HMMT 二月 2014 · 冲刺赛 · 第 22 题

HMMT February 2014 — Guts Round — Problem 22

[ 14 ] Let ! be a circle, and let ABCD be a quadrilateral inscribed in ! . Suppose that BD and AC intersect at a point E . The tangent to ! at B meets line AC at a point F , so that C lies between E and F . Given that AE = 6 , EC = 4 , BE = 2, and BF = 12, find DA .

解析

[ 14 ] Let ω be a circle, and let ABCD be a quadrilateral inscribed in ω . Suppose that BD and AC intersect at a point E . The tangent to ω at B meets line AC at a point F , so that C lies between E and F . Given that AE = 6 , EC = 4 , BE = 2, and BF = 12, find DA . √ Answer: 2 42 By power of a point, we have ED · EB = EA · EC , whence ED = 12. Additionally, 2 by power of a point, we have 144 = F B = F C · F A = F C ( F C + 10), so F C = 8. Note that ∠ F BC = ∠ F AB and ∠ CF B = ∠ AF B , so 4 F BC ∼ 4 F AB . Thus, AB/BC = F A/F B = 18 / 12 = 3 / 2, so AB = 3 k and BC = 2 k for some k . Since 4 BEC ∼ 4 AED , we have AD/BC = AE/BE = 3, so AD = 3 BC = 6 k . By Stewart’s theorem on 4 EBF , we have 2 2 2 2 (4)(8)(12) + (2 k ) (12) = (2) (8) + (12) (4) = ⇒ 8 + k = 8 / 12 + 12 , 2 whence k = 14 / 3 . Thus, √ √ √ 42 DA = 6 k = 6 14 / 3 = 6 = 2 42 . 3