A beam of light strikes $\overline{BC}\,$ at point $C\,$ with angle of incidence $\alpha=19.94^\circ\,$ and reflects with an equal angle of reflection as shown.  The light beam continues its path, reflecting off line segments $\overline{AB}\,$ and $\overline{BC}\,$ according to the rule: angle of incidence equals angle of reflection.  Given that $\beta=\alpha/10=1.994^\circ\,$ and $AB=BC,\,$ determine the number of times the light beam will bounce off the two line segments.  Include the first reflection at $C\,$ in your count.
[AIME 1994 Problem 14.png](https://artofproblemsolving.com/wiki/index.php/File:AIME_1994_Problem_14.png)
At each point of reflection, we pretend instead that the light continues to travel straight. 

Note that after $k$ reflections (excluding the first one at $C$) the extended line will form an angle $k \beta$ at point $B$. For the $k$th reflection to be just inside or at point $B$, we must have $k\beta \le 180 - 2\alpha \Longrightarrow k \le \frac{180 - 2\alpha}{\beta} = 70.27$. Thus, our answer is, including the first intersection, $\left\lfloor \frac{180 - 2\alpha}{\beta} \right\rfloor + 1 = \boxed{071}$.