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Existing of non-intersecting rays

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2

1

$begingroup$

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

share|cite|improve this question

asked 21 mins ago

athosathos

98611340

$endgroup$

add a comment | 

2

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

asked 21 mins ago

athosathos

98611340

$endgroup$

share|cite|improve this question

asked 21 mins ago

athosathos

98611340

share|cite|improve this question

asked 21 mins ago

athosathos

98611340

asked 21 mins ago

athosathos

98611340

asked 21 mins ago

athosathos

98611340

2 Answers
2

active

oldest

votes

2

$begingroup$

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.

share|cite|improve this answer

edited 3 mins ago

answered 10 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 | 

0

$begingroup$

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.

share|cite

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

$begingroup$

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.

share|cite|improve this answer

edited 3 mins ago

answered 10 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 | 

2

$begingroup$

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.

share|cite|improve this answer

edited 3 mins ago

answered 10 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.

share|cite|improve this answer

edited 3 mins ago

answered 10 mins ago

FredHFredH

2,6041021

$endgroup$

share|cite|improve this answer

edited 3 mins ago

answered 10 mins ago

FredHFredH

2,6041021

answered 10 mins ago

FredHFredH

2,6041021

answered 10 mins ago

FredHFredH

2,6041021

0

$begingroup$

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.

share|cite

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 | 

0

$begingroup$

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.

share|cite

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.

share|cite

answered 8 mins ago

CyborgOctopusCyborgOctopus

1407

$endgroup$

share|cite

answered 8 mins ago

CyborgOctopusCyborgOctopus

1407

answered 8 mins ago

CyborgOctopusCyborgOctopus

1407

answered 8 mins ago

CyborgOctopusCyborgOctopus

1407