T.D. Taylor- Finding Gold in the Forest: Self-Contacting Symmetric Binary Fractal Trees and the Golden Ratio

8/3/2019 T.D. Taylor- Finding Gold in the Forest: Self-Contacting Symmetric Binary Fractal Trees and the Golden Ratio

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FINDING GOLD IN THE FOREST: SELF-CONTACTING

SYMMETRIC BINARY FRACTAL TREES AND THE GOLDEN

RATIO

T. D. TAYLOR

Abstract. This paper presents the four self-contacting symmetric binaryfractal trees that scale with the golden ratio. These trees possess remarkablesymmetries in addition to the usual symmetries associated with symmetric bi-nary fractal trees. The trees provide new visual representations of well-knownequations and geometrical ob jects involving the golden ratio. Golden vari-ations of familiar fractals including the Cantor set and the Koch curve are

discussed, along with their connections to the golden fractal trees. The ob-servations about the golden trees presented here provide many interesting andentertaining exercises involving trigonometry, geometric series, fractal treesand the golden ratio. Various tilings of regular polygons with golden trees arealso presented.

1. Introduction and Background

1.1. Self-contacting Symmetric Binary Fractal Trees. Fractal trees were firstintroduced by Mandelbrot in “The Fractal Geometry of Nature” [M]. In general,fractal trees are compact connected subsets of Rn (for some n ≥ 0) that exhibit somekind of branching pattern at arbitrary levels. The class of symmetric binary fractaltrees were more recently studied by Mandelbrot and Frame [MF]. A symmetricbinary fractal tree T (r, θ) is defined by two parameters, the scaling ratio r (a realnumber between 0 and 1) and the branching angle θ (an angle between 0◦ and180◦). The trunk splits into two branches, one on the left and one on the right.Both branches have length equal to r times the length of the trunk and form anangle of θ with the affine hull of the trunk. Each of these branches splits into twomore branches following the same rule, and the branching is continued ad infinitum

to obtain the fractal tree.A self-contacting symmetric binary fractal tree has self-intersection but no actual

branch crossings. For a given branching angle θ, there is a unique scaling ratio rsc(θ)(or just rsc) such that the corresponding tree is self-contacting [MF]. The valuesof rsc as a function of θ have been completely determined [MF]. To determine the

value of rsc, find the smallest scaling ratio such that there is part of the tree besidesthe trunk that is on the affine hull of the trunk [MF], [TT2]. Figure 1 displays threedifferent self-contacting symmetric binary fractal trees. Further details about thegeneral class of self-contacting trees is available in other literature [MF], [TT1],[TT2].

To describe the fractal trees and their scaling properties more precisely, we usea monoid representation. For any r ∈ (0, 1) and any θ ∈ (0◦, 180◦), the generator

Key words and phrases. self-contacting fractal trees, golden ratio, Cantor set, Koch curve,fractal tiling.

1





GOLDEN SELF-CONTACTING TREES 3

address of length i that consists of the first i elements of A. The top points of atree are all points of the tree that have maximal y-value. For self-contacting trees

with θ < 135◦

, the top points form a middle Cantor set [TT2].In general, given a set U that is a subset of a tree T (r, θ) and an address A,

mA(U ) is also a subset of the tree, and we refer to it as a descendant of U . The levelof a descendant is equal to the level of the address map A. A particular class of descendants are subtrees. A level k subtree of a tree T (r, θ) is defined to be mA(T )for some level k address A. The subtree is denoted by S A(r, θ), S A, S b, or just S (the branch b = b(A) acts as the trunk of the subtree).

For an address A = A1A2 · · · , there exists a natural path on the tree that startswith the trunk and goes to P A. The path consists of the trunk along with allbranches b(Ai), where Ai = A1 · · ·Ai. We denote this path p(A), and because itstarts at the trunk we consider it to be level 0.

Each tree is equal to its reflection across the y-axis. Because of this left-rightsymmetry, we often restrict our attention to the right side of the tree.

