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 20 mins ago









      athosathos

      98611340




      98611340






















          2 Answers
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          active

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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











          $endgroup$













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










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



















          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









          $endgroup$













          • $begingroup$
            No . the new point might fall on an existing line
            $endgroup$
            – athos
            6 mins ago











          Your Answer





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          2 Answers
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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          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











          $endgroup$













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










          • $begingroup$
            Thx ! This is a “oh of course “ moment of me
            $endgroup$
            – athos
            7 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











          $endgroup$













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










          • $begingroup$
            Thx ! This is a “oh of course “ moment of me
            $endgroup$
            – athos
            7 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











          $endgroup$



          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














          share|cite



          share|cite








          edited 2 mins ago

























          answered 8 mins ago









          FredHFredH

          2,6041021




          2,6041021












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










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


















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










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
















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




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












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




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











          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









          $endgroup$













          • $begingroup$
            No . the new point might fall on an existing line
            $endgroup$
            – athos
            6 mins ago
















          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









          $endgroup$













          • $begingroup$
            No . the new point might fall on an existing line
            $endgroup$
            – athos
            6 mins ago














          0












          0








          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









          $endgroup$



          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.







          share|cite












          share|cite



          share|cite










          answered 7 mins ago









          CyborgOctopusCyborgOctopus

          1407




          1407












          • $begingroup$
            No . the new point might fall on an existing line
            $endgroup$
            – athos
            6 mins ago


















          • $begingroup$
            No . the new point might fall on an existing line
            $endgroup$
            – athos
            6 mins ago
















          $begingroup$
          No . the new point might fall on an existing line
          $endgroup$
          – athos
          6 mins ago




          $begingroup$
          No . the new point might fall on an existing line
          $endgroup$
          – athos
          6 mins ago


















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