HMMT 二月 2015 · 冲刺赛 · 第 24 题

HMMT February 2015 — Guts Round — Problem 24

[ 14 ] ABCD is a cyclic quadrilateral with sides AB = 10, BC = 8, CD = 25, and DA = 12. A circle ω is tangent to segments DA , AB , and BC . Find the radius of ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT FEBRUARY 2015, 21 FEBRUARY 2015 — GUTS ROUND Organization Team Team ID#

解析

[ 14 ] ABCD is a cyclic quadrilateral with sides AB = 10, BC = 8, CD = 25, and DA = 12. A circle ω is tangent to segments DA , AB , and BC . Find the radius of ω . √ √ 1209 8463 Answer: OR Denote E an intersection point of AD and BC . Let x = EA and 7 7 y +8 25 y = EB . Because ABCD is a cyclic quadrilateral, △ EAB is similar to △ ECD . Therefore, = x 10 x +12 25 128 152 and = . We get x = and y = . Note that ω is the E -excircle of △ EAB , so we may y 10 21 21 finish by standard calculations. x + y +10 EA + AB + BE 35 Indeed, first we compute the semiperimeter s = = = . Now the radius of ω is 2 2 3 (by Heron’s formula for area) √ √ √ [ EAB ] s ( s − x )( s − y ) 1209 8463 r = = = = . E s − AB s − 10 7 7