人教群某几条几何题

tommywong

1.
对于任意三角形如何作这样的内接正三角形，正三角形的顶点与原三角形连线交于一点.


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3.

青青子衿

回复 1# tommywong

3. ……
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tommywong 发表于 2014-8-2 13:28

修改了一下楼主的字母(⊙_⊙?)
\[\begin{gathered}
  \angle CAD = \alpha ,\angle CBD = \beta , \odot {O_{\vartriangle ACD}} =  \odot {O_1}, \odot {O_{\vartriangle BCD}} =  \odot {O_2} \\
  \angle C{O_1}D = 2\angle CAD = 2\alpha ,\angle C{O_2}D = 2\angle CBD = 2\beta \\
  {R_{ \odot {O_1}}} = \frac{{CD}}{{2\sin \alpha }} \\
  {R_{ \odot {O_2}}} = \frac{{CD}}{{2\sin \beta }} \\
  {S_{{{}_{ \odot {O_1}}}}} = \frac{1}{2}\alpha {R_{ \odot {O_1}}}^2 - \frac{1}{2}{R_{ \odot {O_1}}}^2\sin 2\alpha  = \frac{1}{2}\alpha {\left( {\frac{{CD}}{{2\sin \alpha }}} \right)^2} - \frac{1}{2}{\left( {\frac{{CD}}{{2\sin \alpha }}} \right)^2}\sin 2\alpha  \\
  {S_{{{}_{ \odot {O_2}}}}} = \frac{1}{2}\beta {R_{ \odot {O_2}}}^2 - \frac{1}{2}{R_{ \odot {O_2}}}^2\sin 2\beta  = \frac{1}{2}\beta {\left( {\frac{{CD}}{{2\sin \beta }}} \right)^2} - \frac{1}{2}{\left( {\frac{{CD}}{{2\sin \beta }}} \right)^2}\sin 2\beta \\
  {S_{{{}_{ \odot {O_1}}}}} + {S_{{{}_{ \odot {O_2}}}}} \propto {R_{ \odot {O_2}}}^2,{S_{{{}_{ \odot {O_1}}}}} + {S_{{{}_{ \odot {O_2}}}}} \propto C{D^2} \\
\end{gathered} \]
所以最大是取决于\(AC\)或\(BC\)；
最小为\(CD\)垂直于\(AB\)时

青青子衿

回复 1# tommywong

2.
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tommywong 发表于 2014-8-2 13:28

修改了一下楼主的字母(⊙_⊙?)
因为三角形顶点\(A\),内心\(I\),旁心\(I_a\)三点共角平分线，
\[\begin{gathered}
  \therefore \vartriangle AIL \sim \vartriangle A{I_a}R \\
  \therefore \frac{{AI}}{{A{I_a}}} = \frac{{IL}}{{{I_a}R}} = \frac{r}{{{r_a}}} \\
  \therefore \frac{{EI}}{{E{I_a}}} = \frac{{IM}}{{{I_a}S}} = \frac{{IL}}{{{I_a}R}} = \frac{r}{{{r_a}}} = \frac{{AI}}{{A{I_a}}} \Rightarrow \frac{{AI}}{{EI}} = \frac{{A{I_a}}}{{E{I_a}}} \\
  \because \frac{{BF}}{{EF}} = \frac{{A{I_a}}}{{E{I_a}}} \\
  \therefore \frac{{AI}}{{EI}} = \frac{{BF}}{{EF}} \Rightarrow \frac{{BF}}{{EF}}\frac{{EI}}{{AI}} = 1 \\
  \because \frac{{AD}}{{BD}} = 1 \\
  \therefore \frac{{AD}}{{BD}}\frac{{BF}}{{EF}}\frac{{EI}}{{AI}} = 1 \\
\end{gathered} \]
由对\(\vartriangle ABE\)的梅涅劳斯逆定理知命题成立。