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Reflections in a Square



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Ray reflection inside the cubeDissecting a squareEnlarge the Square?Quadrilateral inside a squareThe Challenge SquareThe square and the compassNo ordinary magic squareSide length of square inside a triangleDissect a square into 3:1 rectanglesUgh! The extra SQUARE!Looking for Euler square solver










2












$begingroup$


An ideal billiards table (no friction, ideal reflections off of the walls, no pockets) is shaped like a square. From the bottom-left corner, shoot a point-sized cue ball at some angle.



What is the shortest path that hits all 4 edges of the table at least once each and ends in the top-right corner?



Note 1: hitting a corner ends the path without counting as hitting the walls on either side of it.



Note 2: I'm aware of this puzzle, which is higher-dimensional and more mathematically involved but has a similar answer to this one. I felt that this warranted its own puzzle, though, because I want more people to have the chance to find its elegant, visual solution without being intimidated by the apparent need to use math (and it's not asking for exactly the same thing).










share|improve this question







New contributor




Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$
















    2












    $begingroup$


    An ideal billiards table (no friction, ideal reflections off of the walls, no pockets) is shaped like a square. From the bottom-left corner, shoot a point-sized cue ball at some angle.



    What is the shortest path that hits all 4 edges of the table at least once each and ends in the top-right corner?



    Note 1: hitting a corner ends the path without counting as hitting the walls on either side of it.



    Note 2: I'm aware of this puzzle, which is higher-dimensional and more mathematically involved but has a similar answer to this one. I felt that this warranted its own puzzle, though, because I want more people to have the chance to find its elegant, visual solution without being intimidated by the apparent need to use math (and it's not asking for exactly the same thing).










    share|improve this question







    New contributor




    Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      2












      2








      2





      $begingroup$


      An ideal billiards table (no friction, ideal reflections off of the walls, no pockets) is shaped like a square. From the bottom-left corner, shoot a point-sized cue ball at some angle.



      What is the shortest path that hits all 4 edges of the table at least once each and ends in the top-right corner?



      Note 1: hitting a corner ends the path without counting as hitting the walls on either side of it.



      Note 2: I'm aware of this puzzle, which is higher-dimensional and more mathematically involved but has a similar answer to this one. I felt that this warranted its own puzzle, though, because I want more people to have the chance to find its elegant, visual solution without being intimidated by the apparent need to use math (and it's not asking for exactly the same thing).










      share|improve this question







      New contributor




      Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      An ideal billiards table (no friction, ideal reflections off of the walls, no pockets) is shaped like a square. From the bottom-left corner, shoot a point-sized cue ball at some angle.



      What is the shortest path that hits all 4 edges of the table at least once each and ends in the top-right corner?



      Note 1: hitting a corner ends the path without counting as hitting the walls on either side of it.



      Note 2: I'm aware of this puzzle, which is higher-dimensional and more mathematically involved but has a similar answer to this one. I felt that this warranted its own puzzle, though, because I want more people to have the chance to find its elegant, visual solution without being intimidated by the apparent need to use math (and it's not asking for exactly the same thing).







      logical-deduction geometry






      share|improve this question







      New contributor




      Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question







      New contributor




      Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question






      New contributor




      Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 2 hours ago









      Gilad MGilad M

      111




      111




      New contributor




      Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Suppose we label the corner on the table like this:




          enter image description here




          Now we want to move from $A$ to $D$.



          Now, imagine the table like this:




          enter image description here




          Here,




          We can simulate the path as a single straight line.

          As the line passes the border, it will simulate the result of the ball's reflection hence the table will be mirrored as above.




          Now, to hits all $4$ edges, that means




          The path should pass $overlineAB$, $overlineCD$, $overlineAC$, $overlineBD$.

          In other words, the path must go beyond $2 times 2$ tables.




          To get the shortest path,




          We should choose the nearest $D$ (beyond $2 times 2$ tables).

          One of the options is the $D$ located $4$ above and $4$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.




          Thus,




          This is the shortest path.

          enter image description here


          The overall length will be $sqrt5^2 + 3^2 = sqrt34$ times the table length.




          Which is like this:




          enter image description here







          share|improve this answer











          $endgroup$












          • $begingroup$
            Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
            $endgroup$
            – Ahmed Abdelhameed
            40 mins ago







          • 1




            $begingroup$
            @AhmedAbdelhameed added :)
            $endgroup$
            – athin
            33 mins ago











          Your Answer








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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Suppose we label the corner on the table like this:




          enter image description here




          Now we want to move from $A$ to $D$.



          Now, imagine the table like this:




          enter image description here




          Here,




          We can simulate the path as a single straight line.

          As the line passes the border, it will simulate the result of the ball's reflection hence the table will be mirrored as above.




          Now, to hits all $4$ edges, that means




          The path should pass $overlineAB$, $overlineCD$, $overlineAC$, $overlineBD$.

