Difference between revisions of "1994 AIME Problems/Problem 14"
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− | + | Note that after <math>k</math> reflections (excluding the first one at <math>C</math>) the extended line will form an angle <math>k \beta</math> at point <math>B</math>. For the <math>k</math>th reflection to be just inside or at the point <math>C</math>, we must have <math>k\beta \le 180 - 2\alpha \Longrightarrow k \le \frac{180 - 2\alpha}{\beta} = 70.27</math>. Thus, our answer is, including the first intersection, <math>\left\lfloor \frac{180 - 2\alpha}{\beta} \right\rfloor + 1 = \boxed{071}</math>. | |
== See also == | == See also == |
Revision as of 21:45, 26 March 2011
Contents
[hide]Problem
A beam of light strikes at point with angle of incidence and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments and according to the rule: angle of incidence equals angle of reflection. Given that and determine the number of times the light beam will bounce off the two line segments. Include the first reflection at in your count.
Solution
Solution 1
At each point of reflection, we pretend instead that the light continues to travel straight.
Note that after reflections (excluding the first one at ) the extended line will form an angle at point . For the th reflection to be just inside or at the point , we must have . Thus, our answer is, including the first intersection, .
See also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |