• Kaustubh

MemberDec 15, 2008

## What is the Size of the Largest Hexagon Inside Square?

Here's an easier one for you -

What is the size of the largest regular hexagon that can be constructed inside a square with side-length 's'?

Crack it fast, please.

Solution:

Problem Analysis: To determine the size of the largest regular hexagon that can be constructed inside a square with side-length 's', we need to find the maximum size of the regular hexagon while keeping it fully contained within the square.

Solution: A regular hexagon is a polygon with six equal sides and six equal angles. Let's analyze the problem step by step:

1. Determine the side length of the regular hexagon: Since the hexagon is regular, all six sides are equal in length. Let's denote the side length of the hexagon as 'h'.

2. Find the distance from the center of the hexagon to one of its vertices: The distance from the center of the hexagon to any of its vertices is equal to the side length of the hexagon (h).

3. Determine the distance from the center of the hexagon to one of its sides: The distance from the center of the hexagon to one of its sides is equal to the apothem (a). The apothem is the perpendicular distance from the center of the hexagon to one of its sides.

4. Calculate the length of the apothem: To find the apothem (a), we can divide the hexagon into two congruent triangles. Each of these triangles has a base equal to one side of the hexagon (h) and a height equal to the apothem (a). The angle between the base and the height in each triangle is 30 degrees, as the hexagon has interior angles of 120 degrees. Using trigonometry, we can calculate the apothem (a) as:

a = h sin(30 degrees) = h (1/2) = h / 2

1. Determine the side length of the square: Since the hexagon must be fully contained within the square, the distance from the center of the hexagon to one of its vertices (h) should be less than or equal to half the side length of the square (s/2).

2. Solve for the side length of the hexagon in terms of the square's side length: Using the previous step, we can set up the inequality:

h ≤ s/2

1. Calculate the maximum side length of the hexagon: By substituting a = h / 2 from step 4, we have:

h ≤ s/2 h/2 ≤ s/4 a ≤ s/4

So, the maximum length of the apothem (a) is equal to s/4. Since the distance from the center to the vertex is twice the apothem, the maximum side length of the hexagon is 2 * (s/4), which simplifies to s/2.

Example: Let's take a square with a side length of 10 units (s = 10). To find the size of the largest regular hexagon that can be constructed inside this square, we can use the formula derived above:

Maximum side length of hexagon = s/2 = 10/2 = 5 units

Therefore, the largest regular hexagon that can be constructed inside a square with a side length of 10 units has a side length of 5 units.

Note: The same approach can be applied to squares of different sizes by substituting the corresponding value of 's' in the formula.

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Replies
• MemberDec 15, 2008

Hexagon side will be half of the side of square. i.e s/2 .Am I right ??
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• MemberDec 15, 2008

i think it should be s/3....pls contradict
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• MemberDec 15, 2008

So easy to contradict over my answer.It was asked largest regular hexa gon so i think hexagon with side s/2 > s/3 .Am i right??
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• AdministratorDec 15, 2008

How'd we know if your answer is right or wrong? You have to provide an explanation along with your answer.
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• MemberDec 16, 2008

assuming the side of the regulr polygon to be x
assuming the size of the square to be a

any corner of the hexagon should not be concurrent with the corner of the square ,, as angle =90 which is not in a case of regular hexagon

hence working upon the right agnle trigaingle at one corner of the square which cannot be the part of the hexagon
since the polygon is aregular hexagon .. the internal angles are 120 degrees
thus considering one of the rigth corner tirangles
cos 60=((a-x)/2 )/x
x/2=(a-x)/2
x=a/2
now since x is the base of 6 equilateral triangles formed within the hexagon
the altitue being sqrt(3)/2*a/2=(sqrt(3)*a)/4
area of any triangle =1/2*b*h
1/2* (a/2) *((sqrt(3)*a)/4)
(sqrt(3)a^2)/16
since 6 such triangles are formed
6*(sqrt(3)a^2)/16

:rolling eyes:

i guess there is something i miseed.. its 1"30 AM here , so relatively no fuctionality of brian
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• MemberDec 16, 2008

X cannot be a/2...
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• AdministratorDec 16, 2008

This is why I love this section 😉 !

