Equilateral Pentagons
Equilateral Pentagons E_{5} are polygons with five sides of equal
length. There are an infinite number of distinct E_{5} since the
internal angles can take practically any real value. Next java applet shows some
familiar geometric shapes disected into E_{5}s. Where the disection
is not unique, an animation shows some possibilities. A selector permits to switch
between all the figures.
Theory
Only four intersected equal circles are enough to define a E_{5}.
The center of each circle corresponds to one vertex of E_{5}
namely angles e, a, b and c.
The fifth vertex d is determined by the intersection of circles
C1 and C4 which gives the smaller angle.
Only two angles are enough to determine the positions of the four circles, so
E_{5} can be defined as a twovariablefunction:
E_{5} = F(a, b)
Conditions for Simplification
In order to avoid reflections, translations, rotations (and complications)
of a given E_{5} two conditions can be stated:
 d must be the smaller of the five angles.
(d <= a) && (d <= b) && (d <= c) && (d <= e)

 a must be greater or equal than b.
In the following figures we can see how only one E_{5} is valid
among six variations. Figure 1 (in cyan) satisfies both conditions. Figures 25
(in green) doesn't satisfy first condition while Figure 6 (in yellow) satisfies
first conditions but don't satisfy the second one.
Covex and Concave E_{5}
The angle c's values determine the convexity of E_{5}.
We can conclude that:
 E_{5} is CONVEX if angle c is lower than 180^{o}.
 E_{5} is CONCAVE if angle c is greater than 180^{o}.
SelfIntersecting for E_{5}
The angles a and b values determine the selfintersecting
of E_{5}. We can conclude that:
 E_{5} is not selfintersected if:
2a + b is greater than 180^{o}.
 E_{5} is selfintersected if:
2a + b is lower than 180^{o}.
Some Trigonometry of E_{5}
To find the values of angles c, d and e
as functions of (a, b) some calculations and the
following figure can help:
Here E_{5} is divided into three triangles. Two triangles
(left and right) are isoceles. The triangle at the center is more general.
Some angles have being divided too in this way:
a = a1 + a2
c = c1 + c2 + c4
e = e1 + e2
By trigonometry from yellow triange we found that:
a2 = c1 = p/2  b/2
a1 = a  p/2 + b/2
B = 2 * sin(b/2)
From purple triangle we found that:
D^{2} = 1 + B^{2}  (2)(1)(B) * cos(a + b/2  p/2)
or
D^{2} = 1 + 4*sin^{2}(b/2)  4*sin(b/2)*sin(a + b/2)
Now, having D, the angle d value can be calculated
from green triangle:
d = cos^{1} [ cos(a) + cos(b)  cos(a + b)  1/2 ]

With more calculations (too complicated to be included here) we found angle e is:
e = tan^{1} [

sin(a + b)  sin(a) 
 
cos(a)  cos(a + b)  1 

] + 1/2 * (p  d)

Finally angle c value is (yours better equation is welcome) is:
c = (5/2)*p  a  b  d  e

Minimums and Maximums of the angles
Having three angles c, d and e as functions
of other two (a,b), I have drawn a lot of beatiful
families of curves, they are still drawn on real paper. I hope soon have time to
include them here as electronic media... Also, other information such as area,
angles' relations can be plotted too... Analysing angles plots we get some results:
For Noselfintersected E_{5} these are the ranges for the
angles:
 Minumum  Maximum 

a  60^{o}  120^{o} 
b  28.9^{o}  108^{o} 
c  108^{o}  331^{o} 
d  0^{o}  108^{o} 
e  75.5^{o}  240^{o} 
Equilateral Polygons' ToDo
There is a lot of material (original to be included, and not original to be
mentioned) next to be added as time permits...

An applet for drawing the plots of functions F(a, b).

From the 14 know
convex pentagons tesselations
of the plane, some has equilateral pentagons...
There are also some
tesselations
with more than one type of and/or concaves equilateral pentagons.

Polyominoes are in fact equilateral polygons (The pentominoes are
equilateral dodecagons E_{12}). These dodecagons have only few internal
angles values: 0^{o}, 90^{o}, 180^{o}, 270^{o} and
360^{o}. These discrete angles surely simplifyes in some way the families
of equilateral polygons.

A Math verse reads ...every conic curve can be determined by 5 points...
Therefore exits a conic curve touching the 5 vertices of every E_{5}.
I'm writing a Java applet for drawing conics touching all the general
E_{5} vertices. The relations between conic's and E_{5}'s
parameters should be interesting.

I made some calculations for equilateral hexagons (though complicated).
Three angles are needed for the hexagons. While E_{5} needs surface plots,
E_{6} needs threedimensional plots. Altough, more conditions can be
imposed to the angles to get a reduced set. For instance, families of symmetric
equilateral hexagons having their angles values of the form
2p / n, being n an integer, tile the plane chaotically.

Determine minimum solutions for dissections of E_{M}s into
E_{N}s. wheter M > N, M = N, M < N.
There are systematic solutions by software?
Chaotic Tiling with Equilateral Polygons
Jorge Luis Mireles Jasso Torreón, Coah. México. April 4, 2002
Also in this site:
The Polyominoids
