# Three Dimensional Pythagorean Tree

3-D Pythagorean Tree
The three-dimensional Pythagorean tree is a fractal composed of cubes whose side lengths are structured according to the Pythagorean theorem. It is called a "tree" because, after a few iterations, the fractal acquires a tree-like shape.

# Basic Description

The three-dimensional Pythagorean tree may be easier to understand if we start from the two-dimensional case. The Pythagorean tree is a fractal, so we build it by starting with a base case and continually applying a rule to it.

The base case is a square. The rule we apply to each square is that one of its sides serves as the hypotenuse of a right triangle, the legs of which are the sides of two new squares. Each time we apply the rule, it is called an iteration. The triangles are not actually parts of the tree; rather, the triangles are holes in the tree that determine how the squares are related. Because the triangles are right triangles, the side lengths of the squares adhere to the Pythagorean theorem.

On this page, we will specifically consider sides which form isosceles right triangles, so the iterations are symmetrical.

The three-dimensional case is simply the two-dimensional case extruded, or "pulled out". In other words, we just add depth to the two-dimensional Pythagorean tree. Each square is extruded so that its depth is equal to its length and width, making a cube. An example of this is shown below.

Figure 1
Extrusion from the two-dimensional to the three-dimensional Pythagorean tree.

In the three-dimensional case, we can add surface area and volume to the properties under consideration.

### Making the Pythagorean Tree

Figure 2

Our image was made using Google SketchUp. SketchUp allowed us to make accurate cubes by directly changing side lengths to ensure appropriate proportions. Also, we were able to find endpoints, intersections, midpoints, and lines of symmetry while using Sketchup, making the process of constructing the tree much more simple and the results much more accurate.

Figure 3

First, we create the square that will be used as the base case, or the first iteration. To create the triangle (or the sides of the squares of the next iteration), we copy the base square and paste it above the base square. We connect the center of the new square with two vertices of the first square to make a right isosceles triangle, as shown in Figure 2.

We then remove the excess portions, leaving the base square and the triangle (remember that the triangle is not actually "part" of the tree, we are just using it to create the squares). We then create four smaller squares off the triangle to create the next iteration of squares, since SketchUp cannot create rotated squares. By connecting the points of the four smaller squares, we create a square which shares a single side with the triangle. This process is shown in Figure 3. The same process can be performed on the right side of the triangle.

Repeating, or iterating, this process leads to a more developed Pythagorean tree, which is at this stage two-dimensional. It can be made three-dimensional by simply extruding the squares, as shown in Figure 1, so that their depth is equal to their side lengths. Finally, we remove the triangular portions and obtain the finished three-dimensional Pythagorean tree, shown in Figure 4 below.

Figure 4

# A More Mathematical Explanation

Note: understanding of this explanation requires: *Geometry and Algebra

The three-dimensional Pythagorean tree is a manipulation of the two-dimensional Pythagorean tree into [...]

The three-dimensional Pythagorean tree is a manipulation of the two-dimensional Pythagorean tree into the third-dimension. We simply add depth so that each iteration consists of cubes rather than squares. However, the tree's side lengths still correspond to the Pythagorean theorem:

$a^2+b^2=c^2$.

Figure 5

c will denote the side length of a particular cube (we will call this cube the base cube). a and b will denote the side lengths of the two new cubes of the iteration. As specified, we've chosen the case in which the triangles formed are isosceles, meaning $a = b$. Substituting this in, we get:

$a^2 + a^2 = 2a^2 = c^2$.

We will use this extensively when finding relationships pertaining to the three-dimensional Pythagorean tree. Figure 5 to the right illustrates these relationships. Since, as we've noted, the Pythagorean tree is a fractal, it is self-similar, and each iteration "looks like" this single iteration. It will be helpful to refer to this image as we go through the next few sections, as what is proved for one iteration will be true of all of them.

Below is another instance of the Pythagorean tree (here just a two-dimensional case, to keep things simple), along with a number of preliminary calculations of side lengths (referred to as edge length in Figure 6), surface areas, and volumes of each level of iterations. Side length simply refers to the linear distance of one side of a square. While we can talk about area in the two-dimensional case, only in the three-dimensional case can we talk about surface area, which for a cube is the area of one face times 6. Finally, volume is the side length cubed.

Figure 6

The calculations above seem to suggest certain relationships between the side length, surface area, and volume of a base cube and its successive cubes:

• The values in the first column divided by the values in the second column come out to about 1.415 (for example, $2.12 \div 1.50$). This decimal seems to be around $\sqrt 2$, so it seems that the the side length of the base cube is $\sqrt 2$ times longer than the side length of either of its successive cubes.
• The values in the last 2 columns are equal. These represent the surface area of the base cube and the sum of the surface area of the two successive cubes.
• The values in the fifth column divided by the values in the sixth column come out to about 1.415 as well (for example, $26.97 \div 19.07$). It seems, then, that the volume of the base cube is $\sqrt 2$ times larger than the sum of the volume of its successive cubes.

However, the above values are not quite specific enough. We have just guessed at these values based on a few calculations, using values for a specific case. Hopefully we can show that they are necessarily true. In the following few sections, we will prove that they hold for cubes of any size.

