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HMMT 二月 2000 · ORAL 赛 · 第 9 题

HMMT February 2000 — ORAL Round — Problem 9

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [60] Let v , v , v , v and v be vectors in three dimensions. Show that for some i, j in
    1 2 3 4 5 − → − →
    1 , 2 , 3 , 4 , 5, v · v ≥ 0.
    i j
解析

英文解析

  1. This is a special case of the problem with n + 2 vectors in n dimensions. First it isclear that we can take all the vectors to be of length 1. Then we induct on n. The firstcase is n=1. Here the statement is that given a, b, c real numbers, the at least one ofab , bc , and ac is nonnegative. Without loss of generality, we can assume that a and bare of the same sign, but then ab ≥ 0. Now assume the statement is false for the n-
    − − →
    dimensional case. Choose some vector, say v , and project the other vectors onto then +2 − →

    − − → − → − → − − → − − →
    space perpendicular to v to get v = v − ( v · v ) v . This is essentially takingn +2 i i n +2 n +2 − − →iout the space parallel to v and reducing the problem by one dimension. The onlyn +2 − → − →
    ′ ′
    − → − →
    thing left to check is that if v · v < 0 for all i,j then v · v < 0 for all i,j. This is just ai ji j − → − →
    ′ ′
    − → − → − → − − → − → − − → − − → − − → − → − − → − → − − →
    calculation: v · v = v · v − 2( v · v )( v · v )+( v · v )( v · v )( v · v ) andi j j n +2 i n +2 n +2 n +2 j n +2 i n +2
    i j − − → − − → − → − → − → − − → − → − − → − → − − → − → − − →
    since ( v · v ) = 1, this is just v · v − ( v · v )( v · v ). ( v · v ) and ( v · v )
    n +2 n +2 i j j n +2 i n +2 j n +2 i n +2 − → − →
    ′ ′
    − → − →
    are both negative by assumption so their product is positive and v · v < v · v < 0.
    i ji j
    23