坐标原点\(o\),三角形外心\(O\)，重心\(G\)，垂心\(H\)，内心\(I\)
来自老早以前的纸笔记本上，不保证正确性
upd 2021-02-01 验证正确性的话，上mma，找到TriangleCenter函数，然后自己算一下，比较一下。或者你看这个
upd 2021-12-28 今天看到一篇知乎文章，补充了复数表示下的一些公式
三角形重心\(G\)
向量
\[\vec{oG}=\frac{1}{3}(\vec{oA}+\vec{oB}+\vec{oC})
\]
直角坐标
\[x_G=\frac{1}{3}({x_1+x_2+x_3})
\]
\[y_G=\frac{1}{3}({y_1+y_2+y_3})
\]
复数
\[G=\frac{A+B+C}{3}
\]
三角形外心\(O\)
向量
\[sin(2A)\cdot\vec{OA}+sin(2B)\cdot\vec{OB}++sin(2C)\cdot\vec{OC}=\vec{0}
\]
\[\vec{oO}=\frac{sin(2A)\cdot\vec{oA}+sin(2B)\cdot\vec{oB}++sin(2C)\cdot\vec{oC}}{sin(2A)+sin(2B)+sin(2C)}
\]
直角坐标
\[\begin{aligned}
x_O& =\frac{
\left|\begin{array}{cccc}
\frac{x_2^2-x_3^2}{2}+\frac{y_2^2-y_3^2}{2} & y_2-y_3 \\
\frac{x_3^2-x_1^2}{2}+\frac{y_3^2-y_1^2}{2} & y_3-y_1 \\
\end{array}\right|
}{
\left|\begin{array}{cccc}
x_2-x_3 & y_2-y_3 \\
x_3-x_1 & y_3-y_1 \\
\end{array}\right|
}
\\
&=\frac{[x_1^2(y_2-y_3)+x_2^2(y_3-y_1)+x_3^2(y_1-y_2)]-(y_1-y_2)(y_2-y_3)(y_3-y_1)}
{2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]}
\end{aligned}
\]
\[\begin{aligned}
y_O&=\frac{
\left|\begin{array}{cccc}
x_2-x_3 & \frac{x_2^2-x_3^2}{2}+\frac{y_2^2-y_3^2}{2} \\
x_3-x_1 & \frac{x_3^2-x_1^2}{2}+\frac{y_3^2-y_1^2}{2} \\
\end{array}\right|
}{
\left|\begin{array}{cccc}
x_2-x_3 & y_2-y_3 \\
x_3-x_1 & y_3-y_1 \\
\end{array}\right|
}
\\
&=\frac{-[y_1^2(x_2-x_3)+y_2^2(x_3-x_1)+y_3^2(x_1-x_2)]+(x_1-x_2)(x_2-x_3)(x_3-x_1)}
{2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]}
\end{aligned}
\]
复数
\[O=\frac{(A B-A C) \bar{A}+(B C-A B) \bar{B}+(A C-B C) \bar{C}}{(B-C) \bar{A}+(C-A) \bar{B}+(A-B) \bar{C}}
\]
三角形垂心\(H\)
向量
\[\vec{0}=tan(A)\cdot \vec{HA}+tan(B)\cdot \vec{HB}+tan(C)\cdot \vec{HC}
\]
\[\vec{oH}=\frac{tan(A)\cdot \vec{oA}+tan(B)\cdot \vec{oB}+tan(C)\cdot \vec{oC}}{tan(A)+tan(B)+tan(C)}
\]
直角坐标
\[x_H=\frac{(y_1-y_2)(y_2-y_3)(y_3-y_1)-[x_1x_2(y_1-y_2)+x_2x_3(y_2-y_3)+x_3x_1(y_3-y_1)]}
{x_3(y_1-y_2)+x_1(y_2-y_3)+x_2(y_3-y_1)}
\]
\[y_H=\frac{-(x_1-x_2)(x_2-x_3)(x_3-x_1)+[y_1y_2(x_1-x_2)+y_2y_3(x_2-x_3)+y_3y_1(x_3-x_1)]}
{x_3(y_1-y_2)+x_1(y_2-y_3)+x_2(y_3-y_1)}
\]
复数
\[H=\frac{(B-C)(B+C-A) \bar{A}+(C-A)(C+A-B) \bar{B}+(A-B)(A+B-C) \bar{C}}{(B-C) \bar{A}+(C-A) \bar{B}+(A-B) \bar{C}}
\]
三角形内心\(I\)
向量
\[\vec{0}=a\cdot \vec{IA}+b\cdot \vec{IB}+c\cdot \vec{IC}
\]
\[\vec{oI}=\frac{a\cdot \vec{oA}+b\cdot \vec{oB}+c\cdot \vec{oC}}{a+b+c}
\]
直角坐标
\[x_I=\frac{ax_A+bx_B+cx_C}{a+b+c}
\]
\[y_I=\frac{ay_A+by_B+cy_C}{a+b+c}
\]
复数
\[I=\frac{|B-C| A+|C-A| B+|A-B| C}{|B-C|+|C-A|+|A-B|}
\]
编辑公式不易，转载请注明出处