Finding the smallest possible perimeter of a 4-gon inscribed in a square












0












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"Take the square $ABCD$. Let $E_1$ be a point on the side $AB$ such that $AE_1 : E_1B = 2 : 1$. Find the points $E_2, E_3, E_4$ on the sides $BC, CD$, and $DB$ respectively such that the sum $|E_1E_2| + |E_2E_3| + |E_3E_4| + |E_4E_1|$ is the smallest. Find this sum."





My Attempt:



I first took the two triangles (formed by the diagonal $DB$) individually. I reflected $E_1$ in each of the sides and considered the line joining these two reflected points. Where that line hit the triangle on $DB$, I labelled as $E_4$. I did the same for the second triangle and formed a path.



I don't know if this method is correct and, if it is, I do not know how to find the perimeter of this shape, as the question requires. Any help?










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  • $begingroup$
    Please read tags before using them. The algebraic-geometry tag specifically mentions it is not for use in elementary problems involving both algebra and geometry.
    $endgroup$
    – KReiser
    Dec 17 '18 at 5:14
















0












$begingroup$


"Take the square $ABCD$. Let $E_1$ be a point on the side $AB$ such that $AE_1 : E_1B = 2 : 1$. Find the points $E_2, E_3, E_4$ on the sides $BC, CD$, and $DB$ respectively such that the sum $|E_1E_2| + |E_2E_3| + |E_3E_4| + |E_4E_1|$ is the smallest. Find this sum."





My Attempt:



I first took the two triangles (formed by the diagonal $DB$) individually. I reflected $E_1$ in each of the sides and considered the line joining these two reflected points. Where that line hit the triangle on $DB$, I labelled as $E_4$. I did the same for the second triangle and formed a path.



I don't know if this method is correct and, if it is, I do not know how to find the perimeter of this shape, as the question requires. Any help?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Please read tags before using them. The algebraic-geometry tag specifically mentions it is not for use in elementary problems involving both algebra and geometry.
    $endgroup$
    – KReiser
    Dec 17 '18 at 5:14














0












0








0





$begingroup$


"Take the square $ABCD$. Let $E_1$ be a point on the side $AB$ such that $AE_1 : E_1B = 2 : 1$. Find the points $E_2, E_3, E_4$ on the sides $BC, CD$, and $DB$ respectively such that the sum $|E_1E_2| + |E_2E_3| + |E_3E_4| + |E_4E_1|$ is the smallest. Find this sum."





My Attempt:



I first took the two triangles (formed by the diagonal $DB$) individually. I reflected $E_1$ in each of the sides and considered the line joining these two reflected points. Where that line hit the triangle on $DB$, I labelled as $E_4$. I did the same for the second triangle and formed a path.



I don't know if this method is correct and, if it is, I do not know how to find the perimeter of this shape, as the question requires. Any help?










share|cite|improve this question











$endgroup$




"Take the square $ABCD$. Let $E_1$ be a point on the side $AB$ such that $AE_1 : E_1B = 2 : 1$. Find the points $E_2, E_3, E_4$ on the sides $BC, CD$, and $DB$ respectively such that the sum $|E_1E_2| + |E_2E_3| + |E_3E_4| + |E_4E_1|$ is the smallest. Find this sum."





My Attempt:



I first took the two triangles (formed by the diagonal $DB$) individually. I reflected $E_1$ in each of the sides and considered the line joining these two reflected points. Where that line hit the triangle on $DB$, I labelled as $E_4$. I did the same for the second triangle and formed a path.



I don't know if this method is correct and, if it is, I do not know how to find the perimeter of this shape, as the question requires. Any help?







geometry euclidean-geometry






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edited Dec 17 '18 at 5:14









KReiser

9,72721435




9,72721435










asked Dec 17 '18 at 5:12









helpneededhelpneeded

897




897












  • $begingroup$
    Please read tags before using them. The algebraic-geometry tag specifically mentions it is not for use in elementary problems involving both algebra and geometry.
    $endgroup$
    – KReiser
    Dec 17 '18 at 5:14


















  • $begingroup$
    Please read tags before using them. The algebraic-geometry tag specifically mentions it is not for use in elementary problems involving both algebra and geometry.
    $endgroup$
    – KReiser
    Dec 17 '18 at 5:14
















$begingroup$
Please read tags before using them. The algebraic-geometry tag specifically mentions it is not for use in elementary problems involving both algebra and geometry.
$endgroup$
– KReiser
Dec 17 '18 at 5:14




$begingroup$
Please read tags before using them. The algebraic-geometry tag specifically mentions it is not for use in elementary problems involving both algebra and geometry.
$endgroup$
– KReiser
Dec 17 '18 at 5:14










4 Answers
4






active

oldest

votes


















3












$begingroup$

enter image description here



Whatever $E_2$, $E_3$, $E_4$ you choose,




  1. Reflect square $ABCD$ wrt side $BC$ to get square $A'BCD'$.

    The quadrilateral $E_1E_2E_3E_4$ get mapped to quadrilateral $F_1E_2F_3F_4$.


  2. Reflect square $A'BCD'$ wrt side $CD'$ to get square $A''B'CD'$.

    The quadrilateral $F_1E_2F_3F_4$ get mapped to quadrilateral $G_1G_2F_3G_4$.


  3. Reflect square $A''B'CD'$ wrt side $A''D'$ to get square $A''B''C'D'$.

    The quadrilateral $G_1G_2F_3G_4$ get mapped to quadrilateral $H_1H_2H_3G_4$.



The key is independent of choice of $E_2,E_3,E_4$, the point $H_1$ always mapped to same spot.
In fact, if one choose a coordinate system so that $A = (0,0), B = (1,0), C = (1,1)$ and $D = (0,1)$, one find $E_1 = (frac23,0)$ and $H_1 = (frac83,2)$. In general, we always have $|E_1H_1| = 2sqrt{2}|AB|$.



If one stare at above diagram long enough, one will notice





  • $|E_2E_3| = |E_2F_3|$,


  • $|E_3E_4| = |F_3F_4| = |F_3G_4|$,


  • $|E_4E_1| = |F_4F_1| = |G_4G_1| = |G_4H_1|$.


From this, we can deduce
$$begin{align} &|E_1E_2|+|E_2E_3|+|E_3E_4|+|E_4E_1|\
= &; |E_1E_2| + |E_2F_3| + |F_3G_4|+|G_4H_1|\
ge &; |E_1H_1| = 2sqrt{2}|AB|
end{align}$$



This means the perimeter we seek is bounded from below by $2sqrt{2}|AB|$.
To see this is the actual minimum perimeter, we can
repeatedly "fold" segment $E_1H_1$ back to square $ABC$ to construct
the "optimal" quadrilateral. We find when



$$BE_2 : E_2C = DE_4 : E_4A = 1 : 2quadtext{ and }quad CE_3 : E_3D = 2 : 1$$



the quadrilateral $E_1E_2E_3E_4$ does have $2sqrt{2}|AB|$ as perimeter.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
    $endgroup$
    – Quang Hoang
    Dec 17 '18 at 16:04










  • $begingroup$
    @Quang Hoang I think, my solution is the same.
    $endgroup$
    – Michael Rozenberg
    Dec 17 '18 at 17:34






  • 2




    $begingroup$
    Yes, exactly, but the picture helps clarify the proof a lot though.
    $endgroup$
    – Quang Hoang
    Dec 17 '18 at 17:35










  • $begingroup$
    @Quang Hoang I need to learn how to draw pictures.
    $endgroup$
    – Michael Rozenberg
    Dec 17 '18 at 17:36






  • 1




    $begingroup$
    @helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
    $endgroup$
    – achille hui
    Dec 19 '18 at 0:09





















2












$begingroup$

Let $ABC'D'$, $A''BC'D''$, $A'''B''C'D''$ and $A'''B'''C'''D''$ be squares



and $E_1'in A'''B'''$ such that $A'''E_1':E_1'B'''=2:1.$



Now, let $E_1E_1'cap BC'={E_2'},$ $E_1E_1'cap D''C'={E_3'}$ and $E_1E_1'cap A'''D''={E_4'}.$



Now, easy to see that $$E_1E_2+E_2E_3+E_3E_4+E_4E_1geq E_1E_1'=E_1E_2'+E_2'E_3'+E_3'E_4'+E_4'E_1'$$ and
$$BE_2:E_2C=BE_2':E_2C'=1:2,$$
$$CE_3:E_3D=C'E_3':E_3'D''=2:1$$ and
$$DE_4:E_4A=D''E_4':E_4'A'''=1:2.$$






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    1












    $begingroup$

    Hint: Start by drawing the $ABCD$ square, and the point $E_1$ on $AB$. Now let's choose a random point $E_3$ on $CD$. How do you get the shortest path from $E_1$ to $E_3$ so that it touches $BC$ in some $E_2$? Now repeat the procedure for the $AD$ side as well. Once you have these steps, there is just one more (similar) step to find the final answer.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
      $endgroup$
      – helpneeded
      Dec 17 '18 at 5:54










