Radius of a circle touching a rectangle both of which are inside a square

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
17
down vote

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Given this configuration :



enter image description here



We're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.



If we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $ sqrt2times R-R$)



I really can't understand how to solve it. Any help appreciated.










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  • 1




    This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this.
    – The Count
    Sep 13 at 18:29






  • 2




    The information you have is $R(1-cosalpha) = 2a$ and $R(1-sinalpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$.
    – amsmath
    Sep 13 at 18:29







  • 1




    @amsmath juuuusssssttt beat me to it. Nice!
    – The Count
    Sep 13 at 18:31






  • 1




    If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
    – SMM
    Sep 13 at 18:31







  • 1




    This question could benefit from adding the 20 cm and 10 cm to the image itself.
    – NotThatGuy
    Sep 14 at 13:23














up vote
17
down vote

favorite
2












Given this configuration :



enter image description here



We're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.



If we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $ sqrt2times R-R$)



I really can't understand how to solve it. Any help appreciated.










share|cite|improve this question

















  • 1




    This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this.
    – The Count
    Sep 13 at 18:29






  • 2




    The information you have is $R(1-cosalpha) = 2a$ and $R(1-sinalpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$.
    – amsmath
    Sep 13 at 18:29







  • 1




    @amsmath juuuusssssttt beat me to it. Nice!
    – The Count
    Sep 13 at 18:31






  • 1




    If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
    – SMM
    Sep 13 at 18:31







  • 1




    This question could benefit from adding the 20 cm and 10 cm to the image itself.
    – NotThatGuy
    Sep 14 at 13:23












up vote
17
down vote

favorite
2









up vote
17
down vote

favorite
2






2





Given this configuration :



enter image description here



We're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.



If we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $ sqrt2times R-R$)



I really can't understand how to solve it. Any help appreciated.










share|cite|improve this question













Given this configuration :



enter image description here



We're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.



If we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $ sqrt2times R-R$)



I really can't understand how to solve it. Any help appreciated.







geometry circle rectangles






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asked Sep 13 at 18:19









Mooncrater

291112




291112







  • 1




    This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this.
    – The Count
    Sep 13 at 18:29






  • 2




    The information you have is $R(1-cosalpha) = 2a$ and $R(1-sinalpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$.
    – amsmath
    Sep 13 at 18:29







  • 1




    @amsmath juuuusssssttt beat me to it. Nice!
    – The Count
    Sep 13 at 18:31






  • 1




    If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
    – SMM
    Sep 13 at 18:31







  • 1




    This question could benefit from adding the 20 cm and 10 cm to the image itself.
    – NotThatGuy
    Sep 14 at 13:23












  • 1




    This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this.
    – The Count
    Sep 13 at 18:29






  • 2




    The information you have is $R(1-cosalpha) = 2a$ and $R(1-sinalpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$.
    – amsmath
    Sep 13 at 18:29







  • 1




    @amsmath juuuusssssttt beat me to it. Nice!
    – The Count
    Sep 13 at 18:31






  • 1




    If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
    – SMM
    Sep 13 at 18:31







  • 1




    This question could benefit from adding the 20 cm and 10 cm to the image itself.
    – NotThatGuy
    Sep 14 at 13:23







1




1




This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this.
– The Count
Sep 13 at 18:29




This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this.
– The Count
Sep 13 at 18:29




2




2




The information you have is $R(1-cosalpha) = 2a$ and $R(1-sinalpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$.
– amsmath
Sep 13 at 18:29





The information you have is $R(1-cosalpha) = 2a$ and $R(1-sinalpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$.
– amsmath
Sep 13 at 18:29





1




1




@amsmath juuuusssssttt beat me to it. Nice!
– The Count
Sep 13 at 18:31




@amsmath juuuusssssttt beat me to it. Nice!
– The Count
Sep 13 at 18:31




1




1




If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
– SMM
Sep 13 at 18:31





If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
– SMM
Sep 13 at 18:31





1




1




This question could benefit from adding the 20 cm and 10 cm to the image itself.
– NotThatGuy
Sep 14 at 13:23




