Details about Golden 108° Symmetric Binary Trees

The scaling factor is \(r = r_{sc} = \frac{1}{\phi} = \phi-1\) where \(\phi\) is the golden ratio. It is useful to note that \[ \phi^2 = \phi+1 \Rightarrow 1 = \frac{1}{\phi} + \frac{1}{\phi^2} \Rightarrow 1-\frac{1}{\phi^2} = \frac{1}{\phi} \Rightarrow 1-r^2 = r. \]

 

The dotted regular pentagon in the figure above has its top vertex placed at the origin and has sides of length \(\sqrt{7-4\phi} = \sqrt{5-2\sqrt{5}}.\) The green golden 108° tree is placed with the base of its trunk at this top vertex. Rotate the green tree by 72° clockwise and move the base of its trunk to the next vertex (in the clockwise direction). We claim that the branch tip of the path R²(LR)^∞ for the green tree will coincide with the branch tip of the path L²(RL)^∞ for the red tree.

Let \(\alpha = e^{108^\circ i}\). This complex number represents a counterclockwise rotation of 108°. The complex number \(\alpha^{-1}\) represents to a clockwise rotate of 108°.

The vector for the trunk of the green tree can be taken to be the complex number i. The branch tip for the green R²(LR)^∞ path then corresponds to the complex number \[ \begin{align} i &+ ir\alpha^{-1} + ir^2\alpha^{-2} + ir^3\alpha^{-1} + ir^4\alpha^{-2} + ir^5\alpha^{-1} + \cdots \\ \\ &= i + \frac{ir\alpha^{-1}}{1-r^2} + \frac{ir^2\alpha^{-2}}{1-r^2} \\ \\ &= i + i\alpha^{-1} + i r \alpha^{-2} \\ \\ &= 0.58778525+0.19098301i \end{align} \] The vector for the trunk of the red tree can be represented by the complex number \( k=\cos(18^\circ) + i\sin(18^\circ)\). The vector S from the trunk of the green tree to the trunk of the red tree is given (as a complex number) by \(S = \sqrt{7-4\phi}\left(\cos(36^\circ) - i\sin(36^\circ)\right).\) The branch tip for the red path L²(RL)^∞ is therefore \[ \begin{align} S &+ k + kr\alpha + kr^2\alpha^2 + kr^3\alpha + kr^4\alpha^2 + kr^5\alpha + \cdots \\ \\ &= S + k + \frac{kr\alpha}{1-r^2} + \frac{kr^2\alpha^2}{1-r^2} \\ \\ &= S + k + k\alpha + kr\alpha^2 \\ \\ &= 0.58778525+0.19098301i \end{align} \] So the two branch tips are the same and the claim is verified [See symbolic derivation].

The top edge of the green subtree whose trunk is the last branch of the path RR makes an angle of 54° with the main trunk [Proof]. The same is true of the top edge of the red subtree whose trunk is the last branch of the path LL. These two edges both contain the same point as shown above. Because the two trunks both make the same angle of 126° with the side of the pentagon, and the sum of those four angles is 360°, the green and red edges must form a straight line segment (angle of 180° at the intersection point) since the sum of the interior angles of a pentagon is 540°.

By symmetry the same thing will happen with all the other golden 108° trees placed around the pentagon whose vertices are the trunks of the trees. In particular, this means that the five trees both lie around the outside of an inner pentagon as well as inside of a larger outer pentagon.

The right most point of the top edge of a golden 108° tree will be the branch tip for the path (RL)^∞. With the notation above, this branch tip will correspond to the complex number \[ i + ir\alpha^{-1} + ir^2 + ir^3\alpha^{-1} + \ldots = \frac{i}{1-r^2} + \frac{ir\alpha^{-1}}{1-r^2} = i\phi + i\alpha^{-1} \] The x-coordinate of this point is the real part of the complex number, and therefore \[ x = \text{Re}\left(i\phi+i\left(\cos(108^\circ)-i\sin(108^\circ)\right)\right) = \sin(108^\circ) = \sqrt{\frac{5+\sqrt{5}}{8}} = \frac{1}{2}\sqrt{\phi+2} \] This means that the top edge of the tree can be placed along the side of a regular pentagon of length \(\sqrt{\phi+2}\). Four more copies of the golden 108° tree can then be placed along the other sides of the pentagon. The five points at the base of the trunks are the vertices of another regular pentagon.