HMMT 二月 2015 · 冲刺赛 · 第 24 题
HMMT February 2015 — Guts Round — Problem 24
题目详情
- [ 14 ] ABCD is a cyclic quadrilateral with sides AB = 10, BC = 8, CD = 25, and DA = 12. A circleω 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#
英文原题
[ 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
英文解析
- [ 14 ] ABCD is a cyclic quadrilateral with sides AB = 10, BC = 8, CD = 25, and DA = 12. A circleω 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
y = EB . Because ABCD is a cyclic quadrilateral, △ EAB is similar to △ ECD . Therefore, =25
x 10
x +12 25 128 152
and = . We get x = and y = . Note that ω is the E -excircle of △ EAB , so we mayy 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 = = = = .
s − AB s − 10 7 7E