190, 我总算做出来了。 |
送交者: 零加一中 2019年11月30日09:40:19 于 [灵机一动] 发送悄悄话 |
找到正确的辅助线用了约两天时间。剩下的工作用了大约半天。整理到目前的条理用了一个多小时。最近几天的早锻炼(走路约50分钟)都在考虑这道题目。 Let ∆ABC = S. Choose F in AB so that ED//CF. In ∆ACF, sin ∠A/CF = sin(∠ACF)/AF i.e. CF/sin(45°) = AF/sin(75°) Because ∠B = 30° and ∠CFB = 120°, CF = BF。 BF/AF = sin(45°)/ sin(75°) = sqrt(3) – 1 ∆BCF/∆ACF = BF/AF = sqrt(3) – 1 ∆ACF = S/sqrt(3) (by ∆ACF+∆BCF=S) ∆AED/∆ACF = sqrt(3)/2 = (AD/AF)^2 (by ∆AED=S/2) AD/AB = AD/(AF+BF) = (AD/AF)/(1+BF/AF) = (AD/AF)/sqrt(3) =12^(-1/4) |
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实用资讯 | |