Agnes Scott College
Larry Riddle, Agnes Scott College

Golden 108° Symmetric Binary Tree

108AnglesProof

Let θ = 108°. The red subtree whose trunk starts at the top of the main tree's trunk is a scaled version of the main tree. It has been rotated by θ and therefore the top edge of the main tree is also rotated by θ to become the top edge of the red subtree (and also scaled by r). Therefore angle β = 180° − θ = 72°.

We also have ω + β = θ and so ω = θ − β = 2θ − 180°. Hence δ = 90° − ω = 270° − 2θ. For the golden 108° tree, therefore, δ = 54°.

(Note: These calculations for the angles work for any self-contacting symmetric binary tree with 90° ≤ θ ≤ 135°.)

goldenTriangles

If the two sides of the golden 108° tree are extended until they meet, the result will be a triangle with angles 72°-72°-36°, that is, the golden triangle △ACE in the figure to the right. But the subtree whose trunk is the first R branch is self-similar to the original tree, and thus it lies inside the golden triangle △ACB. Moreover, the subtree whose trunk is the last branch of the path RR is also self-similar to the original tree, and thus it lies inside the golden triangle △ABF. Since the edge BF of the tree makes an angle of 54° with the trunk of the tree, angle BFD is 108°, and therefore segments AF and FD are collinear.

By symmetry, triangles △ACD and △CFD are also golden triangles, and segment BC passes through point F.