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Difference between revisions of "1992 AHSME Problems/Problem 16"

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==Solution 2==
 
==Solution 2==
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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>
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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}=2 \rightarrow \boxed{E}</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 13:31, 6 January 2021

Problem

If \[\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\] for three positive numbers $x,y$ and $z$, all different, then $\frac{x}{y}=$

$\text{(A) } \frac{1}{2}\quad \text{(B) } \frac{3}{5}\quad \text{(C) } \frac{2}{3}\quad \text{(D) } \frac{5}{3}\quad \text{(E) } 2$

Solution 1

$\fbox{E}$ We have $\frac{x+y}{z} = \frac{x}{y} \implies xy+y^2=xz$ and $\frac{y}{x-z} = \frac{x}{y} \implies y^2=x^2-xz \implies x^2-y^2=xz$. Equating the two expressions for $xz$ gives $xy+y^2=x^2-y^2 \implies x^2-xy-2y^2=0 \implies (x+y)(x-2y)=0$, so as $x+y$ cannot be $0$ for positive $x$ and $y$, we must have $x-2y=0 \implies x=2y \implies \frac{x}{y}=2$.

Solution 2

We cross multiply the first and third fractions and the second and third fractions, respectively, for \[x(x-z)=y^2\] \[y(x+y)=xz\] Notice how the first equation can be expanded and rearranged to contain an $(x+y)$ term. \[x^2-xz=y^2\] \[x^2-y^2=xz\] \[(x+y)(x-y)=xz\] We can divide this by the second equation to get \[\frac{(x+y)(x-y)}{y(x+y)}=\frac{xz}{xz}\] \[\frac{x-y}{y}=1\] \[\frac{x}{y}-1=1\] \[\frac{x}{y}=2 \rightarrow \boxed{E}\]

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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All AHSME Problems and Solutions

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