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[几何] $X^2$的轨迹为三角形内切二次曲线

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hbghlyj posted 2025-5-4 09:34 |Read mode
en.wikipedia.org/wiki/Circumconic_and_inconic#Inconic_2
重心坐标$X=(p_{1}+p_{2}t):(q_{1}+q_{2}t):(r_{1}+r_{2}t)$,则$X$的轨迹为直线。
重心坐标$X^2=(p_{1}+p_{2}t)^2:(q_{1}+q_{2}t)^2:(r_{1}+r_{2}t)^2$,则$X^2$的轨迹为内切二次曲线

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