2001 SMT/Algebra Problems/Problem 6

Problem

If for three distinct positive numbers $x$ , $y$ , and $z$ , $\frac{y}{x+z} = \frac{x+y}{z} = \frac{x}{y}$ Then find the numerical value of $\frac{x}{y}$ .

Solution

Using the first equality yields $yz=(x+y)(x+z)$ ; thus $x(x+y+z)=0$ after expanding. We know that $x\ne0$ , so $x+y+z=0$ . Then $x+y=-z$ , so $\frac{x+y}{z} = \frac{-z}{z}=1$ ; this is equal to $\frac{x}{y}$ , so $\frac{x}{y}=\boxed{-1}$ .

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2001 SMT/Algebra Problems/Problem 6

Problem

Solution