SPANISH MATHEMATICAL OLYMPIAD (FINAL ROUND)

that is, $$\tan (2ky)+\tan (my)+\tan (ny)=0$$ provided that $r$, $s$, $t$ are $\tan (ky)$, $\tan (my)$ and $\tan (ny)$ respectively. We consider now the following cases: If $r=\tan y$, $s=\tan 2y$ and $t=\tan 3y$ then $\tan 2y+\tan 5y=0$, and $$y\in \left\{\frac{\pi }{7},\frac{2\pi }{7},\frac{3\pi }{7},\frac{4\pi }{7},\frac{5\pi }{7},\frac{6\pi }{7}\right\}.$$ If $r=\tan 2y$, $s=\tan y$ and $t=\tan 3y$ then $\tan 4y+\tan 4y=0$ and it follows that $$y\in \left\{\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8}\right\},$$ We have discarded $y=\frac{\pi }{2}$, $y=\frac{\pi }{4}$ and $y=\frac{3\pi }{4}$ because $\tan y$ and $\tan (2y)$ must be real numbers. * If $r=\tan 3y$, $s=\tan y$ and $t=\tan 2y$ then $\tan 6y+\tan 3y=0$ and $$y\in \left\{\frac{\pi }{9},\frac{2\pi }{9},\frac{\pi }{3},\frac{4\pi }{9},\frac{5\pi }{9},\frac{2\pi }{3},\frac{7\pi }{9},\frac{8\pi }{9}\right\}.$$