HMMT 二月 2008 · 冲刺赛 · 第 18 题

HMMT February 2008 — Guts Round — Problem 18

[ 9 ] Let ABC be a right triangle with ∠ A = 90 . Let D be the midpoint of AB and let E be a point ◦ on segment AC such that AD = AE . Let BE meet CD at F . If ∠ BF C = 135 , determine BC/AB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

解析

[ 9 ] Let ABC be a right triangle with ∠ A = 90 . Let D be the midpoint of AB and let E be a point ◦ on segment AC such that AD = AE . Let BE meet CD at F . If ∠ BF C = 135 , determine BC/AB . √ 13 Answer: Let α = ∠ ADC and β = ∠ ABE . By exterior angle theorem, α = ∠ BF D + β = 2 ◦ 45 + β . Also, note that tan β = AE/AB = AD/AB = 1 / 2. Thus, 1 tan α − tan α − tan β ◦ 2 1 = tan 45 = tan( α − β ) = = . 1 1 + tan α tan β 1 + tan α 2 4 3 Solving for tan α gives tan α = 3. Therefore, AC = 3 AD = AB . Using Pythagorean Theorem, we 2 √ √ 13 13 find that BC = AB . So the answer is . 2 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND