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[讨论] 三角形外心坐标公式

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发表于 2019-6-26 13:50:32 | 显示全部楼层
本帖最后由 葡萄糖 于 2019-6-26 16:13 编辑
gxqcn 发表于 2012-7-24 09:47
P1,P2,P3 为三角形三顶点,  

d1 = - (P2–P1)·(P1–P3)    // 此为矢量运算:减法、内积;  
d2 = - (P3–P2)·(P2–P1)    // 所得为标量  
d3 = - (P1–P3)·(P3–P2)      

c1 = d2*d3  
c2 = d3*d1  
c3 = d1*d2  
c = c1 + c2 + c3

则,其外接圆半径及圆心为:  
Radius = 1/2 sqrt( (d1+d2)*(d2+d3)*(d3+d1)/c )   
Center = {(c2+c3)P1 + (c3+c1)P2 + (c1+c2)P3}/2c
...


二维情形
\begin{align*}
\begin{pmatrix}
x_{\overset{\,}P}\\
y_{\overset{\,}P}\\
\end{pmatrix}
=
\dfrac{1}{\begin{pmatrix}
1&1&1
\end{pmatrix}\begin{pmatrix}
\Big|\overrightarrow{BC}\Big|^2\left(\Big|\overrightarrow{AB}\Big|^2+\Big|\overrightarrow{AC}\Big|^2-\Big|\overrightarrow{BC}\Big|^2\right)\\
\Big|\overrightarrow{AC}\Big|^2\left(\Big|\overrightarrow{AB}\Big|^2+\Big|\overrightarrow{BC}\Big|^2-\Big|\overrightarrow{AC}\Big|^2\right)\\
\Big|\overrightarrow{AB}\Big|^2\left(\Big|\overrightarrow{BC}\Big|^2+\Big|\overrightarrow{AC}\Big|^2-\Big|\overrightarrow{AB}\Big|^2\right)\\
\end{pmatrix}\,}
\begin{pmatrix}
x_{\overset{\,}A}&x_{\overset{\,}B}&x_{\overset{\,}C}\\
y_{\overset{\,}A}&y_{\overset{\,}B}&y_{\overset{\,}C}\\
\end{pmatrix}
\begin{pmatrix}
\Big|\overrightarrow{BC}\Big|^2\left(\Big|\overrightarrow{AB}\Big|^2+\Big|\overrightarrow{AC}\Big|^2-\Big|\overrightarrow{BC}\Big|^2\right)\\
\Big|\overrightarrow{AC}\Big|^2\left(\Big|\overrightarrow{AB}\Big|^2+\Big|\overrightarrow{BC}\Big|^2-\Big|\overrightarrow{AC}\Big|^2\right)\\
\Big|\overrightarrow{AB}\Big|^2\left(\Big|\overrightarrow{BC}\Big|^2+\Big|\overrightarrow{AC}\Big|^2-\Big|\overrightarrow{AB}\Big|^2\right)\\
\end{pmatrix}
\end{align*}
推广到三维情形
\begin{align*}
\begin{pmatrix}
x_{\overset{\,}P}\\
y_{\overset{\,}P}\\
z_{\overset{\,}P}
\end{pmatrix}
=
\dfrac{1}{\begin{pmatrix}
1&1&1
\end{pmatrix}\begin{pmatrix}
\Big|\overrightarrow{BC}\Big|^2\left(\Big|\overrightarrow{AB}\Big|^2+\Big|\overrightarrow{AC}\Big|^2-\Big|\overrightarrow{BC}\Big|^2\right)\\
\Big|\overrightarrow{AC}\Big|^2\left(\Big|\overrightarrow{AB}\Big|^2+\Big|\overrightarrow{BC}\Big|^2-\Big|\overrightarrow{AC}\Big|^2\right)\\
\Big|\overrightarrow{AB}\Big|^2\left(\Big|\overrightarrow{BC}\Big|^2+\Big|\overrightarrow{AC}\Big|^2-\Big|\overrightarrow{AB}\Big|^2\right)\\
\end{pmatrix}\,}
\begin{pmatrix}
x_{\overset{\,}A}&x_{\overset{\,}B}&x_{\overset{\,}C}\\
y_{\overset{\,}A}&y_{\overset{\,}B}&y_{\overset{\,}C}\\
z_{\overset{\,}A}&z_{\overset{\,}B}&z_{\overset{\,}C}
\end{pmatrix}
\begin{pmatrix}
\Big|\overrightarrow{BC}\Big|^2\left(\Big|\overrightarrow{AB}\Big|^2+\Big|\overrightarrow{AC}\Big|^2-\Big|\overrightarrow{BC}\Big|^2\right)\\
\Big|\overrightarrow{AC}\Big|^2\left(\Big|\overrightarrow{AB}\Big|^2+\Big|\overrightarrow{BC}\Big|^2-\Big|\overrightarrow{AC}\Big|^2\right)\\
\Big|\overrightarrow{AB}\Big|^2\left(\Big|\overrightarrow{BC}\Big|^2+\Big|\overrightarrow{AC}\Big|^2-\Big|\overrightarrow{AB}\Big|^2\right)\\
\end{pmatrix}
\end{align*}
另外,(可以利用 正弦定理+海伦公式 得到)半径为
\[ R=\dfrac{\Big|\overrightarrow{AB}\Big|\Big|\overrightarrow{BC}\Big|\Big|\overrightarrow{AC}\Big|}{2\sqrt{p_{\overset{\,}{\triangle}\,\!}\left(p_{\overset{\,}{\triangle}\,\!}-\Big|\overrightarrow{AB}\Big|\right)\left(p_{\overset{\,}{\triangle}\,\!}-\Big|\overrightarrow{BC}\Big|\right)\left(p_{\overset{\,}{\triangle}\,\!}-\Big|\overrightarrow{AC}\Big|\right)}} \]
毋因群疑而阻独见  毋任己意而废人言
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 楼主| 发表于 2019-6-26 14:42:28 | 显示全部楼层
我在 3# 用的向量法,通用于二维、三维的情形,并已应用于实际编程项目中
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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