Difference between revisions of "Pythagorean Tree"

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The total area of a Pythagorean Tree is actually quite simple to find. Simply put, it is the area of the original square (Iteration 0) multiplied by the number of iterations plus one. Or ''A<sub>n</sub>'' = ''s''<sup>2</sup> (''n'' + 1), where ''s'' is the side length of the original square, and ''n'' is the number of iterations. This method of finding the area does not consider the triangles in between the squares. We consider those triangles as empty spaces.
 
The total area of a Pythagorean Tree is actually quite simple to find. Simply put, it is the area of the original square (Iteration 0) multiplied by the number of iterations plus one. Or ''A<sub>n</sub>'' = ''s''<sup>2</sup> (''n'' + 1), where ''s'' is the side length of the original square, and ''n'' is the number of iterations. This method of finding the area does not consider the triangles in between the squares. We consider those triangles as empty spaces.
  
[[Image:PTree01.JPG|Figure 2: The 0th iteration|thumb|150px|left]]
+
[[Image:PTree01.JPG|Figure 2: The 0th iteration|thumb|80px|left]]
  
 
As you can see in Figure 2, for 0 iterations (the original square), the area is ''s<sup>2</sup>''. Or, when using the formula, ''s''<sup>2</sup> (0 + 1) = ''s''<sup>2</sup>.
 
As you can see in Figure 2, for 0 iterations (the original square), the area is ''s<sup>2</sup>''. Or, when using the formula, ''s''<sup>2</sup> (0 + 1) = ''s''<sup>2</sup>.
  
[[Image:PTree02.JPG|Figure 3: The 1st iteration|thumb|150px|right]]
+
[[Image:PTree02.JPG|Figure 3: The 1st iteration|thumb|200px|right]]
  
 
For the first iteration as shown in Figure 3, the 0th iteration had one square with area ''s''<sup>2</sup>, and the new iteration had two squares. As we mentioned before, the sum of the smaller squares' areas would be equal to the area of the original square, so ''s''<sup>2</sup> + ''s''<sup>2</sup> = 2''s''<sup>2</sup>. We now use our formula to confirm this.  
 
For the first iteration as shown in Figure 3, the 0th iteration had one square with area ''s''<sup>2</sup>, and the new iteration had two squares. As we mentioned before, the sum of the smaller squares' areas would be equal to the area of the original square, so ''s''<sup>2</sup> + ''s''<sup>2</sup> = 2''s''<sup>2</sup>. We now use our formula to confirm this.  

Revision as of 08:09, 26 July 2013

Inprogress.png
Pythagorean Tree, in 2 Dimensions
Le pytho.jpg
Fields: Algebra and Fractals
Image Created By: Enri Kina and John Wallison

Pythagorean Tree, in 2 Dimensions

A Pythagorean Tree is a fractal that is created out of squares. Starting from an initial square, two additional smaller squares are added to one side of the first square such that the space between all three squares is a right triangle. The side of the larger square becomes the hypotenuse of that right triangle.


Basic Description

This animation shows how the angles of the triangle affect the shape of the tree.

The Pythagorean Tree begins with a square that has a right triangle branching off of it. The hypotenuse of the triangle must always be directly connected to the square. When the right triangle is created, the legs of that triangle then become one of the sides of two brand new squares. The lengths of the legs are not changed during this process, so the area of the each new square is less than the area of the original square. The sum of the areas of the two smaller squares is equal to the area of the big square (shown in the More Mathematical Explanation). The interesting thing about the tree is that the other non-right angles of the right triangle can be any value. Side lengths correspond to angles, so change the angle will change the side length too. This relationship between side length and angle is what causes the tilt for most Pythagorean tress.

A More Mathematical Explanation

Note: understanding of this explanation requires: *Basic Algebra and Geometry, Trigonometry

Figure 1: Example pythagorean tree
In this image, CDF i [...]

Figure 1: Example pythagorean tree

In this image, CDF is a right triangle. The original square ABCD has a side length of s and an area of s2. We will derive the relationship between the area of the smaller squares and the area of the larger square. We can rewrite the area of CFHG and DFIJ in terms of s and θ1:

Square CFHG:  a^2 = (s \cos(\theta_1))^2 = s^2 \cos^2(\theta_1)
Square DFIJ:  b^2 = (s \sin(\theta_1))^2 = s^2 \sin^2(\theta_1)

We now have the relationship between the larger square and each of the smaller squares. However, note what happens when we add the areas of the smaller squares.

 \begin{align}
 a^2 + b^2 &= s^2 \cos^2(\theta_1) + s^2 \sin^2(\theta_1) \\
&= s^2(\cos^2(\theta_1) + \sin^2(\theta_1)) = s^2 
\end{align}

We have just proved that the sum of the areas of the smaller squares always adds up to the area of the larger square. Another way of thinking about this property is the Pythagorean theorem. Right triangle CDF has side lengths a, b, and s, with s as the hypotenuse. The Pythagorean theorem states that

 a^2 + b^2 = s^2 .

Since a2, b2, and s2 are the areas of each square, the Pythagorean also shows this interesting property of Pythagorean Trees.

The Area of the Tree for any number of iterations

The total area of a Pythagorean Tree is actually quite simple to find. Simply put, it is the area of the original square (Iteration 0) multiplied by the number of iterations plus one. Or An = s2 (n + 1), where s is the side length of the original square, and n is the number of iterations. This method of finding the area does not consider the triangles in between the squares. We consider those triangles as empty spaces.

Figure 2: The 0th iteration

As you can see in Figure 2, for 0 iterations (the original square), the area is s2. Or, when using the formula, s2 (0 + 1) = s2.

Figure 3: The 1st iteration

For the first iteration as shown in Figure 3, the 0th iteration had one square with area s2, and the new iteration had two squares. As we mentioned before, the sum of the smaller squares' areas would be equal to the area of the original square, so s2 + s2 = 2s2. We now use our formula to confirm this.

 A_2 = s^2 (1 + 1) = 2s^2

A 45-45-90 tree will be used to explain the way the perimeter works, because such a tree shows the most basic way it changes. In a 45-45-90 tree, each and every square in an iteration is congruent. Not so much true for other types of trees, and therefore a bit more difficult to find the perimeter. Referring back to..


This image, we can see that the area of the smaller squares is s2/2. Of course, the area of a square is one of the side lengths squared, so if we take s2/2, and find the square root, we get one of the side lengths, which is s/√2'. Now, if we were to iterate further...

We'd find that the area of the newly iterated squares is s2/4. If you take the square root of that, you'd find that it comes out to a clean s/2. Also known as √2*√2. Iterate further, the denominator in the area becomes 8. Square root of 8? 2√2. It goes like this indefinitely. For each iteration, you just multiply the denominator of the side length by √2. The numerator is always s. Of course, this only applies to 45-45-90. Although that's only for one side, the formula remains unchanged. All that happens is that you multiply the fraction by 4, since all four sides of a square are congruent. The perimeter of the squares changes wildly when different right triangles are used.


History of the Tree.

Albert E. Bosman was the one who created the Pythagorean Tree. He did so in 1942. He was a Dutch Mathematician, who was born in 1891 and died in 1961.




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