HMMT 二月 2007 · 冲刺赛 · 第 28 题

HMMT February 2007 — Guts Round — Problem 28

[ 15 ] Compute the circumradius of cyclic hexagon ABCDEF , which has side lengths AB = BC = 2 , CD = DE = 9, and EF = F A = 12. 5 4 2

解析

[ 15 ] Compute the circumradius of cyclic hexagon ABCDEF , which has side lengths AB = BC = 2 , CD = DE = 9, and EF = F A = 12. 8 ′ ′ Answer: 8 . Construct point E on the circumcircle of ABCDEF such that DE = EF = 12 and √ √ ′ ′ ′ ′ ′ 2 2 2 E F = DE = 9; then BE is a diameter. Let BE = d . Then CE = BE − BC = d − 4 and √ √ ′ ′ 2 ′ 2 2 BD = BE − DE = d − 144. Applying Ptolemy’s theorem to BCDE now yields √ 2 2 9 · d + 2 · 12 = ( d − 4)( d − 144) 4 2 2 Squaring and rearranging, we find 0 = d − 229 d − 432 d = d ( d − 16)( d + 16 d + 27). Since d is a positive real number, d = 16, and the circumradius is 8. 5 4 2