Don't look at me for the right answer fellas. When you are sure, you know you've gotten the right answer.
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• MemberDec 16, 2008

I dotn say I am correct, But I need a proof to say I am wrong...
I guess I can prove x=a/2
length os side of sq =a
max length of side of hexagon =a
since internal angle =120 degress
consider a point a on the side of the sq.. where the corner of a hexagon meets the sq side
internal angles -120 degrees
so sum of two external angles =60 degress
with the congruency of trinagles.. it can be proved that each extranl angle =30 degrees
and
as well....
the side oft he hexagon coincides with that of the side of sq
ie 180-120 degress is internal angle of the rt angle trinagle at the corner of the sq..

hence cos 60 =((a-x)/2)/x
ie x=a/2
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• MemberDec 16, 2008

But try constructing it. U've considered a side of the square which coincides with a side of the hexagon. Now consider any two sides whose common edge lies on one of the sides of the square( but these two sides of hexagon do not lie on a side of the square).

So now, if x=a/2, these tho sides of the hexagon would lie on the side of the square... Think about it and contradict me if i'm wrong.
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• MemberDec 16, 2008

hmmm 😀

I understadn what you are saying...

I guess thats the same point 'a' what i talked about in the previous psot..

i mean even if two sides dont lie on one side of sq.. and opp two sides also dont lie on the sq...

you still ahve two side which can( and need) to lie on the sq...

Vidya..

may be you can try uplaoding an image if what you are trying to depict...

what if finally we both land up doing the same analysis!!! 😁
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• MemberDec 16, 2008

In one of your previous posts, you have mentioned 'a' as a side, not a point...
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• MemberDec 16, 2008

Hey, i am changing my answer.My answer is s/sqrt(3). I will give explanation with figure.You need to wait for that.But you can tell either if it is right or wrong 😒 .
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• MemberDec 16, 2008

The area of the hexagon will be maximum when the diagonal of the hexagon is equal to the side of the square.

ie, if x is the side of the hexagon, then 2x is it's diagonal. So,

2x=a
x=a/2.

so the largest hexagon that can fit into a square of length a is of side a/2.
It must be oriented in such a way that two of it's sides coincide with the opposite sides of the square and two corners of the hexagon coincide with the midpoints of the other two sides of the square.

I hope I'm clear...
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• MemberDec 16, 2008

well i think someone shud draw both s/2 and s/3 figures and post it....well i am sticking to my ans
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• MemberDec 17, 2008

shalini_goel14
Hey, i am changing my answer.My answer is s/sqrt(3). I will give explanation with figure.You need to wait for that.But you can tell either if it is right or wrong 😒 .
hehe!!!😀
we have all very unique answers... no one can say what they have said is 100% the answer..

lets wait for some more entires😒
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• MemberDec 19, 2008

If the side of the square is s, then the side of hexagon would be
s/(2+√2)
. Considering the hexagon made would be regular
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• MemberDec 19, 2008

If the side of the square is s, then the side of hexagon would be
s/(2+√2)
. Considering the hexagon made would be regular.
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• MemberDec 19, 2008

show with diagram!!!
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• MemberDec 21, 2008

Note: For Image see my avatar ( I dont know how to add image)
Y is the side of regular polygon (in red)
And x will be s-y;
X=y/√2 since the white portion forms a right triangle.
And x+y=s (given)
Putting the values
(y/√2)+y=s
Or y=√2s/(1+√2)
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• AdministratorDec 21, 2008

Naveen89
Note: For Image see my avatar ( I dont know how to add image)
Y is the side of regular polygon (in red)
And x will be s-y;
X=y/âˆš2 since the white portion forms a right triangle.
And x+y=s (given)
Putting the values
(y/âˆš2)+y=s
Or y=s/(1+âˆš2)
Adding image is simple. Upload your image to #-Link-Snipped-# and then, copy the complete image path [obtained by right clicking on the image and selecting 'Copy Image Location' option] in the dialog box presented once you click button in CE post editor
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• MemberDec 21, 2008

The_Big_K
Adding image is simple. Upload your image to #-Link-Snipped-# and then, copy the complete image path [obtained by right clicking on the image and selecting 'Copy Image Location' option] in the dialog box presented once you click button in CE post editor
*psst* Flickr does not allow hotlinking 😉
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• AdministratorDec 21, 2008

ash
*psst* Flickr does not allow hotlinking 😉
[off-topic]
It does. All our images are hosted there 😒.