This question could benefit from adding the 20 cm and 10 cm to the image itself.
– NotThatGuy
Sep 14 at 13:23










5 Answers
5






active

oldest

votes

















up vote
22
down vote



accepted










It is just using the pythagorean theorem:

$a=10$ $cm$

$b=20$ $cm$

$(r-a)^2+(r-b)^2=r^2$

$(r-10)^2+(r-20)^2=r^2$

$r^2+100-20r+r^2+400-40r=r^2$

$r^2-60r+500=0$

$r=50$ $cm$

$r=10$ $cm$

The $r=50$ $cm$ is the acceptable answer.



enter image description here






share|cite|improve this answer


















  • 3




    It's worth mentioning that with a slightly looser set of conditions on the question, r=10cm is a valid answer which corresponds to the rectangle occupying the entire top half of a 20x20 square.
    – Philip C
    Sep 14 at 7:12










  • @PhilipC, in that case how a circle with r=10cm can touch all four sides of a 20x20 square?
    – Seyed
    Sep 14 at 10:12










  • A circle of radius 10cm has a diameter of 20cm, i.e. is 20cm wide and 20cm high, the same as the square.
    – Philip C
    Sep 14 at 11:24










  • @PhilipC, yes but how is it going to touch the lower right corner of the rectangle externally?
    – Seyed
    Sep 14 at 11:40











  • @XanderHenderson, Is it better now?
    – Seyed
    Sep 14 at 12:19

















up vote
4
down vote













Place the center of the cricle at $O.$



Let the radius be $R$



The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.



Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$



And the distance from this point equals the $R.$



That should put you on your way to the solution.






share|cite|improve this answer



























    up vote
    3
    down vote













    $(x, y) = (R-20, R-10)$ as a point on the circle $y = sqrtR^2 - x^2$



    $R - 10 = sqrtR^2 - (R-20)^2$



    $(R- 10)^2 = R^2 - (R-20)^2$



    $R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$



    $R^2 - 60R + 500 = 0$



    $(R - 50)(R-10) = 0$



    $R = 50$ is the only sensible option.






    share|cite|improve this answer




















    • $R=10$ is also possible : the 20x10 rectangle covers the top half of a 20x20 square.
      – Eric Duminil
      Sep 14 at 8:10











    • @EricDuminil: But that's not what the diagram shows... The rectangle is clearly entirely outside the circle which it would not be if it covered the entire top half of the square...
      – Chris
      Sep 14 at 8:55










    • @Chris It fits with the title : Radius of a circle touching a rectangle both of which are inside a square and I tend to interpret a drawing as a guideline, not a perfect representation on which you could simply measure the solution. I guess it's a personal preference though.
      – Eric Duminil
      Sep 14 at 9:01










    • @Eric Duminil and Chris. Thanks for your comments. If you allow a solution of R = 10, which touches 2 corners and a side and intersects the rectangle, there are an infinity of solutions with similar characteristics. Example: a 40x40 square with a radius of R = 20 which doesn't contradict the title.
      – Phil H
      Sep 14 at 13:12










    • But if you assume that the circle is inscribed in the square, there are only 2 possibilities at most, right?
      – Eric Duminil
      Sep 14 at 13:50

















    up vote
    3
    down vote













    The point where rectangle touches the circle is $|R-a|$ and $|R-b|$ away from $x$ and $y$ axes, where $a$ and $b$ are lengths of sides of the rectangle and $R$ is the radius of the circle.



    This leads to the equation $$(R-a)^2 + (R-b)^2 = R^2,$$ which has solutions $$R_1,2 = a+bpmsqrt2ab.$$



    One solution corresponds to a bigger rectangle (compared to the circle), one touching the circle on the other side, which is not the case here. Smaller rectangle compared to the circle means that the circle is bigger if the rectangle is kept fixed, so the correct radius is $$R = a+b+sqrt2ab.$$



    Plugging in $a=10$ and $b=20$ gives $R=50$.






    share|cite|improve this answer



























      up vote
      1
      down vote













      You can use trig to get the same answer as those above.



