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100鳳山高中

#4
100彰女填充第二部分第11題
https://math.pro/db/thread-1113-2-2.html

[ 本帖最後由 老王 於 2011-6-19 10:20 PM 編輯 ]
名豈文章著官應老病休飄飄何所似Essential isolated singularity

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#2
將\(\displaystyle  y^2=mx \)代入
\(\displaystyle  x^2-4x+4+mx=3,x^2-(4-m)x+1=0 \)
\(\displaystyle  x_1+x_2=4-m \)
另一方面
\(\displaystyle \frac{y^4}{m^2}-\frac{4y^2}{m}+4+y^2=3,y^4-(4m-m^2)y^2+m^2=0 \)
\(\displaystyle y_1^2+y_2^2=4m-m^2,y_1^2y_2^2=m^2 \)
因為m是正數,以及交點在第一象限,所以
\(\displaystyle y_1y_2=m \)
\(\displaystyle (y_1+y_2)^2=6m-m^2 \)
\(\displaystyle y_1+y_2=\sqrt{6m-m^2} \)
中點在y=x上
\(\displaystyle 4-m=\sqrt{6m-m^2} \)
\(\displaystyle 16-8m+m^2=6m-m^2 \)
\(\displaystyle 2m^2-14m+16=0,m^2-7m+8=0 \)
\(\displaystyle m=\frac{7+\sqrt{17}}{2}, or m=\frac{7-\sqrt{17}}{2} \)
但是m要小於4
所以
\(\displaystyle m=\frac{7-\sqrt{17}}{2} \)
名豈文章著官應老病休飄飄何所似Essential isolated singularity

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#5
假設AB=a,CD=b,AC=BC=AD=BD=1
取AB中點M,CM和DM都垂直AB
所以AB垂直平面CDM
將四面體分成A-CDM和B-CDM
體積為
\(\displaystyle \frac{1}{3} \times \frac{1}{2}b \times \sqrt{1-\frac{a^2}{4}-\frac{b^2}{4}} \times a \)
\(\displaystyle =\frac{1}{12}ab\sqrt{4-a^2-b^2} \)
令\(\displaystyle c=\sqrt{4-a^2-b^2} \)
可知若\(\displaystyle a^2+b^2\rightarrow 4 \),體積會接進0,故無最小值。
\(\displaystyle a^2+b^2+c^2=4 \)
算幾不等式
\(\displaystyle a^2+b^2+c^2 \ge 3\sqrt[3]{a^2b^2c^2} \)
\(\displaystyle abc \le \frac{8\sqrt3}{9} \)
體積最大值
\(\displaystyle \frac{2\sqrt3}{27} \)
名豈文章著官應老病休飄飄何所似Essential isolated singularity

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重做的時候發現第五題也可以用向量來作
因為AC是直徑,所以
\(\displaystyle \vec{AC} \cdot \vec{AB}=|\vec{AB}|^2 \)
以及
\(\displaystyle \vec{AC} \cdot \vec{AD}=|\vec{AD}|^2 \)
就會得到
\(\displaystyle \frac{3}{2}|\vec{AB}|^2+\frac{5}{2}\vec{AB} \cdot \vec{AD}=|\vec{AB}|^2 \)
\(\displaystyle \frac{3}{2}\vec{AB} \cdot \vec{AD}+\frac{5}{2}|\vec{AD}|^2=|\vec{AD}|^2 \)
整理得
\(\displaystyle \vec{AB} \cdot \vec{AD}=-\frac{1}{5}|\vec{AB}|^2=-|\vec{AD}|^2 \)

又AC=2得到
\(\displaystyle 4=\frac{9}{4}|\vec{AB}|^2+\frac{2}{15}\vec{AB} \cdot \vec{AD}+\frac{25}{4}|\vec{AD}|^2 \)
令\(\displaystyle |\vec{AD}|^2=K \)
則\(\displaystyle |\vec{AB}|^2=5K,\vec{AB} \cdot \vec{AD}=-K \)
\(\displaystyle 4=10K,K=\frac{2}{5} \)
於是
\(\displaystyle |\vec{BD}|^2 \)
\(\displaystyle =|\vec{AD}|^2-2\vec{AB} \cdot \vec{AD}+|\vec{AB}|^2 \)
\(\displaystyle =8K=\frac{16}{5} \)
最後得到
\(\displaystyle BD=\frac{4\sqrt{5}}{5} \)
名豈文章著官應老病休飄飄何所似Essential isolated singularity

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