Pythagoras tree

Classic Pythagoras tree

Balanced Pythagoras Tree

A Pythagoras tree is a special type of fractal . The original method for creating a Pythagorean tree is based on the Pythagorean Theorem , in which two more, smaller squares are arranged at right angles on a square . By recursively calling this construction rule, a fractal is generated which, in the borderline case, resembles the shape of a tree. Due to the right angle of the enclosed triangle , the total area of ​​each level remains the same, therefore the area of ​​the basic element (trunk) is exactly as large as the sum of the area of ​​all outer elements (leaves).

construction

Image 1	picture 2
picture 3	Picture 4

A square is constructed from a baseline. A Thales circle is drawn on the top of this basic element (trunk) and this is divided as desired. The resulting point is connected to the basic element (Fig. 1) so that a right-angled triangle is created. From each of the two legs of the triangle a square is constructed (Fig. 2) , a Thales circle is drawn, this is divided, a right-angled triangle is constructed (Fig. 3) and thus expanded to a square again (Fig. 4) . This process can be repeated as often as required.

Calculations on the right-angled and isosceles Pythagoras tree

height

Basic construction of a branch

The following is the side length of the first square (the "stem") . The height of the first (base) branch is, as seen in the first drawing . To determine the maximum height, it is sufficient to place branches of the shape shown on top of each other. Each branch has half the base of the previous branch. This is the height of the second branch , that of the third , etc. The total height is: ${\displaystyle a}$ ${\displaystyle 2a}$ ${\displaystyle a}$ ${\displaystyle {\frac {a}{2}}}$

${\displaystyle Hoehe=2a+a+{\frac {a}{2}}+{\frac {a}{4}}+{\frac {a}{8}}+\ldots =2a\left(1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+\ldots \right)}$

So:

${\displaystyle Hoehe=2a\sum _{i=0}^{\infty }\left({\frac {1}{2}}\right)^{i}=2a{\frac {1}{1-{\frac {1}{2}}}}=4a}$

width

The left branch corresponds to a transverse tree with the base side . Likewise the right branch. In the middle remains a trunk with the width and the two main branches each with the width . In order to: ${\displaystyle {\frac {a}{2}}}$ ${\displaystyle a}$ ${\displaystyle {\frac {a}{2}}}$

${\displaystyle Breite=\underbrace {4{\frac {a}{2}}} _{\begin{array}{c}\scriptstyle {Hoehe\ linker}\\\scriptstyle {querliegender\ Baum}\end{array}}+\underbrace {4{\frac {a}{2}}} _{\begin{array}{c}\scriptstyle {Hoehe\ rechter}\\\scriptstyle {querliegender\ Baum}\end{array}}+a+{\frac {a}{2}}+{\frac {a}{2}}=6a}$

Trunk length

Side length and length of the treetop

To calculate the trunk length - red in the adjacent drawing - only the side lengths of the squares have to be added.

1. Square: ${\displaystyle a}$
2. Square: ${\displaystyle {\frac {\sqrt {2}}{2}}a}$
3. Square: ${\displaystyle {\frac {\sqrt {2}}{2}}{\frac {\sqrt {2}}{2}}a={\frac {1}{2}}a}$
4. Square: ${\displaystyle {\frac {\sqrt {2}}{2}}{\frac {1}{2}}a={\frac {\sqrt {2}}{4}}a}$

etc.

There is always the factor . If you start the numbering at 0, the following applies: ${\displaystyle {\frac {\sqrt {2}}{2}}}$

Side length of the -th square: ${\displaystyle i}$ ${\displaystyle \left({\frac {\sqrt {2}}{2}}\right)^{i}a}$

So the total length of the red line is:

${\displaystyle a\sum _{i=0}^{\infty }\left({\frac {\sqrt {2}}{2}}\right)^{i}=a{\frac {1}{1-{\frac {\sqrt {2}}{2}}}}={\frac {2a}{2-{\sqrt {2}}}}={\frac {2a(2+{\sqrt {2}})}{4-2}}=a\left(2+{\sqrt {2}}\right)\approx 3{,}414214\ a}$

Length of the treetop

Right-left-right branch of a Pythagorean tree

To calculate the length of the treetop - blue in the drawing - first consider the following: You can get to the corners of the tree by walking along the branches alternately to the left and right. In order to calculate the length of the upper horizontal line, the deviation from the stem center line, which results from the growth of a right-left branch, is first calculated.

	Base side	Deviation from the center line ${\displaystyle x_{i}}$
First right-left combination (square, triangle, square, triangle - dashed in the drawing)	${\displaystyle a}$	${\displaystyle x_{1}={\frac {a}{2}}+{\frac {a}{4}}={\frac {3}{4}}a}$
Second right-left combination (dotted in the drawing)	${\displaystyle {\frac {a}{2}}}$	${\displaystyle x_{2}={\frac {3}{4}}{\frac {a}{2}}}$
Third right-left combination	${\displaystyle {\frac {a}{4}}}$	${\displaystyle x_{3}={\frac {3}{4}}{\frac {a}{4}}}$
i-th right-left combination (attention: renumbering: the first is now the zeroth!)	${\displaystyle \left({\frac {1}{2}}\right)^{i}a}$	${\displaystyle x_{i}={\frac {3}{4}}\left({\frac {1}{2}}\right)^{i}a}$

