Amazing $\prod$ . <1>

Note that $\begin{aligned} \prod_{k=1}^K \frac{3}{1 + 2\cos\frac{\pi}{3^k}} & = \; \tfrac32 \prod_{k=2}^K \frac{3}{e^{\frac{\pi i}{3^k}} + 1 + e^{\frac{-\pi i}{3^k}}} \; = \; \tfrac123^K\prod_{k=2}^K \frac{e^{\frac{\pi i}{3^k}}\big(e^{\frac{\pi i}{3^k}} - 1\big)}{e^{\frac{\pi i}{3^{k-1}}}-1} \\ & = \; \tfrac123^K \prod_{k=2}^K \frac{\sin \frac{\pi}{2\times3^k}}{\sin \frac{\pi}{2\times3^{k-1}}} \; = \; \tfrac123^K\frac{\sin\frac{\pi}{2\times3^K}}{\sin\frac{\pi}{6}} \; = \; 3^K \sin \tfrac{\pi}{2\times3^K} \end{aligned}$ From this it is clear that $\prod_{k=1}^\infty \frac{3}{1 + 2\cos\frac{\pi}{3^k}} \; = \; \lim_{K \to \infty} 3^K \sin \tfrac{\pi}{2\times3^K} \; = \; \tfrac12\pi$ making the answer $\lfloor 50\pi \rfloor = \boxed{157}$ .