HMMT 二月 2008 · 冲刺赛 · 第 18 题
HMMT February 2008 — Guts Round — Problem 18
题目详情
- [ 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 .
11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUNDth
英文原题
[ 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
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11 th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND
解析
英文解析
- [ 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 .
√
Answer: Let α = ∠ ADC and β = ∠ ABE . By exterior angle theorem, α = ∠ BF D + β =13
°2
45 + β . Also, note that tan β = AE/AB = AD/AB = 1 / 2. Thus,
tan α −1
tan α − tan β
°
1 = tan 45 = tan( α − β ) = = .2
1 + tan α tan β1
1 + tan α
Solving for tan α gives tan α = 3. Therefore, AC = 3 AD = AB . Using Pythagorean Theorem, we 342
√ √2
13 13
find that BC = AB . So the answer is .
2 2
11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUNDth