How do I prove that a circle has the largest area?
Michael Abbott
How do I prove that the maximum area that a curve of unit length can enclose
is 1/(4 pi)?
I "know" that the shape with the smallest perimiter is a circle ... but I
cannot even begin to think of a proof! Any suggestions, please?
David Tanenbaum
Michael -
Referring to a circle whose "curve length" is a unit, means
the circumference is one unit long. Since the circumference is equal
to Pi*D, then D, the diameter is 1/Pi. The formula for the area of a
circle is (1/4)Pi*D^2 and by substituting 1/Pi for D,
Area = (1/4)Pi*(1/Pi)^2
= 1/4*Pi QED
dlt
Michael Abbott
Yes, I knew that!
I am asking -
How do I prove that of all possible closed curves of unit length, the
circle has the largest area?
All you've done is verify my value, 1/4pi. Sorry I wasn't clearer.
David Tanenbaum wrote in message <345ea324...@news.campus.mci.net>...
Michael Abbott
It's definitely a tricky one.
I think your comments help to reinforce the conviction that the statement
must be true. But I don't think we've even got the germ of proof yet...
However, you've now convinced me that the best curve must be convex. We
can define a convex curve as one where every chord (line segment joining two
points on the perimeter) lies entirely within the area, in which case a
concave curve will have chords lying entirely outside the area (exercise for
the reader).
It's fairly obvious that simply turning the offending segment of
perimeter inside out will generate a curve with the same perimeter but with
larger area. It's not quite inevitable though - I think we still have to
worry (in principle) about self intersection when we turn the segment inside
out.
Unfortunately I don't see how enumerating a handful of patterns is going to
get us much further.
Raymond E. Griffith wrote in message <345F36...@vnet.net>...
>Michael Abbott wrote:
>> How do I prove that of all possible closed curves of unit length, the
>> circle has the largest area?
>
>It's been a long time, so I am very rusty in this area (and it might
>show).
>
>But I would think of it this way--a circle is everywhere convex (i.e.,
>each chord attaching two points on the curve lies entirely within the
>curve). Closed curves which are concave must have less area than closed
>curves which are convex.
>
>Closed curves such as ellipses have areas < the area of a circle
>(formula not in front of me).
>
>Regular polygons are the next closed curves to be investigated. A
>regular polygon with n sides has an area (I think) of
>tan(pi/2 - pi/n)/(4n). This converges to 1/(4pi) as n increases without
>bound.
Michael Abbott
Nope, not helpful. Not true either.
Tony Suranno wrote in message <63qqid$r93$1...@winter.news.erols.com>...
>It would seem that a short trip to a calculus book might help refresh your
>memory as to the proof of a circle having the greatest area of all closed
>curves.
cre...@flash.net
This problem has really caught my interest, mostly because I cannot solve it.
I'll give the two approaches which seem plausible to me, in the hopes that
maybe someone else can make something work. A very simple, beautiful
problem, really.
Prove : Of all curves of length 1, a circle encloses the largest area.
Method 1 : (Not pretty.) We can see the curve must be convex. (I hope I
don't mean concave. It's one of those parity things I can never keep straight
...] Thus it is a function in polar coordinates. Write the integral for the
length as well as the area. Show that the constant function f(theta)=K
maximizes the area given fixed length over all other possible functions.
Perhaps use a Fourier Transform to capture the notion of all possible
functions. Like I said, not pretty.
Method 2 : Use the full rotational symmetry of the circle. It is the only
geometric figure which can go through a rotation of
arbitrary degree and be unchanged [in fact, it is the
only figure that has an uncountably infinite number
of non-distorting automorphisms.] So we have two properties
unique to a circle : maximizes area , full rotational
symmetry. I find it very compelling to try to connect the
two.
In fact, extremum prefer to exists at points of symmetry. We
can see this sort of thing at work in a much simpler example.
