18. Numbers $a, b$ and $c$ are such that $\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=k$. How many possible values of $k$ are there?
18. 2 Consider the equation $\frac{a}{b+c}=\frac{b}{c+a}$. Multiply each side by $(b+c)(c+a)$ to get $a^{2}+a c=b^{2}+b c$ and so $a^{2}-b^{2}+a c-b c=0$. Therefore $(a-b)(a+b+c)=0$. Hence $a=b$ or $a+b+c=0$. Similarly, if we consider the equations $\frac{b}{c+a}=\frac{c}{a+b}$ and $\frac{c}{a+b}=\frac{a}{b+c}$, then $b=c$ or $a+b+c=0$ and $c=a$ or $a+b+c=0$ respectively. Therefore, the possible values of $k$ when all three equations are satisfied simultaneously occur when $a=b=c$, giving $k=\frac{1}{2}$, or when $a+b+c=0$, giving $k=-1$. Hence there are two possible values of $k$.