HMMT 二月 2015 · 几何 · 第 4 题

HMMT February 2015 — Geometry — Problem 4

Let ABCD be a cyclic quadrilateral with AB = 3 , BC = 2 , CD = 2 , DA = 4. Let lines perpendicular ′ ′ to BC from B and C meet AD at B and C , respectively. Let lines perpendicular to AD from A and ′ ′ [ BCC B ] ′ ′ D meet BC at A and D , respectively. Compute the ratio , where [ ̟ ] denotes the area of ′ ′ [ DAA D ] figure ̟ .

解析

Let ABCD be a cyclic quadrilateral with AB = 3 , BC = 2 , CD = 2 , DA = 4. Let lines perpendicular ′ ′ to BC from B and C meet AD at B and C , respectively. Let lines perpendicular to AD from A and ′ ′ [ BCC B ] ′ ′ D meet BC at A and D , respectively. Compute the ratio , where [ ̟ ] denotes the area of ′ ′ [ DAA D ] figure ̟ . 37 ′ Answer: To get a handle on the heights CB , etc. perpendicular to BC and AD , let 76 − − → − − → ̂ ̂ X = BC ∩ AD , which lies on ray BC and AD since AB > CD (as chords AB > CD ). By similar triangles we have equality of ratios XC : XD : 2 = ( XD + 4) : ( XC + 2) : 3, so we have a system of linear equations: 3 XC = 2 XD + 8 and 3 XD = 2 XC + 4, so 9 XC = 6 XD + 24 = 4 XC + 32 84 / 5 32 28 gives XC = and XD = = . 5 3 5 ′ ′ ′ ′ [ BC B C ] BC ( CB + BC ) BC XC + XB It’s easy to compute the trapezoid area ratio = = (where we have ′ ′ ′ ′ [ AA D D ] AD ( AD + DA ) AD XA + XD XC + BC/ 2 32 / 5+1 BC 2 37 similar right triangles due to the common angle at X ). This is just = = . AD XD + AD/ 2 4 28 / 5+2 76 Geometry