HMMT 二月 2022 · 团队赛 · 第 10 题

HMMT February 2022 — Team Round — Problem 10

[60] On a board the following six vectors are written: (1 , 0 , 0) , ( − 1 , 0 , 0) , (0 , 1 , 0) , (0 , − 1 , 0) , (0 , 0 , 1) , (0 , 0 , − 1) . Given two vectors v and w on the board, a move consists of erasing v and w and replacing them with 1 1 √ √ ( v + w ) and ( v − w ). After some number of moves, the sum of the six vectors on the board is u . 2 2 Find, with proof, the maximum possible length of u .

解析

[60] On a board the following six vectors are written: (1 , 0 , 0) , ( − 1 , 0 , 0) , (0 , 1 , 0) , (0 , − 1 , 0) , (0 , 0 , 1) , (0 , 0 , − 1) . Given two vectors v and w on the board, a move consists of erasing v and w and replacing them with 1 1 √ √ ( v + w ) and ( v − w ). After some number of moves, the sum of the six vectors on the board is u . 2 2 Find, with proof, the maximum possible length of u . Proposed by: Daniel Zhu √ Answer: 2 3 Solution: For a construction, note that one can change √ (1 , 0 , 0) , ( − 1 , 0 , 0) → ( 2 , 0 , 0) , (0 , 0 , 0) → (1 , 0 , 0) , (1 , 0 , 0) and similarly for (0 , 1 , 0) , (0 , − 1 , 0) and (0 , 0 , 1) , (0 , 0 , − 1). Then u = (2 , 2 , 2). For the bound, argue as follows: let the vectors be v , . . . , v , n = ( x, y, z ) be any unit vector, and 1 6 P 2 S = ( n · v ) , where the sum is over all vectors on the board. We claim that S is invariant. Indeed, i i we have 2 2 2 2 1 1 n · v + n · w n · v − n · w n · √ ( v + w ) + n · √ ( v − w ) = √ + √ 2 2 2 2 2 2 2( n · v ) + 2( n · w ) = 2 2 2 = ( n · v ) + ( n · w ) . 2 2 2 Also, at the beginning we have S = 2 x + 2 y + 2 z = 2. Therefore we must always have S = 2. Thus, by the Cauchy-Schwarz inequality we have s X X √ √ √ 2 n · u = n · v ≤ ( n · v ) 6 = 12 = 2 3 . i i i √ But since n is arbitrary, this implies that | u | ≤ 2 3; otherwise we could pick n = u/ | u | and reach a contradiction.