details

Yma16,the tilting angle between the coin plane and the hole plane is not critical, any angle greater than 0 degree(including 90 degree) will work. The key is the angle the coin rotate within its plane. Here is what I mean: lets's start with the coin and the hole on the same plane, with one vertex of the coin match with one vertex of the hole. Also, two sides connected with this coin vertex match with two sides connected with the hole vertex. The side length of the coin is x, to be determined. Now rotate the coin 30 degrees counter closewise around the matching vertex,then the altitude of the coin starting from the matching vertex will match wih one side of the hole. Now,as long as this coin altitude equals the hole side, any tilting of the coin plane will let the coin fall through. As the altitude of the coin is 1/2*sqrt(3)x, so let 1/2*sqrt(3)x=1, we get the side of the coin x=2/sqrt(3).

Without drawing it is hard to explain, hope I made it clear.

所有跟帖： 

• 回复：details -yma16- ♀ (767 bytes) () 01/18/2004 postreply 21:31:00

• same thing -atom9- ♀ (768 bytes) () 01/19/2004 postreply 01:20:00

• Thank you! But -yma16- ♀ (222 bytes) () 01/19/2004 postreply 06:16:00

• 回复：Thank you! But -atom9- ♀ (433 bytes) () 01/19/2004 postreply 13:33:00