The maximum deviation of the "last" peak from the first center line is then the sum of the individual deviations:

${\displaystyle Abweichung=\sum _{i=0}^{\infty }{\frac {3}{4}}\left({\frac {1}{2}}\right)^{i}a={\frac {3}{4}}a\sum _{i=0}^{\infty }\left({\frac {1}{2}}\right)^{i}={\frac {3}{4}}a{\frac {1}{1-{\frac {1}{2}}}}={\frac {3}{2}}a}$

A left-right-left-right-… branch therefore deviates from the first center line at most. The same applies to the mirrored right-left-right-left-… branch. The two upper corners therefore have the maximum distance of . This is the length of the top horizontal blue line. ${\displaystyle {\frac {3}{2}}a}$ ${\displaystyle 2{\frac {3}{2}}a=3a}$

The lengths of the other blue lines can be easily calculated. The second blue line corresponds to the upper horizontal line of the main tree, etc.

	Base tree	length
First blue line	${\displaystyle a}$	${\displaystyle 3a}$
Second blue line	${\displaystyle {\frac {\sqrt {2}}{2}}a}$	${\displaystyle 3{\frac {\sqrt {2}}{2}}a}$
Third blue line	${\displaystyle \left({\frac {\sqrt {2}}{2}}\right)^{2}a}$	${\displaystyle 3\left({\frac {\sqrt {2}}{2}}\right)^{2}a}$
i-th blue line (attention: renumbering: the first is now the zero!)	${\displaystyle \left({\frac {\sqrt {2}}{2}}\right)^{i}a}$	${\displaystyle 3\left({\frac {\sqrt {2}}{2}}\right)^{i}a}$

Each blue line is three times longer than the corresponding red line. This means that the total length of the blue line is three times the red line: ${\displaystyle 3a\left(2+{\sqrt {2}}\right)}$

scope

If you want to go around the tree once, you have to walk twice the blue and twice the red line and the line on which the tree stands. The upper blue line is double, so you have to subtract it once:

${\displaystyle Umfang=2\cdot \underbrace {3a\left(2+{\sqrt {2}}\right)} _{\scriptstyle {blau}}\underbrace {-3a} _{\begin{array}{c}\scriptstyle {doppelt\ gerechnete}\\\scriptstyle {obere\ horizontale}\\\scriptstyle {Firstlinie}\end{array}}+2\cdot \underbrace {a\left(2+{\sqrt {2}}\right)} _{\scriptstyle {rot}}\underbrace {+a} _{\scriptstyle {Grundlinie}}=2\cdot a\left(7+4{\sqrt {2}}\right)\approx 25{,}313708\ a}$

Distance to the lawn

Distance between the "leaves" and the "grass"

In order to be able to drive the lawnmower up to the trunk, you have to know how high the clear height is under the foliage of the tree. So: how far is it from the lawn to the first leaves?

With a base side of is the total width . So one side of the tree protrudes beyond the trunk (length of the green line). To calculate the required clearance, consider the third branch - the first horizontally growing branch (or: third square): ${\displaystyle a}$ ${\displaystyle 6a}$ ${\displaystyle {\frac {5}{2}}a}$

The base side of this part of the tree is: . The width of this branch is therefore . With this branch, too, the crown protrudes by the factor , i.e . : . This third branch has a distance from the lawn of . The clear height is then the difference: ${\displaystyle {\frac {a}{2}}}$ ${\displaystyle 6{\frac {a}{2}}=3a}$ ${\displaystyle {\frac {5}{2}}}$ ${\displaystyle {\frac {5}{2}}{\frac {a}{2}}={\frac {5}{4}}a}$ ${\displaystyle {\frac {3}{2}}a}$ ${\displaystyle {\frac {3}{2}}a-{\frac {5}{4}}a={\frac {a}{4}}}$

history

The Pythagoras tree was first constructed by Albert E. Bosman (1891–1961), a Dutch math teacher, in 1942.

Other forms

Since such a tree, which was created strictly according to Pythagoras, looks very unnatural, it is of course possible to deviate from the original form.

Pythagoras tree: Right-angled, isosceles triangles Different colors	Fractal tree: Free angle No squares
Pythagoras tree: Right triangles Different colors	Pythagoras tree: No right triangles Different colors
Pythagoras tree: Random trunk lengths and random trunk division ratios Right triangles Different colors	Pythagoras tree: Isosceles triangles Right triangles Different colors
Pythagoras tree	SW Pythagoras tree

Web links

Commons : Pythagoras Tree  - album with pictures, videos and audio files

• Java applet
• Programming description
• Special Pythagorean trees
• Generator with code

Individual evidence

1. ↑ http://www.wisfaq.nl/show3archive.asp?id=32367&j=2005
2. ↑ Archived copy ( Memento of the original from January 18, 2009 in the Internet Archive ) Info: The archive link was inserted automatically and has not yet been checked. Please check the original and archive link according to the instructions and then remove this notice. 
3. ↑ Archived copy ( memento of the original from January 1, 2017 in the Internet Archive ) Info: The archive link was inserted automatically and has not yet been checked. Please check the original and archive link according to the instructions and then remove this notice.