Exponential Spiral Tilings

The vertices of a spiral tiling can be expressed as a sequence of numbers z–₁, z₀, z₁, z₂, … in the complex plane defined by

for given real constants ρ and θ. Arbitrarily setting z₀ = 1, we have simply z_n = cⁿ. Three vertices z_a, z_b, z_c are co-linear if and only if (z_b – z_a)/(z_c – z_a) is purely real, since this ratio represents the interpolation factor for locating z_b relative to z_a and z_c. Naturally if θ is of the form π/D for some positive integer D, then the vertices z₀, z_D, z_2D, … are all co-linear, because e^kπi is purely real for every integer k. Since the zero vertex is arbitrary, it follows that every sequence of vertices with indices the differ by multiples of D are co-linear, which is just a consequence of the fact that D vertices cover half of the complete angular distance around the center.  This is true for any value of ρ, so we are free to select a value of ρ so that the vertices satisfy some additional condition.

For example, if we have θ = π/2, the value of z₄ is automatically co-linear with z₀ and z₂, as expressed by the fact that the interpolation factor k in the equation

is purely real. Specifically, the interpolation factor is

We are free to choose any real value of k, and in particular we can set k = ρ. In other words, we require that every vertex of the spiral be interpolated (between the vertices preceding it in the sequence by two and four steps) by the same factor ρ by which the lengths of successive radii change. This gives the condition 1 – ρ² = ρ, which has the positive real root

This is the celebrated “golden proportion”, and using this value of ρ, with θ = π/2, the resulting spiral pattern is the diagonals of the well-known progression of perfect squares shown below.

The relation to Fibonacci numbers is obvious, since the diagonal of each square equals the sum of the diagonals of the two subsequent squares.

The analogous construction with θ = π/3 in place of π/2 leads to the following relation between ρ and the interpolation factor for the co-linear vertices z₀, z₆, and z₃.

Rather than simply setting k to some particular value, and thereby determining the value of ρ, let us instead impose the requirement that the vertices z₀, z₅, and z₄ be co-linear. This implies that the ratio (c⁵ – 1)/(c⁴ – 1) is purely real. This ratio can be written in the form

Neglecting the real factor of ρ, the numerator and denominator of this fraction can be plotted on the complex plain as shown below.

In order for the ratio to be purely real, the numerator and denominator must lie on the same ray from the origin, so this imposes the condition

Multiplying through by ρ⁵ and re-arranging, this is equivalent to

The only real root comes from the cubic factor, which gives

This produces the spiral shown below, which can be represented as a tiling of equilateral triangles.

The two polynomials, ρ⁵ + ρ – 1 = 0 and ρ³ + ρ² – 1 = 0, correspond to the two recurrence relations that are evident in this spiral. Letting s_n denote the distance from z_n to z_n+1, we have s₅ + s₁ – s₀ = 0 and s₃ + s₂ – s₀ = 0.  The Perrin sequence 3, 0, 2, 3, 2, 5, 5, 7, 10, 12, etc., satisfies both of these relations.

The same spiral can also be derived by requiring that the vertices z₀, z₁, and z₅ form an equilateral triangle. As discussed in the note on Napoleon’s Theorem, the necessary and sufficient condition for this is the vanishing of the sum of squares of their differences, which is to say

Thus the spiral parameter c must satisfy the condition

Expanding and factoring this expression, we get

which has the root c = ρe^iπ/3 where ρ³ + ρ² – 1 = 0. 

To give another example, suppose we stipulate that the vertices z₀, z₂, and z₃ form an equilateral triangle. This gives the condition

Expanding this and factoring the result, we get

The right-hand polynomial factors as

Up to reflection, each quadratic factor gives a spiral that satisfies the stated conditions. These spirals are characterized by the parameters

Alternatively these can be expressed in the form c = ρe^iθ where

Quadrilateral tilings based on these two solutions are shown below. In each case, the tile is an equilateral triangle with one of the edges carved inward to accommodate the neighboring tiles.

Another interesting spiral is produced by the parameters θ = π/5, and then finding the magnitude ρ of the spiral parameter c = ρe^iθ such that z₀, z₈, and z₇ are co-linear. Thus we seek a value of c such that the ratio (c⁸ – 1)/(c⁷ – 1) is purely real. This constant can be written in the form

Neglecting the constant real factor ρ, the numerator and denominator can be represented in the complex plane as shown below.

The length f is the golden proportion, so we have

The real root of this equation is ρ = 0.84282962349828… We can also express the condition on r as a polynomial with integer coefficients, by clearing the fractions and isolating the square root, and then squaring both sides. This leads to

The resulting spiral tiling is shown in the figure below.

For a final example, let us take θ = 2π/7, and stipulate that z₈ be co-linear with z₀ and z₂. Thus we seek a real ρ such that the interpolation parameter k₀₈₂ = (c⁸ – c⁰)/(c² – c⁰) is purely real, where c = ρe^iπ/7. Solving for the real root of

we get ρ = 0.934431232…, and from this we can compute k₀₈₂ = 0.53385673…  The resulting spiral tiling is shown in the figure below, where we have drawn the lines through the sets of vertices {082}, {193}, and so on. Also, the figure is scaled so that the segment z₀ to z₁ has unit length.

The red numerals represent angular multiples of π/7.  The blue triangles are isosceles, but the green triangles are not, as is clear from the fact that a = ρb, where a and b are the smaller edge lengths of a green triangle with unit base. The interpolation factor can be expressed as

Also, by similar triangles, we have

from which we get the quadratic in b

Hence we have

From the law of sines, and the fact that α + β = 2π/7, we get the angles of the green triangles, α = 0.432477214… and β = 0.46512077…  The green triangles are all similar to each other, as well as to the larger composite triangles, such as z₀z₁z₂, which consists of two smaller green triangles and one blue triangle. Likewise the blue triangles are all similar to the composite triangle z₁z₇ z₉, which consists of two blue and four green triangles. Therefore, by recursively replacing each tile with a set of smaller tiles, we can tile the entire plane with these shapes.

Similar techniques to the ones discussed above can be used to generate n-fold polygonal spirals.

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