Question

If $\frac{x}{x+2y+z}=\frac{y}{y+2z+x}=\frac{z}{z+2x+y}$ and x + y + z $\ne$0, then show that each ratio is equal to $\frac{1}{4}$.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ \frac{x}{x+2y+z}=\frac{y}{y+2z+x}=\frac{z}{z+2x+y}=k \\ \mathrm{By}\mathrm{the}\mathrm{theorem}\mathrm{on}\mathrm{equal}\mathrm{ratios},\mathrm{we}\mathrm{get}: \\ \frac{x+y+z}{x+2y+z+y+2z+x+z+2x+y}=k \\ \Rightarrow \frac{x+y+z}{4x+4y+4z}=k \\ \Rightarrow \frac{x+y+z}{4\left(x+y+z\right)}=k \\ \Rightarrow \frac{1}{4}=k \\ Or, \\ \frac{x}{x+2y+z}=\frac{y}{y+2z+x}=\frac{z}{z+2x+y}=\frac{1}{4} \end{gathered}$$