China Girls' Mathematical Olympiad

There are angles $A,B$, and $C$ in the interval $(0^{\circ },180^{\circ })$ such that $$x=\tan \frac{A}{2},y=\tan \frac{B}{2},z=\tan \frac{C}{2}.$$ By the addition and subtraction formulas, the second equation in the given system becomes $$\begin{array}{rl}1& =\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}\\ & =\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{C}{2}\tan \frac{A+B}{2}\left(1-\tan \frac{A}{2}\tan \frac{B}{2}\right),\end{array}$$ $$(1-\tan \frac{C}{2}\tan \frac{A+B}{2})(1-\tan \frac{A}{2}\tan \frac{B}{2})=0.$$ Note that $\tan \frac{A}{2}\tan \frac{B}{2}=xy\ne 1$, otherwise $z(x+y)=0$, implying $z=0$, which is impossible. Therefore, $\tan \frac{C}{2}\tan \frac{A+B}{2}=1$, or $\frac{A+B}{2}+\frac{C}{2}=90^{\circ }$. In other words, $A,B,C$ are the angles of a triangle. Let $ABC$ denote that triangle.

We rewrite the first equation in the given system as $$\frac{x}{5(x^2+1)}=\frac{y}{12(y^2+1)}=\frac{z}{13(z^2+1)}.$$ Note that, by the double-angle formula, we have $$\frac{x}{x^2+1}=\frac{\tan \frac{A}{2}}{{\tan }^2\frac{A}{2}+1}=\frac{\tan \frac{A}{2}}{{\sec }^2\frac{A}{2}}=\sin \frac{A}{2}\cos \frac{A}{2}=\frac{\sin A}{2},$$ and analogously for the expressions of $y$ and $z$. We conclude that $$\frac{\sin A}{5}=\frac{\sin B}{12}=\frac{\sin C}{13}.$$ By the sine rule, the sides of triangle $ABC$ are in ratio $5:12:13$ with $\sin A=\frac{5}{13}$, $\sin B=\frac{12}{13}$ and $\sin C=1$. Hence $\frac{x}{x^2+1}=\frac{5}{26}$, or $5x^2-26x+5=0$, implying that $x=5$ or $x=\frac{1}{5}$. Likewise, we have $y=\frac{3}{2}$ or $y=\frac{2}{3}$, and $z=1$. Substituting $z=1$ into the second equation in the given system leads to $xy+x+y=1$, implying that $(x,y,z)=(\frac{1}{5},\frac{2}{3},1)$ is the only solution with $x>0$. Hence $(\frac{1}{5},\frac{2}{3},1)$ and $(-\frac{1}{5},-\frac{2}{3},-1)$ are the solutions of the problem.