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:

Clipping example:

geometry trigonometry transformation rotations rectangles
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up vote
1
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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:

Clipping example:

geometry trigonometry transformation rotations rectangles
add a comment |Â
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:

Clipping example:

geometry trigonometry transformation rotations rectangles
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:

Clipping example:

geometry trigonometry transformation rotations rectangles
geometry trigonometry transformation rotations rectangles
asked Sep 5 at 6:35
jippyjoe4
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1 Answer
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The following figure depicts the "original" arrangement when both rectangles are placed in the same coordinate system (red) with their centers in the origin:

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?
1
Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/â¦
â jippyjoe4
Sep 5 at 15:01
@jippyjoe4: In principle, your solution is OK.
â zoli
Sep 6 at 14:31
is that the answer you managed to come up with?
â jippyjoe4
Sep 6 at 21:11
Yes, yes. (+ a few characters)
â zoli
Sep 7 at 7:10
add a comment |Â
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:

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?
1
Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/â¦
â jippyjoe4
Sep 5 at 15:01
@jippyjoe4: In principle, your solution is OK.
â zoli
Sep 6 at 14:31
is that the answer you managed to come up with?
â jippyjoe4
Sep 6 at 21:11
Yes, yes. (+ a few characters)
â zoli
Sep 7 at 7:10
add a comment |Â
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:

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?
1
Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/â¦
â jippyjoe4
Sep 5 at 15:01
@jippyjoe4: In principle, your solution is OK.
â zoli
Sep 6 at 14:31
is that the answer you managed to come up with?
â jippyjoe4
Sep 6 at 21:11
Yes, yes. (+ a few characters)
â zoli
Sep 7 at 7:10
add a comment |Â
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:

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

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?
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/â¦
â jippyjoe4
Sep 5 at 15:01
@jippyjoe4: In principle, your solution is OK.
â zoli
Sep 6 at 14:31
is that the answer you managed to come up with?
â jippyjoe4
Sep 6 at 21:11
Yes, yes. (+ a few characters)
â zoli
Sep 7 at 7:10
add a comment |Â
1
Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/â¦
â jippyjoe4
Sep 5 at 15:01
@jippyjoe4: In principle, your solution is OK.
â zoli
Sep 6 at 14:31
is that the answer you managed to come up with?
â jippyjoe4
Sep 6 at 21:11
Yes, yes. (+ a few characters)
â zoli
Sep 7 at 7:10
1
1
Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/â¦
â jippyjoe4
Sep 5 at 15:01
Please make sure my calculations are correct drive.google.com/file/d/1dtz5LsM1zEP3PW2oXo4pWK5V0Bk1o-zy/â¦
â jippyjoe4
Sep 5 at 15:01
@jippyjoe4: In principle, your solution is OK.
â zoli
Sep 6 at 14:31
@jippyjoe4: In principle, your solution is OK.
â zoli
Sep 6 at 14:31
is that the answer you managed to come up with?
â jippyjoe4
Sep 6 at 21:11
is that the answer you managed to come up with?
â jippyjoe4
Sep 6 at 21:11
Yes, yes. (+ a few characters)
â zoli
Sep 7 at 7:10
Yes, yes. (+ a few characters)
â zoli
Sep 7 at 7:10
add a comment |Â
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