Maximum Angle of Rotation of without Clipping of Rectangle

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Given a rectangle $A$ of width $a_x$ and height $a_y$ and a rectangle $B$ of width $b_x$ and height $b_y$ with the same center as $A$, with $b_x>a_x$ and $b_y>a_y$, calculate the maximum positive angle $alpha$ at which rectangle $B$ can be rotated without any part of it clipping $A$ (meaning that the sides of $B$ never cross the sides of $A$. If $B$ is large enough, then just leave $alpha=360$ degrees since it can be fully rotated without ever clipping.




Here are some images clarifying what I mean:



Maximum angle example:



enter image description here



Clipping example:



enter image description here










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    Given a rectangle $A$ of width $a_x$ and height $a_y$ and a rectangle $B$ of width $b_x$ and height $b_y$ with the same center as $A$, with $b_x>a_x$ and $b_y>a_y$, calculate the maximum positive angle $alpha$ at which rectangle $B$ can be rotated without any part of it clipping $A$ (meaning that the sides of $B$ never cross the sides of $A$. If $B$ is large enough, then just leave $alpha=360$ degrees since it can be fully rotated without ever clipping.




    Here are some images clarifying what I mean:



    Maximum angle example:



    enter image description here



    Clipping example:



    enter image description here










    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite












      Given a rectangle $A$ of width $a_x$ and height $a_y$ and a rectangle $B$ of width $b_x$ and height $b_y$ with the same center as $A$, with $b_x>a_x$ and $b_y>a_y$, calculate the maximum positive angle $alpha$ at which rectangle $B$ can be rotated without any part of it clipping $A$ (meaning that the sides of $B$ never cross the sides of $A$. If $B$ is large enough, then just leave $alpha=360$ degrees since it can be fully rotated without ever clipping.




      Here are some images clarifying what I mean:



      Maximum angle example:



      enter image description here



      Clipping example:



      enter image description here










      share|cite|improve this question














      Given a rectangle $A$ of width $a_x$ and height $a_y$ and a rectangle $B$ of width $b_x$ and height $b_y$ with the same center as $A$, with $b_x>a_x$ and $b_y>a_y$, calculate the maximum positive angle $alpha$ at which rectangle $B$ can be rotated without any part of it clipping $A$ (meaning that the sides of $B$ never cross the sides of $A$. If $B$ is large enough, then just leave $alpha=360$ degrees since it can be fully rotated without ever clipping.




      Here are some images clarifying what I mean:



      Maximum angle example:



      enter image description here



      Clipping example:



      enter image description here







      geometry trigonometry transformation rotations rectangles






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      asked Sep 5 at 6:35









      jippyjoe4

      4247




      4247




















          1 Answer
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          up vote
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          accepted










          The following figure depicts the "original" arrangement when both rectangles are placed in the same coordinate system (red) with their centers in the origin:



          enter image description here



          Turn, about it center, the blue triangle counterclockwise until its upper-right corner meets the bigger rectangle. The angle at which this takes place is the same angle by which we would have turned the white rectangle clockwise to hit the blue rectangle.



          Take the red circle centered at the origin and calculate its radius: $r=frac12sqrta_x^2+a_y^2.$ So, the equation of the red circle is $$x^2+y^2=frac14left(a_x^2+a_y^2 right).$$



          Where does this circle hits the straight $y=y_b$? Now, you have two points: the upper right corner of the blue rectangle and the crossing point mentioned above. Can you calculate the angle between the two red lines?






          share|cite|improve this answer


















          • 1




            Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/…
            – jippyjoe4
            Sep 5 at 15:01










          • @jippyjoe4: In principle, your solution is OK.
            – zoli
            Sep 6 at 14:31










          • is that the answer you managed to come up with?
            – jippyjoe4
            Sep 6 at 21:11










          • Yes, yes. (+ a few characters)
            – zoli
            Sep 7 at 7:10










          Your Answer




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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          The following figure depicts the "original" arrangement when both rectangles are placed in the same coordinate system (red) with their centers in the origin:



          enter image description here



          Turn, about it center, the blue triangle counterclockwise until its upper-right corner meets the bigger rectangle. The angle at which this takes place is the same angle by which we would have turned the white rectangle clockwise to hit the blue rectangle.



