HMMT 二月 2011 · CALCGEO 赛 · 第 13 题

HMMT February 2011 — CALCGEO Round — Problem 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 .

解析

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 R be the circumradius of 4 ABD . Since BC = CD = 2, C is on the midpoint of arc BD on the opposite side 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 = = A sin ∠ IAM sin 2 A BC = 2 R sin 2 BD = 2 R sin A The last two equations follow from the extended law of sines on 4 ABC and 4 ABD , respectively. 2 A r Using AI = 2 = BC gives sin = . However, it is well-known that R ≥ 2 r with equality for an 2 2 R 2 r A 1 equilateral triangle (one way to see this is the identity 1+ = cos A +cos B +cos D ). Hence sin ≤ R 2 4 A ◦ and ≤ 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. A I B D C Remark: Similar but perhaps simpler computations can be made by noting that if AC intersects BD at X , then AB/BX = AD/DX = 2, which follows from the exterior angle bisector theorem; if I is A the A -excenter of triangle ABC , then AI /XI = 2 since it is well-known that C is the circumcenter A A of cyclic quadrilateral BIDI . A Calculus & Geometry Individual Test