Calculating side in quadrilateral in Poincare disc
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If I have a quadrilateral ABCD inside the Poincaré disc such that $angle A=angle B=frac{2pi}{3}$, $AD=BC$ and we know the hyperbolic lengths of sides $AB$ and $CD$, how can I calculate the hyperbolic length of $AD$ in terms of $AB$ and $CD$?
hyperbolic-geometry
asked Dec 7 '18 at 18:35
user164226user164226
31
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add a comment |
$begingroup$
If I have a quadrilateral ABCD inside the Poincaré disc such that $angle A=angle B=frac{2pi}{3}$, $AD=BC$ and we know the hyperbolic lengths of sides $AB$ and $CD$, how can I calculate the hyperbolic length of $AD$ in terms of $AB$ and $CD$?
hyperbolic-geometry
asked Dec 7 '18 at 18:35
user164226user164226
31
$endgroup$
$begingroup$
Hint: Draw the common perpendicular bisector of $AB,CD$ and an obvious Saccheri quadrilateral.
$endgroup$
– user10354138
Dec 7 '18 at 18:51
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Well, @user10354138, the new quadrilaterals gotten by drawing the bisector are not Saccheri quadrilaterals, since they themselves have no symmetry. Do you see a way of using known formulas for this problem?
$endgroup$
– Lubin
Dec 7 '18 at 19:22
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Yes, the quadrilateral formed by AD is not Saccheri, but there is a (actually two) Saccheri screaming out
$endgroup$
– user10354138
Dec 7 '18 at 20:15
add a comment |
$begingroup$
If I have a quadrilateral ABCD inside the Poincaré disc such that $angle A=angle B=frac{2pi}{3}$, $AD=BC$ and we know the hyperbolic lengths of sides $AB$ and $CD$, how can I calculate the hyperbolic length of $AD$ in terms of $AB$ and $CD$?
hyperbolic-geometry
asked Dec 7 '18 at 18:35
user164226user164226
31
$endgroup$
If I have a quadrilateral ABCD inside the Poincaré disc such that $angle A=angle B=frac{2pi}{3}$, $AD=BC$ and we know the hyperbolic lengths of sides $AB$ and $CD$, how can I calculate the hyperbolic length of $AD$ in terms of $AB$ and $CD$?
hyperbolic-geometry
hyperbolic-geometry
asked Dec 7 '18 at 18:35
user164226user164226
31
asked Dec 7 '18 at 18:35
user164226user164226
31
asked Dec 7 '18 at 18:35
user164226user164226
31
asked Dec 7 '18 at 18:35
user164226user164226
31
asked Dec 7 '18 at 18:35
user164226user164226
31
31
$begingroup$
Hint: Draw the common perpendicular bisector of $AB,CD$ and an obvious Saccheri quadrilateral.
$endgroup$
– user10354138
Dec 7 '18 at 18:51
$begingroup$
Well, @user10354138, the new quadrilaterals gotten by drawing the bisector are not Saccheri quadrilaterals, since they themselves have no symmetry. Do you see a way of using known formulas for this problem?
$endgroup$
– Lubin
Dec 7 '18 at 19:22
$begingroup$
Yes, the quadrilateral formed by AD is not Saccheri, but there is a (actually two) Saccheri screaming out
$endgroup$
– user10354138
Dec 7 '18 at 20:15
add a comment |
$begingroup$
Hint: Draw the common perpendicular bisector of $AB,CD$ and an obvious Saccheri quadrilateral.
$endgroup$
– user10354138
Dec 7 '18 at 18:51
$begingroup$
Well, @user10354138, the new quadrilaterals gotten by drawing the bisector are not Saccheri quadrilaterals, since they themselves have no symmetry. Do you see a way of using known formulas for this problem?
$endgroup$
– Lubin
Dec 7 '18 at 19:22
$begingroup$
Yes, the quadrilateral formed by AD is not Saccheri, but there is a (actually two) Saccheri screaming out
$endgroup$
– user10354138
Dec 7 '18 at 20:15
$begingroup$
Hint: Draw the common perpendicular bisector of $AB,CD$ and an obvious Saccheri quadrilateral.
$endgroup$
– user10354138
Dec 7 '18 at 18:51
$begingroup$
Hint: Draw the common perpendicular bisector of $AB,CD$ and an obvious Saccheri quadrilateral.
$endgroup$
– user10354138
Dec 7 '18 at 18:51
$begingroup$
Well, @user10354138, the new quadrilaterals gotten by drawing the bisector are not Saccheri quadrilaterals, since they themselves have no symmetry. Do you see a way of using known formulas for this problem?
$endgroup$
– Lubin
Dec 7 '18 at 19:22
$begingroup$
Well, @user10354138, the new quadrilaterals gotten by drawing the bisector are not Saccheri quadrilaterals, since they themselves have no symmetry. Do you see a way of using known formulas for this problem?
