Find the area of the largest hexagon that can be inscribed in a unit square












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How so I find the area of the largest regular hexagon that can be inscribed in a unit square?
I am using this http://www.drking.org.uk/hexagons/misc/deriv4.html to figure it out but not sure if it is correct or not.






geometry optimization area







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  • $begingroup$
    Looks like that page refers to other cases already treated, for some of the steps.
    $endgroup$
    – coffeemath
    Dec 16 '16 at 16:36










  • $begingroup$
    What are you asking? How to compute the area of the hexagon in the image? Or how to deduce that the hexagon in the picture really is the largest possible?
    $endgroup$
    – Karolis Juodelė
    Dec 16 '16 at 16:37










  • $begingroup$
    @KarolisJuodelė How do I deduce that it is really the largest? If possible, can you tell me what would be the area of the largest hexagon.
    $endgroup$
    – ImVikash_0_0
    Dec 16 '16 at 16:40










  • $begingroup$
    If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$.
    $endgroup$
    – Jack D'Aurizio
    Dec 16 '16 at 16:40






  • 1




    $begingroup$
    Maybe you are looking for the largest inscribed regular hexagon?
    $endgroup$
    – Jack D'Aurizio
    Dec 16 '16 at 16:41
















0












$begingroup$


How so I find the area of the largest regular hexagon that can be inscribed in a unit square?
I am using this http://www.drking.org.uk/hexagons/misc/deriv4.html to figure it out but not sure if it is correct or not.






geometry optimization area







share|cite|improve this question











$endgroup$












  • $begingroup$
    Looks like that page refers to other cases already treated, for some of the steps.
    $endgroup$
    – coffeemath
    Dec 16 '16 at 16:36










  • $begingroup$
    What are you asking? How to compute the area of the hexagon in the image? Or how to deduce that the hexagon in the picture really is the largest possible?
    $endgroup$
    – Karolis Juodelė
    Dec 16 '16 at 16:37










  • $begingroup$
    @KarolisJuodelė How do I deduce that it is really the largest? If possible, can you tell me what would be the area of the largest hexagon.
    $endgroup$
    – ImVikash_0_0
    Dec 16 '16 at 16:40










  • $begingroup$
    If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$.
    $endgroup$
    – Jack D'Aurizio
    Dec 16 '16 at 16:40






  • 1




    $begingroup$
    Maybe you are looking for the largest inscribed regular hexagon?
    $endgroup$
    – Jack D'Aurizio
    Dec 16 '16 at 16:41














0












0








0


1



$begingroup$


How so I find the area of the largest regular hexagon that can be inscribed in a unit square?
I am using this http://www.drking.org.uk/hexagons/misc/deriv4.html to figure it out but not sure if it is correct or not.






geometry optimization area







share|cite|improve this question











$endgroup$




How so I find the area of the largest regular hexagon that can be inscribed in a unit square?
I am using this http://www.drking.org.uk/hexagons/misc/deriv4.html to figure it out but not sure if it is correct or not.






geometry optimization area




geometry optimization area






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '16 at 16:43







ImVikash_0_0

















asked Dec 16 '16 at 16:33









ImVikash_0_0ImVikash_0_0

12




12












  • $begingroup$
    Looks like that page refers to other cases already treated, for some of the steps.
    $endgroup$
    – coffeemath
    Dec 16 '16 at 16:36










  • $begingroup$
    What are you asking? How to compute the area of the hexagon in the image? Or how to deduce that the hexagon in the picture really is the largest possible?
    $endgroup$
    – Karolis Juodelė
    Dec 16 '16 at 16:37










  • $begingroup$
    @KarolisJuodelė How do I deduce that it is really the largest? If possible, can you tell me what would be the area of the largest hexagon.
    $endgroup$
    – ImVikash_0_0
    Dec 16 '16 at 16:40










  • $begingroup$
    If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$.
    $endgroup$
    – Jack D'Aurizio
    Dec 16 '16 at 16:40






  • 1




    $begingroup$
    Maybe you are looking for the largest inscribed regular hexagon?
    $endgroup$
    – Jack D'Aurizio
    Dec 16 '16 at 16:41


