    • $begingroup$
      I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
      $endgroup$
      – Andrei
      Dec 17 '18 at 6:01










    • $begingroup$
      Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
      $endgroup$
      – helpneeded
      Dec 17 '18 at 6:04










    • $begingroup$
      Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
      $endgroup$
      – Andrei
      Dec 17 '18 at 6:23



















    1












    $begingroup$

    Let $A,B,C$ be three corners of the square in rotational order. Draw a linecsegment $PQ$ from point $P$ on $AB$ to point $Q$ on $BC$. Using SOH CAH TOA on right triangle $PQB$ infer that



    $|PB|=|PQ|cosangle QPB$



    $|QB|=|PQ|sinangle QPB$



    $dfrac{|PQ|}{|PB|+|QB|}=dfrac{1}{cosangle QPB+sinangle QPB}ge dfrac{1}{sqrt{2}}$



    We know that $cosangle QPB+sinangle QPB lesqrt{2}$ because $(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)=2$.



    Applying this result to all four sides of the inscribed quadrilateral leads to the perimeter of this quadrilateral being $ge 1/sqrt{2}$ times that of the square. That this bound is sharp is proved from the simple case of the sides of the inscribed quadrilateral being parallel to the angle bisectors of the square (always possible given the symmetries involved).






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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      enter image description here



      Whatever $E_2$, $E_3$, $E_4$ you choose,




      1. Reflect square $ABCD$ wrt side $BC$ to get square $A'BCD'$.

        The quadrilateral $E_1E_2E_3E_4$ get mapped to quadrilateral $F_1E_2F_3F_4$.


      2. Reflect square $A'BCD'$ wrt side $CD'$ to get square $A''B'CD'$.

        The quadrilateral $F_1E_2F_3F_4$ get mapped to quadrilateral $G_1G_2F_3G_4$.


      3. Reflect square $A''B'CD'$ wrt side $A''D'$ to get square $A''B''C'D'$.

        The quadrilateral $G_1G_2F_3G_4$ get mapped to quadrilateral $H_1H_2H_3G_4$.



      The key is independent of choice of $E_2,E_3,E_4$, the point $H_1$ always mapped to same spot.
      In fact, if one choose a coordinate system so that $A = (0,0), B = (1,0), C = (1,1)$ and $D = (0,1)$, one find $E_1 = (frac23,0)$ and $H_1 = (frac83,2)$. In general, we always have $|E_1H_1| = 2sqrt{2}|AB|$.



      If one stare at above diagram long enough, one will notice





      • $|E_2E_3| = |E_2F_3|$,


      • $|E_3E_4| = |F_3F_4| = |F_3G_4|$,


      • $|E_4E_1| = |F_4F_1| = |G_4G_1| = |G_4H_1|$.


      From this, we can deduce
      $$begin{align} &|E_1E_2|+|E_2E_3|+|E_3E_4|+|E_4E_1|\
      = &; |E_1E_2| + |E_2F_3| + |F_3G_4|+|G_4H_1|\
      ge &; |E_1H_1| = 2sqrt{2}|AB|
      end{align}$$



      This means the perimeter we seek is bounded from below by $2sqrt{2}|AB|$.
      To see this is the actual minimum perimeter, we can
      repeatedly "fold" segment $E_1H_1$ back to square $ABC$ to construct
      the "optimal" quadrilateral. We find when



      $$BE_2 : E_2C = DE_4 : E_4A = 1 : 2quadtext{ and }quad CE_3 : E_3D = 2 : 1$$



      the quadrilateral $E_1E_2E_3E_4$ does have $2sqrt{2}|AB|$ as perimeter.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
        $endgroup$
        – Quang Hoang
        Dec 17 '18 at 16:04










      • $begingroup$
        @Quang Hoang I think, my solution is the same.
        $endgroup$
        – Michael Rozenberg
        Dec 17 '18 at 17:34






      • 2




        $begingroup$
        Yes, exactly, but the picture helps clarify the proof a lot though.
        $endgroup$
        – Quang Hoang
        Dec 17 '18 at 17:35










      • $begingroup$
        @Quang Hoang I need to learn how to draw pictures.
        $endgroup$
        – Michael Rozenberg
        Dec 17 '18 at 17:36






      • 1




        $begingroup$
        @helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
        $endgroup$
        – achille hui
        Dec 19 '18 at 0:09


















      3












      $begingroup$

      enter image description here



      Whatever $E_2$, $E_3$, $E_4$ you choose,




      1. Reflect square $ABCD$ wrt side $BC$ to get square $A'BCD'$.

        The quadrilateral $E_1E_2E_3E_4$ get mapped to quadrilateral $F_1E_2F_3F_4$.


