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$\newcommand{\bm}{\mathbf}
\newcommand{\pp}{\partial}$
For a triangle $ABC$, assume the points have potential
$\varphi_a,\varphi_b,\varphi_c$. As shown in Fig. ???,
$AB'\perp AC, AC'\perp AB$, combine the components of $\nabla\varphi$,
we have
$$(\nabla\varphi)^2=\frac{1}{\sin^2 A}\left(\frac{\varphi_{ab}^2}{c^2}+\frac{\varphi_{ac}^2}{b^2}-\frac{2\varphi_{ab}\varphi_{ac}\cos A}{bc}\right)$$
The area of triangle is $$S=\frac{bc\sin A}{2}$$
The energy of this triangle is $$\begin{aligned}
E_{\triangle}&\propto S(\nabla\varphi)^2\
&\propto \frac{bc}{\sin A}\left(\frac{\varphi_{ab}^2}{c^2}+\frac{\varphi_{ac}^2}{b^2}-\frac{2\varphi_{ab}\varphi_{ac}\cos A}{bc}\right)\end{aligned}$$
对$\varphi_a$求导得到:
$$\frac{\pp E_\triangle}{\pp \varphi_a}\propto \left(\frac{b}{c\sin A}-\cot A\right)\varphi_{ab}+\left(\frac{c}{b\sin A}-\cot A\right)\varphi_{ac}\label{a}$$
其中:
$$\frac{b}{c\sin A}-\cot A=\frac{b-c\cos A}{c\sin A}=\frac{a\cos C}{a\sin C}=\cot C$$
同理$$\frac{c}{b\sin A}-\cot A=\cot B$$ 故偏导化为
$$\frac{\pp E_\triangle}{\pp \varphi_a}\propto \varphi_{ab}\cot C+\varphi_{ac}\cot B
\label{comp}$$
Stationary point equation for energy minimum
Let $E$ be the total energy and use $i$ to mark all triangles containing
$A$. $B_i, C_i$ are other points in triangle $i$. In the stationary
point, we have
$$\frac{\pp E}{\pp \varphi_a}=\sum_i\frac{\pp E_i}{\pp \varphi_a}
\propto\sum_i \varphi_{ab_i}\cot C_i+\varphi_{ac_i}\cot B_i=0$$ So,
$$\sum_i\left(\cot C_i+\cot B_i\right)\varphi_{a}=\sum_i\left( \varphi_{b_i}\cot C_i+\varphi_{c_i}\cot B_i\right)$$
Assume we know coordinates of a triangle $ABC$. To calculate $\cot A$,
we define $\bm b=\overrightarrow{AB}, \bm c=\overrightarrow{AC}$.
$$\cot A=\frac{bc\cos A}{bc\sin A}=\frac{\bm{b\cdot c}}{|\bm{b\times c}|}=\frac{\bm{b\cdot c}}{2S}$$