Here's an example -

[CE's 3[sup]rd[/sup] Birthday Cake ]
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• MemberDec 21, 2008

I know it does. But I stress "not allow" as in policy wise 😉
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• MemberDec 21, 2008

but was my answer right?
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• AdministratorDec 21, 2008

How about " sqrt(2-sqrt(3)) s " or ~ 0.518 s?

---------------------------------------------------

@ Ash - I think the Flickr TOS says a 'visible link' should be provided along with the image. I'll remember that. Thanks 😀
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• MemberJun 27, 2012

This is an old thread, and there was already a correct answer. But, it is a nice problem and the previous discussion left me unsure who was correct and why? So, with a little scratching...

If the hexagon has a radius of R and aligned so that a diameter of the hexagon is parallel to the bottom side of the square, then the correct answer is R = S/2. But if you rotate the hexagon a little you will be able to get a slightly larger polygon. By symmetry if you rotate the hexagon by 60 degrees you're right back where you started. If you rotate it 30 degrees you will have a diameter that is now perpendicular to the bottom side and you are still touching the square at R = S/2. It turns out that rotating it by 15 degrees will give you the maximum hexagon. So the correct answer is R = (S/2)/Cos(15) ~ 0.518*S.

You can use the fact that the Cos(15) = Cos(45 - 30) and the formula for the Cos(A - B) = Cos(45)Cos(30) + Sin(45)Sin(30) and much simplification to get the Sqrt(2 - Sqrt(3)) * S as already posted by The_Big_K.
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• MemberDec 11, 2016

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• MemberDec 11, 2016

Gary Truesdale
This is a/2 in sketchup. Not quite right.
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• MemberDec 11, 2016

.518S is sill not quite right.
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• MemberDec 16, 2016

This might be previously answered based on intuition (drawing a hexagon inside a square) and I have to say that the intuition of drawing a hexagon - with its vertices on the sides of the square is wrong (I spent a lot of time doing the same, then realized the trigonometry does not give me the correct result - something very basic must be wrong). You can only draw such a polygon (n sides) for cases where n is a proper multiple (1,2,3..) of 4 (since for a square, n = 4). For a hexagon, such a sketch will bring us to wrong conclusions. And that is to say that the maximum sized hexagon inside the square will be tilted (no sides of the hexagon will coincide with that of the square). Please correct me if you find something wrong.

Therefore, this becomes a problem of optimization. I immediately thought about the <a href="https://mathworld.wolfram.com/BrachistochroneProblem.html" target="_blank" rel="nofollow noopener noreferrer">Brachistochrone Problem -- from Wolfram MathWorld</a>. This problem literally led to the birth of a branch of mathematics called the <a href="https://en.wikipedia.org/wiki/Calculus_of_variations" target="_blank" rel="nofollow noopener noreferrer">Calculus Of Variations</a>
I have grown to be increasingly drawn to this offshoot of calculus, since it was very recently that I learned about the Brachistochrone Problem. Now with this post, I started wondering if this optimization problem could be solved using calculus of variations. I understood the basics (very basic concepts) of the theory (that I learned online, a lot of literature is available), but the point where I am stuck is: how to find the "functional" to be optimized in this case.

During the course of our education, we learn to find the optima of a function (the procedure of finding the first derivative and equating it to 0, afterwards finding the second derivative and deciding if it is a local minima/maxima - is based on the calculus of variations. I did not know that.). It is easy if one is given the textbook problems, with boundary conditions. Here, I am thinking about the possible boundary conditions. A lot of unknowns.

I will continue to fight with this problem. Thus far, this thread has been highly rewarding.
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