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
asked Dec 16 '16 at 16:33
ImVikash_0_0
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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
asked Dec 16 '16 at 16:33
ImVikash_0_0
11
Looks like that page refers to other cases already treated, for some of the steps.
– coffeemath
Dec 16 '16 at 16:36
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?
– Karolis Juodelė
Dec 16 '16 at 16:37
@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.
– ImVikash_0_0
Dec 16 '16 at 16:40
If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$.
– Jack D'Aurizio
Dec 16 '16 at 16:40
1
Maybe you are looking for the largest inscribed regular hexagon?
– Jack D'Aurizio
Dec 16 '16 at 16:41
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
asked Dec 16 '16 at 16:33
ImVikash_0_0
11
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
asked Dec 16 '16 at 16:33
ImVikash_0_0
11
asked Dec 16 '16 at 16:33
ImVikash_0_0
11
edited Dec 16 '16 at 16:43
asked Dec 16 '16 at 16:33
ImVikash_0_0
11
asked Dec 16 '16 at 16:33
ImVikash_0_0
11
asked Dec 16 '16 at 16:33
ImVikash_0_0
11
11
Looks like that page refers to other cases already treated, for some of the steps.
– coffeemath
Dec 16 '16 at 16:36
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?
– Karolis Juodelė
Dec 16 '16 at 16:37
@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.
– ImVikash_0_0
Dec 16 '16 at 16:40
If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$.
– Jack D'Aurizio
Dec 16 '16 at 16:40
1
Maybe you are looking for the largest inscribed regular hexagon?
– Jack D'Aurizio
Dec 16 '16 at 16:41
add a comment |
Looks like that page refers to other cases already treated, for some of the steps.
– coffeemath
Dec 16 '16 at 16:36
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?
– Karolis Juodelė
Dec 16 '16 at 16:37
@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.
– ImVikash_0_0
Dec 16 '16 at 16:40
If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$.
– Jack D'Aurizio
Dec 16 '16 at 16:40
1
Maybe you are looking for the largest inscribed regular hexagon?
– Jack D'Aurizio
Dec 16 '16 at 16:41
Looks like that page refers to other cases already treated, for some of the steps.
– coffeemath
Dec 16 '16 at 16:36
Looks like that page refers to other cases already treated, for some of the steps.
– coffeemath
Dec 16 '16 at 16:36
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?
– Karolis Juodelė
Dec 16 '16 at 16:37
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?
– Karolis Juodelė
Dec 16 '16 at 16:37
@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.
– ImVikash_0_0
Dec 16 '16 at 16:40
@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.
– ImVikash_0_0
Dec 16 '16 at 16:40
If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$.
– Jack D'Aurizio
Dec 16 '16 at 16:40
If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$.
– Jack D'Aurizio
Dec 16 '16 at 16:40
1
1
Maybe you are looking for the largest inscribed regular hexagon?
– Jack D'Aurizio
Dec 16 '16 at 16:41
Maybe you are looking for the largest inscribed regular hexagon?
– Jack D'Aurizio
Dec 16 '16 at 16:41
add a comment |
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
$${A(H)over A(Q)}leq {3sqrt{3}over2}cdot{1over4cos^2{piover12}}=3sqrt{3}-{9over2}doteq 0.6962 .$$
answered Dec 16 '16 at 19:27
Christian Blatter
170k7111324
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1 Answer
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oldest
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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
$${A(H)over A(Q)}leq {3sqrt{3}over2}cdot{1over4cos^2{piover12}}=3sqrt{3}-{9over2}doteq 0.6962 .$$
answered Dec 16 '16 at 19:27
Christian Blatter
170k7111324
add a comment |
up vote
0
down vote
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
$${A(H)over A(Q)}leq {3sqrt{3}over2}cdot{1over4cos^2{piover12}}=3sqrt{3}-{9over2}doteq 0.6962 .$$
answered Dec 16 '16 at 19:27
Christian Blatter
170k7111324
add a comment |
up vote
0
down vote
up vote
0
down vote
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
$${A(H)over A(Q)}leq {3sqrt{3}over2}cdot{1over4cos^2{piover12}}=3sqrt{3}-{9over2}doteq 0.6962 .$$
answered Dec 16 '16 at 19:27
Christian Blatter
170k7111324
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
$${A(H)over A(Q)}leq {3sqrt{3}over2}cdot{1over4cos^2{piover12}}=3sqrt{3}-{9over2}doteq 0.6962 .$$
answered Dec 16 '16 at 19:27
Christian Blatter
170k7111324
answered Dec 16 '16 at 19:27
Christian Blatter
170k7111324
answered Dec 16 '16 at 19:27
Christian Blatter
170k7111324
answered Dec 16 '16 at 19:27
Christian Blatter
170k7111324
170k7111324
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Looks like that page refers to other cases already treated, for some of the steps.
– coffeemath
Dec 16 '16 at 16:36
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?
– Karolis Juodelė
Dec 16 '16 at 16:37
@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.
– ImVikash_0_0
Dec 16 '16 at 16:40
If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$.
– Jack D'Aurizio
Dec 16 '16 at 16:40
1
Maybe you are looking for the largest inscribed regular hexagon?
– Jack D'Aurizio
Dec 16 '16 at 16:41