Problem 1953

Let us denote $x_1 = a = \operatorname{ctg} \alpha$. Then
$$x_2 = \frac{1}{2}\left(x_1 - \frac{1}{x_1}\right) = \frac{1}{2}(\operatorname{ctg} \alpha - \operatorname{tg} \alpha) = \frac{1}{2} \cdot \frac{\cos^2 \alpha - \sin^2 \alpha}{\sin \alpha \cdot \cos \alpha} = \frac{\cos 2\alpha}{\sin 2\alpha} = \operatorname{ctg} 2\alpha,$$
In the same way we easily prove that
$$x_{n+1} = \frac{1}{2}\left(x_n - \frac{1}{x_n}\right) = \frac{1}{2}\left(\operatorname{ctg} 2^{n-1}\alpha - \operatorname{tg} 2^{n-1}\alpha\right) = \frac{1}{2} \cdot \frac{\cos^2 2^{n-1}\alpha - \sin^2 2^{n-1}\alpha}{\sin 2^{n-1}\alpha \cdot \cos 2^{n-1}\alpha} = \operatorname{ctg} 2^n \alpha, \forall n \in \mathbb{N}.$$
The statement of the problem will be satisfied if $x_i \ne 0$, $i = \overline{1, 2017}$, and $x_{2018} = 0$. Therefore, it is sufficient to choose $\alpha$ such that $x_{2018} = \operatorname{ctg}(2^{2017}\alpha) = 0$. Thus we have $2^{2017}\alpha = \frac{\pi}{2}$, and $\alpha = \frac{\pi}{2^{2018}}$.