Why three $60^circ$ corner angles cannot be drawn to make a triangle?

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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 ?







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    curvature$$?
    – Lord Shark the Unknown
    2 days ago










  • That the sum of the angles of a triangle is $pi$ is equivalent to the parallel postulate, hence false in other geometries: math.stackexchange.com/questions/2807341/…
    – Ethan Bolker
    2 days ago














up vote
1
down vote

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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 ?







share|cite|improve this question

















  • 1




    curvature$$?
    – Lord Shark the Unknown
    2 days ago










  • That the sum of the angles of a triangle is $pi$ is equivalent to the parallel postulate, hence false in other geometries: math.stackexchange.com/questions/2807341/…
    – Ethan Bolker
    2 days ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











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 ?







share|cite|improve this question













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 ?









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edited yesterday
























asked 2 days ago









Narasimham

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20.1k51857







  • 1




    curvature$$?
    – Lord Shark the Unknown
    2 days ago










  • That the sum of the angles of a triangle is $pi$ is equivalent to the parallel postulate, hence false in other geometries: math.stackexchange.com/questions/2807341/…
    – Ethan Bolker
    2 days ago












  • 1




    curvature$$?
    – Lord Shark the Unknown
    2 days ago










  • That the sum of the angles of a triangle is $pi$ is equivalent to the parallel postulate, hence false in other geometries: math.stackexchange.com/questions/2807341/…
    – Ethan Bolker
    2 days ago







1




1




curvature$$?
– Lord Shark the Unknown
2 days ago




curvature$$?
– Lord Shark the Unknown
2 days ago












That the sum of the angles of a triangle is $pi$ is equivalent to the parallel postulate, hence false in other geometries: math.stackexchange.com/questions/2807341/…
– Ethan Bolker
2 days ago




That the sum of the angles of a triangle is $pi$ is equivalent to the parallel postulate, hence false in other geometries: math.stackexchange.com/questions/2807341/…
– Ethan Bolker
2 days ago










3 Answers
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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 $K<0$.






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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., $pm2pi$), 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.






    share|cite|improve this answer




























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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.






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        3 Answers
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        3 Answers
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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 $K<0$.






        share|cite|improve this answer

























          up vote
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          down vote













          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 $K<0$.






          share|cite|improve this answer























            up vote
            1
            down vote










            up vote
            1
            down vote









            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 $K<0$.






            share|cite|improve this answer













            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 $K<0$.







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered 2 days ago









            Arthur

            97.9k792173




            97.9k792173




















                up vote
                1
                down vote













                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., $pm2pi$), 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.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  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., $pm2pi$), 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.






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    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., $pm2pi$), 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.






                    share|cite|improve this answer













                    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., $pm2pi$), 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.







                    share|cite|improve this answer













                    share|cite|improve this answer



                    share|cite|improve this answer











                    answered 2 days ago









                    Christian Blatter

                    163k7105305




                    163k7105305




















                        up vote
                        1
                        down vote













                        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.






                        share|cite|improve this answer

























                          up vote
                          1
                          down vote













                          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.






                          share|cite|improve this answer























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            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.






                            share|cite|improve this answer













                            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.







                            share|cite|improve this answer













                            share|cite|improve this answer



                            share|cite|improve this answer











                            answered 2 days ago









                            Takahiro Waki

                            1,985420




                            1,985420






















                                 

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