1.2. The Golden Ratio. The ‘golden ratio’ φ is a remarkable number that arisesin various areas of mathematics, and also in nature and art [L], [HF], [W], [G],[H], [D]. The most basic geometric description of the golden ratio involves a linesegment. Without loss of generality, assume the line segment is [0, 1]. There existsa ∈ (0, 1) such that the ratio of the length of [0, 1] to the length of [0, a] is equal tothe ratio of [0, a] to [a, 1] (see Figure 2). By design, a = 1/φ. Thus φ satisfies

(3)11φ

=

1φ

1 − 1φ

In general, to say that a line segment is “divided according to the golden ratio”means that it is divided into two parts such that the ratio of the original line

segment length to the length of the larger part is the golden ratio.

Figure 2. Dividing the unit interval according to the golden ratio

Another geometric description is given by the golden rectangle, as displayed inFigure 3. Given a rectangle having sides in the ratio 1 : φ, φ is such that partitioningthe original rectangle into a square and a new rectangle results in a new rectanglehaving sides with a ratio 1 : φ. Thus φ also satisfies

(4)1

φ=

φ − 1

1

This implies that φ is the unique positive solution to the quadratic equation x2 −x − 1 = 0, hence

(5) φ =1 +

√5

2.

One can also show that 1/φ is the unique positive solution the the quadratic equa-tion 1 − x − x2 = 0. There are many interesting polynomial equations that can be



4 T. D. TAYLOR

Figure 3. The Golden Rectangle

derived. Because φ + 1 = φ2, any positive power of φ can be expressed in terms of

a linear expression with φ. For example:

φ3 = φφ2

= φ(1 + φ)

= φ + φ2

= φ + φ + 1

= 2φ + 1(6)

Similarly, φ4 = 3φ + 2. In general, it is a nice exercise to show that for k ≥ 2 wehave

(7) φn = F n−1φ + F n−2,

where F n is the nth Fibonacci number (F 0 = 1, F 1 = 1, F 2 = 2, F n = F n−1 +F n−2). For reciprocal powers of φ, we can also find linear equations that involvethe Fibonacci numbers. Using the fact that 1/φ = φ − 1, one can show that forn ≥ 1 we have

(8)1

φn= (−1)n(F n − F n−1φ).

The Fibonacci numbers F n are connected to φ in another way:

φ = limn→∞

F nF n−1

The golden ratio can be found in many classical geometrical figures. We brieflymention figures that will be relevant for the fractal trees. First consider an equilat-eral triangle and the circle that circumscribes it, as in Figure 4a. Join the midpointsof two sides with the line segment AB, and extend this to the circle at C . The ratioof AC to AB is φ. The golden ratio φ also appears in the pentagon and related ge-ometrical figures. Consider a pentagon and a pentagram with the same vertices asin Figure 4b. The interior angles at each vertex of a pentagon are 108◦. Again onecan show that the ratio of AC to AB is φ. A triangle with the angles 72◦, 72◦, 36◦

is commonly called a golden triangle (ABC in Figure 4c), and a triangle withthe angles 36◦, 36◦, 108◦ is a golden gnomon (ABD in Figure 4c). Any trianglein Figure 4b is either a golden triangle or a golden gnomon. A useful trigonometric




A B C

(a) (b)

(c) (d)

Figure 4. (a) Equilateral triangle, (b) Pentagon and Pentagram,(c) Golden Triangles (ABC and BDC ) and Golden Gnomon(ABD), (d) Decagon

relationship can be proved using a golden triangle:

(9) cos 72◦ =−1 + 5

4=

1

2φ

To start the proof of the previous identity, note that the line segment BD bisectsthe angle ∠ABC . Then the original golden triangle is divided into a golden gnomonand a smaller golden triangle, and proceed from there. The golden triangles alsoappear in the decagon (Figure 4d).There are many other interesting aspects of the golden ratio φ. The golden ratioφ can be considered to be the most ‘irrational’ number because it has a continuedfraction representation

(10) φ = [1, 1, 1, · · · ] = 1 +1

1 + 11+ 1

1+···

.

