HMMT 二月 2015 · 几何 · 第 4 题
HMMT February 2015 — Geometry — Problem 4
题目详情
英文原题
- 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
英文解析
- 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 asystem 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