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Determining Equable Shapes

Date: 06/15/99 at 08:11:46
From: Jonathan
Subject: Equable shapes

Could you please list some of the equable shapes and show me how you 
determined them? Can you please write down the formula?

Date: 06/15/99 at 08:18:12
From: Doctor Anthony
Subject: Re: Equable shapes

Consider figures that have perimeters and areas NUMERICALLY equal. I 
emphasize the words 'numerically equal' because one is a length and 
the other is an area so they cannot be 'equal' in the normal sense of 
the word.

A general formula relating the number of sides to the length of side 
of an equable shape (equal area and perimeter)?

The formula given below is for any regular polygon of 'n' sides, each 
side being of length 'x':

   x = 4 / [tan(90 - 180/n)]

For example,

           x = 4 / [tan(90 - 180/4)] = 4 / tan(45) = 4

        Area = x*x = 16

   Perimeter = 4*x = 16

           x = 4 / [tan(90 - 180/6)]  =  4 / tan(60) = 4/sqrt(3)

so      area = perimeter = 24/sqrt(3) 

The formula works for any regular polygon.

For the case of a regular concave polygon - like a star shape -  the 
Koch snowflake is an interesting example.

The Koch Snowflake
To draw this you start with an equilateral triangle of side a. Now 
divide each side into three equal parts and on the middle third of 
each side construct an equilateral triangles pointing outwards from 
the original triangle. The total perimeter is now (4/3)(3a) = 4a. We 
now further subdivide each straight edge into 3 parts and construct 
equilateral triangles on the middle third of each side - again 
pointing outwards from the original figure. This process will enlarge 
the perimeter by a further factor of 4/3. There is no overlapping of 
the extra sides with those already present. The above process is 
repeated n times, at each stage the perimeter being increased by a 
factor of 4/3.

So we have:

   perimeter = (3a)(4/3)(4/3)(4/3) ... to n terms.

Clearly the perimeter will increase in this process, but the area of 
the figure will be less than the area of the circumcircle of the 
original equilateral triangle. So the perimeter and area will converge 
in numerical value.  

If A = area of original triangle then:

The area of the snowflake follows the pattern: 

   A[1 + 1/3 + 4/27 + 16/243 + 64/2187 + ...]

where after the first term we have a GP with common ratio 4/9.  

   = A + A[1/3 + 4/27 + n terms]

   = A + A(1/3)[1-(4/9)^n]/(1-4/9)

   = A + A(1/3)[1-(4/9)^n]/(5/9)

   = A + (3A/5)[1-(4/9)^n]

Now if area and perimeter are to be equal we need solutions to the 

   (3a)(4/3)^n = A + (3A/5)[1-(4/9)^n] = A[1+(3/5)(1-(4/9)^n)]

where  A = (1/2)a^2.sin(60) = a^2.sqrt(3)/4

   (3a)(4/3)^n = a^2.sqrt(3)/4[1+(3/5)(1-(4/9)^n)]

       (4/3)^n = a.sqrt(3)/12[1+(3/5)(1-(4/9)^n)]

We can see that for different values of 'a' there will be different 
values of n. 

Solving for n with a given value of 'a' would be difficult, so choose 
a value of n and see what value of 'a' this requires.

For example, n = 3 gives

        2.3703 = 0.144337a[1+(3/5)(0.912208)]

        2.3703 = 0.144337a[1.547325]

      10.61314 = a

So for any value of n there will be a value of 'a' that makes the 
perimeter and area numerically equal.

This is just an example of the type of figure where perimeter and area 
can be numerically equal. However there will be any number of concave 
polygons where a similar relation could be established.

- Doctor Anthony, The Math Forum   
Associated Topics:
High School Euclidean/Plane Geometry
High School Fractals
High School Geometry

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