HMMT 二月 2003 · 冲刺赛 · 第 38 题

英文解析

38. Given are real numbers x, y . For any pair of real numbers a , a , define a sequence by
    0 1
    a = xa + ya for n ≥ 0. Suppose that there exists a fixed nonnegative integern +2 n +1 nm such that, for every choice of a and a , the numbers a , a , a , in this order,
    0 1 m m +1 m +3
    form an arithmetic progression. Find all possible values of y .
    √
    Solution: 0 , 1 , (1 ± 5) / 2
    Note that x = 1 (or x = 0), y = 0 gives a constant sequence, so it will always havethe desired property. Thus, y = 0 is one possibility. For the rest of the proof, assumey 6 = 0.
    We will prove that a and a may take on any pair of values, for an appropriatem m +1
    choice of a and a . Use induction on m . The case m = 0 is trivial. Suppose that
    0 1
    a and a can take on any value. Let p and q be any real numbers. By settingm m +1
    q − xp 10
    a = (remembering that y 6 = 0) and a = p , we get a = p and a = q .
    m m +1 m +1 m +2
    Therefore, a and a can have any values if a and a can. That completesym +1 m +2 m m +1
    the induction.
    Now we determine the nonzero y such that a , a , a form an arithmetic sequence;
    m m +1 m +3
    that is, such that a − a = a − a . But because a = ( x + y ) a + xya 2
    m +3 m +1 m +1 m m +3 m +1 mby the recursion formula, we can eliminate a from the equation, obtaining them +3
    equivalent condition ( x + y − 2) a + ( xy + 1) a = 0. Because the pair a , a 2
    m +1 m m m +1
    can take on any values, this condition means exactly that x + y − 2 = xy + 1 = 0.2
    2 3 2
    Then x = − 1 /y , and 1 /y + y − 2 = 0, or y − 2 y + 1 = 0. One root of this cubic is
    √
    y = 1, and the remaining quadratic factor y − y − 1 has the roots (1 ± 5) / 2. Since 2
    each such y gives an x for which the condition holds, we conclude that the answer to
    √
    the problem is y = 0 , 1, or (1 ± 5) / 2.