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回复:If it is from Ps Q, We can do like that

cma 2012-01-02 13:48:33 ( reads)

At each point of reflection, we pretend instead that the light continues to travel straight.pathpen = linewidth(0.7); size(250); real alpha = 28, beta = 36; pair B = MP("B",(0,0),NW), C = MP("C",D(...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 kth reflection to be just inside or at the point C, 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}.

 

This is the official solution. However, I found that it should be  kb<=180-a instead of kb<=180-2a. I got 81. What do you think?

跟帖(1)

jinjing

2012-01-02 16:54:15

dear cma,81is not good.