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HMMT 二月 2011 · TEAM1 赛 · 第 5 题

HMMT February 2011 — TEAM1 Round — Problem 5

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 20 ] Let a and b be positive real numbers. Define two sequences of real numbers { a } and { b } for alln npositive integers n by ( a + bi ) = a + b i . Prove thatnn n
    2 2
    | a | + | b | a + bn +1 n +1
    ≥
    | a | + | b | a + bn nfor all positive integers n .
    Page 1 of 3
    Coin Flipping [75]
    In a one-player game, the player begins with 4 m fair coins. On each of m turns, the player takes 4
    unused coins, flips 3 of them randomly to heads or tails, and then selects whether the 4 th one is headsor tails (these four coins are then considered used). After m turns, when the sides of all 4 m coins have been determined, if half the coins are heads and half are tails, the player wins; otherwise, the playerloses.
解析

英文解析

  1. [ 20 ] Let a and b be positive real numbers. Define two sequences of real numbers { a } and { b } for alln npositive integers n by ( a + bi ) = a + b i . Prove thatnn n
    2 2
    | a | + | b | a + bn +1 n +1
    ≥
    | a | + | b | a + bn nfor all positive integers n .
    Solution: Let z = a + bi . It is easy to see that what we are asked to show is equivalent ton +1 n +1 n +1 n +1
    | z + ¯ z | + | z − z ¯ | 2 z z ¯
    ≥
    n n n n
    | z + ¯ z | + | z − z ¯ | | z + ¯ z | + | z − z ¯ |
    Cross-multiplying, we see that it suffices to shown +2 n +1 n +1 n +2 n +2 n +1 n +1 n +2
    | z + z z ¯ + z z ¯ + ¯ z | + | z − z z ¯ + z z ¯ − z ¯ |
    n +2 n +1 n +1 n +2 n +2 n +1 n +1 n +2 + | z + z z ¯ − z z ¯ − z ¯ | + | z − z z ¯ − z z ¯ + ¯ z |
    n +1 n +1 n +1 n +1
    ≥ 2 | z z ¯ + z z ¯ | + 2 | z z ¯ − z z ¯ |
    However, by the triangle inequality,
    n +2 n +1 n +1 n +2 n +2 n +1 n +1 n +2 n +1 n +1
    | z + z z ¯ + z z ¯ + ¯ z | + | z − z z ¯ − z z ¯ + ¯ z | ≥ 2 | z z ¯ + z z ¯ |
    Team Round Aandn +2 n +1 n +1 n +2 n +2 n +1 n +1 n +2 n +1 n +1
    | z − z z ¯ + z z ¯ − z ¯ | + | z + z z ¯ − z z ¯ − z ¯ | ≥ 2 | z z ¯ − z z ¯ |
    This completes the proof.
    Remark: more computationally intensive trigonometric solutions are also possible by reducing theproblem to maximizing and minimizing the values of the sine and cosine functions.
    Coin Flipping [75]
    In a one-player game, the player begins with 4 m fair coins. On each of m turns, the player takes 4
    unused coins, flips 3 of them randomly to heads or tails, and then selects whether the 4 th one is headsor tails (these four coins are then considered used). After m turns, when the sides of all 4 m coins have been determined, if half the coins are heads and half are tails, the player wins; otherwise, the playerloses.