HMMT 二月 2003 · 冲刺赛 · 第 38 题
HMMT February 2003 — Guts Round — Problem 38
题目详情
- [15] Given are real numbers x, y . For any pair of real numbers a , a , define a sequence 0 1 by a = xa + ya for n ≥ 0. Suppose that there exists a fixed nonnegative integer n +2 n +1 n m 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 .
解析
- 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 integer n +2 n +1 n m 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 have the desired property. Thus, y = 0 is one possibility. For the rest of the proof, assume y 6 = 0. We will prove that a and a may take on any pair of values, for an appropriate m 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 setting m m +1 10 q − xp a = (remembering that y 6 = 0) and a = p , we get a = p and a = q . m m +1 m +1 m +2 y Therefore, a and a can have any values if a and a can. That completes m +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 2 that is, such that a − a = a − a . But because a = ( x + y ) a + xya m +3 m +1 m +1 m m +3 m +1 m by the recursion formula, we can eliminate a from the equation, obtaining the m +3 2 equivalent condition ( x + y − 2) a + ( xy + 1) a = 0. Because the pair a , a m +1 m m m +1 2 can take on any values, this condition means exactly that x + y − 2 = xy + 1 = 0. 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 √ 2 y = 1, and the remaining quadratic factor y − y − 1 has the roots (1 ± 5) / 2. Since 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.