This representation is straightforward to derive by using the equation φ = 1 + 1/φ.There is also a nested radical expression for the golden ratio:

(11) φ =

1 +

1 +

√1 + · · ·.

This representation can be derived from the fact that φ2 = φ + 1.

1.3. Fractals and The Golden Ratio. The golden ratio appears in fractal ge-ometry as well as classical geometry, perhaps due to its self-similar nature. Thisself-similar nature can be seen in the golden rectangle where one can divide a goldenrectangle into a square and a smaller golden rectangle, and continue this processwith the smaller golden rectangles ad infinitum . Self-similarity is also apparent



6 T. D. TAYLOR

in a pentagon and pentagram with the same vertices (as in Figure 4b). There issmaller pentagon inside, and one can then construct the smaller pentagram with

the same vertices as this smaller pentagon. This process can be repeated ad infinitum . We have already mentioned that the golden triangle can be divided into agolden gnomon and a smaller golden triangle (as in Figure 4c), and one could alsorepeat this process to obtain smaller and smaller golden triangles within. Besidesgeometric representations of self-similarity, the self-similar nature of the golden ra-tio is also displayed in the continued fraction representation (Equation 10) and thenested radical (Equation 11).

We shall see that a ‘golden’ Cantor set appears in several of our golden trees.The general family of middle Cantor sets {C α}α∈(0,1) includes the classic middlethirds Cantor set. Given α ∈ (0, 1), define the the middle-α Cantor set C α asfollows [K]. Let β = (1 − α)/2. Let f 0, f 1 be the affine maps

(12) f 0(x) = βx and f 1(x) = βx + (1 − β )The middle-α Cantor set is the unique fixed point of the similarities f 0 and f 1. Moreintuitively, it is obtained from removing the middle open set of length α from theunit interval, then continuing to remove middles ad infinitum . It is straightforwardto show that the similarity dimension of a middle-α Cantor set is log 2/ log(1/β ).Cantor sets are noteworthy b ecause of various properties they possess. They areclosed and compact (since they are the intersection of closed subsets of the unitinterval); they are totally disconnected (given any two elements of the set we canfind disjoint neighborhoods of them); they are perfect (since closed and no isolatedpoints); they are nowhere dense; and they are uncountable [SS]. In general, any setthat is nowhere dense and perfect is considered to be a Cantor set.

The following geometric problem was posed [K]. Given β ∈ (0, 1/2), is it possible

to find λ ∈ (0, 1) such that C α∩λC α = {0}? Here λC α denotes the set {λx|x ∈ C α}.There is a critical value β such that for β < β , the sets are small enough to have asolution. For β ≥ β , the sets are too large to have solution. This critical value β

is (3−√5)/2 = 1/φ2, and the corresponding α for this value is such that β /α = φ[K].

There are many other instances of the golden ratio and fractals appearing to-gether in literature, and thus it is not possible to provide a complete survey here.It is worth noting, however, that there is an interesting relationship between thegolden ratio and variations on another well-known fractal, the Sierpinski gasket[BMS].

In the next sections we will discuss the relationship between the golden ratioand self-contacting symmetric binary fractal trees. There are exactly four self-contacting symmetric binary fractal trees for which rsc = 1/φ. The four branchingangles are 60◦, 108◦, 120◦ and 144◦. We shall discuss each tree separately, thoughthere are some common features. Each tree possesses remarkable symmetries inaddition to the usual left-right symmetry of symmetric binary trees. Classicalgeometrical objects such as the pentagon and decagon which are related to thegolden ratio also make an appearance. We shall also see special subsets of thetrees which are other fractals related to the golden ratio, such as a golden middleCantor set and a golden Koch curve. Using the scaling nature of the trees, wefind new representations of well-known facts about the golden ratio. The trees allseem to ‘line up’, and we shall explain what this means in each case. Some of




the observations are given without proof, and generally the proofs are wonderfulexercises involving trigonometry, geometric series, the scaling nature of the fractal

trees, and the many special equations involving the golden ratio.

2. Golden 60

The first golden tree that we present is T (rsc, 60◦). This particular tree has beendiscussed in other literature [W], [P]. We begin by demonstrating that rsc = 1/φ.

Figure 5. T (1/φ, 60◦)

To determine rsc, we find a path which leads to a point on the subtree S R thathas minimal distance to the y-axis (other than the point that is the top of thetrunk). Then we find the scaling ratio that places this point on the y-axis. The

Figure 6. Path p(RL3(RL)∞) to self-contact on T (1/φ, 60◦)

path to such a point is given by p(RL3(RL)∞), as displayed in Figure 6. Let



8 T. D. TAYLOR

P c = (xc, yc) denote P RL3(RL)∞. Then

xc = r sin θ − r

3

sin θ − r

4

sin(2θ) − r

5

sin θ − r

6

sin(2θ) · · ·= r sin θ − r3(sin θ + r sin(2θ))(1 + r2 + · · · )

= r sin θ − r3

1 − r2(sin θ + 2r sin θ cos θ)

=r sin θ

1 − r2[(1 − r2) − r2(1 + 2r cos θ)]

Substituting the value of θ = 60◦:

xc =

√3r

2(1 − r2)[1 − 2r2 − r3]

=

√3r

2(1

−r2)

(1 + r)(1 − r − r2)

The value of rsc that gives the solution to xc = 0 is the root of 1−r−r2 in (0, 1). Asmentioned in Subsection 1.2, the unique positive root of this quadratic is rsc = 1/φ.General Note. For any self-contacting tree, a self-contact point refers to a pointon the tree that is on the x-axis that corresponds to more than one address. Forthe tree T (1/φ, 60◦), a self-contact point is P RL3(RL)∞, because this point is thesame as P LR3(LR)∞. A self-contact point of a subtree is on the affine hull of thetrunk of the subtree and corresponds to more than one address. The two cornerpoints of a tree are P (RL)∞ and P (LR)∞, and corner points of a subtree S A are justthe images of these two points under mA, i.e., the points P A(RL)∞ and P A(LR)∞.

The top points of this tree form a middle Cantor set, as is the case for all otherself-contacting trees with branching angle less than 135◦ [TT2]. Let l be the lengthof the top of the tree, i.e., the length between the leftmost top point P (LR)∞ and

the rightmost top point P (RL)∞. It can easily be shown that l = √3. The top

points are geometrically similar (with a factor of √

3) to the middle Cantor set withα = 1/φ3. The first two iterations of removing open middles are shown in Figure7. At the first iteration, the two remaining line segments each have length l/φ2,because they are on the tops of level 2 subtrees S RL and S LR. The length of thegap at the first iteration must be equal to l/φ3. This is because the gap side is thethird side of an equilateral triangle with the other two sides being the top of level3 subtrees S RLL and S LRR. Thus

l

φ2+

l

φ3+

l

φ2= l

Simplifying this expression gives

2φ + 1 = φ3,

as we established in Equations 6. Thus we have a visual representation on theself-contacting fractal tree of one of the many equations involving φ. The scalingdimension of this middle Cantor set is log 2/(2 log φ).

The curve on the tree between the rightmost top point and and the leftmost toppoint can be considered a ‘golden’ Koch curve (see Figure 8). In the first iteration,

the original line segment of length l =√

3 between the points P (LR)∞ and P (RL)∞ is

replaced with four line segments. Two of these segments have length l/φ2 (because




Figure 7. Golden Cantor Set on T (1/φ, 60◦)

they are the tops of level 2 subtrees) and the other two have length l/φ3 (becausethey are the tops of level 3 subtrees).

Figure 8. Golden Koch Curve on T (1/φ, 60◦)

Another well-known equation involving φ can be visualized on T (1/φ, 60◦). Be-

cause θ = 60◦

, the tops of the subtrees S LRR and S RLLL are collinear. Considerthe triangle consisting of the top of the subtree S LR as one side, the tops of S LRRand S RLLL as another side, and the line segment joining them as in Figure 9a.This triangle is an isosceles triangle because the angles are 120◦, 30◦, 30◦. One sideof the triangle has length l/φ2 and the other side has length equal to l/φ3 + l/φ4

because it consists of the top of the level 3 subtree S LRR together with the top of the level 4 subtree S RLLL. Hence

l

φ2=

l

φ3+

l

φ4.

