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>. | ||
+ | |||
+ | ==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
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
Notice how the first equation can be expanded and rearranged to contain an
term.
We can divide this by the second equation to get
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.