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Prove that the outer circle of the triangle $OMN$ is always tangent to a fixed line.
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Consider a half circle with diameter $AB$.Draw the tangent $Ax$, $By$ with half circle. Take $M$ on $Ax$, $N$ on $By$ such that $AM*BN=R^2$. Prove that the outer circle of the triangle $OMN$ is always tangent to a fixed line.
I will $MN$ is a fixed line and $O;R$ cut $MN$ at OH is a fixed line
I see: $MN=MA+NB$ and $MN^2=MA^2+NB^2+2MA*NB=MA^2+NB^2+2R^2$
So i So I need to prove $MA^2+NB^2$ is a fixed line but i can't. Help me
geometry euclidean-geometry
edited Jul 29 at 11:29
Aretino
21.7k21342
asked Jul 29 at 10:56
Word Shallow
657313
 |Â
show 1 more comment
up vote
0
down vote
favorite
Consider a half circle with diameter $AB$.Draw the tangent $Ax$, $By$ with half circle. Take $M$ on $Ax$, $N$ on $By$ such that $AM*BN=R^2$. Prove that the outer circle of the triangle $OMN$ is always tangent to a fixed line.
I will $MN$ is a fixed line and $O;R$ cut $MN$ at OH is a fixed line
I see: $MN=MA+NB$ and $MN^2=MA^2+NB^2+2MA*NB=MA^2+NB^2+2R^2$
So i So I need to prove $MA^2+NB^2$ is a fixed line but i can't. Help me
geometry euclidean-geometry
edited Jul 29 at 11:29
Aretino
21.7k21342
asked Jul 29 at 10:56
Word Shallow
657313
What is outercircle?
– greedoid
Jul 29 at 11:14
What is a "fixed line"?
– Jens
Jul 29 at 11:15
Sure the "outer circle" refers to the circumcircle of the triangle. A "fixed line" refers to a line that doesn't depend on the choice of the point $M$ on the line $Ax$, etc.
– John Hughes
Jul 29 at 11:17
1
But then $AB$ is obviously
– greedoid
Jul 29 at 11:18
It's certainly obvious that $AB$ contains a point of the circumcircle (namely $O$). It's not obvious to me that $AB$ is the tangent to the circumcircle at $O$, independent of the choice of $M$, but I'm not that good at geometry, so it may be obvious to others.
– John Hughes
Jul 29 at 11:25
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider a half circle with diameter $AB$.Draw the tangent $Ax$, $By$ with half circle. Take $M$ on $Ax$, $N$ on $By$ such that $AM*BN=R^2$. Prove that the outer circle of the triangle $OMN$ is always tangent to a fixed line.
I will $MN$ is a fixed line and $O;R$ cut $MN$ at OH is a fixed line
I see: $MN=MA+NB$ and $MN^2=MA^2+NB^2+2MA*NB=MA^2+NB^2+2R^2$
So i So I need to prove $MA^2+NB^2$ is a fixed line but i can't. Help me
geometry euclidean-geometry
edited Jul 29 at 11:29
Aretino
21.7k21342
asked Jul 29 at 10:56
Word Shallow
657313
Consider a half circle with diameter $AB$.Draw the tangent $Ax$, $By$ with half circle. Take $M$ on $Ax$, $N$ on $By$ such that $AM*BN=R^2$. Prove that the outer circle of the triangle $OMN$ is always tangent to a fixed line.
I will $MN$ is a fixed line and $O;R$ cut $MN$ at OH is a fixed line
I see: $MN=MA+NB$ and $MN^2=MA^2+NB^2+2MA*NB=MA^2+NB^2+2R^2$
So i So I need to prove $MA^2+NB^2$ is a fixed line but i can't. Help me
geometry euclidean-geometry
edited Jul 29 at 11:29
Aretino
21.7k21342
asked Jul 29 at 10:56
Word Shallow
657313
edited Jul 29 at 11:29
Aretino
21.7k21342
edited Jul 29 at 11:29
Aretino
21.7k21342
edited Jul 29 at 11:29
Aretino
21.7k21342
21.7k21342
asked Jul 29 at 10:56
Word Shallow
657313
asked Jul 29 at 10:56
Word Shallow
657313
asked Jul 29 at 10:56
Word Shallow
657313
657313
What is outercircle?
