Unusual Proportions

Suppose $\frac{y}{x-z} = \frac{x+y}{z}= \frac{x}{y}= c \\$ with $c$ constant. So we have, $y=c(x-z) \dots(1) \\ x+y=zc \dots(2) \\x=yc \dots(3) \\$ subtituting $(1)$ in to $(2)$ $\\ x+c(x-z)=zc \dots(4)\\$ subtituting $(3) into (4)$ $\\ yc+c(yc-z)=zc$ $\\ c= \frac{2z-y}{y} \dots(5) \\$ subtituting $(5)$ in to $(3)$ $\\ x= \frac{y(2z-y)}{y} \\ x+y=2z \dots(6) \\$ subtituting $(6)$ in to $(2)$ we have $c=2.$ $\frac{x}{y}= \boxed2$

Adding the numerators and denominators of the first two expressions yields a fraction equal to the original expression.

$\dfrac y{x-z}=\dfrac{x+y}z=\dfrac xy=\dfrac{x+2y}x$

Considering the last two expressions and letting $m=\dfrac xy$ gives $m=1+2/m$ The previous equation results to $m=2$ . Note that $x$ and $y$ are positive numbers.

From y/(x-z) =x/y, we have y^2= x^2-zx (1) From (x+y)/z= x/y, we have zx= xy+y^2 (2) Substituting the value zx in (1) x^2-xy-2y^2=0, x=2y or x= -y. Since x and y are both positive integers, x must be equal to 2y. Thus x/y= 2