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#ed_op#DIV#ed_cl#畫成直角形, 算算看..#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#IMG alt="image file name: 2k89282dd03d.png" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2k89282dd03d.png" border=0#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#BC 若為 2, 那周長就是 14 .#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#因為 AB#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = BC#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# + AC#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#BR#ed_cl#(14-2-AC)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = 2#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# + AC#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#BR#ed_cl#解一元二次方程後, 可得, AC = 35/6, AB = 37/6.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#∠C 的分角線方程 y = x&nbsp;&nbsp;&nbsp; .... (∠C)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#找 ∠B 的分角線#ed_op#BR#ed_cl#sin(∠B) = (35/6) / (37/6) = 35/37#ed_op#BR#ed_cl#cos(∠B) = 2/(37/6) = 12/37#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#tan(∠B/2) = (1 - cos(∠B)) / sin(∠B)#ed_op#BR#ed_cl#= (1 - (12/37)) / (35/37)#ed_op#BR#ed_cl#= 5/7#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#因為身為底邊的 BC =&nbsp;2, 所以對邊 NC =&nbsp;2 * 5/7 = 10/7.#ed_op#BR#ed_cl#故 ∠B 的分角線, 與 y 軸的交點, 為 (0,10/7)#ed_op#BR#ed_cl#因此 ∠B 的分角線方程為 (10/7)x + 2y = 2*(10/7)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#BR#ed_cl#10x + 14y = 20 #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#10x + 14x = 20&nbsp;&nbsp; ← 拿 (∠C) 式來代入, 找兩分角線的交點#ed_op#BR#ed_cl#x = 20/24 = 5/6&nbsp; → 代入 (∠C) 式#ed_op#BR#ed_cl#y = 5/6&nbsp;&nbsp;&nbsp; #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#此交點 (5/6,5/6) 即為圖中內切圓心, 那圓直徑就是 10/6. #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#因此 DE 線方程為 x=5/6&nbsp;&nbsp; ..... (DE)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#又圖中的 O 是 BC 一半,&nbsp;O 坐標 (1,0),#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以 AO 線方程為 (35/6)x + (1)y = (35/6)(1)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#35x + 6y = 35#ed_op#BR#ed_cl#35(5/6) + 6y = 35&nbsp; ← 拿 (DE) 式來代入, 找交點.#ed_op#BR#ed_cl#y = 35/36, 也是 FE 的長度, F (x,y) 即 (5/6, 35/36)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#DF 長 = (直徑-FE) = (10/6 - 35/36) =&nbsp; 25/36#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#DF:FE = 25/36 : 35/36 = 5:7#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

AB+CA=12 ---(1)

AC斜率=(√35sinθ-0)/(6cosθ+1)=√35sinθ/(6cosθ+1)
AB斜率=(√35sinθ-0)/(6cosθ-1)=√35sinθ/(6cosθ-1)

tan2α=√35sinθ/(6cosθ+1)
2tanα/(1-tan2α)=√35sinθ/(6cosθ+1)

tan2β=√35sinθ/(6cosθ-1)
2tanβ/(1-tan2β)=√35sinθ/(6cosθ-1)

MC方程式為:y=(√35/7)tan(θ/2)(x+1) ---(4)
MB方程式為:y=-(√35/7)cot(θ/2)(x-1) ---(5)

BC的中點為O(0,0)，因此OA方程式為:y=(√35/6)xtanθ ---(6)
DE方程式為: x=cosθ ---(7)

DF=(5/42)√35sinθ，FE=(1/6)√35sinθ
DF/FE=(5/42)/(1/6)=5/7

i also think the answer is 5/7.
however i think i have used some 茅招...
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5/7

NO.20