We can show the square is the rectangle containing greatest
area arguing directly from symmetry. Fix the perimeter of
the rectangle, and the area becomes a quadratic in terms of
In article <63rps0$qf9$2...@panther.rmplc.co.uk> "Michael Abbott"
<too.mu...@nowhere.com> writes:>From: "Michael Abbott"
<too.mu...@nowhere.com>>Subject: Re: How do I prove that a circle has the
largest area?>Date: Thu, 6 Nov 1997 06:59:39 -0000
cre...@flash.net
In article <63rps0$qf9$2...@panther.rmplc.co.uk> "Michael Abbott" <too.mu...@nowhere.com> writes:
>From: "Michael Abbott" <too.mu...@nowhere.com>
>Subject: Re: How do I prove that a circle has the largest area?
>Date: Thu, 6 Nov 1997 06:59:39 -0000
AAAAAAAAACK! I did it again! I'm an idiot. Here it is a third time. If I
hit post by accident again, I am going to live in a cave from now on !
This problem has really caught my interest, mostly because I cannot solve it.
I'll give the two approaches which seem plausible to me, in the hopes that
maybe someone else can make something work. A very simple, beautiful
problem, really.
Prove : Of all curves of length 1, a circle encloses the largest area.
Method 1 : (Not pretty.) We can see the curve must be convex. (I hope I
don't mean concave. It's one of those parity things I can never keep straight
...] Thus it is a function in polar coordinates. Write the integral for the
length as well as the area. Show that the constant function f(theta)=K
maximizes the area given fixed length over all other possible functions.
Perhaps use a Fourier Transform to capture the notion of all possible
functions. Like I said, not pretty.
Method 2 : Use the full rotational symmetry of the circle. It is the only
geometric figure which can go through a rotation of
arbitrary degree and be unchanged [in fact, it is the
only figure that has an uncountably infinite number
of non-distorting automorphisms.] So we have two properties
unique to a circle : maximizes area , full rotational
symmetry. I find it very compelling to try to connect the
two.
In fact, extrema prefer to exists at points of symmetry. We
can see this sort of thing at work in a much simpler example.
We can show the square is the rectangle containing greatest
area arguing directly from symmetry. Fix the perimeter of
the rectangle, and the area becomes a quadratic function of
the length of one side. Now suppose a rectangle not a square
is optimal. Then so is the rectangle with length and width swapped.
So the area function has two extrema (turning points). But the area
function is a quadratic, and can't have two extrema. QED.
Of course, I haven't addressed the real proof, but I hope someone
more clever than I can find a beautiful proof. Perhaps something
here can lead them in the right direction.
-- Chris
(I didn't hit Post in the middle! I can stay in my apartment ...]
cre...@flash.net
Concave curves inward, like a cave! Now I'll never forget!
Incidentally, stalagtites hang from the ceiling, because they have
to hang on *tite*. I believe this is due to Gauss.
The joys of dyslexia ...
In article <MPG.ecd8159c...@news.concentric.net>
brow...@concentric.net (Stan Brown) writes:>From: brow...@concentric.net
(Stan Brown)>Subject: convex/concave (Re: How do I prove that a circle has the
largest area?)>Date: Fri, 7 Nov 1997 19:50:24 -0500
>[posted and emailed]
>In article <creiss.61...@flash.net>, cre...@flash.net
>(cre...@flash.net) wrote:
>>Method 1 : (Not pretty.) We can see the curve must be convex. (I hope I
>>don't mean concave. It's one of those parity things I can never keep straight
>>...]
>Convex curves belly outward, like a belly. Concave curves curve inward,
>like a ... er, cave.
>And speaking of caves, stalaGmites Grow up from the floor, and
>stataCtites Come down from the Ceiling.
>(Unless, of course, you believe that stalagmites are miniature WW2 German
>prison guards. :-)
>--
>Stan Brown, Oak Road Systems, Cleveland, Ohio, USA
> http://www.concentric.net/%7eBrownsta/
>Sorry! sig file down for repairs.
nob...@nowhere.on.ca
On Sat, 08 Nov 1997 01:55:45 +0100, Matthew Jackson
<m.ja...@math.canterbury.ac.nz> wrote:
>Raymond E. Griffith wrote:
>>
>> Michael Abbott wrote:
>> >
>> > Yes, I knew that!
