返回题库

HMMT 十一月 2010 · 冲刺赛 · 第 25 题

HMMT November 2010 — Guts Round — Problem 25

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [ 14 ] Triangle ABC is given with AB = 13, BC = 14, CA = 15. Let E and F be the feet of the altitudes from B and C , respectively. Let G be the foot of the altitude from A in triangle AF E . Find AG .
解析
  1. [ 14 ] Triangle ABC is given with AB = 13, BC = 14, CA = 15. Let E and F be the feet of the altitudes from B and C , respectively. Let G be the foot of the altitude from A in triangle AF E . Find AG . √ 396 Answer: By Heron’s formula we have [ ABC ] = 21(8)(7)(6) = 84. Let D be the foot of the 65 84 altitude from A to BC ; then AD = 2 · = 12. Notice that because ∠ BF C = ∠ BEC , BF EC is 14 cyclic, so ∠ AF E = 90 − ∠ EF C = 90 − ∠ EBC = ∠ C . Therefore, we have 4 AEF ∼ 4 ABC , so √ √ ( ) 2 2 2 AG AE 1 56 56 65 − 56 33 2 = ; ( BE )( AC ) = 84 = ⇒ BE = = ⇒ AE = 13 − = = . Then 2 AD AB 2 5 5 5 5 33 / 5 AE 396 AG = AD · = 12 · = . AB 13 65