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发表于 2016-9-3 01:15:53
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接上给出证明:令各点的复数为:\[{Z_A} = 0,{Z_B} = 1,{Z_C} = z,{Z_L} = 0,{Z_M} = 1,{Z_N} = w\] (ABC和LMN分别在两个复平面上.)
于是可知: \[{Z_{{L^ + }}} = \frac{{w - z}}{{w - 1}},{Z_{{M^ + }}} = \frac{z}{w},{Z_{{N^ + }}} = w;\]
并\[A{L^ + },B{M^ + },C{N^ + }\]三线共点于:\[{Z_E} = \frac{{{\mathop{\rm Im}\nolimits} \left( {\frac{w}{{w - z}}} \right)}}{{{\mathop{\rm Im}\nolimits} \left( {\frac{1}{{w - 1}}} \right)}}\left( {\frac{{w - z}}{{w - 1}}} \right)\]
将\[w\]换成它的共轭即得: \[{Z_{{L^ - }}} = \frac{{conj\left( w \right) - z}}{{conj\left( w \right) - 1}},{Z_{{M^ - }}} = \frac{z}{{conj\left( w \right)}},{Z_{{N^ - }}} = conj\left( w \right);\]
及\[A{L^ - },B{M^ - },C{N^ - }\]三线的公共交点:\[{Z_F} = \frac{{{\mathop{\rm Im}\nolimits} \left( {\frac{{conj\left( w \right)}}{{conj\left( w \right) - z}}} \right)}}{{{\mathop{\rm Im}\nolimits} \left( {\frac{1}{{conj\left( w \right) - 1}}} \right)}}\left( {\frac{{conj\left( w \right) - z}}{{conj\left( w \right) - 1}}} \right)\]
\[EF\]的距离:\[{d_{EF}} = |{Z_E} - {Z_F}| \times AB;\]
再将\[a = BC,b = CA,c = AB,l = MN,m = NL,n = LM\],
\[z = \frac{{2b}}{{\left( {a + b - c} \right)}} \times \frac{{{{\left( {a + b + c} \right)}^2}\left( {a + b - c} \right)\left( {b + c - a} \right) - 16S_{ABC}^2}}{{{{\left( {{{\left( {b + c} \right)}^2} - {a^2} - 4i{S_{ABC}}} \right)}^2}}}\]
\[w = \frac{{2m}}{{\left( {l + m - n} \right)}} \times \frac{{{{\left( {l + m + n} \right)}^2}\left( {l + m - n} \right)\left( {m + n - l} \right) - 16S_{LMN}^2}}{{{{\left( {{{\left( {m + n} \right)}^2} - {l^2} - 4i{S_{LMN}}} \right)}^2}}}\]
代入化简即可.
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