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本帖最后由 creasson 于 2021-7-27 13:04 编辑
15#, 结论太复杂了,事实上,有
\[\frac{{A{B^2} + B{C^2} + C{A^2}}}{{8{S_{ABC}}}} = \frac{{R - r}}{{R + r}}\]
记\({R_1},{R_2},{R_3}\)为三绿圆半径,那么
\[\begin{array}{l}
A{B^2} + B{C^2} + C{A^2} = \frac{{8{R_1}{R_2}{R_3}({R_1}{R_2} + {R_2}{R_3} + {R_3}{R_1})}}{{({R_1} + {R_2})({R_2} + {R_3})({R_3} + {R_1})}} \\
{S_{\Delta ABC}} = \frac{{2{R_1}{R_2}{R_3}\sqrt {{R_1}{R_2}{R_3}\left( {{R_1} + {R_2} + {R_3}} \right)} }}{{({R_1} + {R_2})({R_2} + {R_3})({R_3} + {R_1})}} \\
O{I^2} = \frac{{16\left( {\frac{1}{{{R_1}^2}} + \frac{1}{{{R_2}^2}} + \frac{1}{{{R_3}^2}} - \frac{1}{{{R_1}{R_2}}} - \frac{1}{{{R_2}{R_3}}} - \frac{1}{{{R_3}{R_1}}}} \right)}}{{{{\left( {\frac{1}{{{R_1}^2}} + \frac{1}{{{R_2}^2}} + \frac{1}{{{R_3}^2}} - \frac{2}{{{R_1}{R_2}}} - \frac{2}{{{R_2}{R_3}}} - \frac{2}{{{R_3}{R_1}}}} \right)}^2}}} \\
\end{array}\]
\(R,r\)可由笛卡尔定理得。
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