This previous equation simplifies to the familiar equations

1 =1

φ+

1

φ2or φ2 = φ + 1

Not surprisingly, there is a connection between this golden tree and equilateraltriangles. Figure 9b displays the tree together with two other copies of itself, onecopy rotated 120◦ to the right around the origin and the other copy rotated 120◦ tothe left around the origin. The trees contact each other at tip points but there areno branch crossings. This triple tree object is contained in an equilateral trianglewith sides of length l(2 + φ), where l is the length of the top of the tree.

Another regular polygon that naturally relates to T (1/φ, 60◦) is the hexagon,as in Figure 9c. In this figure, each vertex is the bottom of the trunk of a tree.The trees intersect, but there are no branch crossings. The boundary of the center



10 T. D. TAYLOR

(a) (b) (c)

Figure 9. (a) Isosceles triangle on T (1/φ, 60◦), (b) Equilateral tri-angle and T (1/φ, 60◦), (c) Hexagon tiled with copies of T (1/φ, 60◦)

region is a golden Koch snowflake (comprised of golden Koch curves as describedabove).

The golden trees are also special in the sense that branches seem to line uptogether. In the case of T (1/φ, 60◦), consider Figure 10. The affine hull of thebranch b(R) (partially shown in Figure 10a) intersects the tree at the contact pointof the subtree S R and at the top left corner point of the subtree S LLL (besidesthe branch itself). The affine hull of the branch b(LL) (partially shown in Figure10b) is also the affine hull of the branch b(RLLL) and any other branch of theform b(RL)kLL, for k ≥ 0. The proof of these claims is a nice exercise involvingtrigonometry, geometric series and properties of the golden ratio. For the sake of comparison, Figure 11 displays T (1/φ, 60◦) along with the self-contacting trees with

branching angles 58

◦

and 62

◦

.

(a) (b)

Figure 10. Collinearity of branches in T (1/φ, 60◦)

3. Golden 108

The next golden tree we look at is the self-contacting tree with branching angle108◦. As mentioned above in Subsection 1.2, the interior angles at each vertex of a pentagon are 108◦, thus it seems natural that pentagons, and in turn the goldenratio, are relevant for the self-contacting tree with this angle.

The path to a point on the subtree S R with minimal distance to the y-axis is p(RR(RL)∞), as displayed in Figure 13. Let P c = (xc, yc) denote P RR(RL)∞. It




(a) T (rsc, 58◦) (b) T (1/φ, 60◦) (c) T (rsc, 62◦)

Figure 11. Comparison of T (1/φ, 60◦) and two other self-contacting trees with branching angles close to 60◦

Figure 12. T (1/φ, 108◦)

is left to the reader to show that the value of r ∈ (0, 1) that gives the solution toxc = 0 is rsc = 1/φ.

Figure 13. Path p(RR(RL)∞) to self-contact on T (1/φ, 108◦)

As with T (1/φ, 60◦), the top points of T (1/φ, 108◦) form a middle Cantor setwith scaling dimension log 2/(2log φ). Let l be the length of the top of the tree, i.e.,the distance between P (LR)∞ and P (RL)∞. It can be shown that l = 2sin108◦ =

4φ2 − 1/(2φ) ≈ 1.902. The gap between P LR(RL)∞ and P RL(LR)∞ has length

l/φ3. Moreover, one can form a pentagon from the tops of the level 3 subtrees



12 T. D. TAYLOR

S LRR, S LLL, S RRR and S RLL along with the line segment in the gap (as shown inFigure 14a).

a b c

Figure 14. (a) Pentagon formed from the tops of level 3 sub-trees in T (1/φ, 108◦), (b) Outer and inner pentagons aroundT (1/φ, 108◦) and four copies, (c) Another pentagon from five copiesof T (1/φ, 108◦)

There is another relationship between this tree and the pentagon. Consider thetree along with four copies of itself, as shown in Figure 14b. As with the tiling of an equilateral triangle with copies of T (1/φ, 60◦), we have a tiling of the pentagon.The trees contact each other but there are no branch crossings. Finally, Figure 14cdisplays another pentagon that can be associated with five copies of the tree.

