HMMT 二月 2011 · CALCGEO 赛 · 第 13 题

英文解析

13. Let ABCD be a cyclic quadrilateral, and suppose that BC = CD = 2. Let I be the incenter of triangle
    ABD . If AI = 2 as well, find the minimum value of the length of diagonal BD .
    √
    Answer: 2 3 Let T be the point where the incircle intersects AD , and let r be the inradius and Rbe the circumradius of 4 ABD . Since BC = CD = 2, C is on the midpoint of arc BD on the oppositeside of BD as A , and hence on the angle bisector of A . Thus A , I , and C are collinear. We have the following formulas:
    IM r
    AI = =
    sin ∠ IAMAsin
    A2
    BC = 2 R sin
    BD = 2 R sin A2
    The last two equations follow from the extended law of sines on 4 ABC and 4 ABD , respectively.
    A r 2
    Using AI = 2 = BC gives sin = . However, it is well-known that R ≥ 2 r with equality for an
    2 2 Rr A 12
    equilateral triangle (one way to see this is the identity 1+ = cos A +cos B +cos D ). Hence sin ≤
    R 2 4
    °Aand ≤ 30 . Then
    ( )2
    √
    ( )
    √
    A A A 3
    BD = 2 R 2 sin cos = BC · 2 cos ≥ 2 2 · = 2 3
    2 2 2 2
    with equality when 4 ABD is equilateral.
    B DIA
    Remark: Similar but perhaps simpler computations can be made by noting that if AC intersects BDCat X , then AB/BX = AD/DX = 2, which follows from the exterior angle bisector theorem; if I isthe A -excenter of triangle ABC , then AI /XI = 2 since it is well-known that C is the circumcenter a
    A Aof cyclic quadrilateral BIDI .
    Calculus & Geometry Individual Test a