      Draw three lines: One from the center of the circle to the corner shared by the square and the rectangle. Then draw a line from the center of the circle to the corner nearest to the center of the circle. Draw a final line being the diagonal connecting the previously mentioned corners.



      We know the length of the third line by the pythagorean theorem. If we call the side length of the square L, the length of the shorter of the two remaining lines is L/2. The length of the longer, L/sqrt(2).



      Find the angle that the diagonal makes with the longer of the drawn lines allows you to apply the cosine rule.



      The longer line meets the square's corner at a 45 degree angle with respect to either side. Then angle the diagonal makes with the left side of the square has a tangent of 2.



      Apply the cosine rule then solve the resulting quadratic and you get two possible answers, only one of which is plausible.






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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        22
        down vote



        accepted










        It is just using the pythagorean theorem:

        $a=10$ $cm$

        $b=20$ $cm$

        $(r-a)^2+(r-b)^2=r^2$

        $(r-10)^2+(r-20)^2=r^2$

        $r^2+100-20r+r^2+400-40r=r^2$

        $r^2-60r+500=0$

        $r=50$ $cm$

        $r=10$ $cm$

        The $r=50$ $cm$ is the acceptable answer.



        enter image description here






        share|cite|improve this answer


















        • 3




          It's worth mentioning that with a slightly looser set of conditions on the question, r=10cm is a valid answer which corresponds to the rectangle occupying the entire top half of a 20x20 square.
          – Philip C
          Sep 14 at 7:12










        • @PhilipC, in that case how a circle with r=10cm can touch all four sides of a 20x20 square?
          – Seyed
          Sep 14 at 10:12










        • A circle of radius 10cm has a diameter of 20cm, i.e. is 20cm wide and 20cm high, the same as the square.
          – Philip C
          Sep 14 at 11:24










        • @PhilipC, yes but how is it going to touch the lower right corner of the rectangle externally?
          – Seyed
          Sep 14 at 11:40











        • @XanderHenderson, Is it better now?
          – Seyed
          Sep 14 at 12:19














        up vote
        22
        down vote



        accepted










        It is just using the pythagorean theorem:

        $a=10$ $cm$

        $b=20$ $cm$

        $(r-a)^2+(r-b)^2=r^2$

        $(r-10)^2+(r-20)^2=r^2$

        $r^2+100-20r+r^2+400-40r=r^2$

        $r^2-60r+500=0$

        $r=50$ $cm$

        $r=10$ $cm$

        The $r=50$ $cm$ is the acceptable answer.



        enter image description here






        share|cite|improve this answer


















        • 3




          It's worth mentioning that with a slightly looser set of conditions on the question, r=10cm is a valid answer which corresponds to the rectangle occupying the entire top half of a 20x20 square.
          – Philip C
          Sep 14 at 7:12










        • @PhilipC, in that case how a circle with r=10cm can touch all four sides of a 20x20 square?
          – Seyed
          Sep 14 at 10:12










        • A circle of radius 10cm has a diameter of 20cm, i.e. is 20cm wide and 20cm high, the same as the square.
          – Philip C
          Sep 14 at 11:24










        • @PhilipC, yes but how is it going to touch the lower right corner of the rectangle externally?
          – Seyed
          Sep 14 at 11:40











        • @XanderHenderson, Is it better now?
          – Seyed
          Sep 14 at 12:19












        up vote
        22
        down vote



        accepted







        up vote
        22
        down vote



        accepted






        It is just using the pythagorean theorem:

        $a=10$ $cm$

        $b=20$ $cm$

        $(r-a)^2+(r-b)^2=r^2$

        $(r-10)^2+(r-20)^2=r^2$

        $r^2+100-20r+r^2+400-40r=r^2$

        $r^2-60r+500=0$

        $r=50$ $cm$

        $r=10$ $cm$

        The $r=50$ $cm$ is the acceptable answer.