– greedoid
Jul 29 at 11:14
What is a "fixed line"?
– Jens
Jul 29 at 11:15
Sure the "outer circle" refers to the circumcircle of the triangle. A "fixed line" refers to a line that doesn't depend on the choice of the point $M$ on the line $Ax$, etc.
– John Hughes
Jul 29 at 11:17
1
But then $AB$ is obviously
– greedoid
Jul 29 at 11:18
It's certainly obvious that $AB$ contains a point of the circumcircle (namely $O$). It's not obvious to me that $AB$ is the tangent to the circumcircle at $O$, independent of the choice of $M$, but I'm not that good at geometry, so it may be obvious to others.
– John Hughes
Jul 29 at 11:25
 |Â
show 1 more comment
What is outercircle?
– greedoid
Jul 29 at 11:14
What is a "fixed line"?
– Jens
Jul 29 at 11:15
Sure the "outer circle" refers to the circumcircle of the triangle. A "fixed line" refers to a line that doesn't depend on the choice of the point $M$ on the line $Ax$, etc.
– John Hughes
Jul 29 at 11:17
1
But then $AB$ is obviously
– greedoid
Jul 29 at 11:18
It's certainly obvious that $AB$ contains a point of the circumcircle (namely $O$). It's not obvious to me that $AB$ is the tangent to the circumcircle at $O$, independent of the choice of $M$, but I'm not that good at geometry, so it may be obvious to others.
– John Hughes
Jul 29 at 11:25
What is outercircle?
– greedoid
Jul 29 at 11:14
What is outercircle?
– greedoid
Jul 29 at 11:14
What is a "fixed line"?
– Jens
Jul 29 at 11:15
What is a "fixed line"?
– Jens
Jul 29 at 11:15
Sure the "outer circle" refers to the circumcircle of the triangle. A "fixed line" refers to a line that doesn't depend on the choice of the point $M$ on the line $Ax$, etc.
– John Hughes
Jul 29 at 11:17
Sure the "outer circle" refers to the circumcircle of the triangle. A "fixed line" refers to a line that doesn't depend on the choice of the point $M$ on the line $Ax$, etc.
– John Hughes
Jul 29 at 11:17
1
1
But then $AB$ is obviously
– greedoid
Jul 29 at 11:18
But then $AB$ is obviously
– greedoid
Jul 29 at 11:18
It's certainly obvious that $AB$ contains a point of the circumcircle (namely $O$). It's not obvious to me that $AB$ is the tangent to the circumcircle at $O$, independent of the choice of $M$, but I'm not that good at geometry, so it may be obvious to others.
– John Hughes
Jul 29 at 11:25
It's certainly obvious that $AB$ contains a point of the circumcircle (namely $O$). It's not obvious to me that $AB$ is the tangent to the circumcircle at $O$, independent of the choice of $M$, but I'm not that good at geometry, so it may be obvious to others.
– John Hughes
Jul 29 at 11:25
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
HINTS.
- $MON$ is a right triangle.
- The circumcircle of $MON$ has its center at $K$, midpoint of $MN$.
- Radius $OK$ is parallel to $MA$, $NB$ and thus perpendicular to $AB$.
answered Jul 29 at 11:26
Aretino
21.7k21342
Great ............+1
– greedoid
Jul 29 at 11:37
Why $OK$ is parallel to $MA$ ?
– Word Shallow
Jul 29 at 12:22
@WordShallow By the converse of intercept theorem.
– Aretino
Jul 29 at 13:28
add a comment |Â
up vote
1
down vote
Suppose $HO$ is the perpendicular to $MN$ from $O$.
Notice that
$$ AM cdot BN = OAcdot OB = R^2$$
so the triangles $AOM$ and $BON$ are similar.
