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Claim: $\tau \in (0,2\pi/7)$.  Pf: $k=\lceil From problem definition/assumed finite k we have $\tau > 0$. Additionally,$k=\lceil  2\pi/ \tau \rceil > 8 \implies 2\pi/ \tau > 7 \iff \tau < 2\pi/7$ 2\pi/7 \square$  \subsubsection{Lemma 2}