Consider a circle $C$ with diameter $AB=1$. A point $P_0$ is chosen on $C$, $P_0 \ne A$, and starting in $P_0$ a sequence of points $P_1, P_2, \dots, P_n, \dots$ is constructed on $C$, in the following way: $Q_n$ is the symmetrical point of $A$ with respect of $P_n$ and the straight line that joins $B$ and $Q_n$ cuts $C$ at $B$ and $P_{n+1}$ (not necessary different). Prove that it is possible to choose $P_0$ such that: i $\angle {P_0AB} < 1$. ii In the sequence that starts with $P_0$ there are $2$ points, $P_k$ and $P_j$, such that $\triangle {AP_kP_j}$ is equilateral.