HMMT 二月 2009 · 代数 · 第 10 题

英文解析

10. [ 8 ] Let f ( x ) = 2 x − 2 x . For what positive values of a do there exist distinct b, c, d such that ( a, f ( a )),
    ( b, f ( b )), ( c, f ( c )), ( d, f ( d )) is a rectangle?
    √
    Answer: [ , 1]3
    Solution: Say we have four points ( a, f ( a )), ( b, f ( b )), ( c, f ( c )), ( d, f ( d )) on the curve which form 3
    a rectangle. If we interpolate a cubic through these points, that cubic will be symmetric around the center of the rectangle. But the unique cubic through the four points is f ( x ), and f ( x ) has only onepoint of symmetry, the point (0 , 0).
    So every rectangle with all four points on f ( x ) is of the form ( a, f ( a )), ( b, f ( b )), ( − a, f ( − a )), ( − b, f ( − b )),
    and without loss of generality we let a, b > 0. Then for any choice of a and b these points form a parallelogram, which is a rectangle if and only if the distance from ( a, f ( a )) to (0 , 0) is equal to the distance
    2 2 6 4 2
    from ( b, f ( b )) to (0 , 0). Let g ( x ) = x + ( f ( x )) = 4 x − 8 x + 5 x , and consider g ( x ) restricted tox ≥ 0. We are looking for all the values of a such that g ( x ) = g ( a ) has solutions other than a .
    2 3 2
    Note that g ( x ) = h ( x ) where h ( x ) = 4 x − 8 x + 5 x . This polynomial h ( x ) has a relative maximum
    1 5
    of 1 at x = and a relative minimum of 25 / 27 at x = . Thus the polynomial h ( x ) − h (1 / 2) has
    2 6
    2 2
    the double root 1 / 2 and factors as (4 x − 4 x + 1)( x − 1), the largest possible value of a for which
    2 2 2
    h ( x ) = h ( a ) is a = 1, or a = 1. The smallest such value is that which evaluates to 25 / 27 other than
    √ √
    2 3 3
    5 / 6, which is similarly found to be a = 1 / 3, or a = . Thus, for a in the range ≤ a ≤ 1 the
    3 3
    equation g ( x ) = g ( a ) has nontrivial solutions and hence an inscribed rectangle exists. 3