Existing of non-intersecting rays
$begingroup$
Given $n$ points on a plane, it seems intuitive that it’s possible to draw a ray (half-line) from each point s. t. the $n$ rays do not intersect.
But how to prove this?
geometry
asked 35 mins ago
athosathos
98611340
$endgroup$
add a comment |
$begingroup$
Given $n$ points on a plane, it seems intuitive that it’s possible to draw a ray (half-line) from each point s. t. the $n$ rays do not intersect.
But how to prove this?
geometry
asked 35 mins ago
athosathos
98611340
$endgroup$
add a comment |
$begingroup$
Given $n$ points on a plane, it seems intuitive that it’s possible to draw a ray (half-line) from each point s. t. the $n$ rays do not intersect.
But how to prove this?
geometry
asked 35 mins ago
athosathos
98611340
$endgroup$
Given $n$ points on a plane, it seems intuitive that it’s possible to draw a ray (half-line) from each point s. t. the $n$ rays do not intersect.
But how to prove this?
geometry
geometry
asked 35 mins ago
athosathos
98611340
asked 35 mins ago
athosathos
98611340
asked 35 mins ago
athosathos
98611340
asked 35 mins ago
athosathos
98611340
asked 35 mins ago
athosathos
98611340
98611340
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Pick any point $P$ in the plane that is not on a line containing two or more of the given $n$ points. At each point, draw the ray in the direction away from $P$.
One can in fact do better: It is possible to draw lines through all $n$ points that do not intersect. Choose an orientation that is not parallel to any of the lines between any two of the given points, and draw parallel lines in that orientation through each point.
answered 24 mins ago
FredHFredH
2,6041021
$endgroup$
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
23 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
22 mins ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
8 mins ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
6 mins ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
2 mins ago
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Pick any point $P$ in the plane that is not on a line containing two or more of the given $n$ points. At each point, draw the ray in the direction away from $P$.
One can in fact do better: It is possible to draw lines through all $n$ points that do not intersect. Choose an orientation that is not parallel to any of the lines between any two of the given points, and draw parallel lines in that orientation through each point.
answered 24 mins ago
FredHFredH
2,6041021
$endgroup$
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
23 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
22 mins ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
8 mins ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
6 mins ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
2 mins ago
add a comment |
$begingroup$
Pick any point $P$ in the plane that is not on a line containing two or more of the given $n$ points. At each point, draw the ray in the direction away from $P$.
One can in fact do better: It is possible to draw lines through all $n$ points that do not intersect. Choose an orientation that is not parallel to any of the lines between any two of the given points, and draw parallel lines in that orientation through each point.
answered 24 mins ago
FredHFredH
2,6041021
$endgroup$
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
23 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
22 mins ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
8 mins ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
6 mins ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
2 mins ago
add a comment |
$begingroup$
Pick any point $P$ in the plane that is not on a line containing two or more of the given $n$ points. At each point, draw the ray in the direction away from $P$.
One can in fact do better: It is possible to draw lines through all $n$ points that do not intersect. Choose an orientation that is not parallel to any of the lines between any two of the given points, and draw parallel lines in that orientation through each point.
answered 24 mins ago
FredHFredH
2,6041021
$endgroup$
Pick any point $P$ in the plane that is not on a line containing two or more of the given $n$ points. At each point, draw the ray in the direction away from $P$.
One can in fact do better: It is possible to draw lines through all $n$ points that do not intersect. Choose an orientation that is not parallel to any of the lines between any two of the given points, and draw parallel lines in that orientation through each point.
answered 24 mins ago
FredHFredH
2,6041021
edited 17 mins ago
answered 24 mins ago
FredHFredH
2,6041021
answered 24 mins ago
FredHFredH
2,6041021
answered 24 mins ago
FredHFredH
2,6041021
2,6041021
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
23 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
22 mins ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
8 mins ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
6 mins ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
2 mins ago
add a comment |
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
23 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
22 mins ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
8 mins ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
6 mins ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
2 mins ago
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
23 mins ago
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
23 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
22 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
22 mins ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
8 mins ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
8 mins ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
6 mins ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
6 mins ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
2 mins ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
2 mins ago
add a comment |
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