HMMT 二月 2008 · 冲刺赛 · 第 32 题

HMMT February 2008 — Guts Round — Problem 32

[ 18 ] Cyclic pentagon ABCDE has side lengths AB = BC = 5 , CD = DE = 12 , and AE = 14 . Determine the radius of its circumcircle. 3 3 3 5 5 5

解析

[ 18 ] Cyclic pentagon ABCDE has side lengths AB = BC = 5 , CD = DE = 12 , and AE = 14 . Determine the radius of its circumcircle. √ 225 11 ′ ′ ′ Answer: Let C be the point on minor arc BCD such that BC = 12 and C D = 5 , and write 88 ′ ′ ′ ′ AC = BD = C E = x, AD = y, and BD = z. Ptolemy applied to quadrilaterals ABC D, BC DE, and ABDE gives 2 2 x = 12 y + 5 2 2 x = 5 z + 12 yz = 14 x + 5 · 12 Then 2 2 2 2 2 2 ( x − 5 )( x − 12 ) = 5 · 12 yz = 5 · 12 · 14 x + 5 · 12 , 3 from which x − 169 x − 5 · 12 · 14 = 0 . Noting that x > 13 , the rational root theorem leads quickly to √ √ 5 · 12 · 15 √ the root x = 15 . Then triangle BCD has area 16 · 1 · 4 · 11 = 8 11 and circumradius R = = 4 · 8 11 √ 225 11 . 88 3 3 3 5 5 5