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[几何] 一道立体几何

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hbghlyj Posted 2020-5-20 13:19 |Read mode
分别过正方体ABCD-EFGH的六个面中每面的四个顶点作八个球,正方体的八个顶点每点处的三个球面另交于一点,则这八点A'B'C'D'E'F'G'H'共球
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 Author| hbghlyj Posted 2020-5-22 17:33
由于A,B,A',B'为两球的公共点,所以ABA'B'共圆,同理BCB'C',CDC'D',DAD'A'共圆,对球ABCDA'B'C'D'作球极投影,用古镂钱定理,由ABCD共圆得出A'B'C'D'共圆,同理A'B'F'E',B'C'G'F',C'D'H'G',D'A'E'H',E'F'G'H'共圆,由于圆A'B'C'D'的中轴线和圆A'B'F'E'的中轴线都在A'B'的中垂面上,设它们交于O,OA'=OB'=OC'=OD'=OE'=OF'=r,由于OA'=OD'=OE',所以O在圆A'D'E'H'的中轴线上,所以OH'=r,同理OG'=r,所以A'B'C'D'E'F'G'H'共球,球心为O.

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