三角形的心迹

陈九章

陈九章

陈九章

陈九章

陈九章

陈九章

creasson

设 四点电荷分别为1、2、3、4，各点受力为
\[\left\{ \begin{array}{l}
\mathop {{f_A}}\limits^ \to   = \frac{2}{{A{B^3}}}\mathop {BA}\limits^ \to   + \frac{3}{{A{C^3}}}\mathop {CA}\limits^ \to   + \frac{4}{{A{D^3}}}\mathop {DA}\limits^ \to   \\
\mathop {{f_B}}\limits^ \to   = \frac{2}{{A{B^3}}}\mathop {AB}\limits^ \to   + \frac{6}{{B{C^3}}}\mathop {CB}\limits^ \to   + \frac{8}{{B{D^3}}}\mathop {DB}\limits^ \to   \\
\mathop {{f_C}}\limits^ \to   = \frac{3}{{A{C^3}}}\mathop {CA}\limits^ \to   + \frac{6}{{B{C^3}}}\mathop {BC}\limits^ \to   + \frac{{12}}{{C{D^3}}}\mathop {DC}\limits^ \to   \\
\mathop {{f_D}}\limits^ \to   = \frac{2}{{A{D^3}}}\mathop {DA}\limits^ \to   + \frac{8}{{B{D^3}}}\mathop {BD}\limits^ \to   + \frac{{12}}{{C{D^3}}}\mathop {CD}\limits^ \to   \\
\end{array} \right.\]
又设外接球球心坐标$$O = \alpha A + \beta B + \gamma C + \delta D\left( {\alpha  + \beta  + \gamma  + \delta  = 1} \right)$$
\[\left\{ \begin{array}{l}
\mathop {OA}\limits^ \to   = \beta \mathop {BA}\limits^ \to   + \gamma \mathop {CA}\limits^ \to   + \delta \mathop {DA}\limits^ \to   \\
\mathop {OB}\limits^ \to   = \alpha \mathop {AB}\limits^ \to   + \gamma \mathop {CB}\limits^ \to   + \delta \mathop {DB}\limits^ \to   \\
\mathop {OC}\limits^ \to   = \alpha \mathop {AC}\limits^ \to   + \beta \mathop {BC}\limits^ \to   + \delta \mathop {DC}\limits^ \to   \\
\mathop {OD}\limits^ \to   = \alpha \mathop {AD}\limits^ \to   + \beta \mathop {BD}\limits^ \to   + \gamma \mathop {CD}\limits^ \to   \\
\end{array} \right.\]
由对应的向量方向一致知有方程:
\[\left\{ \begin{array}{l}
\frac{2}{{A{B^3}}} = \lambda \beta ,\frac{3}{{A{C^3}}} = \lambda \gamma ,\frac{4}{{A{D^3}}} = \lambda \delta  \\
\frac{2}{{A{B^3}}} = \mu \alpha ,\frac{6}{{B{C^3}}} = \mu \gamma ,\frac{8}{{B{D^3}}} = \mu \delta  \\
\frac{3}{{A{C^3}}} = v\alpha ,\frac{6}{{B{C^3}}} = v\beta ,\frac{{12}}{{C{D^3}}} = v\delta  \\
\frac{2}{{A{D^3}}} = \eta \alpha ,\frac{8}{{B{D^3}}} = \eta \beta ,\frac{{12}}{{C{D^3}}} = \eta \gamma  \\
\end{array} \right.\]
整理一下，即是
\[\left\{ \begin{array}{l}
\frac{2}{{A{B^3}}} = \lambda \beta  = \mu \alpha  \\
\frac{3}{{A{C^3}}} = \lambda \gamma  = v\alpha  \\
\frac{4}{{A{D^3}}} = \lambda \delta  = \eta \alpha  \\