          Take the red circle centered at the origin and calculate its radius: $r=frac12sqrta_x^2+a_y^2.$ So, the equation of the red circle is $$x^2+y^2=frac14left(a_x^2+a_y^2 right).$$



          Where does this circle hits the straight $y=y_b$? Now, you have two points: the upper right corner of the blue rectangle and the crossing point mentioned above. Can you calculate the angle between the two red lines?






          share|cite|improve this answer


















          • 1




            Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/…
            – jippyjoe4
            Sep 5 at 15:01










          • @jippyjoe4: In principle, your solution is OK.
            – zoli
            Sep 6 at 14:31










          • is that the answer you managed to come up with?
            – jippyjoe4
            Sep 6 at 21:11










          • Yes, yes. (+ a few characters)
            – zoli
            Sep 7 at 7:10














          up vote
          1
          down vote



          accepted










          The following figure depicts the "original" arrangement when both rectangles are placed in the same coordinate system (red) with their centers in the origin:



          enter image description here



          Turn, about it center, the blue triangle counterclockwise until its upper-right corner meets the bigger rectangle. The angle at which this takes place is the same angle by which we would have turned the white rectangle clockwise to hit the blue rectangle.



          Take the red circle centered at the origin and calculate its radius: $r=frac12sqrta_x^2+a_y^2.$ So, the equation of the red circle is $$x^2+y^2=frac14left(a_x^2+a_y^2 right).$$



          Where does this circle hits the straight $y=y_b$? Now, you have two points: the upper right corner of the blue rectangle and the crossing point mentioned above. Can you calculate the angle between the two red lines?






          share|cite|improve this answer


















          • 1




            Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/…
            – jippyjoe4
            Sep 5 at 15:01










          • @jippyjoe4: In principle, your solution is OK.
            – zoli
            Sep 6 at 14:31










          • is that the answer you managed to come up with?
            – jippyjoe4
            Sep 6 at 21:11










          • Yes, yes. (+ a few characters)
            – zoli
            Sep 7 at 7:10












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The following figure depicts the "original" arrangement when both rectangles are placed in the same coordinate system (red) with their centers in the origin:



          enter image description here



          Turn, about it center, the blue triangle counterclockwise until its upper-right corner meets the bigger rectangle. The angle at which this takes place is the same angle by which we would have turned the white rectangle clockwise to hit the blue rectangle.



          Take the red circle centered at the origin and calculate its radius: $r=frac12sqrta_x^2+a_y^2.$ So, the equation of the red circle is $$x^2+y^2=frac14left(a_x^2+a_y^2 right).$$



          Where does this circle hits the straight $y=y_b$? Now, you have two points: the upper right corner of the blue rectangle and the crossing point mentioned above. Can you calculate the angle between the two red lines?






          share|cite|improve this answer














          The following figure depicts the "original" arrangement when both rectangles are placed in the same coordinate system (red) with their centers in the origin:



          enter image description here



          Turn, about it center, the blue triangle counterclockwise until its upper-right corner meets the bigger rectangle. The angle at which this takes place is the same angle by which we would have turned the white rectangle clockwise to hit the blue rectangle.



          Take the red circle centered at the origin and calculate its radius: $r=frac12sqrta_x^2+a_y^2.$ So, the equation of the red circle is $$x^2+y^2=frac14left(a_x^2+a_y^2 right).$$



          Where does this circle hits the straight $y=y_b$? Now, you have two points: the upper right corner of the blue rectangle and the crossing point mentioned above. Can you calculate the angle between the two red lines?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Sep 7 at 9:59

























          answered Sep 5 at 12:24









          zoli

          16.4k41643




          16.4k41643







          • 1




            Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/…
            – jippyjoe4
            Sep 5 at 15:01










          • @jippyjoe4: In principle, your solution is OK.
            – zoli
            Sep 6 at 14:31










          • is that the answer you managed to come up with?
            – jippyjoe4
            Sep 6 at 21:11










          • Yes, yes. (+ a few characters)
            – zoli
            Sep 7 at 7:10












          • 1




            Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/…
            – jippyjoe4
            Sep 5 at 15:01










          • @jippyjoe4: In principle, your solution is OK.
            – zoli
            Sep 6 at 14:31










          • is that the answer you managed to come up with?
            – jippyjoe4
            Sep 6 at 21:11










          • Yes, yes. (+ a few characters)
            – zoli
            Sep 7 at 7:10







          1




          1




          Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/…
          – jippyjoe4
          Sep 5 at 15:01




          Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/…
          – jippyjoe4
          Sep 5 at 15:01












          @jippyjoe4: In principle, your solution is OK.
          – zoli
          Sep 6 at 14:31




          @jippyjoe4: In principle, your solution is OK.
          – zoli
          Sep 6 at 14:31












          is that the answer you managed to come up with?
          – jippyjoe4
          Sep 6 at 21:11




          is that the answer you managed to come up with?
          – jippyjoe4
          Sep 6 at 21:11












          Yes, yes. (+ a few characters)
          – zoli
          Sep 7 at 7:10




          Yes, yes. (+ a few characters)
          – zoli
          Sep 7 at 7:10

















           

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