HMMT 二月 2003 · 冲刺赛 · 第 34 题

英文解析

34. OKRA is a trapezoid with OK parallel to RA . If OK = 12 and RA is a positiveinteger, how many integer values can be taken on by the length of the segment in thetrapezoid, parallel to OK , through the intersection of the diagonals?
    Solution: 10
    Let RA = x . If the diagonals intersect at X , and the segment is P Q with P on KR ,
    then 4 P KX ∼ 4 RKA and 4 OKX ∼ 4 RAX (by equal angles), giving RA/P X =
    AK/XK = 1 + AX/XK = 1 + AR/OK = ( x + 12) / 12, so P X = 12 x/ (12 + x ).
    Similarly XQ = 12 x/ (12 + x ) also, so P Q = 24 x/ (12 + x ) = 24 − . This has to be 288
    12+ x
    5 2
    an integer. 288 = 2 3 , so it has (5 + 1)(3 + 1) = 18 divisors. 12 + x must be one ofthese. We also exclude the 8 divisors that don’t exceed 12, so our final answer is 10.