Points closest to the edge of a square












1












$begingroup$


Let S be the square formed by the four vertices (1,1),(1,-1),(-1,1), and (-1,-1). Let the region R be the set of points inside S which are closer to the centre than to any of the four sides. Find the area of region R.



I figure that the area is the area between 4 parabolas, which I can't seem to find.



Some help needed here! (Full proof will be appreciated but hints don't hurt :))






conic-sections coordinate-systems







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you use integration?
    $endgroup$
    – Federico
    Dec 10 '18 at 18:00










  • $begingroup$
    Yep sure! I think you have to calculate the area between 4 curves but please go ahead!
    $endgroup$
    – Pratik Apshinge
    Dec 10 '18 at 18:09










  • $begingroup$
    Have you tried calculating the area inside the parabolas and subtracting from the area of the square?
    $endgroup$
    – saulspatz
    Dec 10 '18 at 18:12










  • $begingroup$
    Split the area into $4$ equal parts, one in each quadrant
    $endgroup$
    – Shubham Johri
    Dec 10 '18 at 18:24
















1












$begingroup$


Let S be the square formed by the four vertices (1,1),(1,-1),(-1,1), and (-1,-1). Let the region R be the set of points inside S which are closer to the centre than to any of the four sides. Find the area of region R.



I figure that the area is the area between 4 parabolas, which I can't seem to find.



Some help needed here! (Full proof will be appreciated but hints don't hurt :))






conic-sections coordinate-systems







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you use integration?
    $endgroup$
    – Federico
    Dec 10 '18 at 18:00










  • $begingroup$
    Yep sure! I think you have to calculate the area between 4 curves but please go ahead!
    $endgroup$
    – Pratik Apshinge
    Dec 10 '18 at 18:09










  • $begingroup$
    Have you tried calculating the area inside the parabolas and subtracting from the area of the square?
    $endgroup$
    – saulspatz
    Dec 10 '18 at 18:12










  • $begingroup$
    Split the area into $4$ equal parts, one in each quadrant
    $endgroup$
    – Shubham Johri
    Dec 10 '18 at 18:24














1












1








1





$begingroup$


Let S be the square formed by the four vertices (1,1),(1,-1),(-1,1), and (-1,-1). Let the region R be the set of points inside S which are closer to the centre than to any of the four sides. Find the area of region R.



I figure that the area is the area between 4 parabolas, which I can't seem to find.



Some help needed here! (Full proof will be appreciated but hints don't hurt :))






conic-sections coordinate-systems







share|cite|improve this question









$endgroup$




Let S be the square formed by the four vertices (1,1),(1,-1),(-1,1), and (-1,-1). Let the region R be the set of points inside S which are closer to the centre than to any of the four sides. Find the area of region R.



I figure that the area is the area between 4 parabolas, which I can't seem to find.



Some help needed here! (Full proof will be appreciated but hints don't hurt :))






conic-sections coordinate-systems




conic-sections coordinate-systems






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 17:40









Pratik ApshingePratik Apshinge

305




305












  • $begingroup$
    Can you use integration?
    $endgroup$
    – Federico
    Dec 10 '18 at 18:00










  • $begingroup$
    Yep sure! I think you have to calculate the area between 4 curves but please go ahead!
    $endgroup$
    – Pratik Apshinge
    Dec 10 '18 at 18:09










  • $begingroup$
    Have you tried calculating the area inside the parabolas and subtracting from the area of the square?
    $endgroup$
    – saulspatz
    Dec 10 '18 at 18:12










  • $begingroup$
    Split the area into $4$ equal parts, one in each quadrant
    $endgroup$
    – Shubham Johri
    Dec 10 '18 at 18:24


















  • $begingroup$
    Can you use integration?
    $endgroup$
    – Federico
    Dec 10 '18 at 18:00










  • $begingroup$
    Yep sure! I think you have to calculate the area between 4 curves but please go ahead!
    $endgroup$
    – Pratik Apshinge
    Dec 10 '18 at 18:09










  • $begingroup$
    Have you tried calculating the area inside the parabolas and subtracting from the area of the square?
    $endgroup$
    – saulspatz
    Dec 10 '18 at 18:12










