# Edit Create an Image Page: Pythagorean Tree

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 Image Title*: Upload a Math Image 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. [[Image:PTree2.gif|This animation shows how the angles of the triangle affect the shape of the tree.|thumb|300px|left]] 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 trees. [[Image:le pyhto.png|Figure 1: Example pythagorean tree|thumb|300px|right]] In this image, ''CDF'' is a right triangle. The original square ''ABCD'' has a side length of ''s'' and an area of ''s''2. 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 ''θ''1: :Square ''CFHG'': Area of $a^2 = (s \cos(\theta_1))^2 = s^2 \cos^2(\theta_1)$ :Square ''DFIJ'': Area of $b^2 = (s \sin(\theta_1))^2 = s^2 \sin^2(\theta_1)$ 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 ''a''2, ''b''2, and ''s''2 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'' = ''s''2 (''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|100px|left]] [[Image:PTree02.JPG|Figure 3: The 1st iteration|thumb|200px|right]] As you can see in Figure 2, for 0 iterations (the original square), the area is ''s2''. Or, when using the formula, ''s''2 (0 + 1) = ''s''2. For the first iteration as shown in Figure 3, the 0th iteration had one square with area ''s''2, 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''2 + ''s''2 = 2''s''2. We now use our formula to confirm this. ::$A_2 = s^2 (1 + 1) = 2s^2$ We can use this formula for only a small amount of iterations. Why? Look at the title image. After a few iterations, the tree starts to overlap and the area of the entire tree starts to converge to a single value. However, our formula implies that if we have infinite iterations, then our tree will have infinite area. Thus while our formula is useful for applying the area relationship per iteration, it is not an all powerful tool we can use to calculate the area for any amount of iterations. ===Perimeter Relationship between iterations=== We will attempt to find a relationship between the perimeters of squares from consequent iterations. Figure 1 will be used as a visual guide. We have the side lengths of all the squares. :Square ''ABCD'': Side length of ''s''. :Square ''CFHG'': Side length of $s \cos (\theta_1)$. :Square ''DFIJ'': Side length of $s \sin (\theta_1)$. The perimeter of a square is just its side length multiplied by 4. Thus we have our perimeters. :Square ''ABCD'': Perimeter of $4s$. :Square ''CFHG'': Perimeter of $4s \cos (\theta_1)$. :Square ''DFIJ'': Perimeter of $4s \sin (\theta_1)$. It seems that unlike area, there is no simple relationship between the perimeters of the squares. Multiplying, adding, dividing, and subtracting any of the perimeters from other values don't seem to work. This is common in mathematics. While some topics have very interesting applications and properties (like the area), other topics have no connections and seem obscure. The lack of a simple relationship between perimeters makes creating a general formula for the tree's perimeter very challenging. However, the one factor that truly makes creating the general formula challenging can be observed in the title image. The perimeter of the tree isn't just the sum of the perimeters of all the squares. The layout of the tree forces only the outer sides of most squares to be counted. We can already see this phenomenon in Figure 3 for the first iteration. If we were to calculate the perimeter of the first iteration, we would not count the inner sides of any of the squares. The situation only worsens as we continue to further iterations. Thus we will not derive a general formula for the tree's perimeter here. Algebra Analysis Calculus Dynamic Systems Fractals Geometry Graph Theory Number Theory Polyhedra Probability Topology Other None Algebra Analysis Calculus Dynamic Systems Fractals Geometry Graph Theory Number Theory Polyhedra Probability Topology Other None Algebra Analysis Calculus Dynamic Systems Fractals Geometry Graph Theory Number Theory Polyhedra Probability Topology Other The Pythagorean Tree is a good example of making interesting shapes with recursive fractals. The relationship for area between iterations gives the tree its name; it is the same equation as the Pythagorean theorem! The lack of a nice equation and relationship for perimeter shows how mathematics is not always beautiful and interesting. Yes, it is.