Difference between revisions of "Pythagorean Tree"

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<math>a^2 = (s * cos\theta)^2</math>
<math>a^2 = (s * cos\theta)^2</math>
<math>b^2 = (s * sin\theta)^2</math>
<math>b^2 = (s * sin\theta)^2</math>

Revision as of 14:24, 5 June 2013

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. The space between the squares in each iteration creates a right triangle. The top line of the square becomes the hypotenuse of the triangle above it.

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 the one that is directly connected to the square. When the right triangle is created, the legs of said triangle then become one of the sides of two brand new squares. Important to note is that the length of the legs is not changed during this creation, so the squares are smaller than the big one. The sum of the areas of the two smaller squares is equal to the area of the big square. The interesting thing about the tree is that the right triangle can have any valid value of the non right angles. When the angles of the triangle are changed, one is made bigger, and the other is made smaller. Length of sides corresponds to measure of angles, so the sides change too. Since the leg is bigger, the square created using that leg is also bigger, creating the illusion of a tilt.

A More Mathematical Explanation

Note: understanding of this explanation requires: *Basic Algebra

Le pyhto.png

In this image, the original square has an area of '"`UNIQ--math-00000000-QIN [...]

Le pyhto.png

In this image, the original square has an area of s^2, meaning its side length is s. This can be put into the Pythagorean theorem: a^2 + b^2 = s^2. This means the sum of the areas of the two branched-off squares will always be equal to the original square. These areas are dependent on the side lengths of the right triangle in the middle, which in turn are dependent the angles. With the hypotenuse s and angle \theta, the length of side b can be found with sin\theta = \frac{b}{s}. This means that b = s(sin\theta). The other side length, a, can be found similarly: cos\theta = \frac{a}{s}, or a = s(cos\theta).

To find the areas of the 2 branched-off squares, square the side lengths:

a^2 = (s * cos\theta)^2

b^2 = (s * sin\theta)^2

Which means:

a^2 + b^2 = s^2, or (s * cos\theta)^2 + (s * sin\theta)^2 = s^2

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