Difference between revisions of "1992 AHSME Problems/Problem 16"
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\text{(E) } 2</math> | \text{(E) } 2</math> | ||
| − | == Solution == | + | == Solution 1== |
<math>\fbox{E}</math> We have <math>\frac{x+y}{z} = \frac{x}{y} \implies xy+y^2=xz</math> and <math>\frac{y}{x-z} = \frac{x}{y} \implies y^2=x^2-xz \implies x^2-y^2=xz</math>. Equating the two expressions for <math>xz</math> gives <math>xy+y^2=x^2-y^2 \implies x^2-xy-2y^2=0 \implies (x+y)(x-2y)=0</math>, so as <math>x+y</math> cannot be <math>0</math> for positive <math>x</math> and <math>y</math>, we must have <math>x-2y=0 \implies x=2y \implies \frac{x}{y}=2</math>. | <math>\fbox{E}</math> We have <math>\frac{x+y}{z} = \frac{x}{y} \implies xy+y^2=xz</math> and <math>\frac{y}{x-z} = \frac{x}{y} \implies y^2=x^2-xz \implies x^2-y^2=xz</math>. Equating the two expressions for <math>xz</math> gives <math>xy+y^2=x^2-y^2 \implies x^2-xy-2y^2=0 \implies (x+y)(x-2y)=0</math>, so as <math>x+y</math> cannot be <math>0</math> for positive <math>x</math> and <math>y</math>, we must have <math>x-2y=0 \implies x=2y \implies \frac{x}{y}=2</math>. | ||
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| + | ==Solution 2== | ||
| + | We cross multiply the first and third fractions and the second and third fractions, respectively, for <cmath>x(x-z)=y^2</cmath> <cmath>y(x+y)=xz</cmath> Notice how the first equation can be expanded and rearranged to contain an <math>(x+y)</math> term. <cmath>x^2-xz=y^2</cmath> <cmath>x^2-y^2=xz</cmath> <cmath>(x+y)(x-y)=xz</cmath> We can divide this by the second equation to get <cmath>\frac{(x+y)(x-y)}{y(x+y)}=\frac{xz}{xz}</cmath> <cmath>\frac{x-y}{y}=1</cmath> <cmath>\frac{x}{y}-1=1</cmath> <cmath>\frac{x}{y}=\boxed{2}</cmath> | ||
== See also == | == See also == | ||
Revision as of 14:30, 6 January 2021
Contents
Problem
If
![\[\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\]](http://latex.artofproblemsolving.com/d/f/8/df8d708f208be3141edf6a0c74b96ec07518a608.png)
for three positive numbers 
and 
, all different, then 
Solution 1
We have 
and 
. Equating the two expressions for 
gives 
, so as 
cannot be 
for positive 
and 
, we must have 
.
Solution 2
We cross multiply the first and third fractions and the second and third fractions, respectively, for ![\[x(x-z)=y^2\]](http://latex.artofproblemsolving.com/5/e/b/5eb16823e18d76db5247d40b70e28a217036b11e.png)
![\[y(x+y)=xz\]](http://latex.artofproblemsolving.com/2/c/5/2c523f62ab69de16ec6a67cdca61ec748a53fc66.png)
Notice how the first equation can be expanded and rearranged to contain an 
term. ![\[x^2-xz=y^2\]](http://latex.artofproblemsolving.com/4/4/e/44e341b30843ce195c4d11901a1f80685772b842.png)
![\[x^2-y^2=xz\]](http://latex.artofproblemsolving.com/4/f/0/4f0b53cb93c903ac3f676d52f5cbe7e7adaa22eb.png)
![\[(x+y)(x-y)=xz\]](http://latex.artofproblemsolving.com/1/2/8/128ab98ec4d2662f7b7ddfa4717de3f91c123198.png)
We can divide this by the second equation to get ![\[\frac{(x+y)(x-y)}{y(x+y)}=\frac{xz}{xz}\]](http://latex.artofproblemsolving.com/9/1/1/911f1111eeea9b3caaff6f06285d450f30162e67.png)
![\[\frac{x-y}{y}=1\]](http://latex.artofproblemsolving.com/0/a/b/0aba277433796620abeb01879ffdcb8cb05c3cc8.png)
![\[\frac{x}{y}-1=1\]](http://latex.artofproblemsolving.com/3/a/0/3a0ce68998a8c4c1aafe8351ae0a65e1d036ac05.png)
![\[\frac{x}{y}=\boxed{2}\]](http://latex.artofproblemsolving.com/0/4/7/047454bd29191dbade41f93a0ebc7a1a9b4e3e9a.png)
See also
| 1992 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