\frac{6}{{B{C^3}}} = \mu \gamma  = v\beta  \\
\frac{8}{{B{D^3}}} = \mu \delta  = \eta \beta  \\
\frac{{12}}{{C{D^3}}} = v\delta  = \eta \gamma  \\
\end{array} \right.\]
求解得到
\[\lambda  \to \frac{{v\alpha }}{\gamma },\mu  \to \frac{{v\beta }}{\gamma },\eta  \to \frac{{v\delta }}{\gamma }\]
以及
\[\left\{ \begin{array}{l}
\frac{2}{{A{B^3}}} = \frac{{v\alpha \beta }}{\gamma } \\
\frac{3}{{A{C^3}}} = v\alpha  \\
\frac{4}{{A{D^3}}} = \frac{{v\alpha \delta }}{\gamma } \\
\frac{6}{{B{C^3}}} = v\beta  \\
\frac{8}{{B{D^3}}} = \frac{{v\beta \delta }}{\gamma } \\
\frac{{12}}{{C{D^3}}} = v\delta  \\
\end{array} \right.\]
再由距离式：
\[\left\{ \begin{array}{l}
{R^2} = O{A^2} = \beta A{B^2} + \gamma A{C^2} + \delta A{D^2} - T \\
{R^2} = O{B^2} = \alpha A{B^2} + \gamma B{C^2} + \delta B{D^2} - T \\
{R^2} = O{C^2} = \alpha A{C^2} + \beta B{C^2} + \delta C{D^2} - T \\
{R^2} = O{D^2} = \alpha A{D^2} + \beta B{D^2} + \gamma C{D^2} - T \\
\alpha  + \beta  + \gamma  + \delta  = 1 \\
T = \alpha \beta A{B^2} + \alpha \gamma A{C^2} + \alpha \delta A{D^2} + \beta \gamma B{C^2} + \gamma \delta C{D^2} \\
\end{array} \right.\]
这里的T是等于$R^2$的（前四式分别乘$\alpha ,\beta ,\gamma ,\delta$相加得到），将之前的式子代入，得到
\[\left\{ \begin{array}{l}
2{R^2}{v^{2/3}}{\alpha ^{2/3}} = {3^{2/3}}\gamma  + {2^{1/3}}{\gamma ^{2/3}}\left( {{2^{1/3}}{\beta ^{1/3}} + 2{\delta ^{1/3}}} \right) \\
2{R^2}{v^{2/3}}{\beta ^{2/3}} = {6^{2/3}}\gamma  + {\gamma ^{2/3}}\left( {{2^{2/3}}{\alpha ^{1/3}} + 4{\delta ^{1/3}}} \right) \\
2{R^2}{v^{2/3}} = {3^{2/3}}\left( {{\alpha ^{1/3}} + {2^{2/3}}{\beta ^{1/3}} + {{22}^{1/3}}{\delta ^{1/3}}} \right) \\
2{R^2}{v^{2/3}}{\delta ^{2/3}} = 2\left( {{2^{1/3}}{\alpha ^{1/3}} + 2{\beta ^{1/3}}} \right){\gamma ^{2/3}} + {22^{1/3}}{3^{2/3}}\gamma  \\
\alpha  + \beta  + \gamma  + \delta  = 1 \\
\end{array} \right.\]
再设定$R = 1$ ,解得（MATHEMATICA可以得到一个很长的表达式）
AB-> 1.4256763368739924361355721322175095867108448151334
AC-> 1.5122540621851561716385500548056689285120031585503
AD-> 1.6031196050911390297710575855509971279275045671404
BC-> 1.6047533133070553394182617889156565156727700850659
BD-> 1.7011767812216463924741294636051178176363249969131
CD-> 1.8044849531123852919849281349929552833395951480575

