# 1992 AHSME Problems/Problem 16

## 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 byProblem 15 Followed byProblem 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. Invalid username
Login to AoPS