没有思路怎么解。。三道我自己感觉有点难的几何最值

elim

\({\small|DF|=(1-u)}\tan({\large\frac{\pi}{6}}+\theta)=\large\frac{(1-u)(1+\sqrt{3})}{\sqrt{3}-u}\)
\({\small\partial_u|DF|=}{\large\frac{\sqrt{3}(u-\sqrt{3})^2-4(\sqrt{3}-1)}{(\sqrt{3}-u)^2}}.\)
\(\therefore\quad\min|CT|=|CT|(u_F)=4\sqrt{3-3^{1/2}}+\sqrt{3}-6=0.23618\ldots\)
\(\qquad u_F=\sqrt{3}-2\sqrt{1-3^{-1/2}}\)

FGNBGHJUOI

elim 发表于 2021-2-11 09:15
\({\small|DF|=(1-u)}\tan({\large\frac{\pi}{6}}+\theta)=\large\frac{(1-u)(1+\sqrt{3})}{\sqrt{3}-u ...

谢谢elim的解答过程，真的详细，辛苦了

王守恩

elim 发表于 2021-2-11 09:15
\({\small|DF|=(1-u)}\tan({\large\frac{\pi}{6}}+\theta)=\large\frac{(1-u)(1+\sqrt{3})}{\sqrt{3}-u ...

求DF的最大值还有点意思。
\(一，设DF的最大值=x,DE=y\)
\(\ \ (1-x)^2+1=(1-y)^2+1+x^2+y^2-2\sqrt{((1-y)^2+1)(x^2+y^2)}\cos60\)
\(二，设∠AEB=x\ \ \ AE=\frac{\cos x}{\sin x}\ \ \ ED=\frac{\sin x-\cos x}{\sin x}\)
\(\ \ DF的最大值=\frac{(\sin x-\cos x)\sin(120-x)}{\sin x\cos(120-x)}\)