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 r and q. Arbitrarily setting z0 = 1, we have simply zn = cn. Three vertices za, zb, zc are co-linear if and only if (zb – za)/(zc – za) is purely real, since this ratio represents the interpolation factor for locating zb relative to za and zc. Naturally if q is of the form p/D for some positive integer D, then the vertices z0, zD, z2D, … are all co-linear, because ekpi 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 r, so we are free to select a value of r so that the vertices satisfy some additional condition.


For example, if we have q = p/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 = r. 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 r by which the lengths of successive radii change. This gives the condition 1 – r2 = r, which has the positive real root



This is the celebrated “golden proportion”, and using this value of r, with q = p/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 q = p/3 in place of p/2 leads to the following relation between r 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 r, 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 r, 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 r5 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, r5 + r – 1 = 0 and r3 + r2 – 1 = 0, correspond to the two recurrence relations that are evident in this spiral. Letting sn denote the distance from zn to zn+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 = reip/3 where r3 + r2 – 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 = reiq 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 q = p/5, and then finding the magnitude r of the spiral parameter c = reiq 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 r, 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 r = 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 q = 2p/7, and stipulate that z8 be co-linear with z0 and z2. Thus we seek a real r such that the interpolation parameter k082 = (c8 – c0)/(c2 - c0) is purely real, where c = reip/7. Solving for the real root of



we get r = 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 p/7.  The blue triangles are isosceles, but the green triangles are not, as is clear from the fact that a = rb, 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 a + b = 2p/7, we get the angles of the green triangles, a = 0.432477214… and b = 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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