HMMT 十一月 2020 · 冲刺赛 · 第 33 题

HMMT November 2020 — Guts Round — Problem 33

[17] In quadrilateral ABCD , there exists a point E on segment AD such that = and ∠ BEC is a ED 9 right angle. Additionally, the area of triangle CED is 27 times more than the area of triangle AEB . If 2 ∠ EBC = ∠ EAB , ∠ ECB = ∠ EDC , and BC = 6, compute the value of AD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMO 2020, November 14–21, 2020 — GUTS ROUND Organization Team Team ID#

解析

[17] In quadrilateral ABCD , there exists a point E on segment AD such that = and ∠ BEC is ED 9 a right angle. Additionally, the area of triangle CED is 27 times more than the area of triangle AEB . 2 If ∠ EBC = ∠ EAB , ∠ ECB = ∠ EDC , and BC = 6, compute the value of AD . Proposed by: Akash Das Answer: 320 Solution: F C B A D E Extend sides AB and CD to intersect at point F . The angle conditions yield 4 BEC ∼ 4 AF D , so ◦ ∠ AF D = 90 . Therefore, since ∠ BF C and ∠ BEC are both right angles, quadrilateral EBF C is cyclic and ◦ ◦ ∠ EF C = ∠ EBC = 90 − ∠ ECB = 90 − ∠ EDF, implying that EF ⊥ AD . F A 2 AE 1 F A 1 EB 1 Since AF D is a right triangle, we have ( ) = = , so = . Therefore = . Since the F D ED 9 F D 3 EC 3 area of CED is 27 times more than the area of AEB , ED = 9 · EA , and EC = 3 · EB , we get that ◦ ◦ ∠ DEC = ∠ AEB = 45 . Since BECF is cyclic, we obtain ∠ F BC = ∠ F CB = 45 , so F B = F C . √ √ 1 Since BC = 6, we get F B = F C = 3 2. From 4 EAB ∼ 4 EF C we find AB = F C = 2, so 3 √ √ 2 2 2 F A = 4 2. Similarly, F D = 12 2. It follows that AD = F A + F D = 320.