Ball questions

yma16 · 发表于 2005-11-20 01:25:58

(1) Find the maximum number of balls of equal size such that you can place each one to touch the rest.

(2) Same as 1. But you can choose any size of a ball. (They do not have to be the same) The second one could be hard. I remember a guy solved the problem for disks (no overlap) about 100 year ago. (and the answer is 4?)

In 1611, Kepler proposed that close packing (either cubic or hexagonal close packing. Do you think (2) is some how having a flavor of that?

husonghu · 发表于 2005-11-20 08:58:46

1. "等径球"的最紧密堆积，最大邻近球数(配位数)为12。

www.ddhw.com

2. 不太懂题意。我想大概可看作“等径球最紧密堆积+较小的球填充空隙”的情况：设大球的半径为R，每个大球与12个大球接触后，周围留下8个正四面体空隙和6个正八面体空隙，前者可填入的球最大半径是0.414R，后者可填入的球最大半径是0.732R。所以大球的最大邻近球数是26 (12大球，8小球，6中球)，小球的最大邻近球数是4个大球，中球的最大邻近球数是6个大球。www.ddhw.com

(2的情况是"非等径球"的最紧密堆积)

本贴由[husonghu]最后编辑于：2005-11-20 3:7:43

yma16 · 发表于 2005-11-21 02:55:05

It requires each one touching the rest of the balls. It is not "等径球"的最紧密堆积.

husonghu · 发表于 2005-11-21 07:43:23

1. 我觉得答案就是4 ---- 4个等径球的球心在正四面体的顶点；
正四面体的棱长是球的直径。

2. 有一个答案是5 ---- 如上述4个等径球放置后，再在中间的空
隙处夹上一个半径为0.414R的球。但我不肯定还能不能更多。

yma16 · 发表于 2005-11-21 19:43:18

For (1), 4 is correct. Can you prove it?

For (2), 5 seems OK. It must be difficult to prove.

husonghu · 发表于 2005-11-22 05:21:08

Haha, the proof part is really hard for me. I would rather not try it at all, as I'm totally clueless at present. See if anybody can help.

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