Besides the pentagon, there is another geometrical figure associated with thegolden ratio that is related this tree. This object is the golden triangle, as shown inFigure 15a. The sides of the triangle in the figure consist of the trunk, the branch

b(L), and the extension of the branch b(LL) to the trunk. Two of the angles are72◦, and thus the triangle is isosceles. This same figure also displays the fact thatcertain branches share the same affine hull. For example, the figure shows that thebranches b((RL)kL), for k ≥ 0, share the same affine hull. So in some sense thebranches do ‘line up’ on the tree.

(a) (b)

Figure 15. (a) Golden triangle and collinearity in T (1/φ, 108◦),(b) Golden gnomons of T (1/φ, 108◦)

Now consider Figure 15b. This figure displays some other interesting geometricalproperties. The affine hull of the branch b(RRR) passes through the top of the trunk




(extension of line segment AC ) and bisects the angle between the trunk and b(R).Thus the point of the subtree S RRR that is on b(R) (at D) has the same distance to

the top of the trunk as does the point of the subtree S RRR that is on the trunk (atB). In addition, the two triangles ABC and ADC are both golden gnomons.The rhombus ABCD is not considered to be a ‘golden rhombus’, however, becausethat name is reserved for a rhombus whose diagonals are in the golden ratio.

(a) (b) (c)

Figure 16. Golden triangles, golden gnomons and pentagon inT (1/φ, 108◦)

We can associate other golden triangles with T (1/φ, 108◦). One can form atriangle with one side going through the top of the tree and the other two sidesgoing through the tops of the level one subtrees (and extended so that they meetat a vertex) as in Figure 16a, the triangle is indeed a golden triangle. In Figure

16b, we have the same golden triangle around the tree, along with the line segmentthat passes through the tops of the subtrees S (RL)kLL and S (RL)kRRR, for k ≥ 0.This line segment divides the original golden triangle into a golden gnomon and asmaller golden triangle.

The final image in Figure 16c shows the original golden triangle with more linesegments, showing more golden triangles and two pentagons. In this figure, thelength of the base is l and the lengths of the sides are φl (because it is a goldentriangle). When a golden triangle is divided into a golden gnomon and smallergolden triangle, the original golden triangle and smaller triangle are geometricallysimilar with a factor of φ. This can be seen using the scaling nature of the trees,because the smaller golden triangle has its base as the top of a level one subtreeand the sides have length equal to the top of the actual tree ( l). The sides on thelarger pentagon have length l/φ2 and the smaller pentagon sides have length l/φ3.

The golden triangle at the bottom of the figure, with base being one side of thelarger pentagon, has side lengths equal to l/φ. Considering the side length of theoriginal golden triangle, we have a visual representation that

l

φ+

l

φ2+

1

φ= φl.

Simplifying this expression gives

2φ + 1 = φ3

as derived in Equation 6.



14 T. D. TAYLOR

For the sake of comparison, Figure 17 displays the tree T (1/φ, 108◦) along withthe self-contacting trees with branching angles 106◦ and 110◦.

(a) T (rsc, 106◦) (b) T (1/φ, 108◦) (c) T (rsc, 110◦)


4. Golden 120

The third golden self-contacting tree has branching angle 120◦. This angle isnot as commonly associated with the golden ratio as 108◦ is. However, we shall

Figure 18. T (1/φ, 120◦)

see that the appearance of equilateral triangles implies that the golden ratio alsoappears. As with the tree with branching angle 108◦, the path to a point on S Rwith minimal distance to the y-axis is p(RR(RL)∞). Let P c = (xc, yc) denotethe point P RR(RL)∞. Because branches of the form b(RRR(RL)k), for k ≥ 0, arevertical if θ = 120◦, the calculations for this tree are simplified. The value of xc isgiven by

xc = r sin θ + r2 sin(2θ)(1 + r2 + r4 · · · )