>> >
>> > I am asking -
>> > How do I prove that of all possible closed curves of unit length, the
>> > circle has the largest area?
>> >
>>
>> It's been a long time, so I am very rusty in this area (and it might
>> show).
>>
>> But I would think of it this way--a circle is everywhere convex (i.e.,
>> each chord attaching two points on the curve lies entirely within the
>> curve). Closed curves which are concave must have less area than closed
>> curves which are convex.
>
>Yes, this is a very interesting question. I can't solve it either
>(yet). But I have some ideas that may lead to the proof.
>
>It seems reasonable to accept that a convex cloed curve will have
>a higher area:perimeter ratio than one which is not convex. My
>argument will be a step to showing that of convex curves, the circle
>has the highestarea:perimeter ratio.
>
Good point, I think. The "most convex" curve is the circle. Then any
change must be to make it less convex.
Michael Abbott
Thank you, that's a much more interesting reply. I agree that this is quite
a subtle problem.
cre...@flash.net wrote in message ...
>In article <63rps0$qf9$2...@panther.rmplc.co.uk> "Michael Abbott"
<mic...@rcp.co.uk> writes:
>>Subject: Re: How do I prove that a circle has the largest area?
Matthew Jackson
Raymond E. Griffith wrote:
>
> Michael Abbott wrote:
> >
> > Yes, I knew that!
> >
> > I am asking -
> > How do I prove that of all possible closed curves of unit length, the
> >
>
> show).
>
> But I would think of it this way--a circle is everywhere convex (i.e.,
> each chord attaching two points on the curve lies entirely within the
> curve). Closed curves which are concave must have less area than closed
> curves which are convex.
Yes, this is a very interesting question. I can't solve it either
(yet). But I have some ideas that may lead to the proof.
It seems reasonable to accept that a convex cloed curve will have
a higher area:perimeter ratio than one which is not convex. My
argument will be a step to showing that of convex curves, the circle
has the highestarea:perimeter ratio.
A convex curve is one which, if it goes around the origin, can be
expressed as f(t), t:0->2pi, where f is the distance of the curve from
the origin at the angle t. The area contained in such a curve is
given by:
2pi
1 /\
- | (f(t))^2 dt
2 \/
0
and the perimeter is given by:
2pi
/\
| sqrt[ (f(t))^2 + (f'(t))^2 ] dt
\/
0
The problem then becomes: Maximize the ratio
2pi
1 /\
- | (f(t))^2 dt
2 \/
0
---------------------------------
2pi
/\
| sqrt[ (f(t))^2 + (f'(t))^2 ] dt
\/
0
subject to the constraint that f(0)=f(2pi).
Speaking for myself, I've never learned to do integral equations, and
I don't think that this is the time to learn. However, as we are
trying to show that this ratio is maximised when the curve is a
circle, we don't need to work quite as hard. The function f will
describe a circle if and only if it is constant, which will be the
case if and only if f'=0. The bottom line then becomes:
2pi
/\
| f(t) dt
\/
0
I don't know if this makes life any easier for people, but I suspect
that it could well lead to a proof.
Good luck everyone,
Matthew Jackson
Darrell Ryan
In article <3463B8...@math.canterbury.ac.nz>,
m.ja...@math.canterbury.ac.nz wrote:
>
> Raymond E. Griffith wrote:
> >
> > Michael Abbott wrote:
> > >
> > > Yes, I knew that!
> > >
> > > I am asking -
> > > How do I prove that of all possible closed curves of unit length, the
> > > circle has the largest area?
>
> Yes, this is a very interesting question. I can't solve it either
> (yet). But I have some ideas that may lead to the proof.
I also do not know how to prove it, but I have a different idea how one
might possibly go about it. For simplicity, the problem can be reduced
to the 1st quadrant (rectangular coordinates). If we have a point on the
y axis (0,y) and another point on the x axis (x,0), find the curve
connecting these two points such that the area enclosed by the curve and
the axes is maximized. Then all you have to do is show this curve to be
the arc of a circle (centered at the origin). Assume the curve is an arc
of the unit circle, in such case x=y=1 for the two points (x,0) and
(0,y). I believe it would be OK to make this assumtion, as long as we
can prove that the area in question is indeed maximized under these
assumptions.
