HMMT 二月 2008 · TEAM2 赛 · 第 8 题

HMMT February 2008 — TEAM2 Round — Problem 8

[ 40 ] Prove that the number of balls b in a juggling sequence j (0) j (1) · · · j ( n − 1) is simply the average j (0) + j (1) + · · · + j ( n − 1) b = . n

解析

[ 40 ] Prove that the number of balls b in a juggling sequence j (0) j (1) · · · j ( n − 1) is simply the average j (0) + j (1) + · · · + j ( n − 1) b = . n Solution: Consider the corresponding juggling diagram. Say the length of an curve from t to 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 that ball is between M n − J and M n + J , where J = max { j (0) , j (1) , . . . , j ( n − 1) } . It follows that 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 that b ( 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) J b 1 − ≤ ≤ b 1 + . nM n nM Since we can take M to be arbitrarily large, we must have j (0) + j (1) + · · · + j ( n − 1) b = , n as desired.