N-Fold Polygonal Spirals It's possible to tile the plane with unequal "30-60-90" triangles by arranging them according to the pattern shown below: The numbers indicate the sequence of triangles according to size, increasing by a factor of s = 1.085196156832515... on each step, where s is the positive real root of Squaring both sides and re-arranging terms, this can be factored as The positive real root comes from the right hand factor, which gives This arrangement can be generated by braid of three right-angled spirals, as illustrated below: Each segment of a spiral contains a vertex of one of the other spirals, in a three-way arrangement, and the vertex cuts the segment into two parts proportional to s3 and the square root of 3 respectively. Thus we need only interpolate between the vertices of one spiral, with an interpolation factor of to find the vertices of another spiral. Of course, this is just one example of infinitely many braided plane-tiling spirals of similar triangles. In general, letting the number zi,j in the complex plane denote the jth vertex (relative to some arbitrary reference vertex zi,0) of the ith spiral in the braid, we have The symbol q represents the angle between the rays from the center of the spiral to two consecutive vertices, and r signifies the factor by which the ray length changes. (In the example above we had q = p/2 and r = 1/s3.)  Given the vertices z0,n of one of the spirals, the vertices z1,n of the inscribed spiral can be interpolated according to for some constant k. Aside from the arbitrary scale factor, we can set z0,0 = 1, and then substitute for the vertices of the 0th spiral to give We can then interpolate the vertices of the next inscribed spiral from these, and so on, leading to the result Now, in order to give a coherent plane-tiling braid of M interlocking spirals, there must be some integer N such that Hence we require Solving for the interpolation factor k, we get Since both k and r are required to be real, we can split this equation into real and imaginary parts, which gives two equations in the two unknowns, k and r. Numerically we can simply determine the value of r that makes the imaginary part of the above expression vanish for the given values of q, M, and N. We could, however, proceed algebraically. For example, our original triple-spiral had M = 5, N = 3, and q = p/2, so the above expression gave Re-arranging terms and eliminating the fractional power, this leads to The real and imaginary parts of this equations give respectively. Equating these two expressions and simplifying, we get The only real root comes from the third factor, which gives For another example, suppose we wish to construct a braid of N = 4 spirals, with a turning angle of q = 2p/5. For these parameters, the only values of M for which k is real in the range 0 to 1 are M = 6 and M = 7. Plots of these two spiral braids are shown below. The right-hand braid of four spirals gives a tiling of the plane with similar isosceles triangles, but there is another way of tiling the plane with this same triangular shape, namely, by setting M = 6 and N = 2, as illustrated below. The shapes are the same, but the 7,4 pattern consists of four inter-laced spirals while the 6,2 pattern (below) consists of only two. Our original example gave a tiling of the plane by a triple-spiral of 30-60-90 triangles, based on the parameters M = 5, N = 3, with the angle q = 2p/4. We can construct another spiral tiling using the same triangular shape, but with six interlaced spirals instead of three. This pattern is shown below. This is a densely populated braid, but we can also construct examples at the other extreme, such as the double-spiral shown below, based on the parameters M = 4, N = 2, and q = 2p/6. In fact, we can even construct some interesting single-spirals by this method. For example, taking M = 10, N = 2, and q = 2p/6, we get two spirals that lie on top of each other, so they constitute just a single spiral, and it corresponds to the golden pentagon related to Perrin’s sequence. This spiral is shown below. Note that the value of k in this example is negative, meaning the “interpolated” points are actually extrapolated backwards along the line from one vertex to the next. Another interesting example is given by the parameter values M = 4, N = 4, and q = 2p/3. The result is a braid of four identical spirals in a pattern consisting of 30-60-90 triangles, and that nicely illustrates one of the best known proofs of Pythagoras’s theorem. (Letting c denote the hypotenuse and a,b the shorter edges, the diagram shows that c2 equals 2ab plus (b-c)2, and hence c2 = a2 + b2.) All the preceding examples were based on angles that evenly divide 2p, but this is not a necessity. We can generate a spiral for any angle of the form 2p/q where q is a real number in the range from 2M/3 to M. For example, the spiral below is based on the parameters M = 6, N = 2, with the angle q = 2p/(5.5). Determining a non-trivial solution of equation (1) is a more delicate operation than one might think. Notice, for example, that only the ratio M/N appears, rather than M and N individually, so one might think 9/6 would give the same result as 6/4, but this isn’t the case, because the root extractions implicit in fractional exponents are not unique. There are six values of 19/6 but only four values of 16/4, so these two expressions are obviously not equivalent. (If they were, it would follow that 11/1 was equivalent to 1n/n for all n, and therefore all the complex numbers on the unit circle would be equivalent.) Likewise the expression (am)1/n is generally not equivalent to (a1/n)m, so exponentiation is not strictly commutative in the complex domain. With this in mind, equation (1) should actually be written as Re-arranging terms, this gives This form helps to clarify the two different constraints that must be satisfied by the equation, given that k and r and both purely real. First, setting aside the magnitudes of the two sides, we require that they point in the same direction in the complex plane. The direction of the left side is simply Mq, so the first task is to ensure that the right side points in this same direction. The typical situation is illustrated below for the case M = 7, N = 4 with q = 2p/5. The real value of (1-k)/(kr) must be such that, when added to eqi, the resulting complex number is at an angle that is 1/N = 1/4 of the angle of eMqi = e7qi. In this example we can read the condition directly from the diagram. Now we need to determine the condition that will ensure the magnitudes of the two sides of equation (2). We have the magnitude so equation (2) implies the condition Simplifying and making the substitution k = 1/(1+r), we get This has only one real root, r = 0.8415406…, and from this we can compute the interpolation parameter k = 0.5430235…, to give the desired spiral. To express the equation as a polynomial with integer coefficients, we could clear the square root, showing that r is a root of the polynomial The first factor is just the cyclotomic (r5 – 1)/(r – 1). Interestingly, if we substitute r = x4 into the second factor, it factors into three polynomials f(x)f(-x)g(x) where f and g are of degrees 10 and 20 respectively. The relevant root (i.e., the fourth root of 0.8415406…) is from the 10th degree polynomial but none of the other roots of this polynomial satisfy the original conditions. Many of the multi-threaded spirals produced by this general method have interesting combinatorial properties. To take one example, consider the 5-fold spiral (i.e., N = 5) with the angle q = 2p/4 and M = 6. Four copies of this spiral outline are shown below, with the tiles colored in different ways to highlight the different symmetries. The all-yellow coloring in the upper left shows how the pattern can be interpreted as a single spiral of triangles. (Removing the largest triangle from the colored tiles leaves a pattern geometrically similar to the original, as does removing the next largest tile, and so on.)  The coloring in the upper right figure shows the five-spiral interpretation. The coloring on the lower left shows how the pattern consists of six spirals and, finally, the coloring on the lower right shows that the pattern also consists of seven spirals. In a sense this can be regarded as simple aliasing, but the cited spirals are all contiguous chains of tiles, so the fact that the pattern has this many different interpretations is not entirely trivial. Return to MathPages Main Menu