In other words, find f(x) such that
INT (0 to 1) f(x) dx
is a maximum, given the initial conditions
f(0)=1
f(1)=0
f'(0)=0
f'(1)=undefined
(do I smell a differential equation?)
Then show that f(x) is equal to x^2+y^2=1 in the 1st quadrant. That part
should be pretty easy.
I don't know if this approach would be valid, it's just an idea to be
kicked around.
....something in the back of my head tells me there is an easier way.
Darrell Ryan dr...@edge.net http://edge.net/~dryan/
-------------------==== Posted via Deja News ====-----------------------
http://www.dejanews.com/ Search, Read, Post to Usenet
Darrell Ryan
David Tanenbaum wrote:
>
> Darrell - What's to constrain the curve between (0,1) and (1,0) to
> the arc of a circle? Have I missed something or couldn't the arc be
> an off axis parabola, for instance....or an ellipse??? dlt
I'm not suggesting that. What I was suggesting is once the curve is
found, then show it is an arc of a circle. Yes, I made some
assumtions but if it can be proven that the area is indeed maximized
if the curve is the arc of a circle, then we have proven what we
wanted to prove.
Are you suggesting an arc of another conic section could also be the
arc of a circle? I don't see how. A circle has constant "curvature"
at all points (recall the definition of curvature). This is not the
case of the other conics. That would be just one (of several) ways to
show it is the arc of a circle. Another thing to keep in mind is this
is just the 1st quadrant, i.e. we are assuming symmetry. Also, the
"unit length" thing comes into play. The only other conic that could
"possibly" qualify is an ellipse (the only other conic that is closed
and thus has finite arc lenght). But of course, an ellipse has
variable curvature, unlike a circle.
For that matter, who's to say the curve f(x) we find must be a conic?
All I am trying to suggest is once we find the curve, if we can show
it is indeed the arc of a circle we have proven what we wanted to
prove. Like I pointed out, I'm sure there are some flaws in my idea.
I just wanted to kick it around a little to see if may lead to a
"correct" idea :-) For instance, I probably shouldn't have assumed
that f'(0)=0 and f'(1)=undefined. That is probably assuming too much.
--
Stan Brown
[posted and emailed]
In article <creiss.67...@flash.net>, cre...@flash.net
(cre...@flash.net) wrote:
>Concave curves inward, like a cave! Now I'll never forget!
>
>Incidentally, stalagtites hang from the ceiling, because they have
>to hang on *tite*. I believe this is due to Gauss.
I doubt it. At least in English there is no such word as "stalagtite".
The American Heritage Dictionary of the English Language (Third Edition)
has an interesting short essay on the common origin of the two real words
stalactite and stalagmite.
Darrell Ryan
For what it's worth, I can prove that a circle of a certain circumference
will have greater area than any regular polygon of the same perimeter.
....still working on the general case for things that are not regular
polygons.
> --
> Darrell Ryan dr...@edge.net http://edge.net/~dryan/
-------------------==== Posted via Deja News ====-----------------------
Magnus Jonsson
After a bit of thinking and finally discussing the matter with my cat,
this is what we came up with.
The proof is probably made in two steps. First you prove that the
largest area of a polygon is when all sides and all angles are equal.
Then prove that the more sides there are in the polygon, starting with a
triangle, the larger is the area. A circle is of an infinite number of
sides. Won't that solve it?
Magnus, art student
Sara, modern philosophy student
cre...@flash.net
In article <3466164B...@edge.net> Darrell Ryan <dr...@edge.net> writes:
>Date: Sun, 09 Nov 1997 14:00:11 -0600
>From: Darrell Ryan <dr...@edge.net>
>Subject: Re: How do I prove that a circle has the largest area?
We need to put this one to rest, fellow travellers!
