1994 AIME Problems/Problem 14

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Problem

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 $AB=AC,\,$ determine the number of times the light beam will bounce off the two line segments. Include the first reflection at $C\,$ in your count.

Solution

Solution 1

At each point of reflection, we pretend instead that the light continues to travel straight.

[Asy_image]

Then each intersection of the extended line with the rotated segments corresponds to a reflection in the original problem. We quickly see that the extended line will intersect each rotation of the angle by $k \beta$ until $k\beta \ge 180 - \alpha \Longrightarrow k \ge \frac{180 - \alpha}{\beta}$ . Thus, our answer is, including the first intersection, $\left\lceil \frac{180 - \alpha}{\beta} \right\rceil = \boxed{083}$ .

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1994 AIME (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1994 AIME Problems/Problem 14

From AoPSWiki

Contents

Problem

Solution

Solution 1

See also