Hans Walser, [20090726a]
Regular K-gon, Rectangles, and Trapezoids
1 Whats about
We start with a regular K-gon and add similar rectangles on every side. The shape of the
width =
1
rectangles depends on K: length
( )
2 sin K
1
2 sin (K)
1
Shape of the rectangles
Then we proceed by adding rectangles, as explained in the next example.
The lengths of the different rectangles are in the proportions 1:1:2:3:5:8…, i. e. in the
proportions of the Fibonacci-numbers.
2 Example
Starting by a regular yellow pentagon ( K = 5 ) in the unit circle we add similar rectanwidth =
1
gles length
0.85065080835 on every side. Then we proceed as indicated in
()
2 sin 5
the following figure.
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Pentagon and rectangles
We get red isosceles trapezoids between the white rectangles. In the first ring we see red
triangles, but we count them as special red trapezoids with upper side zero.
Remark: There is (of course) a link to the golden section = 1+2 5 , since
()
2 sin 5 = 3 .
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Dissecting the rectangles into congruent rhombuses, we see that the lengths of the rectangles are 1, 1, 2, 3, … . These are the Fibonacci Numbers.
Congruent rhombuses
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We can dissect the triangles and trapezoids into congruent triangles and hexagons (both
not regular).
Nice figure
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3 Modification
We can modify the figure by transforming the white rectangles into parallelograms.
This does not change shape or size of the red trapezoids.
Collapsing the rectangles
Most useful is the third of these figures with five sectors.
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All these figures can be dissected into parallelograms (instead of rhombuses) and triangles and hexagons. The triangles and hexagons don’t change their shape or size.
Modification
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Hans Walser: Regular K-gon, Rectangles, and Trapezoids
4 Some calculations
We use the notations of the following figure.
1
1
2 sin (K)
1
2 sin (K)
1
2 sin (K)
a1
2
K
1
2 sin a1
0
a2
a2
1
(K)
a1
1
2 sin (K)
1
a2
a1
a3
a3
a2
1
2 sin (K)
2
a3
a3
a4
1
2 sin (K)
a4
3
1
2 sin (K)
a4
a4
a5
Notations
First we get:
( )
a1 = 2 sin K
a2 = 2 1 a1 sin K = a1
2 sin( K ) ( )
Studying a red trapezoid, we see:
an+2 = 2 1 an+1 sin K + an = an+1 + an
2 sin( K )
( )
This is the usual Fibonacci recursion. Therefore we have:
a1 : a2 : a3 : a4 : a5 :… = 1 :1 : 2 : 3 : 5 :…
For the areas n of the red trapezoids we get by some calculations:
0 : 1 : 2 : 3 : 4 :… = 1 : 3 : 8 : 21 : 55 :…
These are every second Fibonacci number. The areas of the red trapezoids are in a rational relation.
Hans Walser: Regular K-gon, Rectangles, and Trapezoids
5
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Special cases
5.1 K = 1
Division by zero.
5.2
K=2
width =
We use rectangles with length
1
2 sin 2
()
= 12 (half squares).
K=2
The entire Figure is a rectangle, which approximates the so called Golden Rectangle
width = 5 1 0.61803398875 .
with length
2
5.3
K=3
width =
We use rectangles with length
1
2 sin 3
()
= 1 0.57735026919 .
3
K=3
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Both figures fit in the same triangular lattice. In this lattice it is easy to check the Fibonacci property.
Triangular lattice
Hans Walser: Regular K-gon, Rectangles, and Trapezoids
Triangular lattice
Dissections
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Hans Walser: Regular K-gon, Rectangles, and Trapezoids
5.4
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K=4
width =
1
We use rectangles with length
= 1 0.70710678119 . This is the so called
2
2 sin( 4 )
DIN format.
K=4
The figure on the right fits into a square lattice, but not so the figure on the left.
Disections
5.5 K = 5
See example in the introduction.
5.6
K=6
width =
We get “rectangles” with length
1
2 sin 6
()
= 1 , i. e. squares.
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K=6
The figure on the right fits into a regular triangular lattice. The figure on the left not,
since rectangles and regular triangles don’t like each other.
Triangular lattice
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In the dissection of the figure on the left we get squares, equilateral triangles, and regular hexagons. This is the most regular case.
Squares, equilateral triangles, and regular hexagons
Hans Walser: Regular K-gon, Rectangles, and Trapezoids
Pattern in a hexagon
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Hans Walser: Regular K-gon, Rectangles, and Trapezoids
5.7
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K=7
width =
1
We use rectangles with length
1.1523824355 . Now the rectangles are stay2 sin( 7 )
ing out like the hairs of Struwwelpeter.
K =7
5.8
K=8
width =
Rectangles with length
1
2 sin 8
()
= 1 + 22 1.3065629649 .
K=8
Hans Walser: Regular K-gon, Rectangles, and Trapezoids
5.9
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K = 10
width =
Rectangles with length
1
2 sin 10
( )
= 1.6180339887 . We have golden rectangles.
K = 10
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5.10 K = 30
K = 30
The figure approximates circles with the radii 1, 2, 3, 5, and 8, i. e. the Fibonacci numbers.
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Dissection
References
[Deshpande 2009]
Deshpande, M. N. : Proof Without Words: Beyond Extriangles.
MATHEMATICS MAGAZINE. Vol. 82, No. 3, June 2009, p.
208.