Arnold's Trivium problem 69

Question

Does anyone have solution for Arnold's Trivium problem #69? Prove that the solid angle based on a given closed contour is a function of the vertex of the angle that is harmonic outside of the contour.

Answer 1

There might be better / more visual solutions, but: Choose a surface $\Sigma$ bounded by the contour, then your function is $$f(\vec x) = \int_\Sigma \frac{\vec n(\vec z)\cdot (\vec x-\vec z)}{\Vert \vec x-\vec z\Vert^{-2}} dS(\vec z)$$ where $\vec n(\vec z)$ is the unit normal vector to $\Sigma$ at $\vec z\in\Sigma$ (since $\frac{\vec n(\vec z)\cdot (\vec x-\vec z)}{\Vert \vec x-\vec z\Vert^{-2}} dS(\vec z)$ is the solid angle based on the surface element $dS$). The integrant $\frac{\vec n\cdot (\vec x-\vec z)}{\Vert \vec x-\vec z\Vert^{-2}}$ is a harmonic function of $\vec x$ because it's the dipole potential; alternatively, it is harmonic since $1/\Vert\vec x\Vert$ is harmonic and since $$\partial_{\vec n} \frac1{\Vert\vec x\Vert} = \frac{\vec n\cdot \vec x}{\Vert \vec x\Vert^{-2}}.$$ Hence $f$ is harmonic.