  • $begingroup$
    Looks like that page refers to other cases already treated, for some of the steps.
    $endgroup$
    – coffeemath
    Dec 16 '16 at 16:36










  • $begingroup$
    What are you asking? How to compute the area of the hexagon in the image? Or how to deduce that the hexagon in the picture really is the largest possible?
    $endgroup$
    – Karolis Juodelė
    Dec 16 '16 at 16:37










  • $begingroup$
    @KarolisJuodelė How do I deduce that it is really the largest? If possible, can you tell me what would be the area of the largest hexagon.
    $endgroup$
    – ImVikash_0_0
    Dec 16 '16 at 16:40










  • $begingroup$
    If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$.
    $endgroup$
    – Jack D'Aurizio
    Dec 16 '16 at 16:40






  • 1




    $begingroup$
    Maybe you are looking for the largest inscribed regular hexagon?
    $endgroup$
    – Jack D'Aurizio
    Dec 16 '16 at 16:41
















$begingroup$
Looks like that page refers to other cases already treated, for some of the steps.
$endgroup$
– coffeemath
Dec 16 '16 at 16:36




$begingroup$
Looks like that page refers to other cases already treated, for some of the steps.
$endgroup$
– coffeemath
Dec 16 '16 at 16:36












$begingroup$
What are you asking? How to compute the area of the hexagon in the image? Or how to deduce that the hexagon in the picture really is the largest possible?
$endgroup$
– Karolis Juodelė
Dec 16 '16 at 16:37




$begingroup$
What are you asking? How to compute the area of the hexagon in the image? Or how to deduce that the hexagon in the picture really is the largest possible?
$endgroup$
– Karolis Juodelė
Dec 16 '16 at 16:37












$begingroup$
@KarolisJuodelė How do I deduce that it is really the largest? If possible, can you tell me what would be the area of the largest hexagon.
$endgroup$
– ImVikash_0_0
Dec 16 '16 at 16:40




$begingroup$
@KarolisJuodelė How do I deduce that it is really the largest? If possible, can you tell me what would be the area of the largest hexagon.
$endgroup$
– ImVikash_0_0
Dec 16 '16 at 16:40












$begingroup$
If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$.
$endgroup$
– Jack D'Aurizio
Dec 16 '16 at 16:40




$begingroup$
If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$.
$endgroup$
– Jack D'Aurizio
Dec 16 '16 at 16:40




1




1




$begingroup$
Maybe you are looking for the largest inscribed regular hexagon?
$endgroup$
– Jack D'Aurizio
Dec 16 '16 at 16:41




$begingroup$
Maybe you are looking for the largest inscribed regular hexagon?
$endgroup$
– Jack D'Aurizio
Dec 16 '16 at 16:41










1 Answer
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Consider the regular hexagon $H$ with vertices $(pm1,0)$, $left(pm{1over2},pm{sqrt{3}over2}right)$ and area $A(H)={3sqrt{3}over2}$. We have to determine the smallest circumscribed square. Due to symmetry it is sufficient to consider supporting lines tilted by an angle $phiinbigl[0,{piover6}bigr]$ counterclockwise with respect to the vertical, and supporting lines orthogonal to these. The almost vertical lines hit $H$ at $pm(1,0)$, and their orthogonals hit $H$ at $pmleft(-{1over2},-{sqrt{3}over2}right)$. The side-length $s$ of the resulting square $Q$ is then given by
$$s=2maxleft{cosphi, cosleft({piover6}-phiright)right}qquadleft(0leqphileq{piover6}right) ,$$
and is minimal when $phi={piover12}$. It follows that
$${A(H)over A(Q)}leq {3sqrt{3}over2}cdot{1over4cos^2{piover12}}=3sqrt{3}-{9over2}doteq 0.6962 .$$