From here, you obtain that the angle $MON$ is 90, and moreover $MO/MN = AO/NB = BO/NB$ so all the triangles in the figure are similar. In particular, $AO=HO=BO=R$.
answered Jul 29 at 11:22
Exodd
5,3901222
How do you know M,N,H are colinear?
– greedoid
Jul 29 at 11:37
Prove $MN$ is tangent with half circle at $H$.
– Word Shallow
Jul 29 at 11:52
How do you find that?
– greedoid
Jul 29 at 11:52
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
accepted
HINTS.
- $MON$ is a right triangle.
- The circumcircle of $MON$ has its center at $K$, midpoint of $MN$.
- Radius $OK$ is parallel to $MA$, $NB$ and thus perpendicular to $AB$.
answered Jul 29 at 11:26
Aretino
21.7k21342
Great ............+1
– greedoid
Jul 29 at 11:37
Why $OK$ is parallel to $MA$ ?
– Word Shallow
Jul 29 at 12:22
@WordShallow By the converse of intercept theorem.
– Aretino
Jul 29 at 13:28
add a comment |Â
up vote
2
down vote
accepted
HINTS.
- $MON$ is a right triangle.
- The circumcircle of $MON$ has its center at $K$, midpoint of $MN$.
- Radius $OK$ is parallel to $MA$, $NB$ and thus perpendicular to $AB$.
answered Jul 29 at 11:26
Aretino
21.7k21342
Great ............+1
– greedoid
Jul 29 at 11:37
Why $OK$ is parallel to $MA$ ?
– Word Shallow
Jul 29 at 12:22
@WordShallow By the converse of intercept theorem.
– Aretino
Jul 29 at 13:28
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
HINTS.
- $MON$ is a right triangle.
- The circumcircle of $MON$ has its center at $K$, midpoint of $MN$.
- Radius $OK$ is parallel to $MA$, $NB$ and thus perpendicular to $AB$.
answered Jul 29 at 11:26
Aretino
21.7k21342
HINTS.
- $MON$ is a right triangle.
- The circumcircle of $MON$ has its center at $K$, midpoint of $MN$.
- Radius $OK$ is parallel to $MA$, $NB$ and thus perpendicular to $AB$.
answered Jul 29 at 11:26
Aretino
21.7k21342
answered Jul 29 at 11:26
Aretino
21.7k21342
answered Jul 29 at 11:26
Aretino
21.7k21342
answered Jul 29 at 11:26
Aretino
21.7k21342
21.7k21342
Great ............+1
– greedoid
Jul 29 at 11:37
Why $OK$ is parallel to $MA$ ?
– Word Shallow
Jul 29 at 12:22
@WordShallow By the converse of intercept theorem.
– Aretino
Jul 29 at 13:28
add a comment |Â
Great ............+1
– greedoid
Jul 29 at 11:37
Why $OK$ is parallel to $MA$ ?
– Word Shallow
Jul 29 at 12:22
@WordShallow By the converse of intercept theorem.
– Aretino
Jul 29 at 13:28
Great ............+1
– greedoid
Jul 29 at 11:37
Great ............+1
– greedoid
Jul 29 at 11:37
Why $OK$ is parallel to $MA$ ?
– Word Shallow
Jul 29 at 12:22
Why $OK$ is parallel to $MA$ ?
– Word Shallow
Jul 29 at 12:22
@WordShallow By the converse of intercept theorem.
– Aretino
Jul 29 at 13:28
@WordShallow By the converse of intercept theorem.
– Aretino
Jul 29 at 13:28
add a comment |Â
up vote
1
down vote
Suppose $HO$ is the perpendicular to $MN$ from $O$.
Notice that
$$ AM cdot BN = OAcdot OB = R^2$$
so the triangles $AOM$ and $BON$ are similar.
From here, you obtain that the angle $MON$ is 90, and moreover $MO/MN = AO/NB = BO/NB$ so all the triangles in the figure are similar. In particular, $AO=HO=BO=R$.
answered Jul 29 at 11:22
Exodd
5,3901222
How do you know M,N,H are colinear?
