Equilateral Pentagons E5 are polygons with five sides of equal
length. There are an infinite number of distinct E5 since the
internal angles can take practically any real value. Next java applet shows some
familiar geometric shapes disected into E5s. Where the disection
is not unique, an animation shows some possibilities. A selector permits to switch
between all the figures.
Only four intersected equal circles are enough to define a E5.
The center of each circle corresponds to one vertex of E5
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
E5 can be defined as a two-variable-function:
E5 = F(a, b)
Conditions for Simplification
In order to avoid reflections, translations, rotations (and complications)
of a given E5 two conditions can be stated:
In the following figures we can see how only one E5 is valid
among six variations. Figure 1 (in cyan) satisfies both conditions. Figures 2-5
(in green) doesn't satisfy first condition while Figure 6 (in yellow) satisfies
first conditions but don't satisfy the second one.
- 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.
Covex and Concave E5
The angle c's values determine the convexity of E5.
We can conclude that:
- E5 is CONVEX if angle c is lower than 180o.
- E5 is CONCAVE if angle c is greater than 180o.
Self-Intersecting for E5
The angles a and b values determine the self-intersecting
of E5. We can conclude that:
- E5 is not self-intersected if:
2a + b is greater than 180o.
- E5 is self-intersected if:
2a + b is lower than 180o.
Some Trigonometry of E5
To find the values of angles c, d and e
as functions of (a, b) some calculations and the
following figure can help:
Here E5 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
By trigonometry from yellow triange we found that:
c = c1 + c2 + c4
e = e1 + e2
a2 = c1 = p/2 - b/2
From purple triangle we found that:
a1 = a - p/2 + b/2
B = 2 * sin(b/2)
D2 = 1 + B2 - (2)(1)(B) * cos(a + b/2 - p/2)
D2 = 1 + 4*sin2(b/2) - 4*sin(b/2)*sin(a + b/2)
Now, having D, the angle d value can be calculated
from green triangle:
With more calculations (too complicated to be included here) we found angle e is:
d = cos-1 [ cos(a) + cos(b) - cos(a + b) - 1/2 ]
Finally angle c value is (yours better equation is welcome) is:
e = tan-1 [
|sin(a + b) - sin(a)|
|cos(a) - cos(a + b) - 1||
] + 1/2 * (p - d)
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 No-self-intersected E5 these are the ranges for the
| a ||60o||120o|
| b ||28.9o||108o|
| c ||108o||331o|
| d ||0o||108o|
| e ||75.5o||240o|
Equilateral Polygons' To-Do
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
with more than one type of and/or concaves equilateral pentagons.
Polyominoes are in fact equilateral polygons (The pentominoes are
equilateral dodecagons E12). These dodecagons have only few internal
angles values: 0o, 90o, 180o, 270o and
360o. 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 E5.
I'm writing a Java applet for drawing conics touching all the general
E5 vertices. The relations between conic's and E5's
parameters should be interesting.
I made some calculations for equilateral hexagons (though complicated).
Three angles are needed for the hexagons. While E5 needs surface plots,
E6 needs three-dimensional 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 EMs into
ENs. 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
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