09d48.png)
We add 
to both sides, then rearrange:
from which 
or 
- If
then 
then 
- If
then 
then 
~Pega969 (Solution)
~MRENTHUSIASM (Revision)
The second solution gives us 
, and 
, which is not negative, so this solution doesn't work.
Therefore, 
.
EDITING IN PROGRESS
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. 
