来源: constant 于 06-11-09 08:07:29

设有n块相同的砖，在一张桌子的边上一块一块的垒起来（每层一块）。问最远的一块能伸出桌边多远？要证明不能更远了。

Deuss:  (1+1/2+1/3+…+1/n)/2 .

Assume a brick has length 1, the corner of the table is at position 0.
Bricks are numbered top down as 1, 2, ..., n
Let P_k be the center position of the k-th brick,
and Q_k be the weight center of the first k bricks.

To reach farthest, the following relation must hold:
Q_k=P_{k+1}+1/2 (the top k brick's weight center is exactly at the right edge of the (k+1)-th brick, or the table corner).
Then, Q_{k+1}=(Q_k*k+P_{k+1})/(k+1)=Q_k-1/(2(k+1)).
Because Q_n = 0, we have
0 = Q1-(1/2+1/3+…+1/n)/2

The right edge of the 1st brick reaches to
Q1+1/2=(1+1/2+1/3+…+1/n)/2 .

When n->oo, it can reach arbitrarily far.