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楼主 |
发表于 2023-5-26 20:49
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本帖最后由 dodonaomikiki 于 2023-5-26 21:24 编辑
看起来,
难度不大毕竟是初赛,
计算比较繁杂?
来看一哈
\begin{align*}
\Longrightarrow Point \qquad &C( \frac{ 50t }{ t^2+25 } , \frac{ 4t^2-100 }{ t^2+25 } )\\
Cauz \qquad Point \qquad &C,D \qquad are \qquad \qquad symmetrical \qquad about \qquad x-axis\\
\Longrightarrow Point \qquad &D ( \frac{ 50t }{ t^2+25 } , \frac{ 100- 4t^2 }{ t^2+25 } )\\
And \qquad ADE \qquad are \qquad & co-llinear\\
\Longrightarrow &Point \qquad E ( \frac{25}{t} ,0 )\\
\Longrightarrow Area(ABD )&= Area(ABE )- Area(BD E )\\
&=\frac{BE}{2 }(y_A-y_D)\\
&=\frac{ 20(-t^2-5t) }{ t^2+25 }\\
Cauz \qquad t \in 【 -4, 5(1- \sqrt{2} ) 】 \\
\Longrightarrow &\frac{80 }{41} \preceq Area(ABD ) \preceq 10(\sqrt{2} -1) \\
\end{align*} |
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