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# [讨论] 三角形外心坐标公式

 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 $R=\dfrac{\Big|\overrightarrow{AB}\Big|\Big|\overrightarrow{BC}\Big|\Big|\overrightarrow{AC}\Big|}{2\sqrt{p_{\overset{\,}{\triangle}\,\!}\left(p_{\overset{\,}{\triangle}\,\!}-\Big|\overrightarrow{AB}\Big|\right)\left(p_{\overset{\,}{\triangle}\,\!}-\Big|\overrightarrow{BC}\Big|\right)\left(p_{\overset{\,}{\triangle}\,\!}-\Big|\overrightarrow{AC}\Big|\right)}}$

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

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