Mixing
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A problem with two interlinked right triangles is driving me crazy
Clash Royale CLAN TAG#URR8PPP
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This is a problem that is driving me crazy:
$F$ is the center of the circle.
$AhatBC$ is a right angle.
$AhatBF$ is a right angle.
$DhatBF$ is a right angle.
$ChatEF$ is a right angle.
The length $overlineBF$ is known.
The angle $alpha=BhatAC$ is equal to the angle $BhatFD$ and it is known to be different from zero.
Given $alpha$ and $overlineBF$ is it possible to compute $overlineEF$?
The only thing I can write, apart from a bunch of equations deriving from the Pythagorean theorem like $overlineCF^2=overlineCE^2+overlineEF^2$, is
$overlineDF=fracoverlineBFcosalpha$
but I need $overlineEFltoverlineDF$ and not $overlineDF$.
I really can't see the solution...
Assuming some more info: given $alpha$ and $overlineBF$ and $overlineAB$ is it possible to compute $overlineEF$?
geometry trigonometry triangle
asked Aug 3 at 12:50
Alessandro Jacopson
3801421
add a comment |Â
up vote
2
down vote
favorite
This is a problem that is driving me crazy:
$F$ is the center of the circle.
$AhatBC$ is a right angle.
$AhatBF$ is a right angle.
$DhatBF$ is a right angle.
$ChatEF$ is a right angle.
The length $overlineBF$ is known.
The angle $alpha=BhatAC$ is equal to the angle $BhatFD$ and it is known to be different from zero.
Given $alpha$ and $overlineBF$ is it possible to compute $overlineEF$?
The only thing I can write, apart from a bunch of equations deriving from the Pythagorean theorem like $overlineCF^2=overlineCE^2+overlineEF^2$, is
$overlineDF=fracoverlineBFcosalpha$
but I need $overlineEFltoverlineDF$ and not $overlineDF$.
I really can't see the solution...
Assuming some more info: given $alpha$ and $overlineBF$ and $overlineAB$ is it possible to compute $overlineEF$?
geometry trigonometry triangle
asked Aug 3 at 12:50
Alessandro Jacopson
3801421
The points $B,C,D$, and $E$ are all the same, $alpha=0$, and $EF=BF$.
– spiralstotheleft
Aug 3 at 12:55
@spiralstotheleft I updated the question, I forgot to write that $alphaneq0$
– Alessandro Jacopson
Aug 3 at 12:58
1
No, you didn't. If all other conditions are satisfied, they imply $angle DFB=0$.
– spiralstotheleft
Aug 3 at 12:59
@spiralstotheleft I beg your pardon but I am not able to be convinced by your statement.
– Alessandro Jacopson
Aug 3 at 13:03
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is a problem that is driving me crazy:
$F$ is the center of the circle.
$AhatBC$ is a right angle.
$AhatBF$ is a right angle.
$DhatBF$ is a right angle.
$ChatEF$ is a right angle.
The length $overlineBF$ is known.
The angle $alpha=BhatAC$ is equal to the angle $BhatFD$ and it is known to be different from zero.
Given $alpha$ and $overlineBF$ is it possible to compute $overlineEF$?
The only thing I can write, apart from a bunch of equations deriving from the Pythagorean theorem like $overlineCF^2=overlineCE^2+overlineEF^2$, is
$overlineDF=fracoverlineBFcosalpha$
but I need $overlineEFltoverlineDF$ and not $overlineDF$.
I really can't see the solution...
Assuming some more info: given $alpha$ and $overlineBF$ and $overlineAB$ is it possible to compute $overlineEF$?
geometry trigonometry triangle
asked Aug 3 at 12:50
Alessandro Jacopson
3801421
This is a problem that is driving me crazy:
$F$ is the center of the circle.
$AhatBC$ is a right angle.
$AhatBF$ is a right angle.
$DhatBF$ is a right angle.
$ChatEF$ is a right angle.
The length $overlineBF$ is known.
The angle $alpha=BhatAC$ is equal to the angle $BhatFD$ and it is known to be different from zero.
Given $alpha$ and $overlineBF$ is it possible to compute $overlineEF$?
The only thing I can write, apart from a bunch of equations deriving from the Pythagorean theorem like $overlineCF^2=overlineCE^2+overlineEF^2$, is
$overlineDF=fracoverlineBFcosalpha$
but I need $overlineEFltoverlineDF$ and not $overlineDF$.
I really can't see the solution...
