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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,

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

Area = x*x = 16

Perimeter = 4*x = 16

Hexagon:
-------
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 + .....to 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
equation

(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
http://mathforum.org/dr.math/
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Fractals
High School Geometry

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