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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 ?
differential-geometry
asked 2 days ago
Narasimham
20.1k51857
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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 ?
differential-geometry
asked 2 days ago
Narasimham
20.1k51857
1
curvature$$?
– Lord Shark the Unknown
2 days ago
That the sum of the angles of a triangle is $pi$ is equivalent to the parallel postulate, hence false in other geometries: math.stackexchange.com/questions/2807341/…
– Ethan Bolker
2 days ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
On a spherical surface $K=+1$ a curvilinear triangle with $60^circ$ at each vertex cannot be drawn because the spherical excess $ pi - 3 A =0 $ and so its area vanishes. (By virtue of Gauss-Bonnet theorem).
Similarly on a pseudospherical surface $K=-1$ a curvilinear triangle with $60^circ$ at each vertex cannot be drawn because the pseudospherical defect $3 A - pi =0 $ and vanish.
Such a construction is impossible however small the size of geodesic triangle be. It requires that the lines should be concurrent when applied to elliptic/hyperbolic situations like a star made by three concurrent geodesic arcs.
What actually prevents formation of such "triangles" of vanishing enclosed area ?
differential-geometry
asked 2 days ago
Narasimham
20.1k51857
On a spherical surface $K=+1$ a curvilinear triangle with $60^circ$ at each vertex cannot be drawn because the spherical excess $ pi - 3 A =0 $ and so its area vanishes. (By virtue of Gauss-Bonnet theorem).
Similarly on a pseudospherical surface $K=-1$ a curvilinear triangle with $60^circ$ at each vertex cannot be drawn because the pseudospherical defect $3 A - pi =0 $ and vanish.
Such a construction is impossible however small the size of geodesic triangle be. It requires that the lines should be concurrent when applied to elliptic/hyperbolic situations like a star made by three concurrent geodesic arcs.
What actually prevents formation of such "triangles" of vanishing enclosed area ?
differential-geometry
asked 2 days ago
Narasimham
20.1k51857
edited yesterday
asked 2 days ago
Narasimham
20.1k51857
asked 2 days ago
Narasimham
20.1k51857
asked 2 days ago
Narasimham
20.1k51857
20.1k51857
1
curvature$$?
– 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
add a comment |Â
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
add a comment |Â
3 Answers
3
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1
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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$.
answered 2 days ago
Arthur
97.9k792173
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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.
answered 2 days ago
Christian Blatter
163k7105305
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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.
answered 2 days ago
Takahiro Waki
1,985420
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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$.
answered 2 days ago
Arthur
97.9k792173
add a comment |Â
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$.
answered 2 days ago
Arthur
97.9k792173
add a comment |Â
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$.
answered 2 days ago
Arthur
97.9k792173
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$.
answered 2 days ago
Arthur
97.9k792173
answered 2 days ago
Arthur
97.9k792173
answered 2 days ago
Arthur
97.9k792173
answered 2 days ago
Arthur
97.9k792173
97.9k792173
add a comment |Â
add a comment |Â
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.
answered 2 days ago
Christian Blatter
163k7105305
add a comment |Â
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.
answered 2 days ago
Christian Blatter
163k7105305
add a comment |Â
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.
answered 2 days ago
Christian Blatter
163k7105305
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.
answered 2 days ago
Christian Blatter
163k7105305
answered 2 days ago
Christian Blatter
163k7105305
answered 2 days ago
Christian Blatter
163k7105305
answered 2 days ago
Christian Blatter
163k7105305
163k7105305
add a comment |Â
add a comment |Â
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.
answered 2 days ago
Takahiro Waki
1,985420
add a comment |Â
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.
answered 2 days ago
Takahiro Waki
1,985420
add a comment |Â
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.
answered 2 days ago
Takahiro Waki
1,985420
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.
answered 2 days ago
Takahiro Waki
1,985420
answered 2 days ago
Takahiro Waki
1,985420
answered 2 days ago
Takahiro Waki
1,985420
answered 2 days ago
Takahiro Waki
1,985420
1,985420
add a comment |Â
add a comment |Â
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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