Can we find a common differential equation of three lines on a sphere which form a curvilinear scalene triangle with internal angle sum $=\pi ?$ (I.e., no spherical excess.)

By common differential equation we can ensure the same character of curvature for its sides.

( A special case of constant geodesic curvature i.e., circles of a particular radius can always be chosen to cut meridians symmetrically at $ 30^{\circ}$ forming an equilateral triangle of angle sum $\pi $ but a general situation is sought).

The question troubled me for quite some time ...


Earlier answered by achille hui to same question from me.

Three concurrent small circles on a sphere intersect to enclose a curvilinear triangle of three arcs with sum of vertex angles $\pi$ , or zero "spherical excess".

This can be shown in stereographic projection where point of concurrency is the North Pole... which may be shifted to any other point on the sphere.


The picture shows original lines and their three small circles ( Blue,Green,Magenta) passing through North Pole and intersection points determine vertices of a triangle with angle sum $\pi$

A differential equation of small circles is requested. For time being top view is made on inversion basis trig inverting respect to $M$ the bowl or mirror of inversion, obeying relations:

$$ NP\cdot NQ= NS^2= NR^2= NM^2= 4a^2 $$

$$ 1/u^2= 1/\rho^2+ 1/(4a^2),\, u= \frac{2 a \rho}{\sqrt{4a^2+\rho^2} },\, r_1= u \sin\alpha=\frac{4 a^2 \rho}{4a^2+\rho^2 }, \, r=r_1\sec (\theta+ \beta) $$

where $\beta$ is longitude.

Pole view of 3 cutting small circles at NP


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