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    $begingroup$

    Consider the regular hexagon $H$ with vertices $(pm1,0)$, $left(pm{1over2},pm{sqrt{3}over2}right)$ and area $A(H)={3sqrt{3}over2}$. We have to determine the smallest circumscribed square. Due to symmetry it is sufficient to consider supporting lines tilted by an angle $phiinbigl[0,{piover6}bigr]$ counterclockwise with respect to the vertical, and supporting lines orthogonal to these. The almost vertical lines hit $H$ at $pm(1,0)$, and their orthogonals hit $H$ at $pmleft(-{1over2},-{sqrt{3}over2}right)$. The side-length $s$ of the resulting square $Q$ is then given by
    $$s=2maxleft{cosphi, cosleft({piover6}-phiright)right}qquadleft(0leqphileq{piover6}right) ,$$
    and is minimal when $phi={piover12}$. It follows that
    $${A(H)over A(Q)}leq {3sqrt{3}over2}cdot{1over4cos^2{piover12}}=3sqrt{3}-{9over2}doteq 0.6962 .$$






    share|cite|improve this answer









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      1












      $begingroup$

      Consider the regular hexagon $H$ with vertices $(pm1,0)$, $left(pm{1over2},pm{sqrt{3}over2}right)$ and area $A(H)={3sqrt{3}over2}$. We have to determine the smallest circumscribed square. Due to symmetry it is sufficient to consider supporting lines tilted by an angle $phiinbigl[0,{piover6}bigr]$ counterclockwise with respect to the vertical, and supporting lines orthogonal to these. The almost vertical lines hit $H$ at $pm(1,0)$, and their orthogonals hit $H$ at $pmleft(-{1over2},-{sqrt{3}over2}right)$. The side-length $s$ of the resulting square $Q$ is then given by
      $$s=2maxleft{cosphi, cosleft({piover6}-phiright)right}qquadleft(0leqphileq{piover6}right) ,$$
      and is minimal when $phi={piover12}$. It follows that
      $${A(H)over A(Q)}leq {3sqrt{3}over2}cdot{1over4cos^2{piover12}}=3sqrt{3}-{9over2}doteq 0.6962 .$$






      share|cite|improve this answer









      $endgroup$
















        1












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        1





        $begingroup$

        Consider the regular hexagon $H$ with vertices $(pm1,0)$, $left(pm{1over2},pm{sqrt{3}over2}right)$ and area $A(H)={3sqrt{3}over2}$. We have to determine the smallest circumscribed square. Due to symmetry it is sufficient to consider supporting lines tilted by an angle $phiinbigl[0,{piover6}bigr]$ counterclockwise with respect to the vertical, and supporting lines orthogonal to these. The almost vertical lines hit $H$ at $pm(1,0)$, and their orthogonals hit $H$ at $pmleft(-{1over2},-{sqrt{3}over2}right)$. The side-length $s$ of the resulting square $Q$ is then given by
        $$s=2maxleft{cosphi, cosleft({piover6}-phiright)right}qquadleft(0leqphileq{piover6}right) ,$$
        and is minimal when $phi={piover12}$. It follows that
        $${A(H)over A(Q)}leq {3sqrt{3}over2}cdot{1over4cos^2{piover12}}=3sqrt{3}-{9over2}doteq 0.6962 .$$






        share|cite|improve this answer









        $endgroup$



        Consider the regular hexagon $H$ with vertices $(pm1,0)$, $left(pm{1over2},pm{sqrt{3}over2}right)$ and area $A(H)={3sqrt{3}over2}$. We have to determine the smallest circumscribed square. Due to symmetry it is sufficient to consider supporting lines tilted by an angle $phiinbigl[0,{piover6}bigr]$ counterclockwise with respect to the vertical, and supporting lines orthogonal to these. The almost vertical lines hit $H$ at $pm(1,0)$, and their orthogonals hit $H$ at $pmleft(-{1over2},-{sqrt{3}over2}right)$. The side-length $s$ of the resulting square $Q$ is then given by
        $$s=2maxleft{cosphi, cosleft({piover6}-phiright)right}qquadleft(0leqphileq{piover6}right) ,$$
        and is minimal when $phi={piover12}$. It follows that
        $${A(H)over A(Q)}leq {3sqrt{3}over2}cdot{1over4cos^2{piover12}}=3sqrt{3}-{9over2}doteq 0.6962 .$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '16 at 19:27









        Christian BlatterChristian Blatter

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