Find differential equation of an (obviously non-geodesic) line on a sphere such that when three such lines intersect the sum of internal angles of the triangle so formed is $\pi$.
EDIT 1:
DE I get is:
$$ a k_{g{1,2,3}} = \tan \gamma_{1,2,3} $$
where $k_g$ is small circle curvature and $\gamma$ is inclination between NS pole line and plane of small circle.
As @achille hui remarks, it suffices to use a stereographic projection $P$ with center the north pole $N$, to establishing a bijective correspondence between the sphere $(\Sigma)$ (minus $N$) and the (horizontal) plane $(\Pi)$ tangent to the sphere at its south pole $S$; in fact, this sterographic projection
preserves angles.
exchanges straight lines in $(\Pi)$ not passing through $S$ on the plane and non-great-circles as hinted by @Andrew D. Hwang (particular cases: lines passing through the South pole $S$ are mapped onto great circles passing through the north pole $N$).
Thus, it suffices to draw a triangle in $(\Pi)$ and use formulas to send it onto a triangular shape on the sphere with point $M(x,y,-1) \in (\Pi)$ mapped onto its image $M'(x',y',z') \in (\Sigma)$ given by formulas:
$$\begin{cases}x'&=&\dfrac{4x}{x^2+y^2+4}\\y'&=&\dfrac{4y}{x^2+y^2+4}\\z'&=&1-\dfrac{8}{x^2+y^2+4}\end{cases}$$
as a consequence of vector formula $\vec{NM'}=\dfrac{4}{NM^2}\vec{NM}.$
Remark: these triangles on the sphere are not usual spherical triangles, of course.
(nice pumpkin, isn't it ?)
Let $\eta : \mathbb{S}^2 \setminus \{N\} \to \mathbb{R}^2 \times \{-1\} $ be the stereographic projection from unit sphere to the plane $z = -1$ with respect to the north pole $N = (0,0,1)$. i.e. $$\eta: \mathbb{S}^2 \ni (x,y,z) \quad\mapsto\quad (u,v,-1) = \left(\frac{2x}{1-z},\frac{2y}{1-z},-1\right) \in \mathbb{R}^2 \times \{-1\}$$
Let $\gamma : (-\epsilon,\epsilon ) \to \mathbb{S}^2$ be any curve on the unit sphere parametrized by arc-length $s$.
For any point $\gamma(s)$ on $\gamma$, we have
$$
\begin{cases}
u' &= \frac{2x'}{1-z} + \frac{2xz'}{(1-z)^2}\\
v' &= \frac{2y'}{1-z} + \frac{2yz'}{(1-z)^2}
\end{cases}
\quad\implies\quad
\sqrt{u'^2 + v'^2} = \frac{2}{1-z}
$$
This means the unit tangent vector for the image of $\gamma$ under $\eta$ is simply
$$\left( x' + \frac{xz'}{1-z}, y' + \frac{yz'}{1-z}\right)$$
One the plane, a straight line is characterized by having a constant unit tangent vector. In order for $\eta\circ\gamma$ to be a straight line, $\gamma$ need to satisfy:
$$ \begin{align} \frac{d}{ds}\left(x' + \frac{xz'}{1-z}\right) &= 0\\ \frac{d}{ds}\left(y' + \frac{yz'}{1-z}\right) &= 0 \end{align} $$
Conversely, if $(x(s),y(s),z(s))$ is a solution of this ODE on $\mathbb{S}^2$, there will be constants $a, b$ such that $$\begin{cases} x' + \frac{xz'}{1-z} &= -b\\ y' + \frac{yz'}{1-z} &= a\\ \end{cases} \quad\implies\quad \phi' + \frac{\phi z'}{1-z} = 0 \quad\iff\quad \left(\frac{\phi}{1-z}\right)' = 0 $$ where $\phi = ax+by$. This means there is another constant $c$ such that $$\frac{\phi}{1-z} = c \quad\iff\quad ax+by + c(z-1) = 0$$ This is the equation for a plane through $N$ and this solution of ODE lies on the intersection of this plane and the unit sphere (i.e a circle) containing $N$.
What I worked out:
Relation between sphere radius $a,$ curvature $k_g$ and angle $ \gamma$ between planes and diameter through three concurrent circles point, the shifted North pole:
$$ a \cdot k_{g1} = \tan \gamma_1 ; \, a\cdot k_{g2} = \tan \gamma_2 ; \, a\cdot k_{g3} = \tan \gamma_3. $$