Assuming some more info: given $alpha$ and $overlineBF$ and $overlineAB$ is it possible to compute $overlineEF$?
geometry trigonometry triangle
asked Aug 3 at 12:50
Alessandro Jacopson
3801421
edited Aug 3 at 12:57
asked Aug 3 at 12:50
Alessandro Jacopson
3801421
asked Aug 3 at 12:50
Alessandro Jacopson
3801421
asked Aug 3 at 12:50
Alessandro Jacopson
3801421
3801421
The points $B,C,D$, and $E$ are all the same, $alpha=0$, and $EF=BF$.
– spiralstotheleft
Aug 3 at 12:55
@spiralstotheleft I updated the question, I forgot to write that $alphaneq0$
– Alessandro Jacopson
Aug 3 at 12:58
1
No, you didn't. If all other conditions are satisfied, they imply $angle DFB=0$.
– spiralstotheleft
Aug 3 at 12:59
@spiralstotheleft I beg your pardon but I am not able to be convinced by your statement.
– Alessandro Jacopson
Aug 3 at 13:03
add a comment |Â
The points $B,C,D$, and $E$ are all the same, $alpha=0$, and $EF=BF$.
– spiralstotheleft
Aug 3 at 12:55
@spiralstotheleft I updated the question, I forgot to write that $alphaneq0$
– Alessandro Jacopson
Aug 3 at 12:58
1
No, you didn't. If all other conditions are satisfied, they imply $angle DFB=0$.
– spiralstotheleft
Aug 3 at 12:59
@spiralstotheleft I beg your pardon but I am not able to be convinced by your statement.
– Alessandro Jacopson
Aug 3 at 13:03
The points $B,C,D$, and $E$ are all the same, $alpha=0$, and $EF=BF$.
– spiralstotheleft
Aug 3 at 12:55
The points $B,C,D$, and $E$ are all the same, $alpha=0$, and $EF=BF$.
– spiralstotheleft
Aug 3 at 12:55
@spiralstotheleft I updated the question, I forgot to write that $alphaneq0$
– Alessandro Jacopson
Aug 3 at 12:58
@spiralstotheleft I updated the question, I forgot to write that $alphaneq0$
– Alessandro Jacopson
Aug 3 at 12:58
1
1
No, you didn't. If all other conditions are satisfied, they imply $angle DFB=0$.
– spiralstotheleft
Aug 3 at 12:59
No, you didn't. If all other conditions are satisfied, they imply $angle DFB=0$.
– spiralstotheleft
Aug 3 at 12:59
@spiralstotheleft I beg your pardon but I am not able to be convinced by your statement.
– Alessandro Jacopson
Aug 3 at 13:03
@spiralstotheleft I beg your pardon but I am not able to be convinced by your statement.
– Alessandro Jacopson
Aug 3 at 13:03
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
The data $alpha$ and $s:=|BF|$ alone do not determine $t:=|EF|$. In order to see this, let $M$ be the midpoint of $AF$. Then $M$ is the center of a circle with diameter $AF$ and containing $B$, $E$ on its periphery. Since we are not told the angle $angle(BFA)$, or the diameter $|AF|$ of this circle, the knowledge of $s$ ad $alpha$ is insufficient to determine $t$.
answered Aug 3 at 17:50
Christian Blatter
163k7105305
Thank you for your answer, for me $angle(BFA)=AhatBF$ and $AhatBF$ is a right angle. Furthermore I can' t see why $B$ and $E$ should lay on the circumference centered at the midpoint of $AF$ and with diameter equal to $AF$.
– Alessandro Jacopson
2 days ago
Angle $angle(BFA)$ is an angle measured at $F$, and it is obviously not $90^circ$. My point $M$ is the center of aThales circle.
– Christian Blatter
2 days ago
add a comment |Â
up vote
0
down vote
I have simplified your drawing a bit - so, the angle at B is right, and the angle a E is also right. If you choose point p so that the angle there is right, then the distance from F to p is $|BF| cdot cos(v)$, and the distance from D to p is $|BF| cdot sin(v)$, since the angl DBp must be v: the sum of angles in a triangle is 180, BDp shares an angle at D, and both have a right angle. So there you are: $|DF| = |BF| cdot (sin(v) + cos(v))$
answered Aug 3 at 13:32
j4nd3r53n
42039
Thank you but I need to compute $overlineEF$ and not $overlineDF$.