        enter image description here






        share|cite|improve this answer














        It is just using the pythagorean theorem:

        $a=10$ $cm$

        $b=20$ $cm$

        $(r-a)^2+(r-b)^2=r^2$

        $(r-10)^2+(r-20)^2=r^2$

        $r^2+100-20r+r^2+400-40r=r^2$

        $r^2-60r+500=0$

        $r=50$ $cm$

        $r=10$ $cm$

        The $r=50$ $cm$ is the acceptable answer.



        enter image description here







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 14 at 12:19

























        answered Sep 13 at 19:10









        Seyed

        6,02031222




        6,02031222







        • 3




          It's worth mentioning that with a slightly looser set of conditions on the question, r=10cm is a valid answer which corresponds to the rectangle occupying the entire top half of a 20x20 square.
          – Philip C
          Sep 14 at 7:12










        • @PhilipC, in that case how a circle with r=10cm can touch all four sides of a 20x20 square?
          – Seyed
          Sep 14 at 10:12










        • A circle of radius 10cm has a diameter of 20cm, i.e. is 20cm wide and 20cm high, the same as the square.
          – Philip C
          Sep 14 at 11:24










        • @PhilipC, yes but how is it going to touch the lower right corner of the rectangle externally?
          – Seyed
          Sep 14 at 11:40











        • @XanderHenderson, Is it better now?
          – Seyed
          Sep 14 at 12:19












        • 3




          It's worth mentioning that with a slightly looser set of conditions on the question, r=10cm is a valid answer which corresponds to the rectangle occupying the entire top half of a 20x20 square.
          – Philip C
          Sep 14 at 7:12










        • @PhilipC, in that case how a circle with r=10cm can touch all four sides of a 20x20 square?
          – Seyed
          Sep 14 at 10:12










        • A circle of radius 10cm has a diameter of 20cm, i.e. is 20cm wide and 20cm high, the same as the square.
          – Philip C
          Sep 14 at 11:24










        • @PhilipC, yes but how is it going to touch the lower right corner of the rectangle externally?
          – Seyed
          Sep 14 at 11:40











        • @XanderHenderson, Is it better now?
          – Seyed
          Sep 14 at 12:19







        3




        3




        It's worth mentioning that with a slightly looser set of conditions on the question, r=10cm is a valid answer which corresponds to the rectangle occupying the entire top half of a 20x20 square.
        – Philip C
        Sep 14 at 7:12




        It's worth mentioning that with a slightly looser set of conditions on the question, r=10cm is a valid answer which corresponds to the rectangle occupying the entire top half of a 20x20 square.
        – Philip C
        Sep 14 at 7:12












        @PhilipC, in that case how a circle with r=10cm can touch all four sides of a 20x20 square?
        – Seyed
        Sep 14 at 10:12




        @PhilipC, in that case how a circle with r=10cm can touch all four sides of a 20x20 square?
        – Seyed
        Sep 14 at 10:12












        A circle of radius 10cm has a diameter of 20cm, i.e. is 20cm wide and 20cm high, the same as the square.
        – Philip C
        Sep 14 at 11:24




        A circle of radius 10cm has a diameter of 20cm, i.e. is 20cm wide and 20cm high, the same as the square.
        – Philip C
        Sep 14 at 11:24












        @PhilipC, yes but how is it going to touch the lower right corner of the rectangle externally?
        – Seyed
        Sep 14 at 11:40





        @PhilipC, yes but how is it going to touch the lower right corner of the rectangle externally?
        – Seyed
        Sep 14 at 11:40













        @XanderHenderson, Is it better now?
        – Seyed
        Sep 14 at 12:19




        @XanderHenderson, Is it better now?
        – Seyed
        Sep 14 at 12:19










        up vote
        4
        down vote













        Place the center of the cricle at $O.$



        Let the radius be $R$



        The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.



        Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$



        And the distance from this point equals the $R.$



        That should put you on your way to the solution.






        share|cite|improve this answer
























          up vote
          4
          down vote













          Place the center of the cricle at $O.$



          Let the radius be $R$



          The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.



          Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$



          And the distance from this point equals the $R.$



          That should put you on your way to the solution.






          share|cite|improve this answer






















            up vote
            4
            down vote










            up vote
            4
            down vote









            Place the center of the cricle at $O.$



            Let the radius be $R$



            The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.



            Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$



            And the distance from this point equals the $R.$



            That should put you on your way to the solution.






            share|cite|improve this answer












            Place the center of the cricle at $O.$



            Let the radius be $R$



            The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.



            Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$



            And the distance from this point equals the $R.$



            That should put you on your way to the solution.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 13 at 18:34









            Doug M

            40.3k31751




            40.3k31751




















                up vote
                3
                down vote













                $(x, y) = (R-20, R-10)$ as a point on the circle $y = sqrtR^2 - x^2$



                $R - 10 = sqrtR^2 - (R-20)^2$



                $(R- 10)^2 = R^2 - (R-20)^2$



                $R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$



                $R^2 - 60R + 500 = 0$



                $(R - 50)(R-10) = 0$



                $R = 50$ is the only sensible option.






                share|cite|improve this answer




















                • $R=10$ is also possible : the 20x10 rectangle covers the top half of a 20x20 square.
                  – Eric Duminil
                  Sep 14 at 8:10











                • @EricDuminil: But that's not what the diagram shows... The rectangle is clearly entirely outside the circle which it would not be if it covered the entire top half of the square...
                  – Chris
                  Sep 14 at 8:55










                • @Chris It fits with the title : Radius of a circle touching a rectangle both of which are inside a square and I tend to interpret a drawing as a guideline, not a perfect representation on which you could simply measure the solution. I guess it's a personal preference though.
                  – Eric Duminil
                  Sep 14 at 9:01










                • @Eric Duminil and Chris. Thanks for your comments. If you allow a solution of R = 10, which touches 2 corners and a side and intersects the rectangle, there are an infinity of solutions with similar characteristics. Example: a 40x40 square with a radius of R = 20 which doesn't contradict the title.
                  – Phil H
                  Sep 14 at 13:12










                • But if you assume that the circle is inscribed in the square, there are only 2 possibilities at most, right?
                  – Eric Duminil
                  Sep 14 at 13:50














                up vote
                3
                down vote













                $(x, y) = (R-20, R-10)$ as a point on the circle $y = sqrtR^2 - x^2$



                $R - 10 = sqrtR^2 - (R-20)^2$



                $(R- 10)^2 = R^2 - (R-20)^2$



                $R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$



                $R^2 - 60R + 500 = 0$



                $(R - 50)(R-10) = 0$



                $R = 50$ is the only sensible option.






                share|cite|improve this answer




















                • $R=10$ is also possible : the 20x10 rectangle covers the top half of a 20x20 square.
                  – Eric Duminil
                  Sep 14 at 8:10











                • @EricDuminil: But that's not what the diagram shows... The rectangle is clearly entirely outside the circle which it would not be if it covered the entire top half of the square...
                  – Chris
                  Sep 14 at 8:55










                • @Chris It fits with the title : Radius of a circle touching a rectangle both of which are inside a square and I tend to interpret a drawing as a guideline, not a perfect representation on which you could simply measure the solution. I guess it's a personal preference though.
                  – Eric Duminil
                  Sep 14 at 9:01










                • @Eric Duminil and Chris. Thanks for your comments. If you allow a solution of R = 10, which touches 2 corners and a side and intersects the rectangle, there are an infinity of solutions with similar characteristics. Example: a 40x40 square with a radius of R = 20 which doesn't contradict the title.
                  – Phil H
                  Sep 14 at 13:12