      2. Reflect square $A'BCD'$ wrt side $CD'$ to get square $A''B'CD'$.

        The quadrilateral $F_1E_2F_3F_4$ get mapped to quadrilateral $G_1G_2F_3G_4$.


      3. Reflect square $A''B'CD'$ wrt side $A''D'$ to get square $A''B''C'D'$.

        The quadrilateral $G_1G_2F_3G_4$ get mapped to quadrilateral $H_1H_2H_3G_4$.



      The key is independent of choice of $E_2,E_3,E_4$, the point $H_1$ always mapped to same spot.
      In fact, if one choose a coordinate system so that $A = (0,0), B = (1,0), C = (1,1)$ and $D = (0,1)$, one find $E_1 = (frac23,0)$ and $H_1 = (frac83,2)$. In general, we always have $|E_1H_1| = 2sqrt{2}|AB|$.



      If one stare at above diagram long enough, one will notice





      • $|E_2E_3| = |E_2F_3|$,


      • $|E_3E_4| = |F_3F_4| = |F_3G_4|$,


      • $|E_4E_1| = |F_4F_1| = |G_4G_1| = |G_4H_1|$.


      From this, we can deduce
      $$begin{align} &|E_1E_2|+|E_2E_3|+|E_3E_4|+|E_4E_1|\
      = &; |E_1E_2| + |E_2F_3| + |F_3G_4|+|G_4H_1|\
      ge &; |E_1H_1| = 2sqrt{2}|AB|
      end{align}$$



      This means the perimeter we seek is bounded from below by $2sqrt{2}|AB|$.
      To see this is the actual minimum perimeter, we can
      repeatedly "fold" segment $E_1H_1$ back to square $ABC$ to construct
      the "optimal" quadrilateral. We find when



      $$BE_2 : E_2C = DE_4 : E_4A = 1 : 2quadtext{ and }quad CE_3 : E_3D = 2 : 1$$



      the quadrilateral $E_1E_2E_3E_4$ does have $2sqrt{2}|AB|$ as perimeter.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
        $endgroup$
        – Quang Hoang
        Dec 17 '18 at 16:04










      • $begingroup$
        @Quang Hoang I think, my solution is the same.
        $endgroup$
        – Michael Rozenberg
        Dec 17 '18 at 17:34






      • 2




        $begingroup$
        Yes, exactly, but the picture helps clarify the proof a lot though.
        $endgroup$
        – Quang Hoang
        Dec 17 '18 at 17:35










      • $begingroup$
        @Quang Hoang I need to learn how to draw pictures.
        $endgroup$
        – Michael Rozenberg
        Dec 17 '18 at 17:36






      • 1




        $begingroup$
        @helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
        $endgroup$
        – achille hui
        Dec 19 '18 at 0:09
















      3












      3








      3





      $begingroup$

      enter image description here



      Whatever $E_2$, $E_3$, $E_4$ you choose,




      1. Reflect square $ABCD$ wrt side $BC$ to get square $A'BCD'$.

        The quadrilateral $E_1E_2E_3E_4$ get mapped to quadrilateral $F_1E_2F_3F_4$.


      2. Reflect square $A'BCD'$ wrt side $CD'$ to get square $A''B'CD'$.

        The quadrilateral $F_1E_2F_3F_4$ get mapped to quadrilateral $G_1G_2F_3G_4$.


      3. Reflect square $A''B'CD'$ wrt side $A''D'$ to get square $A''B''C'D'$.

        The quadrilateral $G_1G_2F_3G_4$ get mapped to quadrilateral $H_1H_2H_3G_4$.



      The key is independent of choice of $E_2,E_3,E_4$, the point $H_1$ always mapped to same spot.
      In fact, if one choose a coordinate system so that $A = (0,0), B = (1,0), C = (1,1)$ and $D = (0,1)$, one find $E_1 = (frac23,0)$ and $H_1 = (frac83,2)$. In general, we always have $|E_1H_1| = 2sqrt{2}|AB|$.



      If one stare at above diagram long enough, one will notice





      • $|E_2E_3| = |E_2F_3|$,


      • $|E_3E_4| = |F_3F_4| = |F_3G_4|$,


      • $|E_4E_1| = |F_4F_1| = |G_4G_1| = |G_4H_1|$.