– Alessandro Jacopson
Aug 3 at 14:18
Well, I realised that my answer was wrong, too, even if you were after $|DF|$
– j4nd3r53n
Aug 3 at 14:39
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The data $alpha$ and $s:=|BF|$ alone do not determine $t:=|EF|$. In order to see this, let $M$ be the midpoint of $AF$. Then $M$ is the center of a circle with diameter $AF$ and containing $B$, $E$ on its periphery. Since we are not told the angle $angle(BFA)$, or the diameter $|AF|$ of this circle, the knowledge of $s$ ad $alpha$ is insufficient to determine $t$.
answered Aug 3 at 17:50
Christian Blatter
163k7105305
Thank you for your answer, for me $angle(BFA)=AhatBF$ and $AhatBF$ is a right angle. Furthermore I can' t see why $B$ and $E$ should lay on the circumference centered at the midpoint of $AF$ and with diameter equal to $AF$.
– Alessandro Jacopson
2 days ago
Angle $angle(BFA)$ is an angle measured at $F$, and it is obviously not $90^circ$. My point $M$ is the center of aThales circle.
– Christian Blatter
2 days ago
add a comment |Â
up vote
2
down vote
The data $alpha$ and $s:=|BF|$ alone do not determine $t:=|EF|$. In order to see this, let $M$ be the midpoint of $AF$. Then $M$ is the center of a circle with diameter $AF$ and containing $B$, $E$ on its periphery. Since we are not told the angle $angle(BFA)$, or the diameter $|AF|$ of this circle, the knowledge of $s$ ad $alpha$ is insufficient to determine $t$.
answered Aug 3 at 17:50
Christian Blatter
163k7105305
Thank you for your answer, for me $angle(BFA)=AhatBF$ and $AhatBF$ is a right angle. Furthermore I can' t see why $B$ and $E$ should lay on the circumference centered at the midpoint of $AF$ and with diameter equal to $AF$.
– Alessandro Jacopson
2 days ago
Angle $angle(BFA)$ is an angle measured at $F$, and it is obviously not $90^circ$. My point $M$ is the center of aThales circle.
– Christian Blatter
2 days ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The data $alpha$ and $s:=|BF|$ alone do not determine $t:=|EF|$. In order to see this, let $M$ be the midpoint of $AF$. Then $M$ is the center of a circle with diameter $AF$ and containing $B$, $E$ on its periphery. Since we are not told the angle $angle(BFA)$, or the diameter $|AF|$ of this circle, the knowledge of $s$ ad $alpha$ is insufficient to determine $t$.
answered Aug 3 at 17:50
Christian Blatter
163k7105305
The data $alpha$ and $s:=|BF|$ alone do not determine $t:=|EF|$. In order to see this, let $M$ be the midpoint of $AF$. Then $M$ is the center of a circle with diameter $AF$ and containing $B$, $E$ on its periphery. Since we are not told the angle $angle(BFA)$, or the diameter $|AF|$ of this circle, the knowledge of $s$ ad $alpha$ is insufficient to determine $t$.
answered Aug 3 at 17:50
Christian Blatter
163k7105305
edited yesterday
answered Aug 3 at 17:50
Christian Blatter
163k7105305
answered Aug 3 at 17:50
Christian Blatter
163k7105305
answered Aug 3 at 17:50
Christian Blatter
163k7105305
163k7105305
Thank you for your answer, for me $angle(BFA)=AhatBF$ and $AhatBF$ is a right angle. Furthermore I can' t see why $B$ and $E$ should lay on the circumference centered at the midpoint of $AF$ and with diameter equal to $AF$.
– Alessandro Jacopson
2 days ago
Angle $angle(BFA)$ is an angle measured at $F$, and it is obviously not $90^circ$. My point $M$ is the center of aThales circle.
– Christian Blatter
2 days ago
add a comment |Â
Thank you for your answer, for me $angle(BFA)=AhatBF$ and $AhatBF$ is a right angle. Furthermore I can' t see why $B$ and $E$ should lay on the circumference centered at the midpoint of $AF$ and with diameter equal to $AF$.
– Alessandro Jacopson
2 days ago
Angle $angle(BFA)$ is an angle measured at $F$, and it is obviously not $90^circ$. My point $M$ is the center of aThales circle.
– Christian Blatter
2 days ago
Thank you for your answer, for me $angle(BFA)=AhatBF$ and $AhatBF$ is a right angle. Furthermore I can' t see why $B$ and $E$ should lay on the circumference centered at the midpoint of $AF$ and with diameter equal to $AF$.
– Alessandro Jacopson
2 days ago
Thank you for your answer, for me $angle(BFA)=AhatBF$ and $AhatBF$ is a right angle. Furthermore I can' t see why $B$ and $E$ should lay on the circumference centered at the midpoint of $AF$ and with diameter equal to $AF$.
– Alessandro Jacopson
2 days ago
Angle $angle(BFA)$ is an angle measured at $F$, and it is obviously not $90^circ$. My point $M$ is the center of aThales circle.
