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

cma · 2012-01-02 13:48:33Z

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

來源: cma 於 2012-01-02 13:48:33 [檔案] [舊帖] [給我悄悄話] 閱讀數 : (7782 bytes)

回答: If it is from Ps Q, We can do like that 由 jinjing 於 2012-01-02 11:12:38

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

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• dear cma,81is not good. -jinjing- ♀ (266 bytes) () 01/02/2012 postreply 16:54:15