                • But if you assume that the circle is inscribed in the square, there are only 2 possibilities at most, right?
                  – Eric Duminil
                  Sep 14 at 13:50












                up vote
                3
                down vote










                up vote
                3
                down vote









                $(x, y) = (R-20, R-10)$ as a point on the circle $y = sqrtR^2 - x^2$



                $R - 10 = sqrtR^2 - (R-20)^2$



                $(R- 10)^2 = R^2 - (R-20)^2$



                $R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$



                $R^2 - 60R + 500 = 0$



                $(R - 50)(R-10) = 0$



                $R = 50$ is the only sensible option.






                share|cite|improve this answer












                $(x, y) = (R-20, R-10)$ as a point on the circle $y = sqrtR^2 - x^2$



                $R - 10 = sqrtR^2 - (R-20)^2$



                $(R- 10)^2 = R^2 - (R-20)^2$



                $R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$



                $R^2 - 60R + 500 = 0$



                $(R - 50)(R-10) = 0$



                $R = 50$ is the only sensible option.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 13 at 18:48









                Phil H

                2,5512311




                2,5512311











                • $R=10$ is also possible : the 20x10 rectangle covers the top half of a 20x20 square.
                  – Eric Duminil
                  Sep 14 at 8:10











                • @EricDuminil: But that's not what the diagram shows... The rectangle is clearly entirely outside the circle which it would not be if it covered the entire top half of the square...
                  – Chris
                  Sep 14 at 8:55










                • @Chris It fits with the title : Radius of a circle touching a rectangle both of which are inside a square and I tend to interpret a drawing as a guideline, not a perfect representation on which you could simply measure the solution. I guess it's a personal preference though.
                  – Eric Duminil
                  Sep 14 at 9:01










                • @Eric Duminil and Chris. Thanks for your comments. If you allow a solution of R = 10, which touches 2 corners and a side and intersects the rectangle, there are an infinity of solutions with similar characteristics. Example: a 40x40 square with a radius of R = 20 which doesn't contradict the title.
                  – Phil H
                  Sep 14 at 13:12










                • But if you assume that the circle is inscribed in the square, there are only 2 possibilities at most, right?
                  – Eric Duminil
                  Sep 14 at 13:50
















                • $R=10$ is also possible : the 20x10 rectangle covers the top half of a 20x20 square.
                  – Eric Duminil
                  Sep 14 at 8:10











                • @EricDuminil: But that's not what the diagram shows... The rectangle is clearly entirely outside the circle which it would not be if it covered the entire top half of the square...
                  – Chris
                  Sep 14 at 8:55










                • @Chris It fits with the title : Radius of a circle touching a rectangle both of which are inside a square and I tend to interpret a drawing as a guideline, not a perfect representation on which you could simply measure the solution. I guess it's a personal preference though.
                  – Eric Duminil
                  Sep 14 at 9:01










                • @Eric Duminil and Chris. Thanks for your comments. If you allow a solution of R = 10, which touches 2 corners and a side and intersects the rectangle, there are an infinity of solutions with similar characteristics. Example: a 40x40 square with a radius of R = 20 which doesn't contradict the title.
                  – Phil H
                  Sep 14 at 13:12










                • But if you assume that the circle is inscribed in the square, there are only 2 possibilities at most, right?
                  – Eric Duminil
                  Sep 14 at 13:50















                $R=10$ is also possible : the 20x10 rectangle covers the top half of a 20x20 square.
                – Eric Duminil
                Sep 14 at 8:10





                $R=10$ is also possible : the 20x10 rectangle covers the top half of a 20x20 square.
                – Eric Duminil
                Sep 14 at 8:10













                @EricDuminil: But that's not what the diagram shows... The rectangle is clearly entirely outside the circle which it would not be if it covered the entire top half of the square...
                – Chris
                Sep 14 at 8:55




                @EricDuminil: But that's not what the diagram shows... The rectangle is clearly entirely outside the circle which it would not be if it covered the entire top half of the square...
                – Chris
                Sep 14 at 8:55