Has anyone checked some texts on this? It strikes me as 'classical result',
the kind of thing Gauss would have 14 proofs for. (I know, it's no fun just
looking the answer up, but sometimes you just gotta say uncle :)
>David Tanenbaum wrote:
>--
cre...@flash.net
We have to prove it for any closed figure at all; we can't restrict
ourselves to just polygons or conics.
In fact, we are at a bit of a loss because we don't even know
for a fact the curve is differentiable !
I think I'm *this* close to a simple, elementary proof on this.
If I don't get any further I'll post my partial result.
In article <346781...@student.liu.se> Magnus Jonsson
<magj...@student.liu.se> writes:>From: Magnus Jonsson
<magj...@student.liu.se>>Subject: Re: How do I prove that a circle has the
largest area?>Date: Mon, 10 Nov 1997 23:49:27 +0200
Darrell Ryan
cre...@flash.net wrote:
>
> We have to prove it for any closed figure at all; we can't restrict
> ourselves to just polygons or conics.
>
> In fact, we are at a bit of a loss because we don't even know
> for a fact the curve is differentiable !
But since the curve is continuous, it *will* be integrable.
Continuity implies integrability, and under proper restrictions we
know integrals can be interpreted as area, so it seems to me it need
not be differentiable everywhere. In fact, circles are not
differentiable everywhere (vertical tangents exist).
>
> I think I'm *this* close to a simple, elementary proof on this.
> If I don't get any further I'll post my partial result.
I would be interested in seeing whatever you have. Don't be shy!
Someone has informed me that this type of problem can be solved with
calculus of variations, but surely there must be an easier way.
Earle Jones
>This problem has really caught my interest, mostly because I cannot solve it.
>I'll give the two approaches which seem plausible to me, in the hopes that
>maybe someone else can make something work. A very simple, beautiful
>problem, really.
>
>Prove : Of all curves of length 1, a circle encloses the largest area.
---
This thread has received many postings about proving that a circle encloses
the largest area. There's something appealing about the circle -- we just
know that we can maximize our enclosed area with a circle, right?
Now -- here's a problem for you:
A farmer owns a square field, 100 feet on a side. He wants to fence in the
maximum area, taking into account the cost of fencing (let's say that
fencing cost one dollar per foot). What shape field should he enclose?
Why, must be a circle, right?
Case I -- Circle:
Area enclosed = 2500 * pi square feet.
Cost of fence = 100 * pi dollars
That's 25 square feet enclosed per dollar.
Case II -- Square:
Area enclosed = 10,000 square feet
Cost of fence = 400 dollars
That's 25 square feet enclosed per dollar! Just as good as the circle!
Now, see how this grabs you!
Case III -- Square with rounded corners:
Supposed we choose a shape that is a square with the corners rounded.
For example, each side might be a 50-foot straight line, with a quarter
circle (radius = 25 feet) at each corner.
Area enclosed = (25 * 25) (12 + pi) = 9,463.5 square feet
Cost of fence = (50 * pi) + 200 = 357.08 dollars
That's 9,463.5 / 357.08 = 26.5 square feet per dollar! -- better than
either the square or the circle!
----
But (and here is your problem) is this the shape that gives the maximum
enclosed area per unit cost?
Answer: NO!
Because the little quarter circles at the corners should not be 1/4 of the side.
What should the radius of these quarter circles be in order to maximize the
area enclosed per cost?
Have fun!
earle
--
__
__/\_\
/\_\/_/
\/_/\_\ earle
\/_/ jones
We want our Internet back! Get rid of Spam.
See http://www.cauce.org
dauvil
Dear Earle,
This problem belongs to a class of problems known as
isoperimetric problems. You can find the solution at
www.cut-the-knot.com/do_you_know/isoperimetric.html
I believe it is also discussed in the book
"What is Mathematics" by Courant & Robbins.
Hope this helps!
Dan
Darrell Ryan
In article <ejones-ya0236800...@news.wco.com>,
Earle,
In all three of the above cases, we are using a *different* amount of
fencing. These cases have nothing to do with finding the curve that
maximizes area given a *fixed* circumference. Stated another way, given
a *fixed* area, find the curve that encloses this area such that the
perimeter is minimized (not the cost of the perimeter).