PUMaC 2016 · 个人决赛（A 组） · 第 3 题

PUMaC 2016 — Individual Finals (Division A) — Problem 3

On a cyclic quadrilateral ABCD , M is the midpoint of AB and N is the midpoint of CD . Let E be the projection of C onto AB and F the reflection of N about the midpoint of DE . If F is inside quadrilateral ABCD , show that ∠ BM F = ∠ CBD . 1

解析

Denote by φ reflection over AB . Now, let Q = φ ( C ); since 4 P DN ∼ 4 EDC with ratio 2 1 and P, E are the midpoints of CQ, N F , 4 F DN ∼ 4 QDC with ratio so F is the midpoint 2 of QD . Now, let the circle ω with diameter AB intersect AC at X and AD at Y , and let Z = φ ( X ). Since Z ∈ ω , ∠ XZY = ∠ XAY = ∠ CAD = ∠ CBD . Furthermore, XZ ⊥ AB , so it suffices to show Y Z ⊥ M F . Note that BY ⊥ DY, BZ ⊥ QZ . Now, let D , Q be the midpoints of DB, QB , respectively. 1 1 Since 4 DY B, 4 QZB are right triangles and F Q D is the medial triangle of BDQ , Y D = 1 1 1 BD = Q F and ZQ = BQ = D F . Furthermore, ∠ BD F = ∠ BQ F and ∠ BDA = 1 1 1 1 1 1 1 ∠ BCA = ∠ BQA −→ ∠ BD Y = ∠ BQ Z , so ∠ F D Y = ∠ F Q Z . Then, by SAS congruency, 1 1 1 1 ∼ 4 F D Y 4 F Q Z , which implies that F Y = F Z . = 1 1 Let K be the midpoint of Y Z . Since 4 F Y Z is isosceles with base Y Z , F K ⊥ Y Z . Further- more, since Y Z is a chord of ω with center M , M K ⊥ Y Z . Then, F, M, K are collinear and M F ⊥ Y Z , so ∠ BM F = ∠ XZY = ∠ CBD , as desired. Problem written by Bill Huang. 2