HMMT 十一月 2017 · 团队赛 · 第 9 题

HMMT November 2017 — Team Round — Problem 9

[ 60 ] Let A, B, C, D be points chosen on a circle, in that order. Line BD is reflected over lines AB and DA to obtain lines ` and ` respectively. If lines ` , ` , and AC meet at a common point and if 1 2 1 2 AB = 4 , BC = 3 , CD = 2, compute the length DA .

解析

[ 60 ] Let A, B, C, D be points chosen on a circle, in that order. Line BD is reflected over lines AB and DA to obtain lines ` and ` respectively. If lines ` , ` , and AC meet at a common point and if 1 2 1 2 AB = 4 , BC = 3 , CD = 2, compute the length DA . Proposed by: Ashwin Sah √ Answer: 21 Let the common point be E . Then since lines BE and BD are symmetric about line BA , BA is an exterior bisector of ∠ DBE , and similarly DA is also an exterior bisector of ∠ BDE . Therefore A is the E -excenter of triangle BDE and thus lie on the interior bisector of ∠ BED . Since C lies ◦ on line AE , and by the fact that A, B, C, D are concyclic, we get that ∠ ABC + ∠ ADC = 180 , 1 which implies ∠ DBC + ∠ BDC = ( ∠ DBE + ∠ BDE ), so C is the incenter of triangle BDE . Thus 2 π 2 2 2 2 2 2 2 2 2 ∠ ABC = ∠ CDA = , and thus DA = AC − CD = AB + BC − CD = 3 + 4 − 2 = 21. The 2 √ length of DA is then 21.