引用:
原帖由 arend 於 2012-5-1 12:52 AM 發表 
請教計算第一題
我是設FD=x, EB=y , 用餘弦定理,或和角
一直做不出來
試了很多方法,就是差一步
謝謝
\(\displaystyle (5-x)^2+(5-y)^2=16 \)
\(\displaystyle 50-10(x+y)+(x^2+y^2)=16 \).............(1)
令\(\displaystyle \angle {DCF}=\alpha , \angle {BCE}=\beta \)
\(\displaystyle \tan \alpha =\frac{x}{5}, \tan \beta =\frac{y}{5} \)
\(\displaystyle \tan(\alpha+\beta)=\tan(90^o-\angle{ECF})=\cot \angle{ECF} \)
\(\displaystyle \frac{\displaystyle \frac{x}{5}+\frac{y}{5}}{\displaystyle 1-\frac{xy}{25}}=\frac{4}{3} \)
\(\displaystyle 15(x+y)=100-4xy \)...........(2)
再令\(\displaystyle P=x+y,Q=xy \)
(1) => \(\displaystyle 50-10P+P^2-2Q=16 \)
(2) => \(\displaystyle 15P=100-4Q \)
\(\displaystyle 50-10P+P^2-50+\frac{15}{2}P=16 \)
\(\displaystyle 2P^2-5P-32=0 \)
\(\displaystyle P=\frac{5+\sqrt{281}}{4} \)
所求為\(\displaystyle 25-\frac{1}{2}(5x+5y+25-5x-5y+xy)=\frac{1}{2}(25-Q)=\frac{15}{8}P=\frac{75+15\sqrt{281}}{32} \)
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本帖最後由 老王 於 2012-5-1 07:47 PM 編輯 ]