  • $begingroup$
    Split the area into $4$ equal parts, one in each quadrant
    $endgroup$
    – Shubham Johri
    Dec 10 '18 at 18:24
















$begingroup$
Can you use integration?
$endgroup$
– Federico
Dec 10 '18 at 18:00




$begingroup$
Can you use integration?
$endgroup$
– Federico
Dec 10 '18 at 18:00












$begingroup$
Yep sure! I think you have to calculate the area between 4 curves but please go ahead!
$endgroup$
– Pratik Apshinge
Dec 10 '18 at 18:09




$begingroup$
Yep sure! I think you have to calculate the area between 4 curves but please go ahead!
$endgroup$
– Pratik Apshinge
Dec 10 '18 at 18:09












$begingroup$
Have you tried calculating the area inside the parabolas and subtracting from the area of the square?
$endgroup$
– saulspatz
Dec 10 '18 at 18:12




$begingroup$
Have you tried calculating the area inside the parabolas and subtracting from the area of the square?
$endgroup$
– saulspatz
Dec 10 '18 at 18:12












$begingroup$
Split the area into $4$ equal parts, one in each quadrant
$endgroup$
– Shubham Johri
Dec 10 '18 at 18:24




$begingroup$
Split the area into $4$ equal parts, one in each quadrant
$endgroup$
– Shubham Johri
Dec 10 '18 at 18:24










2 Answers
2






active

oldest

votes


















2












$begingroup$

For the point to be closer to the centre $C(0,0)$ of the square than any side,$sqrt{x^2+y^2}<min{|x-1|,x+1,|y-1|,y+1}$.



In the portion of the square in the $1^{st}$ quadrant,



$|x-1|=1-x<x+1, |y-1|=1-y<y+1\thereforesqrt{x^2+y^2}<min{1-x,1-y}$



When $xge y, sqrt{x^2+y^2}<1-x$ and when $x<y, sqrt{x^2+y^2}<1-y$



Simplify these to obtain $y^2<1-2x, xge y$ and $x^2<1-2y, y>x$ as the desired regions in the first quadrant. This area looks something like this:



Graph



$y^2=1-2x, x^2=1-2y$ intersect at $(sqrt2-1,sqrt2-1)$ in the $1^{st}$ quadrant. The area of this region is given by $A=int_{0}^{sqrt2-1} frac{1-x^2}2dx+int_{0}^{sqrt2-1} frac{1-y^2}2-(sqrt2-1)dy$



By symmetry, the total required region looks like this:



enter image description here



The area of this region is just $4A$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you again for yet another answer and the visual! It makes it much easier!
    $endgroup$
    – Pratik Apshinge
    Dec 11 '18 at 6:10



















1












$begingroup$

Due to symmetry, it is enough to calculate the case $xin[0,1]$ and $yin[0,x]$ and multiply the answer by $8$. So let's calculate the parabola first: on the parabola you are at the same distance from $(0,0)$ as from the side $x=1$. Then $$x^2+y^2=(1-x)^2$$You can rewrite this as $$y^2=1-2x$$
Now calculate the intersection with the $x$ axis and with $x=y$. For the $x$ axis you get $y=0$ and $x=1/2$. For the $x=y$ you get $x=pmsqrt 2 -1$, and only the positive solution is of interest in this case. Note that it is less then $1/2$. So $$A_{1/8}=int_0^{sqrt 2 -1} xdx+int_{sqrt 2 -1}^{1/2}sqrt{1-2x} dx$$






share|cite|improve this answer









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    2 Answers
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    2 Answers
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    active

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    2












    $begingroup$

    For the point to be closer to the centre $C(0,0)$ of the square than any side,$sqrt{x^2+y^2}<min{|x-1|,x+1,|y-1|,y+1}$.