          In other words, the path must go beyond $2 times 2$ tables.




          To get the shortest path,




          We should choose the nearest $D$ (beyond $2 times 2$ tables).

          One of the options is the $D$ located $4$ above and $4$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.




          Thus,




          This is the shortest path.

          enter image description here


          The overall length will be $sqrt5^2 + 3^2 = sqrt34$ times the table length.




          Which is like this:




          enter image description here







          share|improve this answer











          $endgroup$












          • $begingroup$
            Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
            $endgroup$
            – Ahmed Abdelhameed
            40 mins ago







          • 1




            $begingroup$
            @AhmedAbdelhameed added :)
            $endgroup$
            – athin
            33 mins ago















          3












          $begingroup$

          Suppose we label the corner on the table like this:




          enter image description here




          Now we want to move from $A$ to $D$.



          Now, imagine the table like this:




          enter image description here




          Here,




          We can simulate the path as a single straight line.

          As the line passes the border, it will simulate the result of the ball's reflection hence the table will be mirrored as above.




          Now, to hits all $4$ edges, that means




          The path should pass $overlineAB$, $overlineCD$, $overlineAC$, $overlineBD$.

          In other words, the path must go beyond $2 times 2$ tables.




          To get the shortest path,




          We should choose the nearest $D$ (beyond $2 times 2$ tables).

          One of the options is the $D$ located $4$ above and $4$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.




          Thus,




          This is the shortest path.

          enter image description here


          The overall length will be $sqrt5^2 + 3^2 = sqrt34$ times the table length.




          Which is like this:




          enter image description here







          share|improve this answer











          $endgroup$












          • $begingroup$
            Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
            $endgroup$
            – Ahmed Abdelhameed
            40 mins ago







          • 1




            $begingroup$
            @AhmedAbdelhameed added :)
            $endgroup$
            – athin
            33 mins ago













          3












          3








          3





          $begingroup$

          Suppose we label the corner on the table like this:




          enter image description here




          Now we want to move from $A$ to $D$.



          Now, imagine the table like this:




          enter image description here




          Here,




          We can simulate the path as a single straight line.

          As the line passes the border, it will simulate the result of the ball's reflection hence the table will be mirrored as above.




          Now, to hits all $4$ edges, that means




          The path should pass $overlineAB$, $overlineCD$, $overlineAC$, $overlineBD$.

          In other words, the path must go beyond $2 times 2$ tables.




          To get the shortest path,




          We should choose the nearest $D$ (beyond $2 times 2$ tables).

          One of the options is the $D$ located $4$ above and $4$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.




          Thus,




          This is the shortest path.

          enter image description here


          The overall length will be $sqrt5^2 + 3^2 = sqrt34$ times the table length.




          Which is like this:




          enter image description here







          share|improve this answer











          $endgroup$



          Suppose we label the corner on the table like this:




          enter image description here




          Now we want to move from $A$ to $D$.



          Now, imagine the table like this:




          enter image description here




          Here,




          We can simulate the path as a single straight line.

          As the line passes the border, it will simulate the result of the ball's reflection hence the table will be mirrored as above.




          Now, to hits all $4$ edges, that means




          The path should pass $overlineAB$, $overlineCD$, $overlineAC$, $overlineBD$.

          In other words, the path must go beyond $2 times 2$ tables.




          To get the shortest path,




          We should choose the nearest $D$ (beyond $2 times 2$ tables).

          One of the options is the $D$ located $4$ above and $4$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.




          Thus,




          This is the shortest path.

          enter image description here


          The overall length will be $sqrt5^2 + 3^2 = sqrt34$ times the table length.




          Which is like this:




          enter image description here








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 33 mins ago

























          answered 45 mins ago









          athinathin

          8,73422877




          8,73422877











          • $begingroup$
            Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
            $endgroup$
            – Ahmed Abdelhameed
            40 mins ago







          • 1




            $begingroup$
            @AhmedAbdelhameed added :)
            $endgroup$
            – athin
            33 mins ago
















          • $begingroup$
            Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
            $endgroup$
            – Ahmed Abdelhameed
            40 mins ago







          • 1




            $begingroup$
            @AhmedAbdelhameed added :)
            $endgroup$
            – athin
            33 mins ago















          $begingroup$
          Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
          $endgroup$
          – Ahmed Abdelhameed
          40 mins ago





          $begingroup$
          Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
          $endgroup$
          – Ahmed Abdelhameed
          40 mins ago





          1




          1




          $begingroup$
          @AhmedAbdelhameed added :)
          $endgroup$
          – athin
          33 mins ago




          $begingroup$
          @AhmedAbdelhameed added :)
          $endgroup$
          – athin
          33 mins ago










          Gilad M is a new contributor. Be nice, and check out our Code of Conduct.









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          Gilad M is a new contributor. Be nice, and check out our Code of Conduct.











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