                @Chris It fits with the title : Radius of a circle touching a rectangle both of which are inside a square and I tend to interpret a drawing as a guideline, not a perfect representation on which you could simply measure the solution. I guess it's a personal preference though.
                – Eric Duminil
                Sep 14 at 9:01




                @Chris It fits with the title : Radius of a circle touching a rectangle both of which are inside a square and I tend to interpret a drawing as a guideline, not a perfect representation on which you could simply measure the solution. I guess it's a personal preference though.
                – Eric Duminil
                Sep 14 at 9:01












                @Eric Duminil and Chris. Thanks for your comments. If you allow a solution of R = 10, which touches 2 corners and a side and intersects the rectangle, there are an infinity of solutions with similar characteristics. Example: a 40x40 square with a radius of R = 20 which doesn't contradict the title.
                – Phil H
                Sep 14 at 13:12




                @Eric Duminil and Chris. Thanks for your comments. If you allow a solution of R = 10, which touches 2 corners and a side and intersects the rectangle, there are an infinity of solutions with similar characteristics. Example: a 40x40 square with a radius of R = 20 which doesn't contradict the title.
                – Phil H
                Sep 14 at 13:12












                But if you assume that the circle is inscribed in the square, there are only 2 possibilities at most, right?
                – Eric Duminil
                Sep 14 at 13:50




                But if you assume that the circle is inscribed in the square, there are only 2 possibilities at most, right?
                – Eric Duminil
                Sep 14 at 13:50










                up vote
                3
                down vote













                The point where rectangle touches the circle is $|R-a|$ and $|R-b|$ away from $x$ and $y$ axes, where $a$ and $b$ are lengths of sides of the rectangle and $R$ is the radius of the circle.



                This leads to the equation $$(R-a)^2 + (R-b)^2 = R^2,$$ which has solutions $$R_1,2 = a+bpmsqrt2ab.$$



                One solution corresponds to a bigger rectangle (compared to the circle), one touching the circle on the other side, which is not the case here. Smaller rectangle compared to the circle means that the circle is bigger if the rectangle is kept fixed, so the correct radius is $$R = a+b+sqrt2ab.$$



                Plugging in $a=10$ and $b=20$ gives $R=50$.






                share|cite|improve this answer
























                  up vote
                  3
                  down vote













                  The point where rectangle touches the circle is $|R-a|$ and $|R-b|$ away from $x$ and $y$ axes, where $a$ and $b$ are lengths of sides of the rectangle and $R$ is the radius of the circle.



                  This leads to the equation $$(R-a)^2 + (R-b)^2 = R^2,$$ which has solutions $$R_1,2 = a+bpmsqrt2ab.$$



                  One solution corresponds to a bigger rectangle (compared to the circle), one touching the circle on the other side, which is not the case here. Smaller rectangle compared to the circle means that the circle is bigger if the rectangle is kept fixed, so the correct radius is $$R = a+b+sqrt2ab.$$



                  Plugging in $a=10$ and $b=20$ gives $R=50$.






                  share|cite|improve this answer






















                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    The point where rectangle touches the circle is $|R-a|$ and $|R-b|$ away from $x$ and $y$ axes, where $a$ and $b$ are lengths of sides of the rectangle and $R$ is the radius of the circle.



                    This leads to the equation $$(R-a)^2 + (R-b)^2 = R^2,$$ which has solutions $$R_1,2 = a+bpmsqrt2ab.$$



                    One solution corresponds to a bigger rectangle (compared to the circle), one touching the circle on the other side, which is not the case here. Smaller rectangle compared to the circle means that the circle is bigger if the rectangle is kept fixed, so the correct radius is $$R = a+b+sqrt2ab.$$



                    Plugging in $a=10$ and $b=20$ gives $R=50$.






                    share|cite|improve this answer












                    The point where rectangle touches the circle is $|R-a|$ and $|R-b|$ away from $x$ and $y$ axes, where $a$ and $b$ are lengths of sides of the rectangle and $R$ is the radius of the circle.



