
Problem 10.1. Find all real numbers $x$ such that $\tan \left(\frac{\pi}{12}-x\right)$, $\tan \frac{\pi}{12}$ and $\tan \left(\frac{\pi}{12}+x\right)$ form (in some order) a geometric progression.


Solution: Denote $a=\tan \frac{\pi}{12}$ and $y=\tan x$. There are three cases to consider:

1. $\tan \left(\frac{\pi}{12}-x\right) \cdot \tan \left(\frac{\pi}{12}+x\right)=\tan ^{2} \frac{\pi}{12}$. Now $\frac{a-y}{1+a y} \cdot \frac{a+y}{1-a y}=$ $a^{2}$. Therefore $a^{2}-y^{2}=a^{2}\left(1-a^{2} y^{2}\right)$, and so $\left(a^{4}-1\right) y^{2}=0$. Since $a \neq \pm 1$ we get that $y=0$, so $\tan x=0$. Obviously all numbers $x$ of the kind $x=k \pi, k \in \mathbb{Z}$ are solutions to the problem.
2. $\tan \frac{\pi}{12} \tan \left(\frac{\pi}{12}+x\right)=\tan ^{2}\left(\frac{\pi}{12}-x\right)$. We obtain $a \frac{a+y}{1-a y}=\left(\frac{a-y}{1+a y}\right)^{2} \Longrightarrow\left(a^{2}+1\right) y\left[a y^{2}+\left(a^{2}-1\right) y+3 a\right]=0$.

The case of $y=0$ is settled in 1 . Let $y_{1}$ and $y_{2}$ be the roots of the equation $a y^{2}+\left(a^{2}-1\right) y+3 a=0$. Since $a=\tan 15 \mathrm{deg}=$ $\tan (45 \mathrm{deg}-30 \mathrm{deg})=2-\sqrt{3}$ we get $y_{1}=y_{2}=\sqrt{3}$, so $\tan x=$ $\sqrt{3}$. Obviously all $x$ of the kind $x=\frac{\pi}{3}+k \pi, k \in \mathbb{Z}$ are solutions of the problem.
3. $\tan \frac{\pi}{12} \tan \left(\frac{\pi}{12}-x\right)=\tan ^{2}\left(\frac{\pi}{12}+x\right)$. The substitution $z=$ $-x$ transforms this case to the previous one. Therefore $x=$ $-\frac{\pi}{3}+k \pi, k \in \mathbb{Z}$.

The required numbers are $x=k \pi$ and $x= \pm \frac{\pi}{3}+k \pi, k \in \mathbb{Z}$.
