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Yahoo!奇摩知識+ - 分類問答 - 科學常識 - 已解決
Yahoo!奇摩知識+ - 分類問答 - 科學常識 - 已解決 
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△ABC 為正三角形 ， D,E兩點在BC 邊上，BD＝2，
Aug 7th 2013, 15:22

第一種情形：
 檢視圖片 
tan a = √3/(k-1)，tan b = [(3√3)/2]/[k-(3/2)] = 3√3/(2k-3)
tan (a+b) = tan 30度
(tan a + tan b)/[1-(tan a*tan b)] = 1/√3
[√3/(k-1) + 3√3/(2k-3)]/{1-[√3/(k-1)][3√3/(2k-3)]} = 1/√3
上式左邊分子分母同乘以(k-1)(2k-3)
[√3(2k-3)+3√3(k-1)]/[(k-1)(2k-3)-9] = 1/√3
交叉相乘得
3(2k-3)+9(k-1) = (k-1)(2k-3)-9
乘開整理得
k^2 - 10k + 6 = 0
k = 5±√19 (負不合)

第二種情形：

檢視圖片 
tan a = [√3(k-3)/2]/[k-(k-3)/2] = √3(k-3)/(k+3)
tan b = [√3(k-2)/2]/[k-(k-2)/2] = √3(k-2)/(k+2)
tan (a+b) = tan 30度
(tan a + tan b)/[1-(tan a*tan b)] = 1/√3
[√3(k-3)/(k+3) + √3(k-2)/(k+2)]/{1-[√3(k-3)/(k+3)][√3(k-2)/(k+2)]} = 1/√3
上式左邊分子分母同乘以(k+3)(k+2)
[√3(k-3)(k+2)+√3(k-2)(k+3)]/[(k+3)(k+2)-3(k-3)(k-2)] = 1/√3
交叉相乘得
3(k-3)(k+2)+3(k-2)(k+3) = (k+3)(k+2)-3(k-3)(k-2)
乘開整理得
2k^2 - 5k - 6 = 0
k = (5±√73)/4 (負不合)

所以正△ABC邊長為 5+√19 或 (5+√73)/4 

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