      From this, we can deduce
      $$begin{align} &|E_1E_2|+|E_2E_3|+|E_3E_4|+|E_4E_1|\
      = &; |E_1E_2| + |E_2F_3| + |F_3G_4|+|G_4H_1|\
      ge &; |E_1H_1| = 2sqrt{2}|AB|
      end{align}$$



      This means the perimeter we seek is bounded from below by $2sqrt{2}|AB|$.
      To see this is the actual minimum perimeter, we can
      repeatedly "fold" segment $E_1H_1$ back to square $ABC$ to construct
      the "optimal" quadrilateral. We find when



      $$BE_2 : E_2C = DE_4 : E_4A = 1 : 2quadtext{ and }quad CE_3 : E_3D = 2 : 1$$



      the quadrilateral $E_1E_2E_3E_4$ does have $2sqrt{2}|AB|$ as perimeter.






      share|cite|improve this answer











      $endgroup$



      enter image description here



      Whatever $E_2$, $E_3$, $E_4$ you choose,




      1. Reflect square $ABCD$ wrt side $BC$ to get square $A'BCD'$.

        The quadrilateral $E_1E_2E_3E_4$ get mapped to quadrilateral $F_1E_2F_3F_4$.


      2. Reflect square $A'BCD'$ wrt side $CD'$ to get square $A''B'CD'$.

        The quadrilateral $F_1E_2F_3F_4$ get mapped to quadrilateral $G_1G_2F_3G_4$.


      3. Reflect square $A''B'CD'$ wrt side $A''D'$ to get square $A''B''C'D'$.

        The quadrilateral $G_1G_2F_3G_4$ get mapped to quadrilateral $H_1H_2H_3G_4$.



      The key is independent of choice of $E_2,E_3,E_4$, the point $H_1$ always mapped to same spot.
      In fact, if one choose a coordinate system so that $A = (0,0), B = (1,0), C = (1,1)$ and $D = (0,1)$, one find $E_1 = (frac23,0)$ and $H_1 = (frac83,2)$. In general, we always have $|E_1H_1| = 2sqrt{2}|AB|$.



      If one stare at above diagram long enough, one will notice





      • $|E_2E_3| = |E_2F_3|$,


      • $|E_3E_4| = |F_3F_4| = |F_3G_4|$,


      • $|E_4E_1| = |F_4F_1| = |G_4G_1| = |G_4H_1|$.


      From this, we can deduce
      $$begin{align} &|E_1E_2|+|E_2E_3|+|E_3E_4|+|E_4E_1|\
      = &; |E_1E_2| + |E_2F_3| + |F_3G_4|+|G_4H_1|\
      ge &; |E_1H_1| = 2sqrt{2}|AB|
      end{align}$$



      This means the perimeter we seek is bounded from below by $2sqrt{2}|AB|$.
      To see this is the actual minimum perimeter, we can
      repeatedly "fold" segment $E_1H_1$ back to square $ABC$ to construct
      the "optimal" quadrilateral. We find when



      $$BE_2 : E_2C = DE_4 : E_4A = 1 : 2quadtext{ and }quad CE_3 : E_3D = 2 : 1$$



      the quadrilateral $E_1E_2E_3E_4$ does have $2sqrt{2}|AB|$ as perimeter.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 17 '18 at 18:38

























      answered Dec 17 '18 at 14:37









      achille huiachille hui

      96k5132259




      96k5132259












      • $begingroup$
        Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
        $endgroup$
        – Quang Hoang
        Dec 17 '18 at 16:04










      • $begingroup$
        @Quang Hoang I think, my solution is the same.
        $endgroup$
        – Michael Rozenberg
        Dec 17 '18 at 17:34






      • 2




        $begingroup$
        Yes, exactly, but the picture helps clarify the proof a lot though.
        $endgroup$
        – Quang Hoang
        Dec 17 '18 at 17:35










      • $begingroup$
        @Quang Hoang I need to learn how to draw pictures.
        $endgroup$
        – Michael Rozenberg
        Dec 17 '18 at 17:36






      • 1




        $begingroup$
        @helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
        $endgroup$
        – achille hui
        Dec 19 '18 at 0:09




















      • $begingroup$
        Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
        $endgroup$
        – Quang Hoang
        Dec 17 '18 at 16:04










      • $begingroup$
        @Quang Hoang I think, my solution is the same.
        $endgroup$
        – Michael Rozenberg
        Dec 17 '18 at 17:34






      • 2




        $begingroup$
        Yes, exactly, but the picture helps clarify the proof a lot though.
        $endgroup$
        – Quang Hoang
        Dec 17 '18 at 17:35










