PUMaC 2011 · 几何（B 组） · 第 3 题

PUMaC 2011 — Geometry (Division B) — Problem 3

[ 4 ] Let ABCD be a trapezoid with AD parallel to BC , AD = 2, and BC = 1. Let M be the midpoint of AD , and let P be the intersection of BD with CM . Extend AP to meet segment CD at point Q . If the ratio CQ/QD = a/b , where a and b are positive integers and gcd( a, b ) = 1, find a + b .

解析

First Solution : Since BC = (1 / 2) AD , we have that BC = M D , and it follows that △ BCP = △ DM P . Thus, CP = P M . Select R on CD such that M R is parallel to AQ . Then, CP = P M = ⇒ CQ = QR and AM = M D = ⇒ QR = RD . Thus, CQ/QD = 1 / 2, so the answer is 1 + 2 = 3 . ∼ Second Solution : As in the first solution, note that △ BCP △ DM P . From this congruence = ′ it follows that BP = P D and CP = P M . Extend AP to meet line BC at point A . Because ′ ′ ∼ CP = P M , we have △ AP M △ A P C . Thus, BC = AM = CA . It follows that point Q is = ′ the centroid of triangle BA D , so CQ/QD = 1 / 2. Thus, our answer is 1 + 2 = 3 . Figure 3: Problem 3 diagram.