HMMT 二月 2003 · 冲刺赛 · 第 34 题

HMMT February 2003 — Guts Round — Problem 34

[12] OKRA is a trapezoid with OK parallel to RA . If OK = 12 and RA is a positive integer, how many integer values can be taken on by the length of the segment in the trapezoid, parallel to OK , through the intersection of the diagonals?

解析

OKRA is a trapezoid with OK parallel to RA . If OK = 12 and RA is a positive integer, how many integer values can be taken on by the length of the segment in the trapezoid, 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 ). 288 Similarly XQ = 12 x/ (12 + x ) also, so P Q = 24 x/ (12 + x ) = 24 − . This has to be 12+ x 5 2 an integer. 288 = 2 3 , so it has (5 + 1)(3 + 1) = 18 divisors. 12 + x must be one of these. We also exclude the 8 divisors that don’t exceed 12, so our final answer is 10.