      • $begingroup$
        @Quang Hoang I need to learn how to draw pictures.
        $endgroup$
        – Michael Rozenberg
        Dec 17 '18 at 17:36






      • 1




        $begingroup$
        @helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
        $endgroup$
        – achille hui
        Dec 19 '18 at 0:09


















      $begingroup$
      Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
      $endgroup$
      – Quang Hoang
      Dec 17 '18 at 16:04




      $begingroup$
      Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
      $endgroup$
      – Quang Hoang
      Dec 17 '18 at 16:04












      $begingroup$
      @Quang Hoang I think, my solution is the same.
      $endgroup$
      – Michael Rozenberg
      Dec 17 '18 at 17:34




      $begingroup$
      @Quang Hoang I think, my solution is the same.
      $endgroup$
      – Michael Rozenberg
      Dec 17 '18 at 17:34




      2




      2




      $begingroup$
      Yes, exactly, but the picture helps clarify the proof a lot though.
      $endgroup$
      – Quang Hoang
      Dec 17 '18 at 17:35




      $begingroup$
      Yes, exactly, but the picture helps clarify the proof a lot though.
      $endgroup$
      – Quang Hoang
      Dec 17 '18 at 17:35












      $begingroup$
      @Quang Hoang I need to learn how to draw pictures.
      $endgroup$
      – Michael Rozenberg
      Dec 17 '18 at 17:36




      $begingroup$
      @Quang Hoang I need to learn how to draw pictures.
      $endgroup$
      – Michael Rozenberg
      Dec 17 '18 at 17:36




      1




      1




      $begingroup$
      @helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
      $endgroup$
      – achille hui
      Dec 19 '18 at 0:09






      $begingroup$
      @helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
      $endgroup$
      – achille hui
      Dec 19 '18 at 0:09













      2












      $begingroup$

      Let $ABC'D'$, $A''BC'D''$, $A'''B''C'D''$ and $A'''B'''C'''D''$ be squares



      and $E_1'in A'''B'''$ such that $A'''E_1':E_1'B'''=2:1.$



      Now, let $E_1E_1'cap BC'={E_2'},$ $E_1E_1'cap D''C'={E_3'}$ and $E_1E_1'cap A'''D''={E_4'}.$



      Now, easy to see that $$E_1E_2+E_2E_3+E_3E_4+E_4E_1geq E_1E_1'=E_1E_2'+E_2'E_3'+E_3'E_4'+E_4'E_1'$$ and
      $$BE_2:E_2C=BE_2':E_2C'=1:2,$$
      $$CE_3:E_3D=C'E_3':E_3'D''=2:1$$ and
      $$DE_4:E_4A=D''E_4':E_4'A'''=1:2.$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Let $ABC'D'$, $A''BC'D''$, $A'''B''C'D''$ and $A'''B'''C'''D''$ be squares



        and $E_1'in A'''B'''$ such that $A'''E_1':E_1'B'''=2:1.$



        Now, let $E_1E_1'cap BC'={E_2'},$ $E_1E_1'cap D''C'={E_3'}$ and $E_1E_1'cap A'''D''={E_4'}.$



        Now, easy to see that $$E_1E_2+E_2E_3+E_3E_4+E_4E_1geq E_1E_1'=E_1E_2'+E_2'E_3'+E_3'E_4'+E_4'E_1'$$ and
        $$BE_2:E_2C=BE_2':E_2C'=1:2,$$
        $$CE_3:E_3D=C'E_3':E_3'D''=2:1$$ and
        $$DE_4:E_4A=D''E_4':E_4'A'''=1:2.$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Let $ABC'D'$, $A''BC'D''$, $A'''B''C'D''$ and $A'''B'''C'''D''$ be squares



          and $E_1'in A'''B'''$ such that $A'''E_1':E_1'B'''=2:1.$



          Now, let $E_1E_1'cap BC'={E_2'},$ $E_1E_1'cap D''C'={E_3'}$ and $E_1E_1'cap A'''D''={E_4'}.$



          Now, easy to see that $$E_1E_2+E_2E_3+E_3E_4+E_4E_1geq E_1E_1'=E_1E_2'+E_2'E_3'+E_3'E_4'+E_4'E_1'$$ and
          $$BE_2:E_2C=BE_2':E_2C'=1:2,$$
          $$CE_3:E_3D=C'E_3':E_3'D''=2:1$$ and
          $$DE_4:E_4A=D''E_4':E_4'A'''=1:2.$$






          share|cite|improve this answer









          $endgroup$



          Let $ABC'D'$, $A''BC'D''$, $A'''B''C'D''$ and $A'''B'''C'''D''$ be squares



          and $E_1'in A'''B'''$ such that $A'''E_1':E_1'B'''=2:1.$



          Now, let $E_1E_1'cap BC'={E_2'},$ $E_1E_1'cap D''C'={E_3'}$ and $E_1E_1'cap A'''D''={E_4'}.$



