Question

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 $\beta=\frac{\alpha}{10}=1.994^\circ$ 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.

Answer

Answer: 71

Concepts:reflection (geometry)angle chasing floor and ceiling functions

Difficulty rating: 2790

Small Hint:

Unfold each reflection by reflecting the next copy of the two-segment angle instead of reflecting the beam

Big Hint:

After $k$ reflections beyond the first one, the relevant boundary ray has turned through $k\beta$

Solution:

Unfold the path at each bounce, so the beam becomes one straight line crossing successive reflected copies of the angle at $B.$ After $k$ reflections beyond the initial reflection at $C,$ the next boundary ray makes angle $k\beta$ with the original one. Using $AB=BC,$ the crossing remains on the finite segments exactly while $k\beta\leq180^\circ-2\alpha.$ Therefore $k\leq\frac{180-2(19.94)}{1.994}\approx70.27,$ so there are $70$ further reflections. Including the first reflection at $C$ gives $71.$

1994 AIME Problem 14

Answer: 71

Difficulty rating: 2790

Small Hint:

Big Hint:

Solution:

Problem 14 in Other Years