=r sin θ

1 − r2[(1

−r2) + 2r cos θ]

Substituting θ = 120◦:

xc =r√

3

2(1 − r2)[1 − r − r2]

Thus rsc is the root of 1 − r − r2 in (0, 1), which of course is 1/φ.One equilateral triangle that can be associated with this tree is the triangle that

is the smallest triangle that bounds the tree, as shown in Figure 20a. Moreover,the medians of this triangle possess interesting features. The vertical median iscoincident with the trunk (naturally). The other two medians are such that for any




Figure 19. Path p(RR(RL)∞) to self-contact on T (1/φ, 120◦)

(a) (b) (c)

Figure 20. T (1/φ, 120◦): collinearity and special triangles

subtree that they intersect, the intersection is either along the trunk of the subtree,or just at one corner point of the subtree (so never crossing a branch). One of themedians shows that the branches b((LR)kRRL), for k ≥ 0, and b((LR)kLLLR),for k ≥ 0, have the same affine hull.

Another equilateral triangle is displayed in Figure 20b. This triangle consists of the branch b(R) (segment AB), the extension of the branch b(RR) to the trunk(segment BC ), and the portion of the trunk that joins these two line segments(segment CA). The length of each side is 1/φ, so the line segment that extendsfrom b(RR) divides the trunk according to the golden ratio. This same figure alsoshows that the point C divides the trunk according to the golden ratio (becauseAC has the same length as AB). The extension of CB meets the tree at two otherpoints besides the branch of the subtree S RR and its self-contact point. These othertwo points are both corner points of the tree.

Now consider Figure 20c. The line through the tip points of maximal height of the subtree S RRR is horizontal, and it forms a special 30-60-90 triangle with thetrunk and the branch b(R). This triangle is BDC in Figure 20c. The linearextension of the branch b(L) meets the point with address RL(LR)∞ and is per-pendicular to the top of the subtree S RLL. Because of the left-right symmetry of the tree, BDE is congruent to BDC . A third congruent triangle is BAE .



16 T. D. TAYLOR

This last fact is useful for deriving other information about this tree using methodsof computational topology (to be discussed in future papers).

(a) (b) (c)

Figure 21. T (1/φ, 120◦): more equilateral triangles and ahexagon

As with T (1/φ, 60◦) and T (1/φ, 108◦), the top tip points of T (1/φ, 120◦) form amiddle Cantor set with scaling dimension log 2/(2log φ). Another common featureof this tree with other golden trees is that the tree T (1/φ, 120◦), along with twocopies of itself (rotated 120◦ and 240◦ around the origin) form a figure that is self-contacting. Figure 21a displays the outer and inner equilateral triangles related tothis triple tree object.

Figure 21b displays other equilateral triangles related to the triple tree. ABI is equilateral, and each side of the triangle is l (where l denotes the length of thetop of the tree). Likewise for

CDE and

HF G.

BC J is equilateral, with sides

of length l/φ. Thus the line segment AD has length l + l/φ + l. One can showthat the ratio of AD to AC is φ. Thus triangles ACH , BDF and IEG aresimilar to ADG with factor 1/φ. The iterated function system which consistsof the three similarities that send the largest equilateral triangle ADG to threetriangles scaled by 1/φ, namely ACH , BDF and IEG, corresponds to avariation on the usual Sierpinski gasket that is also self-similar [BMS].

For the sake of comparison, Figure 22 displays the tree T (1/φ, 120◦) along withthe self-contacting trees with branching angles 118◦ and 122◦.

(a) T (rsc, 118◦) (b) T (1/φ, 120◦) (c) T (rsc, 122◦)





Figure 23. T (1/φ, 144◦)

5. Golden 144

Finally we have the self-contacting tree with branching angle 144◦. The comple-ment of the angle 144◦ is 36◦, which is an interior angle of both golden trianglesand golden gnomons.

Figure 24. Path p(RR) to self-contact on T (1/φ, 144◦)

First we show that the self-contacting ratio for this angle is indeed 1 /φ. A pathto a point on S R with minimal distance to the trunk is given by p(RR).