          Now, easy to see that $$E_1E_2+E_2E_3+E_3E_4+E_4E_1geq E_1E_1'=E_1E_2'+E_2'E_3'+E_3'E_4'+E_4'E_1'$$ and
          $$BE_2:E_2C=BE_2':E_2C'=1:2,$$
          $$CE_3:E_3D=C'E_3':E_3'D''=2:1$$ and
          $$DE_4:E_4A=D''E_4':E_4'A'''=1:2.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 13:27









          Michael RozenbergMichael Rozenberg

          105k1892197




          105k1892197























              1












              $begingroup$

              Hint: Start by drawing the $ABCD$ square, and the point $E_1$ on $AB$. Now let's choose a random point $E_3$ on $CD$. How do you get the shortest path from $E_1$ to $E_3$ so that it touches $BC$ in some $E_2$? Now repeat the procedure for the $AD$ side as well. Once you have these steps, there is just one more (similar) step to find the final answer.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
                $endgroup$
                – helpneeded
                Dec 17 '18 at 5:54










              • $begingroup$
                I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
                $endgroup$
                – Andrei
                Dec 17 '18 at 6:01










              • $begingroup$
                Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
                $endgroup$
                – helpneeded
                Dec 17 '18 at 6:04










              • $begingroup$
                Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
                $endgroup$
                – Andrei
                Dec 17 '18 at 6:23
















              1












              $begingroup$

              Hint: Start by drawing the $ABCD$ square, and the point $E_1$ on $AB$. Now let's choose a random point $E_3$ on $CD$. How do you get the shortest path from $E_1$ to $E_3$ so that it touches $BC$ in some $E_2$? Now repeat the procedure for the $AD$ side as well. Once you have these steps, there is just one more (similar) step to find the final answer.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
                $endgroup$
                – helpneeded
                Dec 17 '18 at 5:54










              • $begingroup$
                I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
                $endgroup$
                – Andrei
                Dec 17 '18 at 6:01










              • $begingroup$
                Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
                $endgroup$
                – helpneeded
                Dec 17 '18 at 6:04










              • $begingroup$
                Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
                $endgroup$
                – Andrei
                Dec 17 '18 at 6:23














              1












              1








              1





              $begingroup$

              Hint: Start by drawing the $ABCD$ square, and the point $E_1$ on $AB$. Now let's choose a random point $E_3$ on $CD$. How do you get the shortest path from $E_1$ to $E_3$ so that it touches $BC$ in some $E_2$? Now repeat the procedure for the $AD$ side as well. Once you have these steps, there is just one more (similar) step to find the final answer.






              share|cite|improve this answer









              $endgroup$



              Hint: Start by drawing the $ABCD$ square, and the point $E_1$ on $AB$. Now let's choose a random point $E_3$ on $CD$. How do you get the shortest path from $E_1$ to $E_3$ so that it touches $BC$ in some $E_2$? Now repeat the procedure for the $AD$ side as well. Once you have these steps, there is just one more (similar) step to find the final answer.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 17 '18 at 5:44









              AndreiAndrei

              12.4k21128




              12.4k21128












              • $begingroup$
                Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
                $endgroup$
                – helpneeded
                Dec 17 '18 at 5:54










              • $begingroup$
                I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
                $endgroup$
                – Andrei
                Dec 17 '18 at 6:01










              • $begingroup$
                Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
                $endgroup$
                – helpneeded
                Dec 17 '18 at 6:04










              • $begingroup$
                Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
                $endgroup$
                – Andrei
                Dec 17 '18 at 6:23


















              • $begingroup$
                Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
                $endgroup$
                – helpneeded
                Dec 17 '18 at 5:54










              • $begingroup$
                I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
                $endgroup$
                – Andrei
                Dec 17 '18 at 6:01










              • $begingroup$
                Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
                $endgroup$
                – helpneeded
                Dec 17 '18 at 6:04










              • $begingroup$
                Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
                $endgroup$
                – Andrei
                Dec 17 '18 at 6:23
















              $begingroup$
              Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
              $endgroup$
              – helpneeded
              Dec 17 '18 at 5:54




              $begingroup$
              Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
              $endgroup$
              – helpneeded
              Dec 17 '18 at 5:54












