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On a spherical surface $K=+1$ a curvilinear triangle with $60^{\circ}$ at each vertex cannot be drawn because the spherical excess $ \pi - 3 A =0 $ and so its area vanishes. (By virtue of Gauss-Bonnet theorem).

Similarly on a pseudospherical surface $K=-1$ a curvilinear triangle with $60^{\circ}$ at each vertex cannot be drawn because the pseudospherical defect $3 A - \pi =0 $ and vanish.

Such a construction is impossible however small the size of geodesic triangle be. It requires that the lines should be concurrent when applied to elliptic/hyperbolic situations like a star made by three concurrent geodesic arcs.

What actually prevents formation of such "triangles" of vanishing enclosed area ?

EDIT1:

An attempt to go further in such a search: By drawing three equal geodesic arcs across cuspidal equator $ (K=-1)$ hypo pseudosphere roughly as per sketch we can/may have a zero area enclosed between them surrounding singularity of cone vertex,of course with additional and appropriate interpretation/condition. Corner angles at cuspidal equator is $\pi/3.$

enter image description here

Two related real constants $(b,m)$ arise in geodesic calculation which should be found from a definite integral.

$$\int_{b}^{m} \frac{dr}{r\sqrt{((r/b)^2-1)(1-m^2+r^2)}} = \frac{\pi}{3}$$

Request your help in their determination.

Narasimham
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    curvature${}{}$? – Angina Seng Aug 04 '18 at 05:59
  • That the sum of the angles of a triangle is $\pi$ is equivalent to the parallel postulate, hence false in other geometries: https://math.stackexchange.com/questions/2807341/are-the-proofs-for-the-properties-of-parallel-lines-and-that-a-triangle-has-180/2807600#2807600 – Ethan Bolker Aug 04 '18 at 15:20

3 Answers3

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If you have a line and on that line make two $60^\circ\!$ angles pointed towards eachother (the way you would make an equilateral triangle in a flat plane), the two legs will curve away from one another and meet at an angle smaller than $60^\circ\!\!$, if they meet at all. This is because $K<0$.

Arthur
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Note that you have the same phenomenon on the sphere $S^2$. I'd say that the case $\kappa=0$ of the Euclidean plane is special, in so far as the angle sum of triangles (resp., the total geodesic curvature along a smooth boundary) is $\pi$ (resp., $\pm2\pi$), independent of the triangle or shape in question. This has to do with the fact that in the Euclidean plane we have the group of scalings at our disposal. A scaling changes the areas, but not the total geodesic curvature.

Christian Blatter
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If we push the regular triangle on Euclidean plane, we easily know the triangle's angles are decreased. If we pull that, their angles are increased(for example the spherical triangle). These facts show such a triangle is impossible.

Takahiro Waki
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