– greedoid
Jul 29 at 11:37
Prove $MN$ is tangent with half circle at $H$.
– Word Shallow
Jul 29 at 11:52
How do you find that?
– greedoid
Jul 29 at 11:52
add a comment |Â
up vote
1
down vote
Suppose $HO$ is the perpendicular to $MN$ from $O$.
Notice that
$$ AM cdot BN = OAcdot OB = R^2$$
so the triangles $AOM$ and $BON$ are similar.
From here, you obtain that the angle $MON$ is 90, and moreover $MO/MN = AO/NB = BO/NB$ so all the triangles in the figure are similar. In particular, $AO=HO=BO=R$.
answered Jul 29 at 11:22
Exodd
5,3901222
How do you know M,N,H are colinear?
– greedoid
Jul 29 at 11:37
Prove $MN$ is tangent with half circle at $H$.
– Word Shallow
Jul 29 at 11:52
How do you find that?
– greedoid
Jul 29 at 11:52
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Suppose $HO$ is the perpendicular to $MN$ from $O$.
Notice that
$$ AM cdot BN = OAcdot OB = R^2$$
so the triangles $AOM$ and $BON$ are similar.
From here, you obtain that the angle $MON$ is 90, and moreover $MO/MN = AO/NB = BO/NB$ so all the triangles in the figure are similar. In particular, $AO=HO=BO=R$.
answered Jul 29 at 11:22
Exodd
5,3901222
Suppose $HO$ is the perpendicular to $MN$ from $O$.
Notice that
$$ AM cdot BN = OAcdot OB = R^2$$
so the triangles $AOM$ and $BON$ are similar.
From here, you obtain that the angle $MON$ is 90, and moreover $MO/MN = AO/NB = BO/NB$ so all the triangles in the figure are similar. In particular, $AO=HO=BO=R$.
answered Jul 29 at 11:22
Exodd
5,3901222
answered Jul 29 at 11:22
Exodd
5,3901222
answered Jul 29 at 11:22
Exodd
5,3901222
answered Jul 29 at 11:22
Exodd
5,3901222
5,3901222
How do you know M,N,H are colinear?
– greedoid
Jul 29 at 11:37
Prove $MN$ is tangent with half circle at $H$.
– Word Shallow
Jul 29 at 11:52
How do you find that?
– greedoid
Jul 29 at 11:52
add a comment |Â
How do you know M,N,H are colinear?
– greedoid
Jul 29 at 11:37
Prove $MN$ is tangent with half circle at $H$.
– Word Shallow
Jul 29 at 11:52
How do you find that?
– greedoid
Jul 29 at 11:52
How do you know M,N,H are colinear?
– greedoid
Jul 29 at 11:37
How do you know M,N,H are colinear?
– greedoid
Jul 29 at 11:37
Prove $MN$ is tangent with half circle at $H$.
– Word Shallow
Jul 29 at 11:52
Prove $MN$ is tangent with half circle at $H$.
– Word Shallow
Jul 29 at 11:52
How do you find that?
– greedoid
Jul 29 at 11:52
How do you find that?
– greedoid
Jul 29 at 11:52
add a comment |Â
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What is outercircle?
– greedoid
Jul 29 at 11:14
What is a "fixed line"?
– Jens
Jul 29 at 11:15
Sure the "outer circle" refers to the circumcircle of the triangle. A "fixed line" refers to a line that doesn't depend on the choice of the point $M$ on the line $Ax$, etc.
– John Hughes
Jul 29 at 11:17
1
But then $AB$ is obviously
– greedoid
Jul 29 at 11:18
It's certainly obvious that $AB$ contains a point of the circumcircle (namely $O$). It's not obvious to me that $AB$ is the tangent to the circumcircle at $O$, independent of the choice of $M$, but I'm not that good at geometry, so it may be obvious to others.
– John Hughes
Jul 29 at 11:25