    In the portion of the square in the $1^{st}$ quadrant,



    $|x-1|=1-x<x+1, |y-1|=1-y<y+1\thereforesqrt{x^2+y^2}<min{1-x,1-y}$



    When $xge y, sqrt{x^2+y^2}<1-x$ and when $x<y, sqrt{x^2+y^2}<1-y$



    Simplify these to obtain $y^2<1-2x, xge y$ and $x^2<1-2y, y>x$ as the desired regions in the first quadrant. This area looks something like this:



    Graph



    $y^2=1-2x, x^2=1-2y$ intersect at $(sqrt2-1,sqrt2-1)$ in the $1^{st}$ quadrant. The area of this region is given by $A=int_{0}^{sqrt2-1} frac{1-x^2}2dx+int_{0}^{sqrt2-1} frac{1-y^2}2-(sqrt2-1)dy$



    By symmetry, the total required region looks like this:



    enter image description here



    The area of this region is just $4A$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you again for yet another answer and the visual! It makes it much easier!
      $endgroup$
      – Pratik Apshinge
      Dec 11 '18 at 6:10
















    2












    $begingroup$

    For the point to be closer to the centre $C(0,0)$ of the square than any side,$sqrt{x^2+y^2}<min{|x-1|,x+1,|y-1|,y+1}$.



    In the portion of the square in the $1^{st}$ quadrant,



    $|x-1|=1-x<x+1, |y-1|=1-y<y+1\thereforesqrt{x^2+y^2}<min{1-x,1-y}$



    When $xge y, sqrt{x^2+y^2}<1-x$ and when $x<y, sqrt{x^2+y^2}<1-y$



    Simplify these to obtain $y^2<1-2x, xge y$ and $x^2<1-2y, y>x$ as the desired regions in the first quadrant. This area looks something like this:



    Graph



    $y^2=1-2x, x^2=1-2y$ intersect at $(sqrt2-1,sqrt2-1)$ in the $1^{st}$ quadrant. The area of this region is given by $A=int_{0}^{sqrt2-1} frac{1-x^2}2dx+int_{0}^{sqrt2-1} frac{1-y^2}2-(sqrt2-1)dy$



    By symmetry, the total required region looks like this:



    enter image description here



    The area of this region is just $4A$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you again for yet another answer and the visual! It makes it much easier!
      $endgroup$
      – Pratik Apshinge
      Dec 11 '18 at 6:10














    2












    2








    2





    $begingroup$

    For the point to be closer to the centre $C(0,0)$ of the square than any side,$sqrt{x^2+y^2}<min{|x-1|,x+1,|y-1|,y+1}$.



    In the portion of the square in the $1^{st}$ quadrant,



    $|x-1|=1-x<x+1, |y-1|=1-y<y+1\thereforesqrt{x^2+y^2}<min{1-x,1-y}$



    When $xge y, sqrt{x^2+y^2}<1-x$ and when $x<y, sqrt{x^2+y^2}<1-y$



    Simplify these to obtain $y^2<1-2x, xge y$ and $x^2<1-2y, y>x$ as the desired regions in the first quadrant. This area looks something like this:



    Graph



    $y^2=1-2x, x^2=1-2y$ intersect at $(sqrt2-1,sqrt2-1)$ in the $1^{st}$ quadrant. The area of this region is given by $A=int_{0}^{sqrt2-1} frac{1-x^2}2dx+int_{0}^{sqrt2-1} frac{1-y^2}2-(sqrt2-1)dy$



    By symmetry, the total required region looks like this:



    enter image description here



    The area of this region is just $4A$.






    share|cite|improve this answer











    $endgroup$



    For the point to be closer to the centre $C(0,0)$ of the square than any side,$sqrt{x^2+y^2}<min{|x-1|,x+1,|y-1|,y+1}$.



    In the portion of the square in the $1^{st}$ quadrant,



    $|x-1|=1-x<x+1, |y-1|=1-y<y+1\thereforesqrt{x^2+y^2}<min{1-x,1-y}$



    When $xge y, sqrt{x^2+y^2}<1-x$ and when $x<y, sqrt{x^2+y^2}<1-y$



    Simplify these to obtain $y^2<1-2x, xge y$ and $x^2<1-2y, y>x$ as the desired regions in the first quadrant. This area looks something like this:



    Graph



    $y^2=1-2x, x^2=1-2y$ intersect at $(sqrt2-1,sqrt2-1)$ in the $1^{st}$ quadrant. The area of this region is given by $A=int_{0}^{sqrt2-1} frac{1-x^2}2dx+int_{0}^{sqrt2-1} frac{1-y^2}2-(sqrt2-1)dy$



