China Mathematical Competition 2003
19
Solution As in the diagram, letA, B, C, D be the centers of the 4 balls in the bottom layer, and A’, B’ , C’ , D’ the centers of the 4 balls in the upper layer. Then A, B, C, D and A’, B‘, C’, D’are the 4 vertices of squares of length 2, D respectively. Now , the circumscribed circles with centers 0 and d of the squares constitute the bases of another cylinder, and the projecting point of A’ on the bottom base is the middle point M of arc AB. In AA’AB , we have A’A
= A’B = AB = 2,
then A’N
N is the middle point of AB. Meanwhile, CYM = OA
=&,where
=a,ON
, so
=1
+
Then the height of the original cylinder is& 2. Remark In order to solve the problem, you must first be clear about the way the balls are placed (each ball in contact with 4 other balls) , then determine the positions of the centers of the balls.
<
S@& Let M, = (0. a l a 2 ~ - a , I ai = 0 or 1, 1 i < n- 1, a, = 1) be a set of decimal fractions, T, and S, be the number and the sum of the elements in M, respectively. Then
lim, S n
=
Solution Since a1 , a2 , anPl all have exactly two possible values, so T, = 2”-’. Meanwhile, the frequency of ai = 1 is the same as that of ai = 0 for 1 i n - 1, and a, = 1. Then ..a,
<<
s, = 21 x 2”-1 -
(h+
-+ 1;2
1 ...+-)+ 1on-I
2”-I X - 1 10”
1 18