Existing of non-intersecting rays












2












$begingroup$


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







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$endgroup$

















    2












    $begingroup$


    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









    $endgroup$















      2












      2








      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









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 35 mins ago









      athosathos

      98611340




      98611340






















          1 Answer
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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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            +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













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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











          $endgroup$













          • $begingroup$
            +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


















          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











          $endgroup$













          • $begingroup$
            +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
















          2












          2








          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











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 17 mins ago

























          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




















          • $begingroup$
            +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




















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