# 1992 AHSME Problems/Problem 16

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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 $$\frac{(x+y)(x-y)}{y(x+y)}=\frac{xz}{xz}$$ $$\frac{x-y}{y}=1$$ $$\frac{x}{y}-1=1$$ $$\frac{x}{y}=2 \rightarrow \boxed{E}$$

## Solution 3

Since $$\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} = \frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n},$$ we can say that $$\frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} = \frac {y+(x+y)+x}{(x-z)+z+y} = \boxed{2}.$$ $\implies \boxed{(E)}$.

$\textbf{Proof of the used property:}$

Let $$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} = ... = \frac{a_n}{b_n} = r.$$

Therefore, \begin{align*} a_1 &= b_1r,\\ a_2 &= b_2r,\\ ...,\\ a_n &= b_nr \end{align*} so $$\frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n} = \frac{b_1r + b_2r + ... + b_nr}{b_1 + b_2 + ... + b_n}$$ $$= \frac{(b_1 + b_2 + ... + b_n)(nr)}{(b_1 + b_2 + ... + b_n)(n)} = r$$ --- NamelyOrange