Bariz - Problem

Source: IMOC 2020 A4

Tags: algebra, polynomial

jasperE3

13.08.2021 02:52

One day, before his work time at Jane Street, Sunny decided to have some fun. He saw that there are some real numbers $a_{-1},\ldots,a_{-k}$ on a blackboard, so he decided to do the following process just for fun: if there are real numbers $a_{-k},\ldots,a_{n-1}$ on the blackboard, then he computes the polynomial $$P_n(t)=(1-a_{-k}t)\cdots(1-a_{n-1}t).$$He then writes a real number $a_n$, where $$a_n=\frac{iP_n(i)-iP_n(-i)}{P_n(i)+P_n(-i)}.$$If $a_n$ is undefined (that is, $P_n(i)+P_n(-i)=0$), then he would stop and go to work. Show that if Sunny writes some real number on the blackboard twice (or equivalently, there exists $m>n\ge0$ such that $am=an$), then the process never stops. Moreover, show that in this case, all the numbers Sunny writes afterwards will already be written before. (usjl)

What a troll problem. Let $\theta_k=\arctan{a_k} \in (-\frac{\pi}{2}, \frac{\pi}{2} )$, the condition implies that for non-negative integer $n$, \[ a_n=i \cdot \frac{P_n(i)-\overline{P_n(i)}}{P_n(i)+\overline{P_n(i)}}=i \cdot \frac{2i \cdot \text{Im} P_n(i)}{2 \cdot \text{Re} P_n(i)}=-\frac{\text{Im} P_n(i)}{\text{Re} P_n(i)}. \]Hence, \begin{align*} \theta_n&=\arctan(a_n)=-\text{arg}(P_n(i))=-\text{arg} \left( \prod_{j=-k}^{n-1} (1-a_j i) \right) \\ &\equiv -\sum_{j=-k}^{n-1} \text{arg} (1-a_j i) \pmod{\pi} \\ &\equiv \sum_{j=-k}^{n-1} \theta_j \pmod{\pi} \end{align*}For non-negative integer $n$ we have $\theta_{n+1}-\theta_n \equiv \theta_n \pmod{\pi}$, i.e. $\theta_{n+1} \equiv 2\theta_n \pmod{\pi}$, so $\theta_n=2^n \theta_0 \pmod{\pi}$. The process stops only if $\theta_n = \frac{\pi}{2}$. If $a_m=a_n$, then $\theta_m \equiv \theta_n \pmod{\pi}$, hence $\theta_k$ is eventually periodic, therefore the process never ends.