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[几何] 来自讨论组群:三次曲线上两点PQ≥1

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kuing posted 2025-1-18 23:39 |Read mode
木目也 22:24
微信截图_20250118232905.png
D怎么说理
曲线C在圆心为B半径为0.5的圆外或相切,但是图象不关于B对称呀

生如夏花 22:50
这图左边是包着圆的
42FB5E2456C40605B08F816C35BF4C51.jpg
哦,我知道你的疑问了
93A594B61356D50CD9CD0600C9F5078C.jpg
就是勾股定理

來ー妢铯圖KK 23:09
显然只需考虑 `x_1`, `x_2\leqslant0` 的情形,则 `x_1+x_2=-1`,不妨设 `y_1\leqslant0\leqslant y_2`,则
\begin{align*}
y_1&=-\sqrt{x_1(x_1+1)(x_1-1)}=-\sqrt{x_1x_2(1-x_1)}\leqslant-\sqrt{x_1x_2},\\
y_2&=\sqrt{x_2(x_2+1)(x_2-1)}=\sqrt{x_1x_2(1-x_2)}\geqslant\sqrt{x_1x_2},\\
\riff&\abs{y_1-y_2}\geqslant2\sqrt{x_1x_2},\\
\riff& PQ^2=(x_1-x_2)^2+(y_1-y_2)^2\geqslant(x_1-x_2)^2+4x_1x_2=(x_1+x_2)^2=1.
\end{align*}

和wwd写的其实差不多

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