$endgroup$
– Lubin
Dec 7 '18 at 19:22
$begingroup$
Yes, the quadrilateral formed by AD is not Saccheri, but there is a (actually two) Saccheri screaming out
$endgroup$
– user10354138
Dec 7 '18 at 20:15
$begingroup$
Yes, the quadrilateral formed by AD is not Saccheri, but there is a (actually two) Saccheri screaming out
$endgroup$
– user10354138
Dec 7 '18 at 20:15
add a comment |
1 Answer
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As @user10354138 has pointed out, your figure is symmetrical, and $AB$, $CD$ have a common perpendicular bisector. I think that in the good situation, you can come to a solution of your problem, but I don’t know how to handle the bad situation.
The good situation occurs when $AD$ and $BC$ intersect, say at a point $U$. In the figure, I’ve drawn the common perpendicular bisector: of $AB$ at $V$ and of $CD$ at $W$. We have two right hyperbolic triangles: $triangle AUV$ and $triangle DUW$. The common angle at $U$ I’ve denoted $theta$. Since we know $angle DAV=120^circ$, we have $angle UAV=60^circ$. If I’ve not mistaken my formulas, the relations among these are: $costheta=cosh asin60^circ$ and $sintheta=sinh a/sinh v=sinh b/sinh w$, where $v$ is the length of $UA$ and $w$ is the length of $UD$. (The first equation says that this good situation occurs when $cosh a<2/sqrt3$, in other words, when $a<0.549$ roughly.) Of course you want $w-v$. I haven’t tried to combine all these for a possible simplification that might suggest a solution in the bad situation. I hope this has helped.
answered Dec 7 '18 at 21:56
LubinLubin
44.1k44585
$endgroup$
$begingroup$
thanks Lubin. But in the case the lines do not intersect we can not do the argument you said
$endgroup$
– user164226
Dec 8 '18 at 22:01
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Precisely. That’s what made the good case good.I have no idea how to handle the other case.
$endgroup$
– Lubin
Dec 8 '18 at 22:12
add a comment |
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$begingroup$
As @user10354138 has pointed out, your figure is symmetrical, and $AB$, $CD$ have a common perpendicular bisector. I think that in the good situation, you can come to a solution of your problem, but I don’t know how to handle the bad situation.
The good situation occurs when $AD$ and $BC$ intersect, say at a point $U$. In the figure, I’ve drawn the common perpendicular bisector: of $AB$ at $V$ and of $CD$ at $W$. We have two right hyperbolic triangles: $triangle AUV$ and $triangle DUW$. The common angle at $U$ I’ve denoted $theta$. Since we know $angle DAV=120^circ$, we have $angle UAV=60^circ$. If I’ve not mistaken my formulas, the relations among these are: $costheta=cosh asin60^circ$ and $sintheta=sinh a/sinh v=sinh b/sinh w$, where $v$ is the length of $UA$ and $w$ is the length of $UD$. (The first equation says that this good situation occurs when $cosh a<2/sqrt3$, in other words, when $a<0.549$ roughly.) Of course you want $w-v$. I haven’t tried to combine all these for a possible simplification that might suggest a solution in the bad situation. I hope this has helped.
answered Dec 7 '18 at 21:56
LubinLubin
44.1k44585
$endgroup$
$begingroup$
thanks Lubin. But in the case the lines do not intersect we can not do the argument you said
$endgroup$
– user164226
Dec 8 '18 at 22:01
$begingroup$
Precisely. That’s what made the good case good.I have no idea how to handle the other case.
$endgroup$
– Lubin
Dec 8 '18 at 22:12
add a comment |
$begingroup$
As @user10354138 has pointed out, your figure is symmetrical, and $AB$, $CD$ have a common perpendicular bisector. I think that in the good situation, you can come to a solution of your problem, but I don’t know how to handle the bad situation.
The good situation occurs when $AD$ and $BC$ intersect, say at a point $U$. In the figure, I’ve drawn the common perpendicular bisector: of $AB$ at $V$ and of $CD$ at $W$. We have two right hyperbolic triangles: $triangle AUV$ and $triangle DUW$. The common angle at $U$ I’ve denoted $theta$. Since we know $angle DAV=120^circ$, we have $angle UAV=60^circ$. If I’ve not mistaken my formulas, the relations among these are: $costheta=cosh asin60^circ$ and $sintheta=sinh a/sinh v=sinh b/sinh w$, where $v$ is the length of $UA$ and $w$ is the length of $UD$. (The first equation says that this good situation occurs when $cosh a<2/sqrt3$, in other words, when $a<0.549$ roughly.) Of course you want $w-v$. I haven’t tried to combine all these for a possible simplification that might suggest a solution in the bad situation. I hope this has helped.
answered Dec 7 '18 at 21:56
LubinLubin
44.1k44585
$endgroup$
$begingroup$
thanks Lubin. But in the case the lines do not intersect we can not do the argument you said
$endgroup$
– user164226
Dec 8 '18 at 22:01
$begingroup$
Precisely. That’s what made the good case good.I have no idea how to handle the other case.