### The Side-Length Relationship

We want to prove that the ratio between the the edge of the base cube and the edge of a successive cube is $\sqrt2 : 1$. In other terms, we want to show that $a\sqrt2 = c$. We will do this by starting with the equation:

$ka = c$.

If we can show that the factor k is equal to $\sqrt 2$, then our proof will have succeeded.

To do this, we will begin with the right-hand side of the equation, $c$. We also have the Pythagorean theorem, which we've modified for our isosceles case:

$2a^2 = c^2$.

We rewrite c:

$c = \sqrt {c^2}$.

Now we can substitute the Pythagorean theorem in:

$c = \sqrt {2a^2} = a \sqrt 2$

This completes the proof. Note that this equation holds for all isosceles Pythagorean trees, whether two-dimensional or three-dimensional, because the depth does not affect side length.

### The Surface Area Relationship

The surface area of a three-dimensional shape is the sum of the area of each of its sides. Since each side of a cube with side length $a$ is $a^2$, and a cube has 6 sides, the expression for the surface area of a cube is $6a^2$.

Like above, we want find a relationship between the base cube and its successive cubes, this time as it pertains to the surface area. As mentioned previously, we suspect that the sum of the surface areas of the successive cubes is equal to the surface area of the base cube. The sum of the surface areas of the successive cubes is $6a^2 + 6 b^2$, and the surface area of the base cube is $6c^2$. We want to show that these are equal. In other terms, in the equation

$k(6a^2 + 6b^2) = 6c^2$

we suspect that the factor k = 1. Like before, we will start by looking at the right side of the equation, $6c^2$. We can immediately use the Pythagorean theorem to substitute in for $c^2$:

\begin{align} 6c^2 &= 6(a^2 + b^2) \\ &= 6a^2 + 6b^2 \\ &= 1(6a^2 + 6b^2) \end{align}

We have shown that the relationship established above holds if k = 1. Therefore, the surface area of the base cube is equal to the surface area of its immediately successive cubes. Note that, for this relationship, we did not have to substitute in the relationship $a = b$. The relationship would, of course, still hold even if we did substitute in $a = b$, but this shows that this particular relationship is not dependent on our doing so. In other terms, the surface area of the successive cubes is equal to the surface area of the base cube even when the triangle formed is not isosceles.

### The Volume Relationship

Lastly, we want to cover the volume relationship between the base cube and its two successive cubes. Volume is simply side length cubed, so the sum of the volumes of two successive cubes is $2a^3$, and the volume of the base cube is $c^3$. Once again, then, we want to find k:

$k(2a^3) = c^3$.

Let's start with the right-hand side of the above equation, $c^3$. We will want to find a way to substitute in the Pythagorean Theorem, so we will have to rewrite:

$c^3 = (c^2)^{3/2}$.

Now we can substitute in the Pythagorean theorem (taking into account that, in our isosceles case, $a = b$):

$c^3 = (2a^2)^{3/2}$.

Then:

\begin{align} c^3 &= 2^{3/2} a^3 \\ &= \sqrt {2} (2a^3) \end{align}.

Therefore, the scale $k = \sqrt 2$, as expected. The sum of the volumes of two successive cubes multiplied by the square root of two is equal to the volume of the base cube.

### Wrapping Up

Figure 7

If we look at the three equations, we notice that each outcome is related to the number $\sqrt2$. This should not surprise us, based on our knowledge of the Pythagorean theorem. If we consider our formulation,

$2a^2 = c^2$,

and take the square root of both sides, we get:

$a \sqrt {2} = c$.

The reader might note that this was our side-length relationship. This makes sense, because the triangle constructed by the sides of the three cubes is an isosceles right triangle, whose side-length relationships should follow accordingly. Likewise, we can take the Pythagorean theorem,

$a^2 + b^2 = c^2$,

and multiply both sides by 6 to obtain

$6a^2 + 6b^2 = 6c^2$.

The left side is the sum of the surface areas of each successive cube (whether or not $a = b$), and the right side is the surface area of the base cube. Therefore, as long as the constructed triangle is a right triangle, the surface area relationship must hold.

It is less obvious how we could find the volume relationship by simply manipulating our previous results. However, let's look at the Pythagorean theorem (with $a = b$) and our first result:

$2a^2 = c^2$

and

$a \sqrt {2} = c$.

Since $a \sqrt{2}$ and $c$ are equivalent, we can multiply each side of the Pythagorean theorem by them, and its equality will still hold. In particular, let's try multiplying the left side by $a \sqrt{2}$ and the right side by $c$. We get:

$2a^2 \cdot a \sqrt {2} = c^2 \cdot c$
$\sqrt {2} (2a^3) = c^3$.

The above corresponds with what was derived as the volume relationship.

As we can see, all of the relationships in an isosceles three-dimensional Pythagorean tree are connected to the Pythagorean theorem.

# Why It's Interesting

The three-dimensional Pythagorean tree is a good example of how recursive fractals can be used to create interesting geometrical structures that resemble things in the real world - in this case, a tree! As we've discovered, the Pythagorean theorem is central to all of the dimensional relationships in the Pythagorean tree, namely side length, surface area, and volume.

# How the Main Image Relates

The main image is a three-dimensional Pythagorean tree created by Ankur Pawar