PUMaC 2017 · 几何（B 组） · 第 8 题

PUMaC 2017 — Geometry (Division B) — Problem 8

Triangle ABC has ∠ A = 90 , AB = 2, and AC = 4. Circle ω has center C and radius CA , 1 while circle ω has center B and radius BA . The two circles intersect at point E , different 2 from point A . Point M is on ω and in the interior of ABC , such that BM is parallel to EC . 2 Suppose EM intersects ω at point K and AM intersects ω at point Z . What is the area of 1 1 quadrilateral ZEBK ? 1

解析

By symmetry, ∠ BEC is a right angle, like ∠ A . Since BM is parallel to EC, this means ◦ ∠ EBM = ∠ BEC = 90 . Thus, M BE is an isosceles right triangle and, ∠ BM E = ∠ BEM = ◦ ∠ M EC = 45 . Since ∠ KEC = ∠ M EC , K, M, E are collinear, we get that triangle KCE is 1 ◦ ◦ an isosceles right triangle as well, so ∠ KCE = 90 . Now note ∠ KZE = 45 , it cuts off an ◦ ◦ arc, KE , with central angle 90 . Thus, KZE is an isosceles right triangle, and ∠ CEZ = 45 . ◦ This means that KZ is perpendicular to CE. From before, we had that ∠ KCE = 90 , so KC is perpendicular to CE as well. Thus, K , C and Z are collinear, and ZEBK is an isosceles trapezoid with base KZ = 2 KC = 2 AC = 8, and base BE = AB = 2, and height EC = AC = 4. So its area is 20 . Problem written by Kai Zheng If you believe that any of these answers is incorrect, or that a problem had multiple reasonable interpretations or was incorrectly stated, you may appeal at http://tinyurl.com/PUMaCappeal2017. All appeals must be in by 1 PM to be considered. 2