– Christian Blatter
2 days ago
Angle $angle(BFA)$ is an angle measured at $F$, and it is obviously not $90^circ$. My point $M$ is the center of aThales circle.
– Christian Blatter
2 days ago
add a comment |Â
up vote
0
down vote
I have simplified your drawing a bit - so, the angle at B is right, and the angle a E is also right. If you choose point p so that the angle there is right, then the distance from F to p is $|BF| cdot cos(v)$, and the distance from D to p is $|BF| cdot sin(v)$, since the angl DBp must be v: the sum of angles in a triangle is 180, BDp shares an angle at D, and both have a right angle. So there you are: $|DF| = |BF| cdot (sin(v) + cos(v))$
answered Aug 3 at 13:32
j4nd3r53n
42039
Thank you but I need to compute $overlineEF$ and not $overlineDF$.
– Alessandro Jacopson
Aug 3 at 14:18
Well, I realised that my answer was wrong, too, even if you were after $|DF|$
– j4nd3r53n
Aug 3 at 14:39
add a comment |Â
up vote
0
down vote
I have simplified your drawing a bit - so, the angle at B is right, and the angle a E is also right. If you choose point p so that the angle there is right, then the distance from F to p is $|BF| cdot cos(v)$, and the distance from D to p is $|BF| cdot sin(v)$, since the angl DBp must be v: the sum of angles in a triangle is 180, BDp shares an angle at D, and both have a right angle. So there you are: $|DF| = |BF| cdot (sin(v) + cos(v))$
answered Aug 3 at 13:32
j4nd3r53n
42039
Thank you but I need to compute $overlineEF$ and not $overlineDF$.
– Alessandro Jacopson
Aug 3 at 14:18
Well, I realised that my answer was wrong, too, even if you were after $|DF|$
– j4nd3r53n
Aug 3 at 14:39
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I have simplified your drawing a bit - so, the angle at B is right, and the angle a E is also right. If you choose point p so that the angle there is right, then the distance from F to p is $|BF| cdot cos(v)$, and the distance from D to p is $|BF| cdot sin(v)$, since the angl DBp must be v: the sum of angles in a triangle is 180, BDp shares an angle at D, and both have a right angle. So there you are: $|DF| = |BF| cdot (sin(v) + cos(v))$
answered Aug 3 at 13:32
j4nd3r53n
42039
I have simplified your drawing a bit - so, the angle at B is right, and the angle a E is also right. If you choose point p so that the angle there is right, then the distance from F to p is $|BF| cdot cos(v)$, and the distance from D to p is $|BF| cdot sin(v)$, since the angl DBp must be v: the sum of angles in a triangle is 180, BDp shares an angle at D, and both have a right angle. So there you are: $|DF| = |BF| cdot (sin(v) + cos(v))$
answered Aug 3 at 13:32
j4nd3r53n
42039
answered Aug 3 at 13:32
j4nd3r53n
42039
answered Aug 3 at 13:32
j4nd3r53n
42039
answered Aug 3 at 13:32
j4nd3r53n
42039
42039
Thank you but I need to compute $overlineEF$ and not $overlineDF$.
– Alessandro Jacopson
Aug 3 at 14:18
Well, I realised that my answer was wrong, too, even if you were after $|DF|$
– j4nd3r53n
Aug 3 at 14:39
add a comment |Â
Thank you but I need to compute $overlineEF$ and not $overlineDF$.
– Alessandro Jacopson
Aug 3 at 14:18
Well, I realised that my answer was wrong, too, even if you were after $|DF|$
– j4nd3r53n
Aug 3 at 14:39
Thank you but I need to compute $overlineEF$ and not $overlineDF$.
– Alessandro Jacopson
Aug 3 at 14:18
Thank you but I need to compute $overlineEF$ and not $overlineDF$.
– Alessandro Jacopson
Aug 3 at 14:18
Well, I realised that my answer was wrong, too, even if you were after $|DF|$
– j4nd3r53n
Aug 3 at 14:39
Well, I realised that my answer was wrong, too, even if you were after $|DF|$
– j4nd3r53n
Aug 3 at 14:39
add a comment |Â
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The points $B,C,D$, and $E$ are all the same, $alpha=0$, and $EF=BF$.
– spiralstotheleft
Aug 3 at 12:55
@spiralstotheleft I updated the question, I forgot to write that $alphaneq0$
– Alessandro Jacopson
Aug 3 at 12:58
1
No, you didn't. If all other conditions are satisfied, they imply $angle DFB=0$.
– spiralstotheleft
Aug 3 at 12:59
@spiralstotheleft I beg your pardon but I am not able to be convinced by your statement.
– Alessandro Jacopson
Aug 3 at 13:03