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 21 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 21 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 21 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 21 mins ago
athosathos
98611340
asked 21 mins ago
athosathos
98611340
asked 21 mins ago
athosathos
98611340
asked 21 mins ago
athosathos
98611340
asked 21 mins ago
athosathos
98611340
98611340
add a comment |
add a comment |
2 Answers
2
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 9 mins ago
FredHFredH
2,6041021
$endgroup$
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
9 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
8 mins ago
add a comment |
$begingroup$
I would do it by induction. Say you already have $n$ points, each with a ray, and all rays are non-intersecting. You then place a new point in the plane. Because all of the rays already in the plane are non-intersecting, it must be that this point is not completely enclosed by rays. Because it is not completely enclosed, it must be possible for it to move continually away from its current position on some straight line path, forever, without hitting a ray. This path itself is a ray from the point which intersects no others! The base case of the induction is obviously true, so the proof is complete.
answered 8 mins ago
CyborgOctopusCyborgOctopus
1407
$endgroup$
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
7 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
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 9 mins ago
FredHFredH
2,6041021
$endgroup$
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
9 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
8 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 9 mins ago
FredHFredH
2,6041021
$endgroup$
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
9 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
8 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 9 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 9 mins ago
FredHFredH
2,6041021
edited 3 mins ago
answered 9 mins ago
FredHFredH
2,6041021
answered 9 mins ago
FredHFredH
2,6041021
answered 9 mins ago
FredHFredH
2,6041021
2,6041021
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
9 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
8 mins ago
add a comment |
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
9 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
8 mins ago
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
9 mins ago
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
9 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
8 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
8 mins ago
add a comment |
$begingroup$
I would do it by induction. Say you already have $n$ points, each with a ray, and all rays are non-intersecting. You then place a new point in the plane. Because all of the rays already in the plane are non-intersecting, it must be that this point is not completely enclosed by rays. Because it is not completely enclosed, it must be possible for it to move continually away from its current position on some straight line path, forever, without hitting a ray. This path itself is a ray from the point which intersects no others! The base case of the induction is obviously true, so the proof is complete.
answered 8 mins ago
CyborgOctopusCyborgOctopus
1407
$endgroup$
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
7 mins ago
add a comment |
$begingroup$
I would do it by induction. Say you already have $n$ points, each with a ray, and all rays are non-intersecting. You then place a new point in the plane. Because all of the rays already in the plane are non-intersecting, it must be that this point is not completely enclosed by rays. Because it is not completely enclosed, it must be possible for it to move continually away from its current position on some straight line path, forever, without hitting a ray. This path itself is a ray from the point which intersects no others! The base case of the induction is obviously true, so the proof is complete.
answered 8 mins ago
CyborgOctopusCyborgOctopus
1407
$endgroup$
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
7 mins ago
add a comment |
$begingroup$
I would do it by induction. Say you already have $n$ points, each with a ray, and all rays are non-intersecting. You then place a new point in the plane. Because all of the rays already in the plane are non-intersecting, it must be that this point is not completely enclosed by rays. Because it is not completely enclosed, it must be possible for it to move continually away from its current position on some straight line path, forever, without hitting a ray. This path itself is a ray from the point which intersects no others! The base case of the induction is obviously true, so the proof is complete.
answered 8 mins ago
CyborgOctopusCyborgOctopus
1407
$endgroup$
I would do it by induction. Say you already have $n$ points, each with a ray, and all rays are non-intersecting. You then place a new point in the plane. Because all of the rays already in the plane are non-intersecting, it must be that this point is not completely enclosed by rays. Because it is not completely enclosed, it must be possible for it to move continually away from its current position on some straight line path, forever, without hitting a ray. This path itself is a ray from the point which intersects no others! The base case of the induction is obviously true, so the proof is complete.
answered 8 mins ago
CyborgOctopusCyborgOctopus
1407
answered 8 mins ago
CyborgOctopusCyborgOctopus
1407
answered 8 mins ago
CyborgOctopusCyborgOctopus
1407
answered 8 mins ago
CyborgOctopusCyborgOctopus
1407
1407
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
7 mins ago
add a comment |
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
7 mins ago
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
7 mins ago
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
7 mins ago
add a comment |
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