HMMT 二月 2008 · TEAM2 赛 · 第 8 题

英文解析

8. [ 40 ] Prove that the number of balls b in a juggling sequence j (0) j (1) · · · j ( n − 1) is simply theaveragej (0) + j (1) + · · · + j ( n − 1)
   b = .
   Solution: Consider the corresponding juggling diagram. Say the length of an curve from tnto f ( t ) is f ( t ) − t . Let us draw only the curves whose left endpoint lies inside [0 , M n − 1].
   For every single ball, the sum of the lengths of the arrows drawn corresponding to thatball is between M n − J and M n + J , where J = max { j (0) , j (1) , . . . , j ( n − 1) } . It followsthat the sum of the lengths of the arrows drawn is between b ( M n − J ) and b ( M n + J ).
   Since the arrow drawn at t has length j ( t ), the sum of the lengths of the arrows drawn is
   M ( j (0) + j (1) + · · · + j ( n − 1)). It follows thatb ( M n − J ) ≤ M ( j (0) + j (1) + · · · + j ( n − 1)) ≤ b ( M n + J ) .
   Dividing by M n , we get
   ( ) ( )
   J j (0) + j (1) + · · · + j ( n − 1) Jb 1 − ≤ ≤ b 1 + .
   n m n n m
   Since we can take M to be arbitrarily large, we must havej (0) + j (1) + · · · + j ( n − 1)
   b = ,
   as desired.n