Find the area of the hyperbolic (geodesic) triangle with vertices (0, 0),(0, 1) and (1, 0).
0
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I'm not really sure where to start for this question. I'm aware of the Gauss-Bonnet theorem $Area = pi -(A+B+C)$ where $A,B,C$ are the interior angles of the triangle. However I am not sure how to go about calculating these angles given the vertices.
hyperbolic-geometry geodesic noneuclidean-geometry
asked Dec 9 '18 at 22:54
martinhynesonemartinhynesone
367
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show 10 more comments
0
$begingroup$
I'm not really sure where to start for this question. I'm aware of the Gauss-Bonnet theorem $Area = pi -(A+B+C)$ where $A,B,C$ are the interior angles of the triangle. However I am not sure how to go about calculating these angles given the vertices.
hyperbolic-geometry geodesic noneuclidean-geometry
asked Dec 9 '18 at 22:54
martinhynesonemartinhynesone
367
$endgroup$
1
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Where do these points live? Apparently in neither the hyperbolic half-plane nor the hyperbolic disk. So what's your model for the geometry?
$endgroup$
– Ted Shifrin
Dec 9 '18 at 22:59
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@TedShifrin could be a doubly ideal triangle in the disc
$endgroup$
– Will Jagy
Dec 9 '18 at 23:03
$begingroup$
@WillJagy: Indeed, could be ... Could be doubly ideal in the half-plane, too, for that matter.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 23:12
1
$begingroup$
LOL, @Will, I think a piece of paper and pencil will suffice. And knowledge of what curves are the geodesics.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 23:29
1
$begingroup$
Yes, you need a semicircle perpendicular to the $x$-axis passing through $(0,1)$ and $(1,0)$. I believe $pi/2$ is correct.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 0:57
|
show 10 more comments
0
0
0
$begingroup$
I'm not really sure where to start for this question. I'm aware of the Gauss-Bonnet theorem $Area = pi -(A+B+C)$ where $A,B,C$ are the interior angles of the triangle. However I am not sure how to go about calculating these angles given the vertices.
hyperbolic-geometry geodesic noneuclidean-geometry
asked Dec 9 '18 at 22:54
martinhynesonemartinhynesone
367
$endgroup$
I'm not really sure where to start for this question. I'm aware of the Gauss-Bonnet theorem $Area = pi -(A+B+C)$ where $A,B,C$ are the interior angles of the triangle. However I am not sure how to go about calculating these angles given the vertices.
hyperbolic-geometry geodesic noneuclidean-geometry
hyperbolic-geometry geodesic noneuclidean-geometry
asked Dec 9 '18 at 22:54
martinhynesonemartinhynesone
367
asked Dec 9 '18 at 22:54
martinhynesonemartinhynesone
367
asked Dec 9 '18 at 22:54
martinhynesonemartinhynesone
367
asked Dec 9 '18 at 22:54
martinhynesonemartinhynesone
367
asked Dec 9 '18 at 22:54
martinhynesonemartinhynesone
367
367
1
$begingroup$
Where do these points live? Apparently in neither the hyperbolic half-plane nor the hyperbolic disk. So what's your model for the geometry?
$endgroup$
– Ted Shifrin
Dec 9 '18 at 22:59
$begingroup$
@TedShifrin could be a doubly ideal triangle in the disc
$endgroup$
– Will Jagy
Dec 9 '18 at 23:03
$begingroup$
@WillJagy: Indeed, could be ... Could be doubly ideal in the half-plane, too, for that matter.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 23:12
1
$begingroup$
LOL, @Will, I think a piece of paper and pencil will suffice. And knowledge of what curves are the geodesics.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 23:29
1
$begingroup$
Yes, you need a semicircle perpendicular to the $x$-axis passing through $(0,1)$ and $(1,0)$. I believe $pi/2$ is correct.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 0:57
|
show 10 more comments
1
$begingroup$
Where do these points live? Apparently in neither the hyperbolic half-plane nor the hyperbolic disk. So what's your model for the geometry?
$endgroup$
– Ted Shifrin
Dec 9 '18 at 22:59
$begingroup$
@TedShifrin could be a doubly ideal triangle in the disc
$endgroup$
– Will Jagy
Dec 9 '18 at 23:03
$begingroup$
@WillJagy: Indeed, could be ... Could be doubly ideal in the half-plane, too, for that matter.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 23:12
1
$begingroup$
LOL, @Will, I think a piece of paper and pencil will suffice. And knowledge of what curves are the geodesics.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 23:29
1
$begingroup$
Yes, you need a semicircle perpendicular to the $x$-axis passing through $(0,1)$ and $(1,0)$. I believe $pi/2$ is correct.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 0:57
1
1
$begingroup$
Where do these points live? Apparently in neither the hyperbolic half-plane nor the hyperbolic disk. So what's your model for the geometry?
$endgroup$
– Ted Shifrin
Dec 9 '18 at 22:59
$begingroup$
Where do these points live? Apparently in neither the hyperbolic half-plane nor the hyperbolic disk. So what's your model for the geometry?
$endgroup$
– Ted Shifrin
Dec 9 '18 at 22:59
$begingroup$
@TedShifrin could be a doubly ideal triangle in the disc
$endgroup$
– Will Jagy
Dec 9 '18 at 23:03
$begingroup$
@TedShifrin could be a doubly ideal triangle in the disc
$endgroup$
– Will Jagy
Dec 9 '18 at 23:03
$begingroup$
@WillJagy: Indeed, could be ... Could be doubly ideal in the half-plane, too, for that matter.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 23:12
$begingroup$
@WillJagy: Indeed, could be ... Could be doubly ideal in the half-plane, too, for that matter.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 23:12
1
1
$begingroup$
LOL, @Will, I think a piece of paper and pencil will suffice. And knowledge of what curves are the geodesics.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 23:29
$begingroup$
LOL, @Will, I think a piece of paper and pencil will suffice. And knowledge of what curves are the geodesics.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 23:29
1
1
$begingroup$
Yes, you need a semicircle perpendicular to the $x$-axis passing through $(0,1)$ and $(1,0)$. I believe $pi/2$ is correct.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 0:57
$begingroup$
Yes, you need a semicircle perpendicular to the $x$-axis passing through $(0,1)$ and $(1,0)$. I believe $pi/2$ is correct.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 0:57
|
show 10 more comments
0
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1
$begingroup$
Where do these points live? Apparently in neither the hyperbolic half-plane nor the hyperbolic disk. So what's your model for the geometry?
$endgroup$
– Ted Shifrin
Dec 9 '18 at 22:59
$begingroup$
@TedShifrin could be a doubly ideal triangle in the disc
$endgroup$
– Will Jagy
Dec 9 '18 at 23:03
$begingroup$
@WillJagy: Indeed, could be ... Could be doubly ideal in the half-plane, too, for that matter.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 23:12
1
$begingroup$
LOL, @Will, I think a piece of paper and pencil will suffice. And knowledge of what curves are the geodesics.
$endgroup$
– Ted Shifrin
Dec 9 '18 at 23:29
1
$begingroup$
Yes, you need a semicircle perpendicular to the $x$-axis passing through $(0,1)$ and $(1,0)$. I believe $pi/2$ is correct.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 0:57