一个巨大的圆形水池，周围布满了老鼠洞。猫追老鼠到水池边，老鼠未来得及进洞就掉入水池里。猫继续沿水池边缘企图捉住老鼠（猫不入水）。已知V猫=4V鼠。问老鼠是否有办法摆脱猫的追逐？
菜菜,我知道你们一定能做出来,以后我也要常来作题,出题,长知识,动脑筋 |
I have posted a variation of the problem at 灵机一动: Suppose the speed of the cat is k times the speed of the mouse. What is the upper limit of k so that the mouse can escape? (Hint: It is larger then pi + 1.) |
如果r为圆池的半径， 老鼠在池中央，它会跑到离猫最远的池边，需要的时间为（r/V鼠) 而猫这时需要的时间为(∏r/V猫） 因为（r/V鼠)>(∏r/V猫）所以猫可以捉到老鼠 对不对呢？ |
Or you all know know the answer already? |
pi^2/(pi-1)? |
Yours is 4.6085.... I have 4.6033..., which is the solution of the equation sqr(k^2 - 1) = pi + acos (1/k). Not nearly as nice as yours. But how did you get yours? |
Sorry, after checking I found I only got an upper bound, not a a supremum. The mouse runs on a small circle with radius r, the pool has radius R, the mouse must run fast than the cat in radians, so that mouse could catch farest postion to dash to the edge of the pool. So rpi, otherwise the mouse always wins) ( on the scrach paper I only wrote r On the other hand, (R-r) or k |
TO solve your equations, we get k = pi + 1. This is the best you can get with the strategy you described. (Remember the hint I give?) It is correct to reach the farthest point on the r circle. (r = R/k actually.) But this is only the first step. The real interesting part is what happens next. This variation is a much more difficult problem than the original. I will post My solution if the problem is put on top, so more people can see it. Otherwise I will post a link to a discussion about the solution. |
I like to see your solution, either way is fine to me. Thanks! |
I think I know where your equation comes from: suppose the cat starts from (0,-1) and runs counter clockwise, and the mouse runs at an angle x northwest. It does not make sense for x>pi/2, because in that case, the cat may run the opposite way. It turns out that when x=pi/2, k is the largest. Let o be the center, b be the point where the mouse lands on the circle, let y be the angle between ob and north, then set the running time to be equal, we have: sqrt(1-1/k^2) = (arccos(1/k)+pi)/k |
That is basically the idea. The actual strategy is a little nore complicated. Because the cat can stop or change direction, the mouse needs to modify her strategy accordingly. Furthermore, since the cat can also change his strategy according to what the mouse does, we need to prove tha the mouse can win no matter what the cat does. Apparently our monitor does not like the problem. So here is the link to a discussion: |
for the acute triangle probability question. If is not too much writing, would you post it? |
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