    By symmetry, the total required region looks like this:



    enter image description here



    The area of this region is just $4A$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 10 '18 at 18:37

























    answered Dec 10 '18 at 18:18









    Shubham JohriShubham Johri

    5,082717




    5,082717












    • $begingroup$
      Thank you again for yet another answer and the visual! It makes it much easier!
      $endgroup$
      – Pratik Apshinge
      Dec 11 '18 at 6:10


















    • $begingroup$
      Thank you again for yet another answer and the visual! It makes it much easier!
      $endgroup$
      – Pratik Apshinge
      Dec 11 '18 at 6:10
















    $begingroup$
    Thank you again for yet another answer and the visual! It makes it much easier!
    $endgroup$
    – Pratik Apshinge
    Dec 11 '18 at 6:10




    $begingroup$
    Thank you again for yet another answer and the visual! It makes it much easier!
    $endgroup$
    – Pratik Apshinge
    Dec 11 '18 at 6:10











    1












    $begingroup$

    Due to symmetry, it is enough to calculate the case $xin[0,1]$ and $yin[0,x]$ and multiply the answer by $8$. So let's calculate the parabola first: on the parabola you are at the same distance from $(0,0)$ as from the side $x=1$. Then $$x^2+y^2=(1-x)^2$$You can rewrite this as $$y^2=1-2x$$
    Now calculate the intersection with the $x$ axis and with $x=y$. For the $x$ axis you get $y=0$ and $x=1/2$. For the $x=y$ you get $x=pmsqrt 2 -1$, and only the positive solution is of interest in this case. Note that it is less then $1/2$. So $$A_{1/8}=int_0^{sqrt 2 -1} xdx+int_{sqrt 2 -1}^{1/2}sqrt{1-2x} dx$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Due to symmetry, it is enough to calculate the case $xin[0,1]$ and $yin[0,x]$ and multiply the answer by $8$. So let's calculate the parabola first: on the parabola you are at the same distance from $(0,0)$ as from the side $x=1$. Then $$x^2+y^2=(1-x)^2$$You can rewrite this as $$y^2=1-2x$$
      Now calculate the intersection with the $x$ axis and with $x=y$. For the $x$ axis you get $y=0$ and $x=1/2$. For the $x=y$ you get $x=pmsqrt 2 -1$, and only the positive solution is of interest in this case. Note that it is less then $1/2$. So $$A_{1/8}=int_0^{sqrt 2 -1} xdx+int_{sqrt 2 -1}^{1/2}sqrt{1-2x} dx$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Due to symmetry, it is enough to calculate the case $xin[0,1]$ and $yin[0,x]$ and multiply the answer by $8$. So let's calculate the parabola first: on the parabola you are at the same distance from $(0,0)$ as from the side $x=1$. Then $$x^2+y^2=(1-x)^2$$You can rewrite this as $$y^2=1-2x$$
        Now calculate the intersection with the $x$ axis and with $x=y$. For the $x$ axis you get $y=0$ and $x=1/2$. For the $x=y$ you get $x=pmsqrt 2 -1$, and only the positive solution is of interest in this case. Note that it is less then $1/2$. So $$A_{1/8}=int_0^{sqrt 2 -1} xdx+int_{sqrt 2 -1}^{1/2}sqrt{1-2x} dx$$






        share|cite|improve this answer









        $endgroup$



        Due to symmetry, it is enough to calculate the case $xin[0,1]$ and $yin[0,x]$ and multiply the answer by $8$. So let's calculate the parabola first: on the parabola you are at the same distance from $(0,0)$ as from the side $x=1$. Then $$x^2+y^2=(1-x)^2$$You can rewrite this as $$y^2=1-2x$$
        Now calculate the intersection with the $x$ axis and with $x=y$. For the $x$ axis you get $y=0$ and $x=1/2$. For the $x=y$ you get $x=pmsqrt 2 -1$, and only the positive solution is of interest in this case. Note that it is less then $1/2$. So $$A_{1/8}=int_0^{sqrt 2 -1} xdx+int_{sqrt 2 -1}^{1/2}sqrt{1-2x} dx$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 18:27









        AndreiAndrei

        11.8k21026




        11.8k21026






























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