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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 33 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 33 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 33 mins ago

athosathos

98611340

$endgroup$

share|cite|improve this question

asked 33 mins ago

athosathos

98611340

share|cite|improve this question

asked 33 mins ago

athosathos

98611340

asked 33 mins ago

athosathos

98611340

asked 33 mins ago

athosathos

98611340

1 Answer
1

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 15 mins ago

answered 21 mins ago

FredHFredH

2,6041021

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for being slightly faster than me :)
  $endgroup$
  – Severin Schraven
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thx ! This is a “oh of course “ moment of me
  $endgroup$
  – athos
  20 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
  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
  3 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @FredH Ah ok makes perfect sense now!
  $endgroup$
  – Μάρκος Καραμέρης
  28 secs 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 15 mins ago

answered 21 mins ago

FredHFredH

2,6041021

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for being slightly faster than me :)
  $endgroup$
  – Severin Schraven
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thx ! This is a “oh of course “ moment of me
  $endgroup$
  – athos
  20 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
  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
  3 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @FredH Ah ok makes perfect sense now!
  $endgroup$
  – Μάρκος Καραμέρης
  28 secs 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 15 mins ago

answered 21 mins ago

FredHFredH

2,6041021

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for being slightly faster than me :)
  $endgroup$
  – Severin Schraven
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thx ! This is a “oh of course “ moment of me
  $endgroup$
  – athos
  20 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
  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
  3 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @FredH Ah ok makes perfect sense now!
  $endgroup$
  – Μάρκος Καραμέρης
  28 secs 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 15 mins ago

answered 21 mins ago

FredHFredH

2,6041021

$endgroup$

share|cite|improve this answer

edited 15 mins ago

answered 21 mins ago

FredHFredH

2,6041021

answered 21 mins ago

FredHFredH

2,6041021

answered 21 mins ago

FredHFredH

2,6041021