Difference between revisions of "1992 AHSME Problems/Problem 16"
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== Solution == | == Solution == | ||
| − | <math>\fbox{E}</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>. |
== See also == | == See also == | ||
Revision as of 02:52, 20 February 2018
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
We have 
and 
. Equating the two expressions for 
gives 
, so as 
cannot be 
for positive 
and 
, we must have 
.
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 | ||
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