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Is there any way to prove for any n-sided polygon that if it's regular, it has the highest possible area/circumference ratio?
I have been thinking about this for days, and I can't find an answer. I have thought about calculus, but the problem is that there are a variable number of variables. You end up with a formula containing sums, which you can't take the derivative of (that's true, is it?)
Maybe with a geometric approach? But I have no idea how to do that...
Good question. I'll give it some thought.
what do you mean by a regular polygon?
One where all sides are the same length and all angles have the same measure.Quote:
Originally posted by Gandalf_Grey_
what do you mean by a regular polygon?
twanvl:
Are you asking if the maximum area/circumference ratio for an n-sided polygon is achieved when it is a regular polygon? That's how I understood your question.
Why don't you post what you have so far so that we can look at it? You can take the derivative of a sum, just proceed as usual.
I would think that you would need an expression for the Area and circumference for an n-sided polygon (an irregular polygon not a regular one). Divide one by the other, and then take the complete differential. Set each of the partial derivatives to zero and solve the set of equations (there should be n variables I think). The equations will be nonlinear so they might be a little tricky.
You might want to try to look at a triangle first before doing the general case of an n-sided polygon.
I could do about 2/3 of this proof. Proceeding thus:
1) Show the polygon (whatever it may be) is convex.
2) Show all sides are the same length.
3) Show all the angles are equal.
1) and 2) aren't all that tricky. Anybody post back if you want the details. I'll work on 3).
DavidHooper:
It appears that you're showing a way to check if the polygon is regular. The way I understood the question was, does a regular polygon give you the maxium Area/Circumference ratio for an n-sided polygon? That is a maximization problem.
I began working on the expression for a triangle, it's a little nasty. Do you know if in fact the maximum Area/Circumference ratio for an n-sided polygon is a regular polygon?
wy125: What I was thinking for the derivative:
make a function f(x,q) where x is a list of sides, and q is a list of angles (around the center of the polygon). The function f should calculate the area/circumgerence for that polygon. Once you have the function f, all that's left is to derive it, and solve f'=0 (and f''<0).
The first problem is that you must make sure that all items in q are >0, and sum(q)=2p.
I think the easiest way to do this is give the function a list q, of ratios of the total angle(2p), so that q[n] = 2p*q[n] / sum(q)
the function need only be derived to q[1] or q[1] and x[1], since you can 'rotate' the polygon defined in the lists.
I hope this makes sense to anyone :)
DavidHooper: 1 is really easy, for if it's not equal you can always increase the area/circumference ratio, by flipping the part of the polygon that is facing inwards outwards. Example:
I don't know about 2 or 3 though.Code:_
\__/ \__/
can be changed to
\__ __/
\_/
wy125: It sounds too logical to be false. It is at least true for a n=infinite (a circle), beacause #1 above.
Okay, I did some work on this and I think I can show that a regular polygon does not always give the greatest Area/Circumference ratio. By the way, the fact that there is no constraint on the material in any way, for any given regular polygon with a corresponding A/C ratio you could choose an irregular polygon of greater size that has a greater ratio. So, it's clear that the problem is not properly constrained, and as it is stated there is no maximum.
Here is my proof that a regular polygon does not always give the optimum (this is partially constrained so we can find an answer, but if our hypothesis is correct it won't matter if it it has more constraints or not):
(1) Let us assume for the sake of argument that it is true that regular polygons do give the greatest Area/Circumference Ratio.
(2) We will use a triangle as our polygon. Constrain the polygon so that at least two of the sides are equal while the other side can take on any value that completes the third side of the triangle.
****If our assumption is correct the value we obtain for the third side will equal the other two equal sides *****
(see attached picture for illustration of the triangle)
Starting from here, let's use this notation:
x is the length of the equal sides of the triangle
y is the length of the third side of the triangle
a is the value of the angle between the sides of length x
As this is set up, a, the angle between the two sides of length x can be determined using the law of cosines. Doing so gives,
a = acos(1 - y^2/(2x^2))
The area, A, of this triangle is given by
A = 1/2*x*y*sin[acos(a)]
The circumference, C, is given by
C = 2x + y
And the ratio of the area to cirumference, R, is given by
R = A/C
Putting this ratio only in terms of x and y we obtain:
R(x,y) = 1/2*x^2*sin[acos(1-y^2/(2*x^2))]
---------------------------------------------
2x + y
Now if we want to determine what value of y in terms of x gives the maximum ratio, R, we take the derivative with respect to y and set equal to zero and then solve for y. As previously stated, if our hypothesis is correct, that a regular polygon always gives the maxium, then we should find that y = x.
