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[几何] 求证反演后的圆过反演圆的圆心。

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abababa posted 2024-2-26 19:01 |Read mode

将$\odot A$作为反演圆。以下带撇的字母表示对应的反演点。设直线$BCD$与$\odot A$切于点$B$,将直线$BCD$反演,得到$\odot(BC'D')$,此圆过点$A$且仍与$\odot A$切于点$B$。

将直线$C'D'$再反演,得到一个过$C,D$两点的圆,求证这个圆也过点$A$。

这个性质要怎么用反演的观点来证明?

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爪机专用 posted 2024-2-26 22:03
无穷远点的反演就是圆心吧?那任意直线的反演都过圆心吧?
I am majia of kuing

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hbghlyj posted 2024-2-26 22:49
补充一下$1/\infty=0$
The Riemann Sphere - A world where you can divide by zero

這篇文章裡有很多圖:
1_IYAmHJ8b51r7qdCr5EMxaw.webp
例如上面的,雙曲線的反演經過0
其它例子:
1_J78J_RLOJKGjT-ZQGQVN9w.png
1_oOi3RVPLOiy7SFtaGF1G1g.png

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original poster abababa posted 2024-2-27 09:54
爪机专用 发表于 2024-2-26 22:03
无穷远点的反演就是圆心吧?那任意直线的反演都过圆心吧?
是的,昨天我都迷糊了。

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