The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Limit of Perimeter/Area Ratio for Polygons

Date: 03/23/2000 at 19:56:02
From: Christy Walsh
Subject: Perimeter minimization

I need to find a formula for the perimeter of a regular polygon as a 
function of its area. It needs to be shown that as the number of sides 
of a polygon of given area increases, the perimeter of the figure 
decreases and also the limit of the perimeter is the circumference of 
the circle of that given area. Part of the argument is that if we take 
the limit as the number of sides, n, of a regular polygon increasing
without bound, the polygon will become a circle.

Date: 03/24/2000 at 08:26:40
From: Doctor Luis
Subject: Re: Perimeter minimization

Hi Christy,

Obtaining the area of a regular polygon is not terribly hard. 
Essentially you have to note that you can obtain a regular polygon by 
stacking a bunch of identical triangles side to side. (Draw lines from 
the vertices of your polygon toward the center so you can see this.)

If you can figure out the area of each of those triangles, you'll be 
able to obtain the area of the polygon by simply multiplying by n, the 
number of sides.

Here's one of those triangles, where O is the center of the polygon, h 
is the height of the triangle, and B = 2*b is the length of the base. 
You can see that the area of the triangle is

     T = (1/2)*(2*b)*h = b*h

          / | \
       l /  |  \ l
        /  h|   \ 
       /    |    \
     M   b     b   N

Thus, the area of the polygon is:

     A = n*T = nbh

and its perimeter is

     P = n*(2b) = 2nb

Now, because the angle MON is 2pi/n (in radians), we have

     tan((1/2)2pi/n) = b/h
     tan(pi/n) = b/h

If we keep A fixed, we can express b in terms of n only by eliminating 

        A = nb(b/tan(pi/n))

        b^2 = ------------

This allows us to express P in terms of n only.
                 / A*tan(pi/n)
     P =  2n    / -------------
              \/       n

            /          sin(pi/n)          1
       =   / 4A * pi * ---------- * ------------
         \/              pi/n         cos(pi/n)

You can see that in the limit as n -> +infinity, we'll have

     sin(pi/n) / (pi/n) -> 1

              cos(pi/n) -> 1

so naturally we have

            1/cos(pi/n) -> 1

Finally, we get that

       tan(pi/n)/(pi/n) -> 1

All this goes to show that

     P -> sqrt(4pi*A) as n -> +infinity

Keep in mind that

     A = nbh = (pi*h^2)*tan(pi/n)/(pi/n)

Since h -> R, the radius of the circle, you'll get

     A -> pi*R^2*1 = (pi)R^2

and so

     P -> sqrt(4pi*pi*R^2) = 2pi*R

This just goes to show that you get the correct formulas for A and P 
for an "infinite-sided" polygon. Showing that P is decreasing is a bit 
trickier, as you have to show that the function (n/pi)tan(pi/n) is 
decreasing for n > 3. (Why 3?)

If looking at a graph of (x/pi)tan(pi/x) is not convincing enough, I 
think the shortest way you can show this is by looking at the Taylor 
series for the tangent. You have something like:

     tan(x) = x + (1/3)x^3 + (2/15)x^5 + (17/315)x^7 + ...

This is the Taylor series expansion about the point x = 0. The nice 
thing about the tangent series is that all the coefficients are 
positive, a fact which comes in handy when trying to prove 
inequalities. (You can prove they are all positive by examining the 
sign of the derivatives of the tangent at x = 0. Verifying it for a 
few cases should be enough to convince you, and you can then prove 
by induction.)

Supposing that m > n, we get that pi/m < pi/n, so we get the following 
string of inequalities

        (1/3)(pi/m)^2 <    (1/3)(pi/n)^2
       (2/15)(pi/m)^4 <   (2/15)(pi/n)^4
     (17/315)(pi/m)^6 < (17/315)(pi/n)^6
           ...                ...

Adding all these up, you'd get

       (1/3)(pi/m)^2 + (2/15)(pi/m)^4 + (17/315)(pi/m)^6 + ... 
     < (1/3)(pi/n)^2 + (2/15)(pi/n)^4 + (17/315)(pi/n)^6 + ...  
       1 + (1/3)(pi/m)^2 + (2/15)(pi/m)^4 + (17/315)(pi/m)^6 + ... 
     < 1 + (1/3)(pi/n)^2 + (2/15)(pi/n)^4 + (17/315)(pi/n)^6 + ...  


You should now be able to recognize the tangent expansion on both 
sides of the inequality. (I make use of the fact that a series can be 
thought of as the limit of a sequence of partial sums, and also that 
if the each term of one converging sequence is less than the 
corresponding term of another converging sequence, then the limit of 
the former will also be less than the limit of the latter.)

In other words,

     (m/pi)tan(pi/m) < (n/pi)tan(pi/n)

Since this expansion certainly holds for 0 < pi/m < pi/n < pi/3, and 
since originally we said that m > n, it follows that the function 
(n/pi)tan(pi/n) is certainly decreasing for n > 3, as we suspected.

Hence, if (n/pi)tan(pi/n) is decreasing, then certainly

     P = sqrt(4pi*A*(n/pi)tan(pi/n))

is decreasing.

Although this last argument is somewhat advanced, I find it easier to 
understand than some of the other methods I used to prove that 
(n/pi)tan(pi/n) is decreasing for n > 3.

Alternatively, you could also show that the derivative of

     f(x) = (x/pi)*tan(pi/x)

is always negative for x > 3, although I find the algebra to be very 
tedious and less intuitive than the method I presented to you. Since 
we are restricting ourselves to integral values of x, you could also 
try to prove that f(x) is decreasing by induction, but that proof is 
just too long and the algebra is very involved.

If you don't really care about being rigorous, you can always point 
to the graph and say that it is obvious that f(x) is decreasing for 
x > 3  :)

I hope this helped!

- Doctor Luis, The Math Forum   
Associated Topics:
College Trigonometry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.