The derivative of R with respect to x is a little nasty, it's
dR/dx = 4*x^2 - 2xy - y^2
----------------------------------
4(4x^2-y^2)^0.5 (2x + y)
Setting this equal to 0 gives us the following quadratic equation in terms of y
y^2 + 2xy - 4x^2 = 0
This gives two equations for y
y = -2x +/- SQRT(4x^2 +16x^2)
------------------------------------
2
y1 = (5^0.5 - 1)x
y2 = (-5^0.5 - 1)x
only y1 is our solution since it's positive.
So we see that the maximum ratio is not given by a regular polygon since y != x.
Also note that if you substitue the expression for y back into the expression for the ratio, you'll notice that it is a function of x. That tells us that we need to constrain the problem further since the ratio is a function of the 'size' of the polygon. Like I wrote early, we can always pick a polygon that is bigger than a regular one that yields a greater ratio.
Hope this all came across okay. I can't explain my thoughts very well sometimes.
That is indeed a problem. The best way to rephrase the question, would be:Quote:
Okay, I did some work on this and I think I can show that a regular polygon does not always give the greatest Area/Circumference ratio. By the way, the fact that there is no constraint on the material in any way, for any given regular polygon with a corresponding A/C ratio you could choose an irregular polygon of greater size that has a greater ratio. So, it's clear that the problem is not properly constrained, and as it is stated there is no maximum.
Prove for any n, that if an n-sided polygon has the highest poisble area for a given circumference or the lowest posible circumference for a given area, that polygon is regular.
Your equation doesn't prove anything about this, because both the area and the circumference change. Sorry, I should have been more clear when describing the problem. :(
Note, a area/(circumference^2) ratio will probably work too.
wy125:
Yes I am showing a way to check if the polygon is regular. A way to show that the polygon with maximum area/circumference is regular!Quote:
It appears that you're showing a way to check if the polygon is regular. The way I understood the question was, does a regular polygon give you the maxium Area/Circumference ratio for an n-sided polygon? That is a maximization problem.
This is how I proceeded, considering twanvl's ratio throughout:
1) Show the polygon (whatever it may be) is convex. If it is concave, then the area can be increased without increasing the circumference by flipping the 'cave'.
2) Show all sides are the same length. If two adjacent sides are different lengths then the area can be increased without increasing the length of the sides, but by making the sides the same length.
3) Show all the angles are equal. Not finished yet...
There exists a proof that a circle has the largest area for a given perimeter and the smallest perimeter for a given area. I think these two properties are equivalent.
I think the proof is esoteric. Due to that proof, I think the proof about regular polygons is likely to be valid, but difficult to prove. It might be more difficult than the circle proof.
It does not seem possible to get a general analytical equation for either the area or the perimeter of an arbitrary polygon. Therefore, an approach using derivatives is not possible. I am pretty sure the circle proof does not use that approach either.
It is my guess that the proof involves considering what happens when you distort a regular polygon. You can try to show that a distortion which does not change the perimeter must decrease the area. Or perhaps show that a distortion which keeps the same area, must increase the perimeter.
It might be worthwhile to try to prove the theorem for a polygon with say 6 sides, and see if you can generalize the proof to any polygon. If you cannot prove it for a six sided polygon, you are not going to prove in the general case anyway.
This was what I suggested, just I can't put it as neatly as Guv.Quote:
It is my guess that the proof involves considering what happens when you distort a regular polygon. You can try to show that a distortion which does not change the perimeter must decrease the area. Or perhaps show that a distortion which keeps the same area, must increase the perimeter.
a square with a total of 16 for the perimeter
4m x 4m = 16²m (4+4+4+4 =16)
3m x 5m = 15²m (3+3+5+5 = 16)
so the regular polygon has highest area.
alkatran: I wasn't talking about rectangles, I meant all kinds of polygons, including pentagons, hexagons, decagons, etc.
I have found that it is posible to prove it the way Guv proposed. The solution is rather lengthy:
1. You devide a regular polygon into triangles.
2. You take two adjacent triangles, formed by the points A, B and C, on the outer side of the polygon, and it's center M. The triangles are ABM and BCM
3. You can also divide the triangles as ACM and ABC.
4. AM and CM are constant for these two triangles, we are only going to change BM and the angle AMB. So the area triangle ACM is constant, and we only look at ABC
5. AB+BC is also constant. I'll call the sides of this triangle a=BC,b=AC and c=AB. The sum of a and c is also constant, I'll call it r.
6. A function for the area of ABC can now be made. It has only one variables, a. (b is also constant, and c=r-a)
7. You can take the derivative of this function, and solve where it's 0.
8. That happens to be at a = r/2. c = r-a => c = r/2 = a, thus the area is the highest.
I did not include the functions's here, since they are quite complex. My math teacher agrees that this solution is correct.
if its less for a square and less for a circle, then its probably less for all of em, but ur solution is good too :D
btw since the polygons are made of triangles, and changing shape of triangle decreases it, u know its the same for all of em.
but I'm only in grade 10, so what do i know?.... yet