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[几何] 极圆

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hbghlyj Posted 2020-6-5 23:03 |Read mode
Last edited by hbghlyj 2020-7-16 09:46定义 一个三角形的极圆(pole circle of triangle)是以垂心为圆心,半径由$r^2=-4R^2\cos A\cos B\cos C$给出的圆.由此,仅对钝角三角形,才有实的极圆存在,对钝角三角形,我们可以立即建立如下的定理:
定理 在关于极圆的反演下,三角形的每个顶点与对应的高线足互反;以一条高为弦的圆,经过反演不变,所以与极圆正交;外接圆与九点圆互反.
在极圆上取点P,设PA,PB,PC再次交圆于Q,R,S,则ARS,QBS,QRC共线
R,S处的切线与BC共点于U,S,Q处的切线与CA共点于V,Q,R处的切线与AB共点于W
P,Q处的切线与BC共点于D,P,R处的切线与CA共点于E,P,S处的切线与AB共点于F
AU,BV,CW共点于T
AU,BE,CF共点于X,AD,BV,CF共点于Y,AD,BE,CW共点于Z
等面三面角.png

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 Author| hbghlyj Posted 2020-6-6 00:20
Last edited by hbghlyj 2020-6-6 13:16耐心梳理一下,共点共线好像只有下面这些,如果您发现更多欢迎告诉我!!
BC,YS,ZR共点于J,CA,ZQ,XS共点于K,AB,XR,YQ共点于L,
BC,YR,ZS共点于M,XQ,CA,ZS共点于N,XQ,YR,AB共点于O
JKLT共线
AU,YE,ZF共点于$A_1$,XD,BV,ZF共点于$B_1$,XD,YE,CW共点于$C_1$
Q处的切线,RS,LT共点于$A_2$,R处的切线,SQ,LT共点于$B_2$,S处的切线,QR,LT共点于$C_2$,
BC,YP,ZQ共点于$J_1$,CA,ZP,XR共点于$K_1$,AB,XP,YS共点于$L_1$
BC,ZP,YQ共点于$J_2$,CA,XP,ZR共点于$K_2$,AB,YP,XS共点于$L_2$
$type 配极.ggb (53.14 KB, Downloads: 3316)

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 Author| hbghlyj Posted 2020-6-6 13:08
由于垂心组A,B,C,H这四个点地位是对称的,所以还可以画三个一样的图,这四个图叠起来又会发生很多故事呢

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