              $begingroup$
              I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
              $endgroup$
              – Andrei
              Dec 17 '18 at 6:01




              $begingroup$
              I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
              $endgroup$
              – Andrei
              Dec 17 '18 at 6:01












              $begingroup$
              Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
              $endgroup$
              – helpneeded
              Dec 17 '18 at 6:04




              $begingroup$
              Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
              $endgroup$
              – helpneeded
              Dec 17 '18 at 6:04












              $begingroup$
              Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
              $endgroup$
              – Andrei
              Dec 17 '18 at 6:23




              $begingroup$
              Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
              $endgroup$
              – Andrei
              Dec 17 '18 at 6:23











              1












              $begingroup$

              Let $A,B,C$ be three corners of the square in rotational order. Draw a linecsegment $PQ$ from point $P$ on $AB$ to point $Q$ on $BC$. Using SOH CAH TOA on right triangle $PQB$ infer that



              $|PB|=|PQ|cosangle QPB$



              $|QB|=|PQ|sinangle QPB$



              $dfrac{|PQ|}{|PB|+|QB|}=dfrac{1}{cosangle QPB+sinangle QPB}ge dfrac{1}{sqrt{2}}$



              We know that $cosangle QPB+sinangle QPB lesqrt{2}$ because $(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)=2$.



              Applying this result to all four sides of the inscribed quadrilateral leads to the perimeter of this quadrilateral being $ge 1/sqrt{2}$ times that of the square. That this bound is sharp is proved from the simple case of the sides of the inscribed quadrilateral being parallel to the angle bisectors of the square (always possible given the symmetries involved).






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Let $A,B,C$ be three corners of the square in rotational order. Draw a linecsegment $PQ$ from point $P$ on $AB$ to point $Q$ on $BC$. Using SOH CAH TOA on right triangle $PQB$ infer that



                $|PB|=|PQ|cosangle QPB$



                $|QB|=|PQ|sinangle QPB$



                $dfrac{|PQ|}{|PB|+|QB|}=dfrac{1}{cosangle QPB+sinangle QPB}ge dfrac{1}{sqrt{2}}$



                We know that $cosangle QPB+sinangle QPB lesqrt{2}$ because $(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)=2$.



                Applying this result to all four sides of the inscribed quadrilateral leads to the perimeter of this quadrilateral being $ge 1/sqrt{2}$ times that of the square. That this bound is sharp is proved from the simple case of the sides of the inscribed quadrilateral being parallel to the angle bisectors of the square (always possible given the symmetries involved).






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let $A,B,C$ be three corners of the square in rotational order. Draw a linecsegment $PQ$ from point $P$ on $AB$ to point $Q$ on $BC$. Using SOH CAH TOA on right triangle $PQB$ infer that



                  $|PB|=|PQ|cosangle QPB$



                  $|QB|=|PQ|sinangle QPB$



                  $dfrac{|PQ|}{|PB|+|QB|}=dfrac{1}{cosangle QPB+sinangle QPB}ge dfrac{1}{sqrt{2}}$



                  We know that $cosangle QPB+sinangle QPB lesqrt{2}$ because $(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)=2$.



                  Applying this result to all four sides of the inscribed quadrilateral leads to the perimeter of this quadrilateral being $ge 1/sqrt{2}$ times that of the square. That this bound is sharp is proved from the simple case of the sides of the inscribed quadrilateral being parallel to the angle bisectors of the square (always possible given the symmetries involved).






                  share|cite|improve this answer











                  $endgroup$



                  Let $A,B,C$ be three corners of the square in rotational order. Draw a linecsegment $PQ$ from point $P$ on $AB$ to point $Q$ on $BC$. Using SOH CAH TOA on right triangle $PQB$ infer that



                  $|PB|=|PQ|cosangle QPB$



                  $|QB|=|PQ|sinangle QPB$



                  $dfrac{|PQ|}{|PB|+|QB|}=dfrac{1}{cosangle QPB+sinangle QPB}ge dfrac{1}{sqrt{2}}$



                  We know that $cosangle QPB+sinangle QPB lesqrt{2}$ because $(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)=2$.



                  Applying this result to all four sides of the inscribed quadrilateral leads to the perimeter of this quadrilateral being $ge 1/sqrt{2}$ times that of the square. That this bound is sharp is proved from the simple case of the sides of the inscribed quadrilateral being parallel to the angle bisectors of the square (always possible given the symmetries involved).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 17 '18 at 15:36

























                  answered Dec 17 '18 at 15:03









                  Oscar LanziOscar Lanzi

                  12.8k12136




                  12.8k12136






























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