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

來源: cma 2012-01-02 13:48:33 [] [舊帖] [給我悄悄話] 本文已被閱讀: 次 (7782 bytes)
回答: If it is from Ps Q, We can do like thatjinjing2012-01-02 11:12:38

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?

所有跟帖: 

dear cma,81is not good. -jinjing- 給 jinjing 發送悄悄話 (266 bytes) () 01/02/2012 postreply 16:54:15

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