creasson

设 \[\mathop {BA}\limits^ \to   = \frac{{\left( {s + t} \right)\left( {1 - st} \right)}}{{s{{\left( {1 - it} \right)}^2}}}\mathop {BC}\limits^ \to  ,\left( {s = \tan \frac{B}{2},t = \tan \frac{A}{2}} \right)\]
\[\mathop {BP}\limits^ \to   = \frac{{\left( {u + v} \right)\left( {1 - uv} \right)}}{{u{{\left( {1 - iv} \right)}^2}}}\mathop {BC}\limits^ \to  ,\left( {u = \tan \frac{{\angle PBC}}{2},v = \tan \frac{{\angle BPC}}{2}} \right)\]
由\[\mathop {PA}\limits^ \to   = \mathop {BA}\limits^ \to   - \mathop {BP}\limits^ \to   = \left( {\frac{{\left( {s + t} \right)\left( {1 - st} \right)}}{{s{{\left( {1 - it} \right)}^2}}} - \frac{{\left( {u + v} \right)\left( {1 - uv} \right)}}{{u{{\left( {1 - iv} \right)}^2}}}} \right)\mathop {BC}\limits^ \to  \]
以及\[\mathop {PB}\limits^ \to   = \lambda \exp \left( {i\angle APB} \right)\mathop {PA}\limits^ \to  ,\left( {\lambda  \in R} \right)\]
知\[ - \frac{1}{\lambda } = \exp \left( { - i\angle APB} \right)\left( {1 - \frac{{\left( {s + t} \right)\left( {1 - st} \right)u{{\left( {1 - iv} \right)}^2}}}{{s{{\left( {1 - it} \right)}^2}\left( {u + v} \right)\left( {1 - uv} \right)}}} \right)\]
取虚部，展开即为：
\[\begin{array}{l}
sv{\left( {1 + {t^2}} \right)^2}\sin \theta \left( {1 - {u^2}} \right) + 2\left( {s + t} \right)\left( {1 - st} \right)\left( {v - t} \right)\left( {1 + vt} \right)u\cos \theta  \\
  + t\left( {1 - s - t - st + v + sv + tv - stv} \right)\left( {1 + s + t - st - v + sv + tv + stv} \right)u\sin \theta  = 0 \\
\end{array}\]
其中 ，\[s = \tan \frac{B}{2},t = \tan \frac{A}{2},v = \tan \frac{{mA + n\frac{\pi }{3}}}{2},\theta  = mC + n\frac{\pi }{3}\]
这是关于$u$的二次方程，可解。于是$P$点轨迹可明确写为$$F\left( {m,n} \right) $$的形式
也可以求出\[\tan \theta  = \frac{{2(s + t)\left( {1 - st} \right)u(v - t)(1 + tv)}}{{s{{(1 + {t^2})}^2}(1 - {u^2})v + tu(1 - s - t - st + v + sv + tv - stv)(1 + s + t - st - v + sv + tv + stv)}}\]
也可以写成这样的形式：
\[\left\{ \begin{array}{l}
\tan \left( {mC + n\frac{\pi }{3}} \right) = \frac{{2(s + t)\left( {1 - st} \right)u(v - t)(1 + tv)}}{{s{{(1 + {t^2})}^2}(1 - {u^2})v + tu(1 - s - t - st + v + sv + tv - stv)(1 + s + t - st - v + sv + tv + stv)}} \\
\tan \left( {mA + n\frac{\pi }{3}} \right) = \frac{{2v}}{{1 - {v^2}}} \\
\end{array} \right.\]

creasson

如果$m + n = 2$且$$\angle B = {\rm{6}}{{\rm{0}}^ \circ }$$,那么
\[\tan \left( {mB + n\frac{\pi }{3}} \right) =  - \tan \left( {mA + \frac{{n\pi }}{3} + mC + \frac{{n\pi }}{3}} \right)\]
此时方程恰可以化为多项式方程，这是可解的情形

creasson

进一步的计算结果是:
\[\mathop {BP}\limits^ \to   = \frac{{2t\left( {s + t} \right)\left( { - 1 + st} \right)\left( {1 + pq} \right) + \left( {s - t + {s^2}t + 6s{t^2} + {t^3} - {s^2}{t^3} + s{t^4}} \right)p + \left( {s + t - {s^2}t - 2s{t^2} - {t^3} + {s^2}{t^3} + s{t^4}} \right)q}}{{s{{\left( {1 + {t^2}} \right)}^2}\left( {1 - iq} \right)}}\mathop {BC}\limits^ \to  \]
其中\[s = \tan \frac{B}{2},t = \tan \frac{A}{2},p = \tan \left( {mC + \frac{{n\pi }}{3}} \right),q = \tan \left( {mA + \frac{{n\pi }}{3}} \right)\]