(a) T (rsc, 142◦) (b) T (1/φ, 144◦) (c) T (rsc, 146◦)

Figure 25. T (1/φ, 144◦) and two other self-contacting trees withbranching angles close to 144◦



18 T. D. TAYLOR

The x-coordinate of the point P RR is r sin θ + r2 sin(2θ). For this point to be onthe trunk, sin θ + r2 sin(2θ) = 0, hence 1 + 2r cos θ = 0. Solving for rsc gives

rsc(144◦) =−1

2cos144◦= 1/φ.

The self-contact point P RR divides the trunk according to the golden ratio. Thepoint P RR((LR)∞ also divides the trunk according to the golden ratio.

As with the other golden trees, the branches line up in a certain fashion. Figure25b shows how the branches b(RL)kLL and b(RL)kRRR, for k ≥ 0, share the sameaffine hull. For the sake of comparison, the trees T (rsc, 142◦) and T (rsc, 146◦) aredisplayed in Figures 25a and c.

Consider T (1/φ, 144◦) in Figure 26a. The line segment BD through the origin(at D) and the point with address R (at C ) goes through all points of the formP R(LR)k for k ≥ 0 and P R(LR)∞. This line segment BD has length 1. The linesegment AB goes through the all points of the form P (RL)k for k

≥0 and P (RL)∞.

The triangle ABD is a golden triangle. The line segment AC that goes from thetop of the trunk to the point P R divides ABD into a golden gnomon (ACD)and a smaller golden triangle (ABC ).

(a) (b)

Figure 26. Golden figures on T (1/φ, 144◦)

Figure 26b shows branches of the form Rk. They form a spiral of golden triangles.

(a) (b)

Figure 27.




The tip points of T (1/φ, 144◦) form a golden Koch curve. The initial line seg-ment is shown in Figure 23a, with the first iteration shown in Figure 23b. A line

segment of length l is replaced with four line segments each of length l/φ2

(becausethey correspond to level 2 subtrees). The triangle formed from the two inner linesegments (and a line segment missing from the figure) is a golden triangle becausethe angles are 36◦, 72◦, 72◦.

Figure 28a shows the tree along with four copies, rotated around the bottom of the trunk with angles that are multiples of 72◦. The tip points of the trees form agolden Koch snowflake. Figure 28b shows a decagon associated with the tree andfour copies. Finally, Figure 28 shows a pentagon with each vertex acting as thebottom of a trunk of a tree and such that the five trees have the share the samepoint as the top of the trunk.

Golden Koch Snowflake Decagon Pentagon

Figure 28. More golden figures related to T 1/φ, 144◦)

6. Conclusions

This paper has presented four self-contacting symmetric binary fractal trees thatscale with the golden ratio. The four possible branching angles are 60◦, 108◦,120◦, and 144◦. There are no other branching angles which yield self-contactingsymmetric binary fractal trees that scale with the golden ratio. The reason forthis can be seen in Figure 29, where a plot of the self-contacting scaling ratio as afunction of branching angle.

Figure 29. Plot of self-contacting ratio rsc as a function of branching angle [MF]



20 T. D. TAYLOR

Various geometrical figures, other fractals, and equations associated with thegolden ratio can also be associated with these special trees. Future work includes

a search through other classes of trees such as asymmetric binary (same angle,different scaling ratio for each side) for more connections with the golden ratio.

ACKNOWLEDGEMENTS

The author gratefully acknowledges the Natural Sciences and Engineering Re-search Council of Canada, the Killam Foundation, and St. Francis Xavier Uni-versity for financial support. The initial research for this paper was done duringdoctoral research under the supervision of D. Pronk of Dalhousie University.

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Department of Mathematics, Statistics & Computer Science, St. Francis Xavier

University, Antigonish, Nova Scotia, B2G 2W5, Canada

E-mail address: [email protected]

Documents

T.D. Taylor- Finding Gold in the Forest: Self-Contacting Symmetric Binary Fractal Trees and the Golden Ratio