, 2012

Solution:

Drawing the graphs of the functions $y=x-\frac{\pi}{2}$ and $y=\tan x$, we may observe that the graphs intersect exactly once in each of the intervals $\left(\frac{(2 k-1) \pi}{2}, \frac{(2 k+1) \pi}{2}\right)$ for each $k=1,2, \cdots$. Hence, the 200th intersection has $x$ in the range $\left(\frac{399 \pi}{2}, \frac{401 \pi}{2}\right)$. At this intersection, $y=x-\frac{\pi}{2}$ is large, and thus, the intersection will be slightly less than $\frac{401 \pi}{2}$. We have that $\left\lfloor\frac{401 \pi}{4}\right\rfloor=\left\lfloor 100 \pi+\frac{\pi}{4}\right\rfloor=\left\lfloor 314.16+\frac{\pi}{4}\right\rfloor=314$.