$endgroup$
– Lubin
Dec 8 '18 at 22:12
add a comment |
$begingroup$
As @user10354138 has pointed out, your figure is symmetrical, and $AB$, $CD$ have a common perpendicular bisector. I think that in the good situation, you can come to a solution of your problem, but I don’t know how to handle the bad situation.
The good situation occurs when $AD$ and $BC$ intersect, say at a point $U$. In the figure, I’ve drawn the common perpendicular bisector: of $AB$ at $V$ and of $CD$ at $W$. We have two right hyperbolic triangles: $triangle AUV$ and $triangle DUW$. The common angle at $U$ I’ve denoted $theta$. Since we know $angle DAV=120^circ$, we have $angle UAV=60^circ$. If I’ve not mistaken my formulas, the relations among these are: $costheta=cosh asin60^circ$ and $sintheta=sinh a/sinh v=sinh b/sinh w$, where $v$ is the length of $UA$ and $w$ is the length of $UD$. (The first equation says that this good situation occurs when $cosh a<2/sqrt3$, in other words, when $a<0.549$ roughly.) Of course you want $w-v$. I haven’t tried to combine all these for a possible simplification that might suggest a solution in the bad situation. I hope this has helped.
answered Dec 7 '18 at 21:56
LubinLubin
44.1k44585
$endgroup$
As @user10354138 has pointed out, your figure is symmetrical, and $AB$, $CD$ have a common perpendicular bisector. I think that in the good situation, you can come to a solution of your problem, but I don’t know how to handle the bad situation.
The good situation occurs when $AD$ and $BC$ intersect, say at a point $U$. In the figure, I’ve drawn the common perpendicular bisector: of $AB$ at $V$ and of $CD$ at $W$. We have two right hyperbolic triangles: $triangle AUV$ and $triangle DUW$. The common angle at $U$ I’ve denoted $theta$. Since we know $angle DAV=120^circ$, we have $angle UAV=60^circ$. If I’ve not mistaken my formulas, the relations among these are: $costheta=cosh asin60^circ$ and $sintheta=sinh a/sinh v=sinh b/sinh w$, where $v$ is the length of $UA$ and $w$ is the length of $UD$. (The first equation says that this good situation occurs when $cosh a<2/sqrt3$, in other words, when $a<0.549$ roughly.) Of course you want $w-v$. I haven’t tried to combine all these for a possible simplification that might suggest a solution in the bad situation. I hope this has helped.
answered Dec 7 '18 at 21:56
LubinLubin
44.1k44585
answered Dec 7 '18 at 21:56
LubinLubin
44.1k44585
answered Dec 7 '18 at 21:56
LubinLubin
44.1k44585
answered Dec 7 '18 at 21:56
LubinLubin
44.1k44585
44.1k44585
$begingroup$
thanks Lubin. But in the case the lines do not intersect we can not do the argument you said
$endgroup$
– user164226
Dec 8 '18 at 22:01
$begingroup$
Precisely. That’s what made the good case good.I have no idea how to handle the other case.
$endgroup$
– Lubin
Dec 8 '18 at 22:12
add a comment |
$begingroup$
thanks Lubin. But in the case the lines do not intersect we can not do the argument you said
$endgroup$
– user164226
Dec 8 '18 at 22:01
$begingroup$
Precisely. That’s what made the good case good.I have no idea how to handle the other case.
$endgroup$
– Lubin
Dec 8 '18 at 22:12
$begingroup$
thanks Lubin. But in the case the lines do not intersect we can not do the argument you said
$endgroup$
– user164226
Dec 8 '18 at 22:01
$begingroup$
thanks Lubin. But in the case the lines do not intersect we can not do the argument you said
$endgroup$
– user164226
Dec 8 '18 at 22:01
$begingroup$
Precisely. That’s what made the good case good.I have no idea how to handle the other case.
$endgroup$
– Lubin
Dec 8 '18 at 22:12
$begingroup$
Precisely. That’s what made the good case good.I have no idea how to handle the other case.
$endgroup$
– Lubin
Dec 8 '18 at 22:12
add a comment |
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$begingroup$
Hint: Draw the common perpendicular bisector of $AB,CD$ and an obvious Saccheri quadrilateral.
$endgroup$
– user10354138
Dec 7 '18 at 18:51
$begingroup$
Well, @user10354138, the new quadrilaterals gotten by drawing the bisector are not Saccheri quadrilaterals, since they themselves have no symmetry. Do you see a way of using known formulas for this problem?
$endgroup$
– Lubin
Dec 7 '18 at 19:22
$begingroup$
Yes, the quadrilateral formed by AD is not Saccheri, but there is a (actually two) Saccheri screaming out
$endgroup$
– user10354138
Dec 7 '18 at 20:15