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Author |
isee
Posted 2022-2-25 16:51
Last edited by isee 2022-2-25 18:20(1)
注意圆周角相等,由三弦定理有\[ CA\cdot \sin BCD=CD\cdot \sin BCA+BC \cdot\sin ACD,\]即
\[ AC \cdot \sin BCD=CD\cdot \sin ADB+BC \cdot\sin ABD.\]
即\[ AC =\frac {3\cdot CD+2\cdot BC}4.\]
于是
\begin{align*}
\lambda&=\frac {AC^2}{BC\cdot CD}\\[1ex]
&=\frac {(2\cdot BC+3\cdot CD)^2}{16\cdot BC\cdot CD}\\[1ex]
&=\frac {BC}{4\cdot CD}+\frac {9\cdot CD}{16\cdot BC}+\frac {3}4\\[1ex]
&\geqslant 2\sqrt {\frac 14\cdot \frac 9{16}}+\frac {3}4\\[1ex]
&=\frac 32.
\end{align*}
取“=”时,$AB\cdot CD=BC\cdot DA$即调和四边形$ABCD$.
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(2)
实际在四边形$ABCD$中
\[36^\circ=\angle A=\frac 12\angle B=\frac 13\angle C=\frac 14\angle D.\]
连接$AC$,记$\angle BAC=x,\angle DAC=y$,则$x+y=36^\circ$.
在$\triangle ABC$中由正弦定理易知
\[BC=\frac {AC\cdot \sin x}{\sin B},\]
同理在$\triangle ACD$中有
\[CD=\frac {AC\cdot \sin y}{\sin D}=\frac {AC\cdot \sin y}{\sin B}.\]
从而
\begin{align*}
\lambda&=\frac {AC^2}{BC\cdot CD}\\[1ex]
&=\frac {AC^2}{\frac {AC\cdot \sin x}{\sin B}\cdot \frac {AC\cdot \sin y}{\sin B}}\\[1ex]
&=\frac {\sin^2 72^\circ}{\sin x\sin y}\\[1ex]{}
\xlongequal[\text{积化和差}]{}&=\frac {2\sin^2 72^\circ}{-\cos 36^\circ+\cos (x-y)}\\[1ex]
&\geqslant \frac {2\sin^2 72^\circ}{-\cos 36^\circ+\color{red}{1}}\\[1ex]
&=\frac {2\sin^2 72^\circ}{2\sin^2 18^\circ}\\[1ex]
&=\tan^2 72^\circ=\frac 1{\tan^2 18^\circ}\\[1ex]
&=5+2\sqrt 5.
\end{align*}
取“=”时,$x=y=18^\circ$,即$AB$为圆的直径.
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关于 sin 18度 的求法 1,求法 2 |
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