                    This leads to the equation $$(R-a)^2 + (R-b)^2 = R^2,$$ which has solutions $$R_1,2 = a+bpmsqrt2ab.$$



                    One solution corresponds to a bigger rectangle (compared to the circle), one touching the circle on the other side, which is not the case here. Smaller rectangle compared to the circle means that the circle is bigger if the rectangle is kept fixed, so the correct radius is $$R = a+b+sqrt2ab.$$



                    Plugging in $a=10$ and $b=20$ gives $R=50$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 14 at 6:13









                    Danijel

                    776417




                    776417




















                        up vote
                        1
                        down vote













                        You can use trig to get the same answer as those above.



                        Draw three lines: One from the center of the circle to the corner shared by the square and the rectangle. Then draw a line from the center of the circle to the corner nearest to the center of the circle. Draw a final line being the diagonal connecting the previously mentioned corners.



                        We know the length of the third line by the pythagorean theorem. If we call the side length of the square L, the length of the shorter of the two remaining lines is L/2. The length of the longer, L/sqrt(2).



                        Find the angle that the diagonal makes with the longer of the drawn lines allows you to apply the cosine rule.



                        The longer line meets the square's corner at a 45 degree angle with respect to either side. Then angle the diagonal makes with the left side of the square has a tangent of 2.



                        Apply the cosine rule then solve the resulting quadratic and you get two possible answers, only one of which is plausible.






                        share|cite|improve this answer
























                          up vote
                          1
                          down vote













                          You can use trig to get the same answer as those above.



                          Draw three lines: One from the center of the circle to the corner shared by the square and the rectangle. Then draw a line from the center of the circle to the corner nearest to the center of the circle. Draw a final line being the diagonal connecting the previously mentioned corners.



                          We know the length of the third line by the pythagorean theorem. If we call the side length of the square L, the length of the shorter of the two remaining lines is L/2. The length of the longer, L/sqrt(2).



                          Find the angle that the diagonal makes with the longer of the drawn lines allows you to apply the cosine rule.



                          The longer line meets the square's corner at a 45 degree angle with respect to either side. Then angle the diagonal makes with the left side of the square has a tangent of 2.



                          Apply the cosine rule then solve the resulting quadratic and you get two possible answers, only one of which is plausible.






                          share|cite|improve this answer






















                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            You can use trig to get the same answer as those above.



                            Draw three lines: One from the center of the circle to the corner shared by the square and the rectangle. Then draw a line from the center of the circle to the corner nearest to the center of the circle. Draw a final line being the diagonal connecting the previously mentioned corners.



                            We know the length of the third line by the pythagorean theorem. If we call the side length of the square L, the length of the shorter of the two remaining lines is L/2. The length of the longer, L/sqrt(2).



                            Find the angle that the diagonal makes with the longer of the drawn lines allows you to apply the cosine rule.



                            The longer line meets the square's corner at a 45 degree angle with respect to either side. Then angle the diagonal makes with the left side of the square has a tangent of 2.



                            Apply the cosine rule then solve the resulting quadratic and you get two possible answers, only one of which is plausible.






                            share|cite|improve this answer












                            You can use trig to get the same answer as those above.



                            Draw three lines: One from the center of the circle to the corner shared by the square and the rectangle. Then draw a line from the center of the circle to the corner nearest to the center of the circle. Draw a final line being the diagonal connecting the previously mentioned corners.



                            We know the length of the third line by the pythagorean theorem. If we call the side length of the square L, the length of the shorter of the two remaining lines is L/2. The length of the longer, L/sqrt(2).



                            Find the angle that the diagonal makes with the longer of the drawn lines allows you to apply the cosine rule.



                            The longer line meets the square's corner at a 45 degree angle with respect to either side. Then angle the diagonal makes with the left side of the square has a tangent of 2.



                            Apply the cosine rule then solve the resulting quadratic and you get two possible answers, only one of which is plausible.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Sep 13 at 21:35









                            TurlocTheRed

                            1007




                            1007



























                                 

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